问:
Do Girsanov transformations constitute a group?
Just to briefly sum up, I have a process $\phi \in \Lambda^2_{loc}$ , a Brownian motion $B_t$ and the Doleans-Dade exponential defined as usual:

$Z_t:=\exp \left(\int_0^t \phi_s d B_s - \frac{1}{2} \int_0^t \phi^2_s ds \right).$

If we assume that Novikov's condition is fulfilled

$ \mathbb{E} \left[\exp \left(\frac{1}{2} \int_0^T \phi^2_s ds \right) \right]<+\infty $,

then the process

$\tilde{B}_t := B_t - \int_0^t \phi_s ds$

is still a Brownian motion, but under the equivalent probability measure defined via the Radon-Nikodym derivative

$ \frac{d \mathbb{Q}}{d \mathbb{P}}=Z_T$.

My question is: do Girsanov's transformations constitute a group (with respect to the sum) ? That is, given a Girsanov transformation, is its inverse a Girsanov transformation as well? And what about the composition, that is the sum of two Girsanov transformations?

For the moment, I reasoned as follows. The $0$ is clearly represented by $\phi \equiv 0$. Given a transformation, its inverse is

$\tilde{B}_t := B_t + \int_0^t \phi_s ds$.

Obviously, $-\phi \in \Lambda^2_{loc}$ and Novikov's condition is still fulfilled. However, I am not sure about the DD exponential. Can we still define a Girsanov change of measure with

$Z^*_t:=\exp \left(-\int_0^t \phi_s d B_s - \frac{1}{2} \int_0^t \phi^2_s ds \right)?$

Regarding the product, I just considered

$d \tilde{B}_1(t) = d B_1(t) + \phi_1(t) dt $

$d \tilde{B}_2(t) = d \tilde{B}_1(t) + \phi_2(t) dt $,

which leads to

$d \tilde{B}_2(t) = d B_1(t) + (\phi_1(t) + \phi_2(t)) dt$?

How could I proceed? Thanks a lot for any help!!

答:
I might have an answer to this.

As far as I can tell it is sufficient that $\tilde{\mathbb{P}}=\mathcal{E(A)}\mathbb{P}$ is a probability to preform a transformation. This in turn is equivalent to that $\mathcal{E(A)}$ is a martingale.

One can show that $X$ is a martingale under a measure $P$ iif $XM$ is martingale under $Q$ given that $M=\frac{P}{Q}$, in turn implying that $M^{-1}$ is a martingale.

Update

Also Novikov actually says that $\mathcal{E(A)}$ is a martingale given that the $\mathcal{A}$ satisfies the condition which is what you asked right?


问:
Notation difficulty in AM-GM inequality proof
The proof goes as follows:

If $a_1, \dots, a_n$ are positive numbers then:
$$
(a_1 + \dots + a_n)/n \geq \sqrt[n]{a_1 \dots a_n}
$$
The left-hand side we denote by $A_n$ and the right-hand side by $G_n$.

The inequality is trivial when $n = 1$. Assume the inequality is true if $n = k$.

Applying this inequality to $b_1 = a_{k+1}$ and
$b_2 = b_3 = b_k = A_{k+1}$, we get:
$$
\frac{a_{k+1} + (k-1)A_{k+1}}{k} \geq (a_{k+1}A_{k+1}^{k-1})^{1/k}
$$
Now let:
$$
A = \frac{a_{k+1} + (k-1)A_{k+1}}{k} \quad and \quad G = (a_{k+1}A_{k+1}^{k-1})^{1/k}
\tag{1}
$$

Applying the inequality for the case in which n = 2, we get:
$$
A_{k+1} = \frac{A_k + A}{2} \geq (A_k A)^{1/2} \geq (G_k G)^{1/2} = (G_{k+1}^{k+1}A_{k+1}^{k-1})^{1/2k}
\tag{2}
$$
That is $$A_{k+1} \geq (G_{k+1}^{k+1}A_{k+1}^{k-1})^{1/2k}$$ and thus $A_{k+1} \geq G_{k+1}$.

Specifically once we define $A$ and $G$ in (1) I fail to get the meaning of the subscripts for $A$ and $G$ in (2). I mainly tried to solve my difficulty using $G_k$, but every time I cannot unfold $G_k G$ to get $(G_{k+1}^{k+1}A_{k+1}^{k-1})^{1/2k}$.

