问:
Is this Stochastic Process bounded from above?
I'm dealing with a problem in my thesis that involves proving the boundedness of a Stochastic Process. This question is related to another one I did recently, but the results found in the other one sadly didn't help me sufficiently.

Let $\text{d}X_t= f(t,X_t) \text{d}t +\text{d}W_t$ be a stochastic process defined for the times $t\ge0,$ such that $X_0=0$ and $f(t,x)<a<0$ for all $t\ge0,x>k$ for a certain constant $k$.

Prove that $$\mathbb{P}[\exists C>0: X_t<C \ \forall t\ge 0]=1.$$

In other words, I wish to show that if the function $f$ is negative when $X_t$ goes above a certain value $k$, then the stochastic process is bounded from above, because the drift part is going to be negative.

I know that $Y_t=at+W_t$ is bounded from above if $a<0$, but what I can't manage to deal with is the fact that we cannot control the moments when $X_t$ goes above $k$: every time it goes above $k$ it might reach a higher peak since the bound on $Y_t$ is not $\omega$-wise uniform, and the sequence of those peaks might be unbounded.

Here's a simulation of the SDE $\text{d}Y_t=-Y_t\text{d}t+\text{d}W_t$, which seems to fit these hypotheses, as the time taken is quite long.

EDIT: my SDE, in particular, is $$\text{d}X_t=X_t\frac{D(X_t)-S}{D(X_t)+S}\text{d}t+X_t\sigma\text{d}t,$$ with $D$ being $D(x)=D_0\exp(-\alpha x),$ for some $D_0>S>0$, and $X_0>0$. This SDE is equivalent to $$\text{d}Y_t=\left(\frac{D(\exp(Y_t))-S}{D(\exp(Y_t))+S}-\frac{\sigma^2}2\right)\text{d}t+\sigma\text{d}t,$$ thanks to the substitution $Y_t=\log(X_t)$: this second equation is the one I was referring to. Actually, $S$ is not a constant but a bounded function, which breaks the Markovianity, but I think that if the result is not true for constant $S$ the it's not true also for bounded $S$.

Any help would be immensely appreciated.

答:
So, let's come back to our initial equation but with some modifications to suit our analysis on the impact of volatility.

$$ dX_t = \left( f(t,X_t)-\frac{\sigma^2}{2} \right)dt +\sigma dW_t $$
with intial condition $X_0=x_0$

Here are some additional assumptions we make on $f$ (which are also confirmed by our real $f$):

$f$ is bounded (1)

$f$ is decreasing on second variable (2)

$f$ is locally lipschitz (3)

Let $\tilde{X}$ denote the solution of our SDE when there is no volatility, that is, $\tilde{X}$ is the solution of the following ODE:
$$d\tilde{X}_t = \left( f(t,\tilde{X}_t) \right)dt$$
with intial value $\tilde{X}_0=x_0$.

The condition of local Lipschitz (3) and boundedness of $f$ (1) guarantee the existence and unicity of $\tilde{X}$

After Ito's formula, we have:
$$d(X_t-\tilde{X}_t)^2=2(X_t-\tilde{X}_t)\left[ f(t,X_t)- f(t,\tilde{X}_t) \right]dt + \sigma^2 \left[ 1-\left( X_t-\tilde{X}_t\right)\right]dt+\underbrace{2\sigma (X_t-\tilde{X}_t) dW_t}_{=:dM_t}$$

Let's have some simple analysis:

$(X_t)$ is bounded in $L^2$ ( as a direct consequence of the boundedness of $f$ )

$(M_t)$ is a local martingale and in fact, it's even a $L^2$ martingale ( thanks to the boundedness of (X_t) )

The first weight of $dt$ in the $RHS$ of the above equation is nonpositive (due to the monotonicity of $f$)
So we can imply the following inequality for all $t \ge 0$:

$$g(t):= \mathbb{E}\left( (X_t-\tilde{X}_t)^2 \right) \le \sigma^2t- \int_{0}^t \sigma^2 \mathbb{E}(X_s-\tilde{X}_s)ds$$

Thus,
$$ 0 \le g(t) \le 2\sigma^2 t +\int_{0}^t \sigma^2 g(s)ds $$
So after Gronwalls, we imply that:
$$ 0 \le g(t) \le 2\sigma^2 t e^{\sigma^2t}$$
and that is :
$$ \mathbb{E}( X^{\sigma}_t-\tilde{X}_t)^2 \le 2\sigma^2 t e^{\sigma^2t}$$

The last equation implies most of the convergence we know:

Strong convergence in Euler Maruyama's sense.

(As a consequence) Finite-dimensional convergence (in law and in $L^2$)

Convergence in the weak topology of continuous processes (taking value in $\mathcal{C}[0,T]$ ) ( tightness is free due to the boundedness of $f$)

Discussion:

The condition of local Lipschitz is unnecessary as its existence and unicity can be derived by the same method ( the decreasing monotonicity of $f$ is necessary)

The second assumption (2) plays the central role in our analysis.

We can give a convergence speed if we want.

Some modifications can be made to have the same result in a more general setting of SDE.

We can craft some sort of almost sure convergence as follows:

Choosing a sequence $(\sigma_n , n \ge 1)$ such that: $$\sum_{n \ge 1} {\sigma_n}^2 <\infty $$
(for example $\sigma_n= \frac{1}{n}$ )

Then, there is a sequence of positive real number $(\epsilon_n , n \ge 1)$ that is decreasing to $0$ such that:
$$ \sum_{n \ge 1} \frac{\sigma_n^2}{\epsilon_n^2} <\infty $$
Thus, by Borel-Catelli, we imply that $\forall t \ge 0$:
$$ X^{\sigma_n}(t) \xrightarrow{n \rightarrow +\infty} \tilde{X}(t)  \text{ almost surely}$$.
Then thanks to the boundedness of $f$ and Komogolrov's continuity theorem, we imply the "almost sure" convergence, that is almost surely,
$$ \forall t  X^{\sigma_n}(t) \rightarrow \tilde{X}(t)$$

The SDE can be viewed as an ODE driven by a Holder-continous noise. But I don't have any knowledge on this field to give any adequate proof.


问:
Positive defnite matrices
Hei guys,

I would like to check with you a prove I have

Let the matrix $X_1 = \begin{bmatrix}
    A_1 & B_1\\
    B_1 ^ T & A_1\\
  \end{bmatrix}$ be positive definite. Based on this I would like to show that the matrix $X_2 = \begin{bmatrix}
    X_1 & B_2\\
    B_2 ^ T & A_1\\
  \end{bmatrix}$, where $B_2 = \begin{bmatrix}
    0\\
    B_1\\
  \end{bmatrix}$ is also positive definite.

The prove I have looks like this:

Based on the assumptions on $X_1$ and the Schur Lemma we get that $A_1, A_1 - B_1 ^ T \times A_1 ^ {-1} \times B_1$ and $A_1 - B_1 \times A_1 ^ {-1} \times B_1 ^ T$ are positive definite.

In order for $X_2$ to be positive definite one has to show (based on the same lemma) that $X_1 - B_2 \times A_1 ^ {-1} \times B_2 ^ T$ is positive definite.
This is equivalent with showing that the matrix:

$\begin{bmatrix}
    A_1 & B_1\\
    B_1 ^ T & A_1 - B_1 \times A_1 ^ {-1} \times B_1 ^ T\\
  \end{bmatrix}$ is positive definite.

For me this is true on if the matrix $- B_1 \times A_1 ^ {-1} \times B_1 ^ T$ is positive definite.

I would like to ask you if is possible $X_2$ to be positive definite with the $- B_1 \times A_1 ^ {-1} \times B_1 ^ T$ being positive definite?

Thanks,
Bogdan.

答:
The result seems to follow from the Sylvester criterion for positive definiteness and determinant identity for Schur complement.
 You need $A_1-B_1 A^{-1} B^T_1$ to be positive definite (which is positive definite, because $X_1$ is, as you say), but $-B_1 A^{-1}_1B^T_1$ is not positive definite if $X_1$ is positive definite. In other words, $-B_1 A^{-1}_1B^T_1$ is never positive definite if $A_1$ is. </p>

ContentLicense: CC BY-SA 3.0

Title: Positive defnite matrices


问:
Supremum of a set to the power of -1
$$
A\subseteq(0,\infty)
$$
$$
\sup(A)=3, \\ \inf(A)=1
$$
$$
B=\left\{a^{-1}:a\in A\right\}
$$

How would I go about formally proving that 1 is the supremum of B?

答: Since we are considering inverses of real numbers it should be clear that $1 = \inf(A) $would be a good choice for an upper bound for $B$. It is clear that the supremum will be positive since, for example, take the inverse of any element in $A \subset (0,\infty)$, then we must have that $\sup(B) >$ than that. To prove this is the least upper bound, suppose there was $0< x < 1$ $(\implies \frac{1}{x} > 1)$ such that $x > a^{-1}=\frac {1}{a}$ for all $a \in A$. But then $a > \frac{1}{x}$ so that $1  < \frac{1}{x} < a$ since $x$ is fixed, contradicting that $1$ is the infimum of $A$.

ContentLicense: CC BY-SA 4.0

Title: Supremum of a set to the power of -1


问:
Probability that $10$ random books from a collection of $20$ history and $20$ science fiction contains at least one book of each type; and ...

Let's say Mary is a first grade school teacher and today brings to school 40 different books, 20 of which are books about History and the other 20 books are about Science Fiction.

Question A.  Mary chooses randomly 10 books from her stack of 40 books and gives those to Peter as a reward for his performance on yesterdays Math test. Calculate the probability that Peter receives at least one book about History and at least one book about Science Fiction.

Question B. Mary decides to share all the 40 books to the 20 students of her class. Each book is given randomly to some student and independently of the other books. Obviously some students will get more books than the others (and some may not get any book at all). Calculate the probability Jane and George not getting any book at all.

Regarding question A, I treat the $40$ books as two groups of similar objects (20 Hist and 20 SF).  I count all of the possible combinations (0 Hist books 10 SF books, 1 Hist book 9 SF books, 2 Hist books 8 SF books etc etc ...) and then I find the ways those combinations can be arranged (by groups of similar objects $\frac{n!}{n_1!n_2!}$ where $n = 10$ and $n_1$ is the number of Hist books and $n_2$ is the number of SF books). In the end I divide to find the probability.

For question B, I treat the problem as distributing $40$ non-discrete objects (the books) to $20$ discrete boxes (students). The final result is $\dfrac{C(57,40)}{C(59,40)}$.

答:
Keep in mind that both the books and the students are distinct.

Question A:  Mary can choose $10$ of the $40$ books in
$$\binom{40}{10}$$
ways.

If Peter is to receive at least one book about history and at least one book about science fiction, Mary cannot choose only history books or only science fiction books.  There are
$$\binom{20}{10}$$
ways for Mary to choose only history books.  There are also
$$\binom{20}{10}$$
ways for Mary to choose only science fiction books.  Thus, there are
$$\binom{40}{10} - \binom{20}{10} - \binom{20}{10}$$
ways for Mary to select $10$ of the $40$ books for Peter while giving Peter at least one history book and at least one science fiction book.

Thus, the probability that Mary gives Peter at least one history book and at least one science fiction when randomly selecting $10$ of the $40$ books for Peter is
$$\frac{\dbinom{40}{10} - \dbinom{20}{10} - \dbinom{20}{10}}{\dbinom{40}{10}} = 1 - \frac{2 \cdot \dbinom{20}{10}}{\dbinom{40}{10}}$$
where the term that is being subtracted from $1$ is the probability that Peter receives only history books or only science fiction books.

Your method would work, except that you must make sure not to give Peter only history books or only science fiction books.  Thus, the number of favorable cases is
$$\sum_{k = 1}^{9} \binom{20}{k}\binom{20}{10 - k}$$
where the first binomial counts the number of ways of selecting $k$ of the $20$ history and the second binomial counts the number of ways of selecting $10 - k$ of the $20$ science fiction books.  Notice that the summation starts at $1$ and ends at $9$ since Peter receives at least one history book and at least one science fiction book.

Question B:  Since there are $20$ possible recipients for each of the $40$ books, Mary could distribute the $40$ books in $20^{40}$ ways (if she did not worry about a fair allocation of the books).  If neither Jane nor George were to receive any books, then there would be $18$ possible recipients for each of the $40$ books, which means Mary could distribute the $40$ books to the other $18$ students in the class in $18^{40}$ ways.  Hence, the probability that neither Jane nor George receives a book is
$$\frac{18^{40}}{20^{40}}$$


问:
Repeated application of narcissistic function
Over on Code Golf.SE, I've asked this challenge about the looping behaviour of a function related to narcissistic numbers. In the question, I define the function as, for a natural number $x = d_1d_2...d_n$ where $d_i$ is a single digit between $0$ and $9$ and $x$ has $n$ digits:

$$f(x) = \sum_{i=1}^n d_i^n$$

In this case, a number is narcissistic if it is a fixed point of $f(x)$ i.e. $f(x) = x$. However, the question is about repeated application of this function. For example, repeatedly applying $f$ to $x = 127$ leads to the following chain of numbers:

$$127,352,160,217,352,160,217,...$$

and the triplet $352, 160, 217$ repeats ad infinitum from here. I have two hypotheses which I believe to be correct but have no idea how to prove:

For all natural numbers $x > 0$, repeated application of $f$ eventually leads to a repeating loop (counting a fixed point as a "loop" of 1 element).

The length of this loop is always less than or equal to $14$. For example, $x = 147$ is the lowest number that yields $14$, with a loop of

$$537059, 681069, 886898, 1626673, 1665667, 2021413, 18829, 124618, 312962, 578955, 958109, 1340652, 376761, 329340$$

My brute forcing attempts have shown that this is true for all $x < 10^5$, and is currently (at time or writing) at around $x = 64000$ without having found a counter example to either hypothesis.

This is a program which takes an input $x$ and outputs the repeated application of $f$ to $x$ until a repeated value is found, the loop of $x$ and the length of said loop, if you'd like to test out some other numbers.

Body:
After checking (hopefully) all possibilities, for number bases 3 to 10, I have found the following loops:

Base-3: https://pastebin.com/H6pZtJC1

Base-4: https://pastebin.com/rK7LVnZB

Base-5: https://pastebin.com/XYTXuA4k

Base-6: https://pastebin.com/vFL1hdjs

Base-7: https://pastebin.com/75g0bWP0

Base-8: https://pastebin.com/2MTqi6q5

Base-9: https://pastebin.com/055Kw8dC

Base-10: https://pastebin.com/8extqeGy

Count of narcissistic loops in
different number bases:

                          Base
        3  4  5  6  7  8  9 10
    + ------------------------
  1 |   6 12 18 31 60 63 59 89
  2 |   -  -  2  6 17 11 14 16
  3 |   -  -  -  -  7  4 11  7
  4 |   -  -  1  2  6  5  8  5
  5 |   -  -  -  1  -  1  5  4
  6 |   -  -  1  1  3  2  1  5
  7 |   -  -  -  -  2  2  2  3
  8 |   -  -  -  -  2  -  1  -
  9 |   -  -  -  -  2  -  -  2
 10 |   -  -  -  -  -  1  -  2
 11 |   -  -  -  -  -  -  -  -
 12 |   -  -  -  -  -  -  1  2
 13 |   -  -  -  -  -  1  -  1
 14 |   -  -  -  -  -  -  -  2
 15 |   -  -  -  -  -  1  -  -
 16 |   -  -  -  -  -  -  1  -
 17 |   -  -  -  -  -  -  -  -
 18 |   -  -  -  -  -  -  -  1
 19 |   -  -  -  -  -  -  -  -
 ...|
 31 |   -  -  -  -  1  -  -  -
  ^ Loop length

The largest singleton is in base-10:

$115132219018763992565095597973971522401=1^{39}+1^{39}+5^{39}+1^{39}+3^{39}+2^{39}+2^{39}+1^{39}+9^{39}+0^{39}+1^{39}+8^{39}+7^{39}+6^{39}+3^{39}+9^{39}+9^{39}+2^{39}+5^{39}+6^{39}+5^{39}+0^{39}+9^{39}+5^{39}+5^{39}+9^{39}+7^{39}+9^{39}+7^{39}+3^{39}+9^{39}+7^{39}+1^{39}+5^{39}+2^{39}+2^{39}+4^{39}+0^{39}+1^{39}$

Here's the loop of 18 in base-10:

10074069541108119620821,18935428129475061827932,28415252997720554102092,
27233895488449729771663,28470768372157693427350,10181976920394277400584,
27852314084047558510219,10688610940325001073897,19553092616993382187366,
36662895609794210663524,27211241280778321507795,10180397389391156436853,
28388673910517178592951,29032929335610790615729,44371031395529147765128,
18399103996082548276483,37841895989779904859691,64484670033935168320603

Here's the loop of 31 in base-7:

105343,115515,253446,414526,412436,321136,260316,541235,210236,255345,
303626,543325,212326,255466,1044166,5015551,2441041,263452,400006,300652,
351614,366436,1144156,3215311,462622,554356,556516,1122466,4620541,3215551,
2004031


问:
Check if the following set is convex
I am wondering about the set

$$\{(x,y,z)\in\mathbb R^3\mid xyz\geq a\text{ and } x,y,z\geq 0\}$$

for some $a>0$ and whether it is convex or not. I have tried to show that the function $f(x,y,z)=xyz$ is convex over $\{(x,y,z)\in\mathbb R^3\mid x,y,z\geq 0\}$ by considering its Hessian matrix but I think the function is not convex as the Hessian seems to not be positive semi-definite.

I was wondering if there is maybe an example of two points such that any convex combination is not in the set (which I doubt). Otherwise, I would be interested in some hints on how to solve the problem.

答: This is the epigraph of the function $g:(0,\infty)\times (0,\infty)\to\Bbb R$, $g(x,y)=\frac a{xy}$. Therefore, it is convex if and only if $g$ is convex. The hessian of $g$ is $\nabla^2g(x,y)=\frac a{xy}\begin{pmatrix}\frac {2}{x^2}&\frac{1}{xy}\\ \frac{1}{xy}&\frac {2}{y^2}\end{pmatrix}$, which is positive definite by Sylvester's criterion. Therefore $g$ is convex, and so is your set.

Title: Check if the following set is convex


问:
Huffman Algorithm
We use Huffman’s algorithm to obtain an encoding of alphabet ${a,b,c,d}$ with frequencies $f_{a}, f_{b}, f_{c}, f_{d}.$ In each of the following cases, either give an example of frequencies $(fa, fb, fc, fd)$ that would yield the specified code, or explain why the code cannot possibly be obtained (no matter what the frequencies are).

(a) Code: ${0,10,110,111}$

(b) Code: ${1,00,01,110} $

(c) Code: ${00,01,10,11}$

I mostly solved part $b$, but could not be able to solve $a$ and $c$. Would be welcomed to know your solutions on this!

答:
In Huffman coding, we build the tree bottom-up by considering pairs of alphabets with the least frequency/probability. So the two least frequent alphabets get assigned the longest code length. In part $(b)$, $d$ has the longest code $110$ so some other alphabet must also have the code $111$ which is not the case.

In part $(a),c,d$ have the largest code length so they have the two least probabilities, say $p_c\le p_d$. We start building the tree by forming a dummy vertex labelled $v_1$ with left child $c$, right child $d$ and $p_1=p_c+p_d$. Then $v_1$ has the code $11$. Now $b$ has the code $10$, i.e. $b$ is the left child and $v_1$ is the right child of a dummy vertex $v_2$. So $p_2=p_b+p_1$ and $p_b\le p_1$. Since out of $\{a,b,v_1\},b,v_1$ have the longest codes, we also have $p_b,p_1\le p_a$.

Now $v_2$ has the code $1$. $a$ has the code $0$, i.e. $a$ is the left child and $v_2$ is the right child of the root vertex $R$. So $p_a\le p_2.$ So the constraints we have are$$\begin{align*}&p_c\le p_d\le p_b\le p_a\\&p_b\le p_c+p_d\le p_a\\&p_a\le p_b+p_c+p_d\\&p_a+p_b+p_c+p_d=1\end{align*}$$

You can check that $p_a=0.41,p_b=0.29,p_c=0.15,p_d=0.15$ is a solution and it generates the tree in the picture:

For part $(c)$, all alphabets have the same code length so one solution is when they are all equiprobable, i.e. $p_a=p_b=p_c=p_d=0.25$. Can you check if you get the correct tree and codes?

Edit: Your tree should look like this:

Note that instead of $a,b,c,d$ we could have $b,a,c,d$ or any other order of the four alphabets in the bottom four vertices since they have the same probability, but we have to make the tree according to the codes given. The below tree also corresponds to the given probabilities but the code for $b$ is $00$, which is not desired.

Title: Huffman Algorithm


问:
How do I show the optimal strategy for the game 'guess who' is to pick subsets close to half the size of the current set
Consider the game guess who. In the game, the chooser chooses a person out of a set of $n$ people. The guesser asks questions such as, 'does the person wear glasses'. By asking such questions the number of people is reduced until the guesser can guess exactly who the person is. The guesser wants to minimize the number of guesses. Intuitively, assuming an equal probability that any given person is chosen, they should ask questions that if true, cut the number of people in half, or as close to half as possible. More generally, I suspect if there is some distribution over the people, the best strategy is to always pick one of the subsets of the current remaining set of people that is as close to having a $50\%$ chance of containing the chosen person.

First, I would like to know if this is correct, and second, I would like to prove this. Currently I am trying to prove this using an argument via entropy however I have been unsuccessful so far.

答: You can prove by induction that with $n$ questions you can identify an individual out of $2^n$.  You can then show that if one set is larger than $2^{n-1}$ you might need another question, so you have to split that evenly.  If you have a number of people that is not a power of $2$ you have some flexibility in the splits.  For example, if there are $25$ people you can split them anywhere from $9-16$ to $12-13$ and succeed in five questions at worst.


问:
Can someone find a (two-sided) inverse function to the characteristic function χ? given...
so i have given the folowing info:

In general, if $A$ is a set, for $S \in \mathcal P(A)$ define $\chi(S) : A \to \{0,1\}$ by $$\chi(S)(a) = \begin{cases} 1 & \textrm{if } a \in S, \\ 0 & \textrm{if } a \notin S. \end{cases}$$
Then $\chi : \mathcal P(A) \to 2^A$ is bijective.

But I would like to know the what the inverse is to $\chi$. I'm quite new with the definition "characteristic function", so I don't even now how to start. I normally calculate the inverse by replacing x with y and vice versa, but know I don't have a clue. Please help me.

Thank you in advance

答: Given $f \in 2^A$, define $\chi^{-1}(f) = \{a \in A \mid f(a)=1\} \in \mathcal P(A)$.


问:
Which quantifier to use for the domain of tangent in set-builder notation?
I am trying to write out the domain of the function $\tan x$ in set-builder notation. I have seen this written out as $$\left\{x \in \mathbb{R} \mid (\exists n \in \mathbb{Z})\left[x \neq \frac{\pi}{2} + \pi n\right]\right\},$$ with the existential quantifier. My question is that since the domain excludes an infinite number of asymptotes, would it not be more correct to use the universal quantifier to write the domain like so? $$\left\{x \in \mathbb{R} \mid (\forall n \in \mathbb{Z})\left[x \neq \frac{\pi}{2} + \pi n\right]\right\}$$

答: Yes, the second domain is correct. Note that the first domain includes $\pi/2$ because for $n=1,n\pi+\pi/2=3\pi/2\ne\pi/2$.


答:
You are correct, you should have a universal quantifier. The first set you are describing is just $\mathbb R$.

(Indeed, given any $x$, I can find an $n$ such that $x\neq\frac\pi2 +\pi n$. Just take $n=x$ and it is trivially true.)

Although personally I would just write the domain more concisely as
$$\mathbb R\smallsetminus \{(2n+1)\tfrac\pi2 : n\in\mathbb Z\},$$
or just as
$$\mathbb R\smallsetminus (\pi\mathbb Z+\tfrac\pi2)$$
if you're comfortable with Minkowski set operations.

Note.  If you want to use an existential quantifier, you would have to rephrase the condition as a negation, i.e., as  $\lnot\big[(\exists n\in\mathbb Z)(x=\tfrac\pi2+\pi n)\big]$.

Title: Which quantifier to use for the domain of tangent in set-builder notation?

Body: I am trying to write out the domain of the function $\tan x$ in set-builder notation. I have seen this written out as $$\left\{x \in \mathbb{R} \mid (\exists n \in \mathbb{Z})\left[x \neq \frac{\pi}{2} + \pi n\right]\right\},$$ with the existential quantifier. My question is that since the domain excludes an infinite number of asymptotes, would it not be more correct to use the universal quantifier to write the domain like so? $$\left\{x \in \mathbb{R} \mid (\forall n \in \mathbb{Z})\left[x \neq \frac{\pi}{2} + \pi n\right]\right\}$$


问:
Prove $x^ny(x - y) + y^nz(y - z) + z^nx(z - x) \ge 0$

Prove the inequality with $x, y,z$ is the sides of a triangle and
  $n\in \mathbb Z \land n\ge2$ $${x^n}y(x - y) + {y^n}z(y - z) + {z^n}x(z - x) \ge 0 \tag 1$$

I can prove the inequality with $n=2$:
$$(1)\iff x(y-z)^2(x+z-x) + y(x-y)(x-z)(x+y-z) \ge 0$$
or apply Ravi Substitution: $$(1)\iff \frac {a^2}b+\frac {b^2}c + \frac {c^2}a \ge a+b+c\  $$
(alway true, by AM-GM)

But I can't prove the general problem above

Please forgive my grammar

Body:
you can also induct further: if $n=k$ is OK,then:$n=k+1$, WLOG,there is two cases:

case I: $x\ge z \ge y$,

${x^{k+1}}y(x - y) + {y^{k+1}}z(y - z) + {z^{k+1}}x(z - x) \ge z{x^{k}}y(x - y)+z{y^{k}}z(y - z)+ z{z^{k}}x(z - x)=z\left({x^{k}}y(x - y) + {y^{k}}z(y - z) + {z^{k}}x(z - x)\right) \ge 0$

case II: $x\ge y \ge z$

${x^{k+1}}y(x - y) + {y^{k+1}}z(y - z) + {z^{k+1}}x(z - x) \ge z{x^{k}}y(x - y)+z{y^{k}}z(y - z)+ z{z^{k}}x(z - x)=z\left({x^{k}}y(x - y) + {y^{k}}z(y - z) + {z^{k}}x(z - x)\right) \ge 0$


问:
About a sum of binomial coefficients
I need to calculate the following sum: $\sum_{k=0}^{[\frac{n}{2}]}$ $n\choose 2k$ $\frac{1}{4^{2k}}$.
I guess it has to do something with identity $\sum_{k=0}^{n}$ $n \choose 2k $ $=2^{n-1}$, but I am not sure should I use some substitution for counter of this sum or I am on wrong way?