答:
Notice that we earlier defined$$A_k=\frac{a_1+a_2+\cdot\cdot\cdot+a_k}k\\G_k=(a_1a_2...a_k)^{1/k}$$Thus$$\frac{A_k+A}2=\frac{\frac{a_1+a_2+\cdot\cdot\cdot+a_k}k+\frac{a_{k+1}+(k-1)A_{k+1}}k}2=\frac{(k+1)A_{k+1}+(k-1)A_{k+1}}{2k}=A_{k+1}$$The inductive step gives $A_k\ge G_k$ and $A\ge G$ and thus $(A_kA)^{1/2}\ge(G_kG)^{1/2}$.

Further, $G_kG=(a_1a_2...a_k)^{1/k}\left(a_{k+1}A^{k-1}_{k+1}\right)^{1/k}=\left[\underbrace{(a_1a_2...a_{k+1})}_{G_{k+1}^{k+1}}\left(A_{k+1}^{k-1}\right)\right]^{1/k}$ from where we get our result.

ContentLicense: CC BY-SA 4.0


问:
$AA^{*}-I$ is negative semi-definite
I need to show $(AA^{*}-I)$ is negative semi-definite where A is a linear operator.

I can show $AA^{*}+I$ is positive definite since $\langle x, (AA^{*}+I)x  \rangle=\|A^{*}x\|^2+\|x\|^2>0$ for $x \neq 0$  but I am not sure how to show $\langle x, (AA^{*}-I)x  \rangle \leq 0$.

Any ideas?

p.s. I think it was assumed that $\|AA^{*}-I\| <1$ if that helps.

答:
In general, $AA^{*}-I$ is not negative semi-definite:

Let $A$ be the $2 \times 2$ diagonal matrix $A= diag(2,2).$ Then

$$AA^*-I= diag (3,3),$$

wich is positive definite !

Title: $AA^{*}-I$ is negative semi-definite


问:
Compute the value of $\iint |x+y|\, dxdy$, given the $x=-1$ to $x=1$ and $y=-1$ to $y=1$.
I have no clue how to go about solving it. I tried to split up the given double integral in two parts that is from $-1$ to $0$ and $0$ to $1$, but my answer came out to be $2$ and the given  answer is $8/3$. I would like to know where i am going wrong. thanks

Body:
Split as follows

$$\int_{-1}^1\int_{-1}^1|x+y|dxdy
=- \int_{-1}^1\int_{-1}^{-x}(x+y)dy dx + \int_{-1}^1\int_{-x}^1(x+y)dy dx
= \frac43+\frac43=\frac83
$$


问:
Lie algebra (invariant)-action on bilinear form
Let $(\pi, V)$ be a rational representation of $G$ a Lie algebra, and let $B$ be a bilinear form defined on $V$.
Then there is an action of $G$ on $B$ given by $$ g \cdot B(u,v) = B(g^{-1}u, g^{-1}v).$$
We can choose to write $B(u,v) = v^t \Gamma u$, so that succinctly $ g \cdot \Gamma = g^{-1} \Gamma g^{-1}$.

I want to show that the action of $X \in \mathfrak{g}$ on $B$ is given by $$ X \cdot B(u,v) = - B(d\pi(X)u,v) - B(u, d\pi(X)v).$$
This form looks related to the fact that the differential should be a derivation, but I am having trouble making this clear (this might be related to my other question here).

Otherwise, as in another of my questions, I am familiar with the definition of the differential via both the exponential map and vector fields, however I am unsure how to apply it here to obtain the desired action on the bilinear form.

Furthermore, is it true that if $B$ is $G$-invariant then it is also $\mathfrak{g}$-invariant? And/or if this is true/not true, is it clear from the solution to my first question?

答:
To be clear, the action of $G$ is on the vector space of all bilinear forms

$$ \operatorname{Bil}(V) = \{B \colon V \times V \to \mathbb{C} \mid B \text{ bilinear}\}.$$

For a hand-wavy derivation of the action of $\mathfrak{g}$, suppose that $G$ is embedded inside $\operatorname{GL}_n$, so that we can consider a small perturbation $\pi(I + tX)$ for some $X \in \operatorname{Mat}_n$: by the definition of the differential we should have

$$ \pi(I + tX)(v) = v + t d \pi_I(X)(v) + O(t^2),$$

the term with the $t$ in front gives us the action of $X \in \mathfrak{g}$ on $V$, namely $X \cdot v = d \pi_I(X) v$.