答:
Hint: Compute, using the binomial theorem $$\frac{(x+1)^n+(1-x)^n}{2}.$$

For your particular question, you can not use that one because both are a particular case.

Title: About a sum of binomial coefficients


问:
Example of a representation of $U(1)$ with $n>2$
This is one of those ideas that it seems intuitively clear at first, but then it starts to blur out. I see the comment "There is no such thing as the $n$-dimensional representation of $U(1)$." in this post and the explanation in Peter Woit's Quantum Theory, Groups and Representations: An Introduction

Figure 2.2: Visualizing a representation $π : U(1) → U(n),$ along with its differential.
The spherical figure in the right-hand side of the picture is supposed to
indicate the space $U(n) ⊂ GL(n, C)$ ($GL(n, C)$ is the $n \times n$ complex matrices,
$C^{n^2},$ minus the locus of matrices with zero determinant, which are those that
can’t be inverted). It has a distinguished point, the identity. The representation
$π$ takes the circle $U(1)$ to a circle inside $U(n).$ Its derivative $π'$ is a linear map
taking the tangent space $iR$ to the circle at the identity to a line in the tangent
space to $U(n)$ at the identity.

I understand how

$$R(U(1)) =\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \in GL(2,\mathbb C)$$

which can be expressed as

$$R(U(1)) =\begin{bmatrix}e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix}$$

in the basis of eigenvectors $\left\{ \begin{bmatrix}i \\ 1  \end{bmatrix} , \begin{bmatrix}-i \\ 1  \end{bmatrix}\right\}$

But what is the meaning of representations in $GL(n,\mathbb C)$ with $n>2$? Is the idea to introduce tensorial basis vectors like $e_1 \otimes e_2$? I don't think so, since the block matrix is a direct sum, while tensor products build irreducible representations. What is, in the end, the sphere in the diagram pasted above try to symbolize beyond a single circle embedded on its surface?

Or, alternatively, what is the meaning of each entry (beyond $n=2$) in $e^{-im_n\theta}$:

$$R(U(1)) =\begin{bmatrix}e^{im_1\theta} & 0 & 0 & 0 \\ 0 & e^{im_2\theta} & 0 & 0 \\0&0&\ddots &0\\ 0 & 0 & 0 & e^{im_n\theta} \end{bmatrix}$$

Can I get an example of such a representation with $n>2$ to understand the concept?

答:
Some remarks on language

A representation of a group $G$ is indeed what you describe: a linear map $R$ from $G$ to a group $GL(n, \mathbb{C})$ of matrices. Most of the time, the vector space on which these matrices act is also called the representation, and properties of the space, like its dimension, are in everyday language usage treated as if they are properties of the representation. This can be confusing, but you get used to it. A clear example of this latter phenomenon is where you have two representations of some $G$ and make a third representation by taking their direct sum.

You hopefully know what a direct sum of vector spaces is. The direct sum of representations is then the direct sum of the corresponding vector spaces, where the new representation acts on this direct sum by acting on one summand by the first representation, and on the other by the second. The corresponding matrices will be block matrices, much like in your example, where the blocks are all $1$-by-$1$.

There is a habit of referring to the vector space $V$ on which the matrices $R(G)$ act as if that thing is 'the representation' we were talking about, rather than the map $R$ itself - some people are quite cautious to avoid it too openly, but I, personally, do it all the time. Especially in below answer.

Secondly. The philosophy behind the word 'representation' is that you have this very abstract element $g$ in the very abstract group $G,$ and you represent it by the very concrete linear transformation $R(g)$. In the representation you can understand everything. $R(g)$ is just a reflection, or a rotation, or otherwise some other very concrete description of where every point in the vector space goes when we set $R(g)$ loose on it. So it immensely helps us to visualize or understand what $G$ is doing. At the same time we will not go so far as to say that $R(G)$ is $G$, or that $R(g)$ is the element $g$; it merely represents it, here on this concrete vector space. We could have taken a different representation $R'$ where $R'(g)$ would look equally concrete but still quite different.

Your question

Judging from your comment I think your issue is with this last point. Are there really different representations of the same group? More concretely:

if $G$ itself happens to be a matrix group such as $SU(n)$, are there other representations, beyond the 'standard' one where $R$ is just the inclusion of $G$ into $GL(n, \mathbb{C})$ used to define the group?

The answer is yes, but $U(1) \cong SO(2)$ is a bit of an unfortunate example to see this because all the different representations 'look' more similar to each other than you have in the case of non-commutative groups such as $SU(2)$. I will give some examples for $SU(2)$ instead of $U(1)$ for this reason, and come back to the $U(1)$ case later, when I address your second question:

What are representations trying to accomplish?

Here the circle group $U(1)$ are actually the best example to work with. (So win some, lose some)

Boring vs interesting answers

So we first want to know if there are $SU(2)$ reps in dimensions other than $2.$ A way of getting a boring answer is taking the direct sum of $m$ copies of the standard representation. This is $2m$-dimensional, so we're done. But also it tells us nothing new, it is exactly the kind of almost-tautology you were talking about. So we should refine our question, perhaps to something like this:

Are there $SU(2)$ reps in dimension other than $2$ that are not a direct sum of smaller ones?

Such representations are called indecomposable. A much more famous concept is that of being irreducible. Irreducible means that when the matrices $R(G)$ act on a space $V$, there is no proper, non-zero subspace $W \subset V$ such that each $R(g)$ maps vectors from $W$ back into $W$. It follows that if such a $W$ does exist, and $V$ hence is not irreducible, the subspace $W$ is also a representation of $G$ in its own right (a 'subrepresentation'). This does not automatically mean that $V$ is then also not indecomposable - the existence of $W$ does not automatically imply the existence of some other representation $U$ such that $V \cong U \oplus W$. So irreducible implies indecomposable but not necessarily the other way around.

The good news is that when the group is compact as a topological space, then the two concepts irreducible and indecomposable are equivalent, and for every subrepresentation $W$ of some representation $V$ there is a complementary representation $U$ such that $V \cong U \oplus W$. This is good news for us since all $SU(n)$ are compact, as well as the circle group and all finite groups. So for many practical purposes we can treat indecomposeable and irreducible as interchangeable.

(Note that this is quite magical: representations as we defined them seem to be an entirely algebraic concept and suddenly topology swoops in and starts having all sorts of unexpected consequences. It are unexpected links like this why I love this subject.)

So a reformulation of the question could be:

Are there any irreducible $SU(2)$ representations in dimensions other than $2$?

An answer that is somewhat intermediate on the boring-interesting scale is the trivial representation. Here the matrix $R(g)$ is just the identity matrix for every $g$. The representation forgets all the interesting information about the group $G$ but the map $R$ is a homomorphism - so it still counts.

We have a trivial representation in every dimension, but only one of them, the $1$-dimensional one, is irreducible, the reason being that every subspace of a trivial representation is a (trivial) subrepresentation. Of course trivial representations exist for every group.

When we get to the second part (about the purpose of representation theory) we'll see that we cannot simply ignore the one-dimensional irrep, and strictly speaking, it is an answer for our question, but of course it is a very unsatisfactory one. So I'll give some more interesting examples below.

Higher dimensional irreducible representations of $SU(2)$

$SU(2)$ has an irreducible representation in dimension $n$ for each $n$. This sounds quite magical, the $1$ and $2$-dimensional examples being the only ones sounds much more intuitive. I don't really have a conceptual explanation of why this intuition is wrong so instead I'll just give you explicit descriptions of these irreps and leave the (harder) task of checking that the lower-dimensional ones (or any other rep) do indeed not sit 'inside' the higher-dimensional ones.

What do they look like? The $n$-dimensional irreducible representation of $SU(n)$ consists of all homogeneous polynomials of degree $n-1$ in two variables $X$ and $Y$. So concretely the $4$-dimensional rep consists of all linear combinations of the polynomials $X^3, X^2Y, XY^2$ and $Y^3$. The $6$-dimensional representation consists of polynomials of degree $5,$ and so has basis $X^5, X^4Y, X^3Y^2, XY^4, Y^5,$ etc.

So how does $SU(2)$ act on this space? A typical vector in this space is a polynomial, so we can think about it as function. Let's call it $f$. The way we described it, $f$ takes two inputs: $X$ and $Y$ (or two complex numbers that we denote $X$ and $Y$ in our description of the internal workings of the function, if you like). However for our purposes it is better to think of $f$ as taking a single input, the row vector $(X, Y)$. Now elements $g \in SU(2)$ act on the set of row vectors by right multiplication; whatever the values of $X$ and $Y$, the vector $(X, Y)g$ is again a row-vector of length $2$ and hence can be fed into $f$.

This is what we use to make our representation $R$. Recall that $R(g)$ should map vectors to vectors, that is to say: polynomials to polynomials. Now $R(g)f$ is the polynomial that when given the vector $(X, Y)$ as input, gives as output the number $f((X, Y)g)$, so the same output the function $f$ would have given if, instead of $(X, Y)$ it was fed $(X, Y)g$: the result of letting $g$ act on $(X, Y)$. The fact that $f$ was capable of producing this same answer when given a different input should not distract us: the point is that the new polynomial $R(g)f$ gives this answer already when given only the vector $(X, Y)$.

Why is this a representation? Well, for matrices $g_1, g_2 \in SU(2)$ we want to see that $R(g_1)R(g_2)f = R(g_1g_2)f$. To see that this is indeed the case we feed both functions the vector $(X, Y)$.

$R(g_1)R(g_2)f$ is perhaps better written $R(g_1)(R(g_2)f)$. So feeding $(X, Y)$ to this thing is the same as feeding the vector $(X, Y)g_1$ into the function $R(g_2)f$. But we know what we get when we feed any vector into $R(g_2)f$: just the outcome of feeding the function $f$ the vector we get by letting $g_2$ act from the right on our vector. So in this case that means that we get $f((X, Y)g_1g_2)$.

It is not hard to see that this is also the outcome of feeding $(X, Y)$ into $R(g_1g_2)f$.

So $[R(g_1)R(g_2)f](X, Y) = [R(g_1g_2)f](X, Y)$ for all $f$ and all $X$ and $Y$, hence $R(g_1)R(g_2)f = R(g_1g_2)f$ for all $f,$ and hence $R(g_1)R(g_2) = R(g_1g_2)$ as we were hoping.

This shows that $R$ is indeed a representation. That it is irreducible (when restricting to the spaces described above) is a different issue that needs separate checking. I leave that to you.

Now if we accept that all these $n$-dimensional representations are irreducible then the next step, seeing that they are different, is really easy. I mean: just look at them! Every one has a different dimension!

This is why $SU(2)$ is a better example than $U(1)$: $U(1)$ also has infinitely many different irreducible representations, but they are all $1$-dimensional, so you have to think a bit harder about what it means to be the same or to be different before we can use this as an example.

PART II: what is the objective of representation theory?

Of course there are many, but I single out one that is important and relevant to this story.

I wrote above:

For every subrepresentation $W$ of some representation $V$ [of a compact Lie group $G$] there is a complementary representation $U$ such that $V \cong U \oplus W$.

Continuing this line of thinking we conclude:

Every representation $V$ of a compact Lie group $G$ can be decomposed as a direct sum $V = V_1 \oplus V_2 \oplus ...$ of irreducible subrepresentations.

The goal of (a subset of) representation theory is then to find, for each compact $G$, all its irreducible representations and to understand their properties so well that whenever we encounter some representation $V$ of $G$ in some natural setting or application, all we need to do is find the decomposition of $V$ into irreps (as above) and then apply our pre-computed knowledge about these irreps to understand everything about $V$ that we could possible want.

The only remaining question is then the one from your original post:

When and where would we find 'natural' examples of representations 'in the wild'?

I mean, it is nice to have this machinery, but only if you can ever use it.

Physics provides a lot of examples, but I don't really understand them so won't comment on them. A second class of examples come from, as you mentioned, tensor products.

I wrote a separate answer, some time ago, on how to decompose a tensor product of three copies of the 'standard' 3-dimensional representation of $SO(3)$ into irreducibles. One interesting thing is that the one-dimensional, trivial irrep shows up there in the decomposition. That is why I wrote above that you cannot simply ignore it as a pathological case. The answer is here.

However, as I wrote, the best examples come from the case where $G = U(1)$.

The case of $U(1)$

$U(1)$ has infinitely many irreducible representations, indexed by elements of $\mathbb{Z}$. However they are all one-dimensional, so for each $g \in U(1)$ the matrix $R(g)$ is one-by-one and lies in the group $U(1)$ itself. In fact the representation $R$ that is indexed by the number $n$ (positive, zero or negative) acts by $R(g) = g^n$, or equivalently, $R(e^{i\theta}) = e^{in\theta}$.

The case where $n = 0$ is the trivial representation.

Now for the big, 'natural' representation $V$. For this we take the space of all $2\pi$-periodic functions on $\mathbb{R}$. Periodic means that we might as well consider them functions on the circle. The circle group $U(1)$ acts on the circle by rotating it and by a process that is completely analogous to what we did with $SU(2)$ acting on row-vectors of length $2$ we turn the action of $U(1)$ on the circle into a representation $R$ of $U(1)$ on the (huge, infinite dimensional) space of all functions on the circle.

Now as I said this representation is huge, but since $U(1)$ is compact we can decompose it as a direct sum of very well-understood irreducible, and in this case: one-dimensional, representations. As a result we can take a single vector $f$ in the huge space (that is to say: a single periodic function) and decompose it into an infinite sum of vectors (hence: functions) each of which lives in one of the one-dimensional spaces on which the $U(1)$ action is really nice and simple and well-understood.

You almost certainly know this decomposition: it is the Fourier-series of $f$, one of the most useful concepts in all of mathematics!!

So here you see one perspective on representation theory (or harmonic analysis, as this branch of representation theory is sometimes called for exactly this reason):

representation theory is the generalization of Fourier theory to the case where the underlying group is no longer necessarily $U(1)$.

Part III: what about the picture? (edited in later)

Your first comment gets it better than the second. The sphere in the picture is supposed to denote the group $U(n)$ of $n \times n$ matrices into which the map $R$ maps. It is a metaphor/analogy/etc: there is no $n$ for which this thing looks like a sphere. However the sphere in the picture is much smaller and thinner (of lower dimension) than the $3$-dimensional space surrounding it, which in 'reality' correspond to $U(n)$ being some geometric object floating around in the $2n^2$-dimensional space of all complex $n$-by-$n$ matrices while being itself much smaller and of lower dimension.

For instance: the space of all $2$-by-$2$ matrices has $4$ complex dimensions, but (hence) from a geometric perspective $8$ real dimensions. The group $U(2)$ sitting inside has only $4$ real dimensions: as a Lie group it is isomorphic to the group of non-zero quaternions under multiplication, so it looks like a real $4$D space with one point removed. Not exactly sphere-like, but it fits the analogy of the picture in that it is a somewhat weird shape of lower dimension floating around in a very straightforward space of higher dimension.

Now when you think of Lie groups as geometric objects sitting in the bigger space of all $n$-by-$n$ matrices, then the image of a map $R$ from $U(1)$ to such a Lie group is just a circle sitting somewhere inside that geometric object. Different representations of $U(1)$ on the same space (hence different maps $R$ into the same matrix group) would correspond to different circles on this geometric object (depicted as a sphere in the picture but not a sphere in reality).

To get an example of such a circle one should not take the example of all periodic functions: the space of such functions is infinite-dimensional, so the space of all 'matrices' (linear transformations) on that space is also infinite dimensional and then we have the group of unitary ones sitting inside there as some also infinite dimensional 'smaller' object - it is really hard to picture.

Instead take maps $R$ from $U(1) \cong SO(2)$ into $SO(3)$. The latter group is a $3$-dimensional object of somewhat hard to picture shape (a real projective space of dimension $3$) sitting nicely inside the $9$-dimensional space of all real $3$-by-$3$ matrices. Now every element of $SO(3)$, viewed as some transformation of $\mathbb{R}^3$, is a rotation around some axis. Conversely: if we fix an axis, we can look at the set of all rotations around that axis and it is not hard to see that it is isomorphic to the circle group. So any isomorphsim $R$ from the circle group $U(1)$ to this set of rotations can be viewed as representation of $U(1)$ on real three-dimensional space.

Geometrically speaking, different choices of axis correspond to different circles in the 3D-blob $SO(3)$ sitting in $9$-space. All these circles pass through the same point: the identity matrix.

Different or not?

Now these different representations, depicted by different circles, are actually equivalent ('the same') when viewed as representations. Somewhat informally speaking this is what you get if their corresponding circles in the geometric object $SO(3)$ can be shifted onto one another without tearing or stretching.

Now how about an example of non-equivalent (actually different) representations on the same space? Here we return to the one-dimensional irreducible representations of $U(1)$.

One-dimensional representations of $U(1)$ revisited

Each such $R$ is a map from the Lie group $G = U(1)$ to the matrix group $U(1) \subset GL(1, \mathbb{C})$. The latter is the analogue of the sphere in the picture: it is a one-dimensional sphere (circle) sitting inside the bigger $2$-dimensional space (complex plane) of all 1-by-1 matrices.

Now each representation draws a circle on this circle. How can they be different? This is where the 'different speeds' from your comment come in.

The representation that sends $e^{i\theta}$ to $e^{i\theta}$ is just the ordinary way of mapping a circle to itself. The representation that maps $e^{i\theta}$ to $e^{2i\theta}$ wraps the circle twice around the target circle. The representation that maps $e^{i\theta}$ to $e^{-i\theta}$ wraps the circle once around the target circle, but in the opposite direction. Etc.

You can see how they are inequivalent: if you wrap a rubber band twice around a cylinder you have no way to get back to the situation where it is wrapped around only once.

So what about the decomposition?

What is missing in this picture is how to see the decomposition of a (say) two-dimensional representation into two one-dimensional irreducible ones. For this I recommend to think about $U(2)$ not as a sphere but as a torus. Equally inaccurate, but much more helpful.

A $2$D-representation of $U(1)$ then corresponds with a circle drawn on this torus, that perhaps winds around and trough the hole in some complicated way. The nice decomposition then corresponds to saying: "Wait, if I only look in the 'through the hole'-direction it just makes one simple circle and if I look in the 'around the hole' direction it makes two circles before returning home, so this 'complicated' representation is just the direct sum of the one that sends $e^{i\theta}$ to $e^{i\theta}$ and the one that sends $e^{i\theta}$ to $e^{2i\theta}$!"

In the geometric picture, the subrepresentations correspond to different 'directions' and the decomposition corresponds to understanding the circle by projecting down to those directions.


问:
Decide for which real numbers a, b, c the vectors are linearly independent
$$
    v_1=\begin{bmatrix}
    a \\
    a^2 \\
    a^3 \\
    \end{bmatrix}\
$$

$$
    v_2=\begin{bmatrix}
    b \\
    b^2 \\
    b^3 \\
    \end{bmatrix}\
$$

$$
    v_3=\begin{bmatrix}
    c \\
    c^2 \\
    c^3 \\
    \end{bmatrix}\
$$

When I try to create a matrix and try to solve for $a$, $b$ and $c$ I just simply get that $a=b=c=0$, however this would not make the vectors linearly independent.

答:
As you attempted, you would first want to juxtapose column vectors $v_1, v_2, v_3$ to an order $3$ matrix
\begin{align*}
A = \begin{pmatrix}
a & b & c \\
a^2 & b^2 & c^2 \\
a^3 & b^3 & c^3
\end{pmatrix}.
\end{align*}

Then note that $v_1, v_2, v_3$ are linearly independent if and only if $\det(A) \neq 0$. To find $\det(A)$, using the Vandermonde determinant:
\begin{align*}
\det(A) = abc\begin{vmatrix}
1 & 1 & 1 \\
a & b & c \\
a^2 & b^2 & c^2
\end{vmatrix} = abc(b - a)(c - a)(c - b).
\end{align*}

Therefore $\det(A) \neq 0$ if and only if $abc \neq 0$ and $a, b, c$ are all distinct.


问:
Expected conditional covariance identity
I am trying to understand the identity in 2.6 of the following article, in which we have random variables $X \in \mathbb{R}, Y\in \mathbb{R}$ and $Z \in \mathbb{R}^{p-1}$. They define the conditional mean function
$$
\mu^* := \mu^*(X,Z) = \mathbb{E}[Y|X,Z]
$$
and some estimate of $\mu^*$, denoted $\mu := \mu(X,Z)$. They then claim that
\begin{align*}
            \mathbb{E} [\text{Cov}(\mu^*, \mu|Z)] = \mathbb{E}[Y(\mu - \mathbb{E}[\mu|Z])]
        \end{align*}
which I would like to prove. Using the law of iterated covariances, we have for random variables $A,B,C$
\begin{align*}
      \text{Cov}(A,B) = \mathbb{E}(\text{Cov}(A,B|C)) - \text{Cov}(\mathbb{E}(A|C), \mathbb{E}(B|C)),
            \end{align*}
and so we can rewrite
\begin{align*}
\mathbb{E} [\text{Cov}(\mu^*, \mu|Z)]
&= \text{Cov}(\mu, \mu^*) + \text{Cov}((\mathbb{E}(\mu^*|Z), \mathbb{E}(\mu|Z)))
\end{align*}

Then, considering the terms separately. We have
\begin{align*}
\text{Cov}(\mu, \mu^*)
&= \mathbb{E} ( \mathbb{E}[Y|X,Z] \mu) - \mathbb{E}[\mathbb{E}[Y|X,Z]] \mathbb{E}[\mu]\\
&=\mathbb{E}[Y \mu] - \mathbb{E}[Y]\mathbb{E}[\mu]
\end{align*}
For the second term, we have
\begin{align*}
\text{Cov}((\mathbb{E}(\mu^*|Z), \mathbb{E}(\mu|Z)))
&=\mathbb{E}( \mathbb{E}[\mu^*|Z] \mathbb{E}[\mu|Z]) -
\mathbb{E}[\mathbb{E}[\mu^*|Z]]\mathbb{E}[\mathbb{E}[\mu|Z]]\\
&=\mathbb{E}( \mathbb{E}[Y|Z] \mathbb{E}[\mu|Z]) -
\mathbb{E}[\mathbb{E}[Y|Z]]\mathbb{E}[\mathbb{E}[\mu|Z]]\\
&=\mathbb{E}( Y \mathbb{E}[\mu|Z]) -
Y\mathbb{E}[\mathbb{E}[\mu|Z]]\\
&=\mathbb{E}( Y \mathbb{E}[\mu|Z]) -
Y\mathbb{E}[\mu]\\
\end{align*}

So putting these together yields
\begin{align*}
\mathbb{E}[Y \mu] - \mathbb{E}[Y]\mathbb{E}[\mu] +\
\mathbb{E}( Y \mathbb{E}[\mu|Z]) -
Y\mathbb{E}[\mu]
\end{align*}
which doesn't give the right answer. Can someone point out where I've gone wrong, or suggest an easier proof?

答:
Your error occurs in simplification of the expression
$$\mathbb{E}( \mathbb{E}[Y|Z] \mathbb{E}[\mu|Z]) -
\mathbb{E}[\mathbb{E}[Y|Z]]\mathbb{E}[\mathbb{E}[\mu|Z]].$$
The first factor in the rightmost term simplifies to $\mathbb{E}[Y]$, not $Y$. Fixing this should yield the claimed result.

A quicker proof applies the definition of covariance, but conditioned on $Z$:
$$
\operatorname{Cov}(\mu,\mu^*\mid Z)=\mathbb{E}(\mu^*\mu\mid Z) - \mathbb{E}(\mu^*\mid Z)\mathbb{E}(\mu\mid Z).$$
Taking expectation of both sides yields
$$\mathbb{E}[\operatorname{Cov}(\mu,\mu^*\mid Z)]=\mathbb{E}[Y\mu]-\mathbb{E}[Y\mathbb{E}[\mu\mid Z)],$$
using the same arguments as you've made, but avoiding having to add and subtract $\mathbb{E}[Y]\mathbb{E}[\mu]$.


答:
If you have an $f\in 2^A$, then $f$ is a function from $A$ to $2=\{0,1\}$. So you need to associate $f$ with a subset $F\subseteq A$. As the comment above said, take the inverse image of 1 under $f$, that is, $F=\{a\in A : f(a)=1\}$.
So, you have a function from $2^A$ to $\mathcal P (A)$.

Title: Can someone find a (two-sided) inverse function to the characteristic function χ? given...

Body:
so i have given the folowing info:

In general, if $A$ is a set, for $S \in \mathcal P(A)$ define $\chi(S) : A \to \{0,1\}$ by $$\chi(S)(a) = \begin{cases} 1 & \textrm{if } a \in S, \\ 0 & \textrm{if } a \notin S. \end{cases}$$
Then $\chi : \mathcal P(A) \to 2^A$ is bijective.