Now let $\rho(g)(B)(u, v) = B(\pi(g^{-1})u, \pi(g^{-1})v)$, so considering a perturbation again we get
$$\begin{aligned} (\rho(I + tX)B)(u, v)
&= B(\pi(I + tX){^-1} u, \pi(I + tX)^{-1} v) \\
&= B(\pi(I - tX + O(t^2))u, \pi(I - tX + O(t^2))v) \\
&= B(u - t d\pi_I Xu + O(t^2), v - t d\pi_I Xv + O(t^2)) \\
&= B(u, v) - t B(d\pi_I X u, v) - t B(u, d \pi_I X v) + O(t^2).\end{aligned}$$

Here we have approximated $(I + tX)^{-1} = I - tX + t^2 X^2 - t^3 X^3 + \cdots$ by $I - tX + O(t^2)$, and the last inequality uses the bilinearity of $B$. So we find that the action of $X \in \mathfrak{g}$ is $$(X \cdot B)(u, v) = - B(d\pi_I(X) u, v) - B(u, d \pi_I(X), v).$$

Any "real" proof of this result is essentially upgrading this simplified result to be more rigorous (in fact for affine algebraic groups, this is basically the real proof, once one knows how to interpret $t^2 = 0$ appropriately). But if you are working over differentiable manifolds you can make this rigorous using your definition of the differential.

Another way to see this is to first understand how the Lie algebra acts on the dual $V^*$, and on tensor products; then one uses the canonical isomorphism $V^* \otimes V^* \to \operatorname{Bil}(V)$ taking $\varphi \otimes \psi$ to the bilinear form $(u, v) \mapsto \varphi(u) \psi(v)$. Of course then we have that

$$ X \cdot (\varphi \otimes \psi) = -d\pi_I(X)^t \varphi \otimes \psi - \varphi \otimes d\pi_I(X)^t \psi,$$

which is the same bilinear form.


问:
How to solve an indefinite integral of this form?
I have these two integrals. $$\int \frac{x^3}{(x^4+1)^{3/2}} dx$$$$\int x^2 (x^3+2)^{5/2} dx$$ I'm wondering how I can solve them. In the first one I place the $(x^4+1)$ in $dx$, so it becomes $d(x^4+1)$ and then I multiply everything by $1/4.$ But after that I get stuck and I don't know what to do with the $x^3$ and the $x^2$. Please help. Thank you.

答:
Hints:

First integral: substitute $t=x^4+1.$

Second integral: substitute $t=x^3+2.$


问:
If f is continuous on [a,b] and f(a)=f(b) then show that there exists x,y in (a,b) such that f(x)=f(y)
If f is continuous on $[a,b]$ and $f(a)=f(b)$ then show that there exists $x,y \in (a,b)$ such that $f(x)=f(y)$

It looks obvious if I imagine the graph. But I am not able to prove it.
I am trying to employ intermediate value property, but not able to reach to the conclusion.

答: Without using IVP: from the statement it's clear that $f$ cannot be strictly monotone in $(a,b)$. If for all $x,y \in (a,b)$ we have $x\neq y \Rightarrow f(x) \neq f(y)$ (otherwise we are done), then that tells $f$ is injective and continuous in $(a,b)$. This link says that a continuous and injective function is strictly monotone, raising a contradiction.


问:
Quadratics and their values depending on values of variable
Source: Challenge and Thrill of Pre-College Mathematics, Page 201, Question 18:

"Let $$p(x)=ax^2+bx+c$$ be such that $p(x)$ takes real values on real values of x, and non-real values on non-real values of $x$. Is it true that $a=0$ ? If yes, then prove the same."

My attempt: Consider $x=y+zi$, a complex number. We can make cases on
$a=0, a≠0$ and prove that $a≠0$ doesn't work.

$a=0$ :- $p(x)=a(y^2-z^2+2yzi)+b(y+zi)+c$, and likewise for $a≠0$.

This is only a rough idea I have, but it doesn't help much when I tried to  actually solve the question. Could anyone help me elaborate, or provide comments on what I have done so far?

答:
$p(0),p(1)+p(-1),p(1)-p(-1)$ are all real, giving $a,b,c\in\Bbb R$. This implies that $\forall x\in\Bbb R,p(x)\in\Bbb R$.

For $z\ne0$, we require the imaginary part of $p(y+iz)$ to be non-zero. Thus$$z(2ay+b)\ne0\implies2ay+b\ne0\forall y\in\Bbb R$$If $a\ne0$, then for $y=-b/2a\in\Bbb R,2ay+b$ will be zero. Thus $a=0,b\ne0$.