But I would like to know the what the inverse is to $\chi$. I'm quite new with the definition "characteristic function", so I don't even now how to start. I normally calculate the inverse by replacing x with y and vice versa, but know I don't have a clue. Please help me.

Thank you in advance


问:
Interpretation of a certain integral
What does $\int d(\sin x)$ actually mean?I mean to say when we have $\int f(x)dx$,we interpret it as the values $f(x)$ s are being multiplied by tiny $dx$ and are summed together.But how do we interpret the above from this point of view.Because there is $d(\sin x)$,not $dx$ and we take $\sin x$ in the Y axis.But in integration we take dx to be in the X axis.

答:
@tommik's answer is correct in that we do use this notation as shorthand in, for example, change of variables. However, I think you are looking for an intuition behind what the notation means. The intuition of taking values $f(x)$ and multiplying them by small $\Delta x$ and then summing them up is the Riemann Integral. When we start to have integrals like $\int d(f(x))$ we are moving into the realm of the Riemann-Stieltjes Integral.

The way that I like to think about this (which offers a reasonable generalization to integrating with respect to functions that are not continuous or random variables, as well as to measures) is that we are weighting our divisions not by their length, but by some arbitrary function. So in the example of $\int d(\sin x)$, we are integrating the (constant) function $1$, but instead of weighting the subdivisions of the domain by the length, we are weighting them by the sines of the endpoints (specifically the small change in sine on that interval that @ShubhamJohri mentions in a comment). When our integrator is differentiable, this is equivalent to integrating the derivative with respect to $x$, which helps justify the notational choices.

Title: Interpretation of a certain integral


答:
The goal of Guess Who is to guess faster than your opponent, and not just to minimize the expected number of guesses. This leads to some situations in which you need to gamble on an unequal split.

For example, suppose Alice and Bob play an $n=4$ game with Alice going first. Alice goes for a 2-2 split, guaranteeing that she eliminates two possibilities. So her next question will definitely be enough to determine Bob's person. If Bob then also goes for a 2-2 split, he'll remain a turn behind Alice, and so has no hope of winning. But if he goes for a 3-1 split, he has a $\frac{1}{4}$ probability of getting ahead of her.

For larger numbers, presumably the second player needs to go for risky splits until one of them pays off, at which point the first player needs to start taking risks.

Title: How do I show the optimal strategy for the game 'guess who' is to pick subsets close to half the size of the current set

Body:
Consider the game guess who. In the game, the chooser chooses a person out of a set of $n$ people. The guesser asks questions such as, 'does the person wear glasses'. By asking such questions the number of people is reduced until the guesser can guess exactly who the person is. The guesser wants to minimize the number of guesses. Intuitively, assuming an equal probability that any given person is chosen, they should ask questions that if true, cut the number of people in half, or as close to half as possible. More generally, I suspect if there is some distribution over the people, the best strategy is to always pick one of the subsets of the current remaining set of people that is as close to having a $50\%$ chance of containing the chosen person.

First, I would like to know if this is correct, and second, I would like to prove this. Currently I am trying to prove this using an argument via entropy however I have been unsuccessful so far.


问:
Antilog of Shannon entropy (exponential entropy)
In the source article at the bottom, the antilog of entropy is introduced as the exponential of Shannon entropy:

$$ H(N) = -\sum_N w_i \ln w_i $$

$$ e^{H(N)} = \prod_N w_i ^{w_i} $$

Could someone show the steps in between $e^{H(N)}$ and how to get to $\prod_N w_i ^{w_i} $?

and can the last measure also be applied to point-wise probabilities of a random variable like how Shannon entropy normally is used for, i.e. $p(x)$ instead of $w_i$ so that, for random variable $X$,

$$e^{H(X)} = \prod_N p(x) ^{p(x)}?$$

Straathof, S. M. (2007). Shannon's entropy as an index of product variety. Economics Letters, 94(2), 297-303.

答: By the laws of indices/logarithms, $$\exp H(N)=  \exp\Big(-\sum_{N} w_i\log w_i\Big) = \prod_{N} \exp(-w_i\log w_i) = \prod_{N} \exp(\log(w_i^{-w_i})) = \prod_N {w_i}^{-w_i},$$ since $\exp$ and $\log$ are inverses of each other.


问:
What does the Y-axis represent in probability density function?
When representing the probability density function, what does the Y-axis represent intuitively? Like, what do the values in the Y-axis mean? Do they have a cardinal meaning or they just represent an ordinal one as in the actual value doesn't matter as much as the value being bigger or smaller than the other values</p>

答: The values that the density function takes on are in comparison to the uniform distribution on the space. If the density is larger than 1, you are more likely to find the random variable in that small region than if you had the uniform law. Similarly, if your density is less than 1 at a point, you are less likely to find the random variable in that small region around the point.</p>


问:
Some inequality with complex variables and a concavity of a complex function.
I am doing some project. I have to calculate the estimates of an operator. But I was stuck on a part. I need to show the following form of inequality to derive a conclusion what I want to show.

$|\sqrt{z_{1}} - \sqrt{z_{2}}| < C|\sqrt{z_{1} - z_{2}}|$ for a constant $C \in \mathbb{R}$ and $z_{1}, z_{2} \in \mathbb{C}$

Does anyone give me some comments related to this inequality? $C$ might depend on the values of $z_{1}$ and $z_{2}$.

Here, $z_{1}$ and $z_{2}$ are values of complex functions and so I am not sure how I can deal with this property. I know, for the real-valued, it is trivial because $f(x) = \sqrt{x}$ is a concave function. But, how can I define a concavity of a complex function?

I scanned Functions of One Complex Variables 1 by J. B. Conway, but I couldn't find any related theorems to this case.

Thanks.

答:
I purpose a strategy of finding where the inequality has equality and then just testing a point to see if the inequality holds in a particular connected region.

For example,

If $\sqrt{z}-\sqrt{w}=\sqrt{z-w}$ then $z-2\sqrt{zw}+w=z-w$ and so $w^2=wz$ thus $w=z$ or $w=0$. Given a region $\Omega\subseteq\mathbb{C}^2$ such that $w\not=z$ and $w\not =0$ for all $w,z\in \Omega$, if $\Omega$ is connected then the inequality will hold entirely in $\Omega$ or hold nowhere in $\Omega$.


问:
Defining Constraints for the CSP of Skyscraper puzzle
The Skyscraper puzzle has the goal of positioning Skyscrapers of $n$ different heights on an $n \times n$-Field so that the following requirements are met:

Every field contains a skyscraper.

All skyscrapers in a row/column have different heights. The field may have partially pre-set skyscrapers.

Numbers on the side of the field set the additional requirement that the row/column that the number belongs to have exactly that amount of skyscrapers visible from that position (See picture for better visualization). A skyscraper is considered visible from a position if all the skyscrapers inbetween the position and that skyscraper are smaller than it. If there is no number, then there is no such requirement for the row/column.

I have to formulate this into a CSP now. So far I have the following:

Variables: $X = \{x_{ij}\} \text{ with } i,j = \{1,...,n\}$

Domains: $D = \{D_{ij} = \{1,...,n\}\} \text{ with } i,j = \{1,...,n\}$

Now I am struggling with the constraints. Demanding all skyscrapers in a row/column to be different seems easy.

$\forall x_{ij}: x_{ij} \neq x_{i,l} (l \neq j) \land x_{ij} \neq x_{k,j}(k \neq i)$

I am struggling on the last two constraints though. Some of the fields may already have a pre-set skyscraper in them and I do not know how I can express that mathematically. Furthermore, the numbers which add further constraints to the rows and columns seem to be very hard to formulate elegantly. My only approach is something I haven't even managed to figure out how to write down exactly and it would entail constructing a set of sets which all contain fields. Those sets of fields represent a row/column. Then that set of sets is what represents the collection of rows/columns of which our field is made. Then I'd have to somehow double this list, associating each of these rows/columns with a direction in which it is read and then associate all of these elements with all possible constraint values. And then, a subset containing at most one of these elements per row/column + direction is what forms our constraints.

How can I formulate these two constraints in a more elegant form? Especially the second one, how can I represent that some rows may have such a constraint? How can I represent mathematically how many skyscrapers are visible in a row/column without having to chain many logical operations together?

Thanks for reading!

答:
For each row or column, you can use an alldiff constraint:
\begin{align}
\operatorname{alldiff}(x_{i,1},\dots,x_{i,n}) &&\text{for $i\in\{1,\dots,n\}$} \\
\operatorname{alldiff}(x_{1,j},\dots,x_{n,j}) &&\text{for $j\in\{1,\dots,n\}$} \\
\end{align}

For fixed values $x_{i,j}=c_{i,j}$, you can limit the domain: $D_{i,j}=\{c_{i,j}\}$

The other constraints can be modeled via reification.  Let binary decision variable $\ell_{i,j}$ indicate whether $x_{i,j}$ is taller than all cells to its left and impose
\begin{align}
\ell_{i,1} = 1 &&\\
\operatorname{reify}\left(\ell_{i,j}, \bigwedge_{k=1}^{j-1} (x_{i,j} > x_{i,k})\right) &&\text{for $j > 1$}\\
\end{align}
The three "left" clues are then
\begin{align}
\sum_{j=1}^n \ell_{1,j} &= 3\\
\sum_{j=1}^n \ell_{2,j} &= 3\\
\sum_{j=1}^n \ell_{3,j} &= 4\\
\end{align}
Similarly, introduce binary variables $r_{i,j}$, $t_{i,j}$, and $b_{i,j}$, for right, top, and bottom, respectively, along with the associated constraints.


问:
Quantified formula values and "replacement notation"
Inside my notes I have the following:

I do not see why the notation v(u/d) is necessary; why can we not simply say, for the ∀ case, that $∀xA(x)^v = 1$ if $A(d)^v = 1$ for every d ∈ D? What makes the whole notion of replacing u's with individuals d's necessary, when u's are arbitrary variables anyways and we could just specify the "for all individuals" to begin with?

答:
For the purposes of meta-logic you really want to be clear on the distinction between the language we use to describe some possible domain, and any domain itself.

Here, the $d$ is a mathematical symbol we use to point to an object from some domain. The $u$ is a symbol of our logic language... it only denotes $d$ if we define a valuation that says so ... though other valuations could have $u$ to point to other objects in the domain.

An expression like $A(d)$ is really inappropriate: it mixes a mathematical symbol into a string of symbol from our language of logic.

One way in which this distinction can become pretty important is that we typically assume that our language uses finite strings of symbols, meaning that there is a countable number of variables and constants. However, a domain could be uncountable, such as the real numbers.


问:
Find $\lim_{x \to 4}\frac{x^2-5x+4}{\sqrt{x}-2}$ without L'Hôpital's rule
I am learning limits now and I have a problem:

Find $\lim_{x \to 4} \frac{x^2-5x+4}{\sqrt{x}-2}$

I have tried to rationalize the denominator and I have got something like this:

$\lim_{x \to 4} \frac{(x-4)(x-1)(\sqrt{x}+2)}{x-4}$

I don't know how to continue from here, can you give me a tip or show me a solution to this?

答:
Tip:

Check your factorization of $x^2-5x+4$.

Another tip now that you corrected the factorization:

cancel $x-4$ from numerator and denominator.


问:
I need a function with the following behavior
Probably a simple question, but my math is a bit rusty...
I need a function which looks similar to logarithmic (fast rise at the beginning, then slows down).
$f(0) = 0$,
$f(b) = a$ ($a = 100$, $b = 2500$ but should be changeable, in between $500$-$5000$)

I don't need a horizontal asymptote because I'll cut off the values when $x$ reaches $b$.

Simply, I need it for my site for calculating a score as a percentage based on number of voters, which rises in a logarithmic fashion until $x$ reaches certain number ($500$-$5000$) and then it bumps the $100$ percent value. (thus giving new users instant gratification, and preventing score spamming for those with more votes).

I played with logarithmic functions but it's hard for me to get something which can be easily parametrized to easily change the parameter $b$ while keeping the flow.

Any ideas?

答: Just in case someone wonders, I used a work-around -- tables of a couple of key values, while interpolating points in between. But thanks for your answers, actually they almost did the trick but tables are way more fitting for my purpose which is more of a numerical nature :)</p>


问:
Factorised groups and actions
Let G be a group that can be written as a product of two of its subgroups, that is
$$G = MN = \{mn~~|~~m\in M, n\in N\}.$$
Now let $k$ be a field, let $k(N)$ be the dual space of $N$ (here, the dual space means the collection of functions from $N$ to $k$) and $kM$ the group algebra of $M$.

We have that $M$ has an action on $N$ and $N$ also has an action on $M$. These actions induce an action of $k(N)$ on $kM$ and an action of $kM$ on $k(N)$.

I want to know how these actions are possible.

答:
I am not sure what $G$ has to do with this, but let's say $M$ and $N$ are two groups which each have a group action on the other (with no assumption about how $M\curvearrowright N$ respects $N$'s group operation, or vice-versa).

I will use $k^N$ to refer to the algebra of functions $N\to k$ (where $k$ is a field) with pointwise addition and pointwise multiplication of functions as its operations.

Since $M$ acts on $N$, we can say $M$ acts on $k^N$ by algebra automorphisms. Given $f:N\to k$ and any element $m\in M$, the function $m\cdot f$ is defined by $(m\cdot f)(n)=f(m^{-1}n)$ for all $n\in N$. You could call this a sort of "contragredient" action I guess. You can check it respects the pointwise operations on $k^N$.

Given a linear action of a group $M$ on a vector space $V$, that automatically means there is a representation of $kM$ on $V$, by linear extension. That is, $(\sum c_m m)v:=\sum c_m(mv)$. For $V=k^N$ that means

$$ \big((\sum_{m\in M} c_m m)f\big)(n):=\sum_{m\in M} c_m f(m^{-1}n) $$

for all $n\in N$ and $f:N\to k$.

Title: Factorised groups and actions


答:
Let $$ u_n=\sum_{0\leqslant 2k\leqslant n}\binom{n}{2k}\frac{1}{4^{2k}} $$
and $$ v_n=\sum_{0\leqslant 2k+1\leqslant n}\binom{n}{2k+1}\frac{1}{4^{2k+1}} $$
then
$$ u_n+v_n=\sum_{k=0}^n\binom{n}{k}\frac{1}{4^k}=\left(1+\frac{1}{4}\right)^n=\left(\frac{5}{4}\right)^n $$
because adding the odd and the even terms of the sum gives the whole sum, and
$$ u_n-v_n=\sum_{k=0}^n (-1)^k\binom{n}{k}\frac{1}{4^k}=\left(1-\frac{1}{4}\right)^n=\left(\frac{3}{4}\right)^n $$
by the same argument, noticing that $(-1)^k=1$ if $k$ is even and $(-1)^k=-1$ if $k$ is odd. Thus $u_n=\frac{(u_n+v_n)+(u_n-v_n)}{2}=\frac{5^n+3^n}{2^{2n+1}}$.

Title: About a sum of binomial coefficients

Body:
I need to calculate the following sum: $\sum_{k=0}^{[\frac{n}{2}]}$ $n\choose 2k$ $\frac{1}{4^{2k}}$.
I guess it has to do something with identity $\sum_{k=0}^{n}$ $n \choose 2k $ $=2^{n-1}$, but I am not sure should I use some substitution for counter of this sum or I am on wrong way?


答:
One thing that comes to my mind is to let $x = y^2$ (so if $y \to 2$ then $x \to 4$). Then

$$\frac{x^2-5x+4}{\sqrt{x}-2} = \frac{y^4 - 5y^2 + 4}{y - 2}.$$

Now you can use long division or synthetic division to get

$$ \frac{y^4 - 5y^2 + 4}{y - 2} = y^3 + 2y^2 - y - 2, \quad (y\neq 2). $$

Title: Find $\lim_{x \to 4}\frac{x^2-5x+4}{\sqrt{x}-2}$ without L'Hôpital's rule

Body:
I am learning limits now and I have a problem:

Find $\lim_{x \to 4} \frac{x^2-5x+4}{\sqrt{x}-2}$

I have tried to rationalize the denominator and I have got something like this:

$\lim_{x \to 4} \frac{(x-4)(x-1)(\sqrt{x}+2)}{x-4}$

I don't know how to continue from here, can you give me a tip or show me a solution to this?


Body:
$$\lim_{x \to 4} \frac{x^2-5x+4}{\sqrt{x}-2}=\lim_{x \to 4}\frac{(x-4)(x-1)}{\sqrt{x}-2}=\lim_{x \to 4}\frac{(\sqrt x-2)(\sqrt x+2) (x-1)}{\sqrt{x}-2}$$

$$=\lim_{x \to 4}(\sqrt x+2) (x-1)=4\times 3=12$$


问:
How to solve a system of equations Cov(X,Z) + Cov(Y,Z) + Cov(X,Y) = -Var(Z)/2?
Please Don't vote down here, I can't delete the first post about my initial questions since there are errors in the equations below.

I have 2 covariance matrices known $X$ and $Y$ . I am looking for a way to find a combination of
vectors random variables $Z$ (not null) that could verify :

$$2\,\text{Cov}(X,Z) + 2\,\text{Cov}(Y,Z) + 2\,\text{Cov}(X,Y) = -\text{Var}(Z)$$

If I can manage to find these combinations of vectors, I could write :

$$2\,\text{Cov}(X+Y,Z) + 2\,\text{Cov}(X,Y) = - \text{Var}(Z)$$ and so :

$$\text{Var}(X+Y+Z) = \text{Var}(X) + \text{Var}(Y) + 2\,\text{Cov}(X,Y) + \text{Var}(Z) + 2\,\text{Cov}(X+Y, Z) = \text{Var}(X) + \text{Var}(Y)$$

Any help/track/suggestion is welcome.

EDIT 1:  Sorry, I think that I have made an error in my initial post,  A further checking if the question is well formulated is welcome.

I am going to inspire me from the first answer to solve this problem but I would be glad to get also any help.

答:
If $Z = a X + b Y$, $$\eqalign{\text{Cov}(X+Y,Z) &= a \text{Var}(X) + (a+b) \text{Cov}(X,Y) + b \text{Var}(Y) \cr &= a (\text{Var}(X) + \text{Cov}(X,Y)) + b (\text{Cov}(X,Y) + \text{Var}(Y))}$$
which you want to be $0$.

EDIT: For your new equation, if $Z = a X + b Y + W$ (where $W$ is independent of $X$ and $Y$, you want
$$ a^2 \text{Var}(X) + 2 a b \text{Cov}(X,Y) + b^2 \text{Var}(Y) + 2( \text{Cov}(X,Y) + \text{Var}(X)) a + 2 (\text{Cov(X,Y)} + \text{Var}(Y)) b + 2 \text{Cov}(X,Y) + \text{Var}(W)= 0$$
Call the left side $f(a,b) + \text{Var}(W)$.
It turns out that $f(a,b)$ is minimized at $a=-1$, $b=-1$, with $f(-1,-1) =  - \text{Var}(X) - \text{Var}(Y)$.

So one solution is $a=-1, b=-1$, $\text{Var}(W) = \text{Var}(X) + \text{Var}(Y)$.  Or you can change $a$ and $b$ as long as $f(a,b) \le 0$ and take $\text{Var}(W) = -f(a,b)$.


问:
Which functions can be divided by a polynomial to give a polynomial remainder?
For which functions $f\colon\mathbb{R} \to \mathbb{R}$, given a real polynomial $p(x)$, can we find a real polynomial $h(x)$ such that $deg(h(x))<deg(p(x))$ and:

$f(x)=q(x)p(x)+h(x)$ for some function $q\colon\mathbb{R} \to \mathbb{R}$?

I know that this is true when $f$ is a polynomial function, but does this hold for all functions on $\mathbb{R}$? Additionally given $f$ and $p(x)$ are $q$ and $h(x)$ unique?

答: If you also require that $q(x)$ be polynomial, then $f(x)$ must be.  The polynomials are closed under multiplication and addition.  The process of polynomial division will find $q,h$.  If you do not require $q(x)$ to be polynomial, you can use any $h(x)$ you want and as long as $p(x)$ is not zero you can solve for $q(x)$


问:
How to recognize if a Möbius transformation includes an 'inversion' operation?
For my current project, I am looking into Möbius transformations from the unit disk onto itself. Such Möbius transformations can be found, for example, by specifying three points and their images on the unit circle. Assume that I have the four parameters $a$, $b$, $c$ and $d$ of a given Möbius transformation:

$$M(z) = \frac{az+b}{cz+d}$$

Sometimes, the Möbius transformation causes an inversion, i.e. a mapping from the inside of the unit disk to the outside of the unit disk and vice versa.

Now my question: Is there a way to recognize whether a Möbius transformation induces an inversion directly through the parameters $a$, $b$, $c$, and $d$?

答:
If the Möbius transformation maps $\frac12+\frac12i$ outside the unit disc, then it indicates that the transformation is an inversion. Now
$$|M(\tfrac12+\tfrac12i)|^2 = \frac{(a+b)^2 + b^2}{(c+d)^2+d^2},$$
and when this is $>1$, you have that $\frac12+\frac12i$ is mapped outside the disc. Thus you could take your condition to be that
$$(a+b)^2+b^2>(c+d)^2+d^2.$$

Edit: As suggested in runaway44's comment, we can obtain an easier condition by seeing where $0$ is mapped. Indeed,
$$|M(0)| = b/d,$$
so we just need $|b| > |d|$.


问:
Minimum average lifespan using an exponential law distribution?
It is assumed that the life of a certain equipment follows an exponential law. The manufacturer wants to ensure that this equipment has a probability of less than $0.001$ of failing over 1 year. What is the minimum average lifespan in years of this electronic equipment?

I am not sure at all how to do this. I thought finding the minimum $\lambda$ in using the inequality $P(X < 1) < 0.001$, but it seems not right. Can you help me with that?

In using the distribution function, I got $λ=0.00100005$ years

答:
Your confusion stems from the relationship between the parameter of the distribution (specifically, the choice of parametrization) and the expected value of the random variable as a function of that parameter.

There are two common ways to parametrize an exponential distribution.  One is by rate, and the parameter is typically (but not always) denoted by $\lambda$:  $$\Pr[X \le x] = 1 - e^{-\lambda x}, \quad x, \lambda > 0.$$  With this parametrization, the expectation of $X$ is $$\operatorname{E}[X] = \frac{1}{\lambda}.$$  We call this parametrization by "rate" because in exponential lifetime models for which this distribution applies, $\lambda$ can be conceptualized as a rate at which failure (or more broadly, the event of interest) occurs; thus with a higher rate, the random lifetime $X$ tends to be shorter.

The second parametrization is by scale, and the parameter is often (but not always) denoted by $\theta$ or $\mu$:  $$\Pr[X \le x] = 1 - e^{x/\theta}, \quad x, \theta > 0.$$  The expectation of $X$ is unsurprisingly $$\operatorname{E}[X] = \theta,$$ as the obvious relationship between the two parametrizations is $\theta \lambda = 1$.  This is called a scale parametrization because the value of $\theta$ might be regarded as a scale factor of the random variable $X$.  If we plot the CDF for different choices of $\theta$, we find for instance that doubling $\theta$ will "stretch" the CDF horizontally by a factor of $2$.

All that said, it is straightforward to solve the criterion $$\Pr[X \le 1] \le 0.001$$ for either type of parameter; for instance, $$0.001 \le \Pr[X \le 1] = 1 - e^{-\lambda}$$ implies $$\lambda \le \log \frac{1000}{999} \approx 0.0010005,$$ and since $\theta \lambda = 1$, we also get $$\theta \ge \left(\log \frac{1000}{999}\right)^{-1} \approx 999.4999166,$$ which is the expectation.

A final note.  This represents the smallest expected (average) lifetime because while one could choose to have a more strict quality control  by choosing a $1$-year failure probability that is less than $0.001$, which would then increase the expected lifetime, and still satisfy the criterion.  But one cannot choose a larger failure probability.


问:
Show that $M$ is isometric to a subset of $\ell_{\infty}(D)$.
I just finished this problem from N.L. Corothers' real analysis book and believe I have the right idea, but finished the proof with the wrong conclusion. Let $(M,d)$ be a metric space.

Let $D$ be dense in $M$. Show that $M$ is isometric to a subset of $\ell_{\infty}(D)$.

The hint is to use the following theorem: Let $D$ be dense in $M$, let $N$ be complete, and let $f: D \to N$ be uniformly continuous. Then, $f$ extends uniquely to a uniformly continuous map $F: M \to N$, defined on all of $M$. Moreover, if $f$ is an isometry, then so is the extension $F$.

let $f: D \to \ell_{\infty}(D)$ be given by $f(x) := (x,x,x,...)$ a constant sequence of $x$. Now since $f$ is uniformly continuous, $f$ is an isometry, and $\ell_{\infty}(D)$ is complete: by the given theorem $f$ extends uniquely to a uniformly continuous map $F: M \to \ell_{\infty}(D)$. Since $f$ is an isometry, so is $F$. Thus $M$ is isometric to $F(M) \subset \ell_{\infty}(D)$.

Is this correct? Any criticism is welcome.

答:
Choose any point $x_0 \in D$. Define function $f_x(y)$, $x, y \in D$ as $f_x(y) = d(x, y) - d(x_0, y)$. From triangle inequality, $d(x, y) - d(x_0, y) \leq d(x_0, x)$ and thus $f_x$ is bounded, so in $\ell_\infty(D)$.

For $x_1, x_2 \in D$, $$|f_{x_1}(y) - f_{x_2}(y)| =\\ |d(x_1, y) - d(x_0, y) - d(x_2, y) + d(x_0, y)| =\\ |d(x_1, y) - d(x_2, y)| \leq\\ d(x_1, x_2)$$, so $\|f_{x_1} - f_{x_2}\| \leq d(x_1, x_2)$.
And $|f_{x_1}(x_1) - f_{x_2}(x_1)| = d(x_1, x_2)$. Combining, we get $\|f_{x_1} - f_{x_2}\| = d(x_1, x_2)$, so $F: D \to \ell_\infty(D)$, $F(x) = f_x$ is isometry. By theorem from hint, we can extend $F$ to isometry $M \to \ell_\infty(D)$.


问:
Convergence in probability for each $P_k$ implies convergence in probability in $\sum_k \lambda_k P_k$
Suppose we have $X_n \to X$ convergence in $P_k$-probability for every $k$. And we are given a probability measure $R= \sum_{k\ge 1} \lambda_k P_k$, where $\sum \lambda_k = 1, \lambda_k \ge 0$.

Then how do we show that $X_n \to X$ converges in $R$-probability?

I can see that this easily holds if the number of positive $\lambda_k$ is finite, but I cannot come up with an argument in the infinite case. I would greatly appreciate any help.

答:
Fix $\varepsilon, \delta > 0$. One has
$$R(|X_n - X| \geq \varepsilon) = \sum \lambda_k P_k(|X_n - X| \geq \varepsilon) $$
Since $\sum \lambda_k = 1$ and $\lambda_k \geq 0$ there is $K$ such that for every $n$ $$\sum_{k \geq K} \lambda_k P_k(|X_n - X| \geq \varepsilon) \leq \sum_{k \geq K} \lambda_k \leq \frac{\delta}{2}$$
Then since $X_n \to X$ in $P_k$-probability for $k < K$ we can pick an $N$ such that for each $k < K$, $$P_k(|X_n - X| \geq \varepsilon) \leq \frac{\delta}{2K}$$ for $n > N$. Hence for $n > N$, $$R(|X_n - X| \geq \varepsilon) \leq \frac{\delta}{2} + K \frac{\delta}{2K} = \delta$$ as desired.


问:
Second-order ODE with substitution
I’m struggling with this question:

Use the substitution $y(t) = z(t)\,e^{-t}$ to transform the ordinary differential equation
  $$\frac{d^2 y}{dt^2} + 2\,\frac{dy}{dt} + y = t^2 e^{-t}$$
  into an ordinary differential equation for $z(t)$. Hence obtain the general solution for $y(t)$. What is the specific solution which satisfies the initial conditions, $y(0) = 1$ and $y'(0) = 0$?

I have an exam in two hours, and I can’t get around this question.

So far I managed to get the first order and second order differentials for $z(t)$, and found that $z''(t) = t^2$, but I don’t know what to do next.

How should I proceed?

答:
Let $v= \frac{dz}{dt}$. Then $\frac{dv}{dt}=t^2$ and $ v = \frac{t^3}{3} + c.$

Substituting for $v$, $\frac{dz}{dt} = t^3/3 + c$ so $z(t)=t^4/12 + ct + d$.

Title: Second-order ODE with substitution


问:
Prove that $\lim_{x \to \infty} x^2 \neq10^{10}$ using a proof by contradiction style epsilon delta.
Prove that $\lim_{x \to \infty} x^2 \neq10^{10}$ using a proof by contradiction style epsilon delta.

My Attempt:

We can show pretty straightforwardly that $\lim_{x \to \infty} x^2 = \infty$, hence we need to show a contradiction in this function having two limits. By definition

$(1)$ for all $ M_1 > 0$ there exists an $N_1 > 0 $ s.t. $x > N_1 \rightarrow x^2 > M_1$

$(2)$ for all $ \epsilon_2 > 0$ there exists an $N_2 > 0 $ s.t. $x > N_2 \rightarrow |x^2 - 10^{10}| < \epsilon_2$

Let $N = max(N_1, N_2)$. Then for $x > N \rightarrow x^2 > M$. However $M > L + \epsilon_2$ which would make $|x^2 - 10^{10}| > \epsilon_2$. Hence by contridiction, the limit does not hold.

Is this proof correct?

答:
Yes your proof is correct, although you need to take $N=\max(N_1,N_2)$ so that you can use both (1) and (2).

Also I would use $\Rightarrow$ in place of $\to$, since that is usually meant to denote a limit in these kinds of proofs.

I think you can shorten it just by saying that if $\lim x^2=10^{10}$, given $\epsilon > 0$, we should be able to find $N$ such that $|x^2-10^{10}|<\epsilon$ for all $x>N$, but when $x>10^5+\epsilon$, it is clear that this is false.


答:
Maybe to be more clear we could proceed as follows, suppose

$$\lim_{x \to \infty} x^2 =10^{10}$$

that is

$$\forall  \epsilon> 0 \quad \exists N > 0 \quad \forall x > N \quad |x^2 - 10^{10}| < \epsilon$$

but for $x^2 > 10^{10}+\epsilon$ we obtain a contradiction.

Body:
Prove that $\lim_{x \to \infty} x^2 \neq10^{10}$ using a proof by contradiction style epsilon delta.

My Attempt:

We can show pretty straightforwardly that $\lim_{x \to \infty} x^2 = \infty$, hence we need to show a contradiction in this function having two limits. By definition

$(1)$ for all $ M_1 > 0$ there exists an $N_1 > 0 $ s.t. $x > N_1 \rightarrow x^2 > M_1$

$(2)$ for all $ \epsilon_2 > 0$ there exists an $N_2 > 0 $ s.t. $x > N_2 \rightarrow |x^2 - 10^{10}| < \epsilon_2$

Let $N = max(N_1, N_2)$. Then for $x > N \rightarrow x^2 > M$. However $M > L + \epsilon_2$ which would make $|x^2 - 10^{10}| > \epsilon_2$. Hence by contridiction, the limit does not hold.

Is this proof correct?


答:
You can change your representation of convergence of probability to see the hidden pattern.

Please note that the notation of convergence in probability of $X_n \rightarrow X$ ( in $\mathbb{R}$) is equivalent to say that:
$$ \mathbb{E} \left( \min( |X_n-X|, 1) \right)  \xrightarrow{n \rightarrow +\infty} 0$$
Besides, under your new measure, we have:
$$\mathbb{E^{R}} \left( \min( |X_n-X|, 1) \right)  =\sum_{k \ge 0} \lambda_k \mathbb{E}^k \left( \min( |X_n-X|, 1) \right)$$
So you see why.

Title: Convergence in probability for each $P_k$ implies convergence in probability in $\sum_k \lambda_k P_k$

Body:
Suppose we have $X_n \to X$ convergence in $P_k$-probability for every $k$. And we are given a probability measure $R= \sum_{k\ge 1} \lambda_k P_k$, where $\sum \lambda_k = 1, \lambda_k \ge 0$.

Then how do we show that $X_n \to X$ converges in $R$-probability?

I can see that this easily holds if the number of positive $\lambda_k$ is finite, but I cannot come up with an argument in the infinite case. I would greatly appreciate any help.


问:
Additive and Bijective function on the real line
I was trying to solve the following functional equation:

$$f\big(f(x-y)\big)=f(x)-f(y)\text.$$

And I concluded that $f$ must be additive and bijective. The question is:

Let $f:\mathbb{R} \to \mathbb{R}$ be an additive bijective function. Is it true that $f$ has to be of the form $f(x)=cx$ for some $c\in\mathbb{R}\setminus\{0\}$?

What if we impose the condition that $f\big(f(x)\big)=f(x)$. Is the identity map the only solution?

Edit: if $f$ is surjective and $f\big(f(x)\big)=f(x)$, then $f$ is the identity map (almost by definition).

答:
While there are uncountably many wild additive bijections on $ \mathbb R $ (as @MarcoVergamini has indicated in a comment above), the original equation you encountered has only one injective/surjective solution: the identity function. It's straightforward to check that the identity function is a bijective solution. To see that it's the only injective/surjective solution, let $ x = y = 0 $ in
$$ f \big( f ( x - y ) \big) = f ( x ) - f ( y ) \tag { * } \label 0 $$
to get $ f \big( f ( 0 ) \big) = 0 $. Then, put $ x = f ( 0 ) $ and $ y = 0 $ in \eqref{0} to see that $ f ( 0 ) = 0 $. By setting $ y = 0 $ in \eqref{0} you have $ f \big( f ( x ) \big) = f ( x ) $. This means $ f ( x ) = x $ for every $ x $ in the range of $ f $. So, if $ f $ is supposed to be surjective, it follows that $ f $ must be the identity function. Also, if $ f $ is supposed to be injective, $ f \big( f ( x ) \big) = f ( x ) $ directly implies $ f ( x ) = x $ for all $ x $.

But there seems to be a problem with your claim that solutions of \eqref{0} must be bijective. Just similar to the construction of wild additive bijections, you can use a Hamel basis $ \mathcal H $ of $ \mathbb R $ over $ \mathbb Q $ to construct wild solutions of \eqref{0}: choose a nonempty subset $ \mathcal I $ of $ \mathcal H $, map every member of $ \mathcal I $ to itself and every member of $ \mathcal H \setminus \mathcal I $ to an arbitrary member of $ \mathcal I $, and finally extend this mapping linearly (over $ \mathbb Q $) to a function $ f : \mathbb R \to \mathbb R $. It's straightforward to check that this $ f $ will satisfy \eqref{0}. Therefore, unless any further restriction on $ f $ is given, it cannot be concluded solely from \eqref{0} that $ f $ must be injective/surjective.

But additivity of $ f $ is secured by \eqref{0}. $ f \big( f ( x ) \big) = f ( x ) $ can be used to rewrite \eqref{0} as $ f ( x - y ) = f ( x ) - f ( y ) $. Substituting $ x + y $ for $ x $ and rearranging the terms, that equation is transformed to the Cauchy functional equation. So, this part of your claim is definitely true.


问:
Differential equation - new independent variables
Hi I have a question on;

Transform the differential equation:

y∙f’x - x∙f’y =xy∙f

with new independent variables:
u = x^2 + y^2 and v= e^(-(x^2)/2)

I got xy∙(v∙f’v +f)= 0, but the solution is v∙f’v+ f. Why can I ignore xy in front of the parenthese?

答:
If your PDE is

$$
y f_x(x,y)-x f_y(x,y) = x y f(x,y)
$$

making the change of coordinates

$$
\cases{
r= x^2+y^2\\
s = e^{-\frac{x^2}{2}}
}
$$

we obtain

$$
\sqrt{\ln s}\sqrt{r+2\ln s}\left(f(r,s)+sf_s(r,s)\right)=0
$$

so we have the possibilities

$$
\cases{
s=1\\
r+2\ln s = 0\\
f(r,s)+sf_s(r,s)=0
}
$$

but

$$
\cases{
s=1\\
r+2\ln s = 0
}\Rightarrow \cases{x=0\\ y=0}
$$

which represents a particular solution.


答:
There is a problem when you write “However $M>L+\varepsilon_2$”. Why? I had not mentioned $M$ nor $L$ before. Why is $M$ greater than $L+\varepsilon_2$?

Besides, it is not clear to me that you can use the fact that $\lim_{x\to\infty}x^2=\infty$.

You can do it as follows. In order to prove that we don't have $\lim_{x\to\infty}x^2=10^{10}$, I will take $\varepsilon=1$. Take $M>0$. Now, take some $x$ which is both greater than $M$ and greater than $10^6$. Then$$x>10^6\implies x^2>10^{12}\implies|x^2-10^{10}|\geqslant1.$$So, I have proved that there is a number $\varepsilon>0$ such that, for every $M>0$, there is some $x>M$ such that $|x^2-10^{10}|\geqslant\varepsilon$, or$$(\exists\varepsilon>0)(\forall M>0)(\exists x>M):|x^2-10^{10}|\geqslant\varepsilon,$$which the same thing as asserting that we don't have $\lim_{\to\infty}x^2=10^{10}$.

Title: Prove that $\lim_{x \to \infty} x^2 \neq10^{10}$ using a proof by contradiction style epsilon delta.


问:
Is antilog the same as exponential function?
An article has implied that antilog and exponentiation $e$ just mean the same thing. All I know about antilog though is that it means an inverse logarithm. I don't know if it has a mathematical symbol or anything else, and want to check whether antilog does in fact mean $e$, or if some simple examples can be given that might help understand the earlier link, or in what situations an antilog is possible or not, given that the function in the link contains a logarithm but is not entirely a logarithmic function.

答: Yes the natural log of $e^{x}$ is $x$ and $e^{\ln{x}}=x$ so it is reasonable to say that taking the exponential (to base $e$) is the same as the antilog (to base $e$). The inverse of log to base $10$ (for example) is to raise $10$ to the power of $x$.


问:
What is the Point, and Meaning, of the Mean Value of a Function?
Ok, so I know that the mean value of a function, $f(x)$, on the interval $[a,b]$ is given by (or defined by?)
$$\frac{1}{b-a}\int_a^bf(x)~dx$$
but I have $2$ basic questions about this:

$1$: From a purely mathematical point of view, does this have any practical use? Does it give us any extra weapon to add to our mathematical arsenal?

$2$: How excatly can I interpret this as a mean value? I've seen the geometrical interpretation (looking at the area under the graph within the given interval), but I still don't understand how it links to a mean value. For a finite set of values I divide the sum of the values by however many values I have to obtain the mean, but in this case there are infinitely many values, so how is this a mean value?

Thanks for your help.

答:
Let $f:[a,b] \to \Bbb R$ be integrable. Consider the equidistant partition of $[a,b]$ into $n$ subintervals: $$\mathcal{P}_n: \quad a < a + \frac{b-a}{n} < a + 2\frac{(b-a)}{n} < \cdots < a + (n-1)\frac{(b-a)}{n} < b.$$The length of every subinterval is $(b-a)/n$. Then by definition of integral, we have $$\lim_{n \to +\infty} \sum_{k=1}^n f\left(a+k\frac{(b-a)}{n}\right) \frac{b-a}{n} = \int_a^b f(x)\,{\rm d}x,$$ and hence $$\lim_{n\to +\infty} \frac{1}{n} \sum_{k=1}^n f\left(a+k\frac{(b-a)}{n}\right) = \frac{1}{b-a} \int_a^bf(x)\,{\rm d}x.$$The thing inside the limit is the arithmethic mean of the values of $f$ on right-endpoints of the intervals in the partition. And the integral is a limit of means, i.e., a "continuous" mean of $f$ on the entire interval $[a,b]$. Of course, you can do the same with left-endpoints.

Such means are very frequent in Measure Theory, where one replaces the Riemann integral by a Lebesgue integral, and so on. For instance, Lebesgue's differentiation theorem gives the most general statement of what happens when, say, you have intervals of the form $[x_0,x_0+h]$ and want to see what happens with $$\lim_{h \to 0} \frac{1}{h} \int_{x_0}^{x_0+h} f(x)\,{\rm d}x.$$The above is the mean of $f$ on the interval $[x_0,x_0+h]$ and, under suitable assumptions, this limit equals $f(x_0)$, as one might expect.

One tool for proving theorems of this type is the so-called Hardy-Littlewood maximal operator, which is defined in terms of means. And so on and so on.

Bottom line: keep studying analysis and you will see this appear everywhere.


问:
Function of arc every few steps along cone
So I have no way how to properly explain this.
I have no idea how to solve this problem.

Given the following:

In a 3D world, we have a normalized vector $V(x_v, y_v, z_v)$, a starting point $S(x_s, y_s, z_s)$ and a radius r, as illustrated in the following image, and the vector $V$ is on the same plane as the entire orange circle with middle point $S$ and radius r, and given that the dotted line below is always parallel to the "floor", so the two points of our blue triangle on the orange circle have an equal y value.:

Now given that the angle of alpha is 60 degrees, s.t. the blue triangle becomes an equilateral triangle with sides of length r.

What I want, is the following, and I honestly have no clue how:

Given n concentric arcs between the two lines of the blue triangle, equally spaced s.t. $r = n \cdot d$.

Can I somehow get the function of each of those arcs somehow?

I know this question was asked so incredibly vague that it might not even make sense, but I tried my hardest to explain this the best i could.

答:
The radius of each arc is $r_k=\frac kn r$ where $k$ ranges from $1$ to $n$.  The points can be calculated by  taking intervals of $\theta$ from $0$ to $\alpha$ as $x=r_k \cos (\frac {\pi-\alpha}2+\theta),y=r_k \sin (\frac {\pi-\alpha}2+\theta)$, the standard parameterization of a circle.

Added:  the vertex of your angle is assumed to be the origin of the coordinates.  For 3D you can restrict $\theta$ to $[0,\alpha /2]$, which will give you the point in the $xy$ plane with $x \ge 0$.  Now you can revolve it around the $y$ axis.  The $y$ values are all the same.  Given the $x$ we calculated for a given $\theta$ the point is $(x',y,z)$ with $x'=x \cos \phi, z=x \sin \phi$ as $\phi$ ranges from $0$ to $2\pi$  The shape is no longer an arc, which is what led me to the 2D solution.  It is now a spherical cap.


问:
On the primality of integers of the form $p^2+k$
I am not able to find an answer to the following question:

For which positive even integers $k$ is the integer
  $$p^2+k$$ prime, where $p$ is a prime number $\gt5$?

答:
$p^2 \equiv  1(\mod 3)$

$p^2+ k \not \equiv 0(\mod 3) \implies k $ is of form $3n$ or $3n+1$.

$p^2 \equiv  1(\mod 2) \implies k \equiv 0( \mod 2)$ (since it is the stronger condition)

You can get $k$ in $(\mod 6)$,using CRT.

Similarily, you have $p^2 \equiv \pm 1\mod 5 \implies k \equiv 2,3$ or $0(\mod 5)$ . You can use CRT again to get $k \equiv x(\mod 30)$, you get a stronger condition.

Note that $\gcd(3,4)


问:
If $f$ is R-integrable, prove that if $\int_a^b |f(x)|dx = 0$, then $f(x) = 0$ for all $x\in[a,b]$. If this is not true, provide a counter-example.
I've thought of multiple examples, none of which give $\int_a^b |f(x)|dx = 0$. So at this point I am thinking that a proof is in order. This is all that I have so far:

Let $F(x)=\int_{a}^{x}|f(t)|dt$. By assumption, $0=\int_{a}^{b}|f(t)|dt = \int_{a}^{x}|f(t)|dt+\int_{x}^{b}|f(t)|dt$ means, by the fundamental theorem of calculus, that $|F(x)-F(a)|+|F(x)-F(b)|=0 \Longrightarrow |F(x)-F(a)|=-|F(x)-F(b)|$.

This is the point where I start to get confused. Should I investigate cases or is there a simpler way of approaching this?

答: If$$f(x)=\begin{cases}1&\text{ if }x=a\\0&\text{ if }x>a,\end{cases}$$then $f$ is Riemann-integrable and $\int_a^b|f(x)|\,\mathrm dx=0$, but $f$ is not the null function.


问:
Distortion of the Unknot
In Mikhail Gromov's "Filling Riemannian Manifolds" he defines the distortion of a knot $K$ embedded in $S^3$ as $$\delta (K) := \inf_{\gamma \in K} \sup_{x,y \in \gamma} \frac{d_{\gamma}(x,y)}{||x-y||} \geq 1. $$

It is clear to me why this is well defined, since this is independent of the representation of the knot K you choose. The theorem I'm having trouble proving (and couldn't find a proof of) is as follows -

$$\delta(K) = \frac{\pi}{2} $$ if and only if $K$ is the unknot.

I've successfully proven one direction. If $K$ is the unknot then the furthest apart two points can be is $\pi r$ where r represents the radius of the embedding. In the bottom, you get something like $$ \sqrt{2r^2 - 2r\cos(2 \pi(x-y))  }. $$ By inspection, this is minimized for $r =1 $, that is the standard embedding of the unknot. Further elementary reasoning shows that $ 2 - 2(-1) = 4 \Rightarrow \sqrt{4} = 2. $ Hence  $$\delta(K) = \frac{\pi}{2}$$ whenever $K$ is the unknot.

The other direction has me stumped. I know that Sullivan and Denne showed that $$ \delta(K) \geq \frac{5\pi}{3}$$ whenever $K$ is a knotted curve. Moreover, Pardon has some bounds for $p,q$ Torus knots in the form of $$ \delta{K} \geq \frac{1}{160}\min(p,q). $$ However, Gromov was able to prove this fact independent of the more modern results, I was hoping someone could help me prove this fact.

答:
I think I got to the bottom of this - it's a little shaky, but it's what I got so far.

To prove the converse, set $ \delta(K) = \frac{\pi}{2}.$ Without loss of generality, suppose $\gamma([0,1]) = K$ satisfies $$ |\gamma| := \int_{0}^{1} |\gamma '(t)| \mathrm{d} t = 2\pi $$ for all $x,y \in K$ then $$ ||x-y||_{\mathbb{R}^3} \leq 2. $$ Thus $$ f(x) := \frac{x-y}{2} $$ is non-increasing whenever $y \in K$ such that $$\pi = d_{\gamma}(x,y) = \sup_{x' \in K} d_{\gamma}(x,z). $$ Note that such a $y \in K$ exists since $K$ is compact and $$ d_{\gamma} (x,y) := \int_{t_0}^{t_1} |\gamma'(t)| \mathrm{d} t $$ where $\gamma(t_0) = x$ and $\gamma(t_1) = y$ and $\gamma$ is a continuous mapping of a compact set. So, $d_{\gamma}(x, -)$ is a continuous function over the compact set $K$.

Therefore, $$2 = d_{\mathbb{R}^3}(f(x),f(y)) \leq d_{\mathbb{R}^3}(f(x),f(z)) $$ for any $z \in K.$ That is, our map $f$ is distance non-increasing; a basic fact in geometry implies that $f$ is an isometry from $K$ to the unit circle. It goes without saying that the unit circle is an embedding of the unkot.

Title: Distortion of the Unknot


问:
Equivalent definitions of limiting point of function.

Let $(X,d_X)$, $(Y,d_Y)$ be metric spaces, let $E$ be a subset of $X$ and let $f:X\to Y$ be a function. Let $x_0\in X$ be an adherent point of $E$ and $L\in Y$. Prove the following are equivalent:

a. For every sequence $(x^{(n)})_{n=1}^\infty$ in $E$, converging to $x_0$, the sequence $f((x^{(n)}))_{n=1}^\infty$ converges to $L$.

b. For every open set $V\subseteq Y$ which contains $L$, there exists an open set $U\subseteq X$ containing $x_0$ such that $f(U\cap E)\subseteq V$

Assuming a. I want to prove by contradiction,

Soo suppose $L\in V$ which is open in $Y$ and assume there is no open set $U\subseteq X$ containing $x_0$ such that $f(U\cap E)\subseteq V$

Define the sequence $(x_n)_{n=1}^\infty$ by $x_n\in B_{\frac{1}{n}}(x_0)$

By archimedean property for any $\epsilon >0$ there will exist an $n$ large enough that $d_X(x^{(n)},x+0)<\epsilon$, so this sequence converges to $x_0$.

Since $V$ is open and $L\in V$, there exists an open ball $B_r(L)\subseteq V$

By assuming $a.$ however we know that $f((x^{(n)}))_{n=1}^\infty$ converges thus there exists an $m\in \mathbb{N}$ such that if $n\geq m$, $d_Y(f(x^{(n)}),L)<r$, so it is in $B_r(L)$, or in other words, I know there is a ball small enough $B_{\frac{1}{m}}(x_0)$ such that all point of my sequence within this ball have image in $B_r(L)$.

How do I go from saying I know all the points which belong to my sequence which are in $B_{\frac{1}{n}}(x_0)$ have image in $B_r(L)$ to know the entire open ball is, regardless of whether the point is in my sequence or not?

答: You need to choose the points $x_n$ a bit more carefully. You’ve assumed that there is no open nbhd $U$ of $x_0$ such that $f[U\cap E]\subseteq V$, and each $B_{1/n}(x_0)$ is an open nbhd of $x_0$, so for each $n\in\Bbb Z^+$ there must be at least one point $x_n\in B_{1/n}(x_0)\cap E$ such that $f(x_n)\notin V$. Now $\langle x_n:n\in\Bbb Z^+\rangle$ is a sequence converging to $x_0$ with the added property that $f(x_n)\notin V$ for each $n\in\Bbb Z^+$. This clearly implies that the sequence $\langle f(x_n):n\in\Bbb Z^+\rangle$ does not converge to $L$, contradicting the assumption (a).


问:
Wich is the theorem of existence and uniqueness?
I have to enounce and prove the theorem of existence and uniqueness of the problem  $$\begin{cases} u_t +v\cdot\nabla u =f(x,t)-\gamma u, & u(x,t)\in\mathbb{R}^n \times (0,\infty), \\ u(x,0)=g(x), & x\in\mathbb{R}^n,  \end{cases} $$ where $\gamma >0$, and $v\in\mathbb{R}^n$ are constants, $g\in C^1(\mathbb{R}^n)$ and $f\in C(\mathbb{R}^n\times [0,\infty))$ but I don´t know where find the theorem and an idea of the proof. Please help.

答:
You don't have to find the theorem, because it just says that there is one solution to the problem. To prove uniqueness the usual method is to suppose there are two solutions $u_1$ and $u_2$, and prove they are equal. To do this, find the equation satisfied by the function $w = u_1-u_2$, and show that equation has only the solution with $w$ identically zero. The process of doing that will give you clues about how to show existence.

If you need more hint, the $w$ equation is $w_t+v\cdot\nabla w = -\gamma w$, with $w(x,0) = 0$.

If you need more hint, recognize that the left hand side of the $w$ equation is a directional derivative in the direction of the vector $(v,1)$ in the $(x,t)$ space.