问:
Random walk with reflective barriers
Consider a random walk starting at $x_0 \in [0,1]$. At any discrete time step $k \in \mathbb{N}$, the new position is given by: $$x_k = \min\{\max\{x_{k-1}+C_k,0\},1\}$$ where $C_k \sim U(-a,a)$ are i.i.d. uniform random variables with support in $[-a,a]$. In other words, we are moving at random in $[0,1]$ and we have a certain probability to be ''absorbed'' at $0$ or $1$ and then eventually reflected back.

I am trying to find the stationary distribution of this process. Based on simulations, I have the feeling that the stationary distribution must be something like
$$\pi(x) = \alpha \delta_0(x)+ \alpha \delta_1(x)+(1-2\alpha){1}_{[0,1]}(x) $$
where $\alpha$ is the stationary probability of being at the extreme points and ${1}_{[0,1]}$ is the indicator function on $[0,1]$.

Is there any way to find $\pi(x)$ rigorously?

答:
Consider the discrete version.  Suppose there are $N$ states.
You can move from any state $x$ to any state from $x-m$ to $x+m$ with probability $1/(2m+1)$.

Adjustments are needed for states $1$ and $N$.

Build the transition matrix.  Find its dominant eigenvector and plot it.

Play around with various $N$ and $m$.

Title: Random walk with reflective barriers


问:
How can I determine the cardinality of a set of polymorphic functions?
It seems obvious to me that the set of functions with the signature $\forall A. A \rightarrow A$ is "once-inhabited", i.e. there is only one such polymorphic function which "works" for any set $A$, which is the identity function. (I think perhaps mathematicians would consider this to be actually a whole family of functions; I'm coming from CS where we tend to talk about single polymorphic functions, but feel free to explain how a mathematician would phrase the problem). However I don't know how to prove it except in a very hand-wavy manner.

A slightly more complex example would be to show that $$\forall A B C. (A \rightarrow B) \rightarrow (B \rightarrow C) \rightarrow (A \rightarrow C)$$ is also once-inhabited, and that the only such function with this signature is the function composition operator.

答:
If you know about $\lambda$-calculus, you can make this precise by looking for normalized $\lambda$-terms that have type $\forall A. A \rightarrow A$ in the system F type system (which is described in details on https://fr.wikipedia.org/wiki/Système_F -- the English page is not as complete, for example it doesn't have the rules for typing).

In this basic case, looking at the inference rules, you quickly get that $\Lambda A. \lambda x:A. x$ is the only normalized term having the type $\forall A. A \rightarrow A$.

The Wikipedia page also explains how to make a boolean type (a type inhabited by exactly 2 terms), a natural number type, product types, and so forth.

ContentLicense: CC BY-SA 3.0


问:
Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$
I'm trying to prove rigorously that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$.  Where $f$ is the pdf of the random variable $X$.

I can't find a proof on the wikipedia article, or if it's there then it's disguised enough that I can't recognize it.  Basically what I've got is a sort of semi-rigorous (and probably incorrect) proof of the equality, but maybe somebody could help me flesh out the details.

Proof:

Beginning from the definition that $E[X]:=\int_{\Omega}X\;dP$.

Given a random variable $X:\Omega\rightarrow\mathbb{R}$, we have the induced measure on $(\mathbb{R}, \mathscr{B}(\mathbb{R}))$ given by $P(\{X\in A\})$.  Then by the Radon-Nikodym theorem there exists a measurable function $f:\mathbb{R}\rightarrow [0,\infty)$ such that $$P(\{X\in A\}) = \int_Afd\mu,$$

where $d\mu$ is the Lebesgue measure.  From here it gets a bit hand-wavy.  Basically since this measure was induced by the random variable $X$, then on $\mathbb{R}$ this random variable is simply given by the identity function $g(x)=x$.  And thus by definition we write the expected value of $g$ with respect to our Radon-Nikodym produced measure in the form $$\int_{-\infty}^{\infty}x\;d\Big(\int_Afd\mu\Big).$$

Now by the Fundamental Theorem of Calculus, this becomes $$\int_{-\infty}^{\infty}xf(x)\;dx.$$

What does everyone think about this?

答:
If $X\geqslant0$ almost surely, Fubini yields
$$
E[X]=\int_\Omega X\mathrm dP=\int_\Omega\int_0^\infty\mathbf 1_{t\leqslant X}\mathrm dt\mathrm dP=\int_0^\infty\int_\Omega\mathbf 1_{t\leqslant X}\mathrm dP\mathrm dt=\int_0^\infty P[X\geqslant t]\mathrm dt.
$$
Now, for each $t$,
$$
P[X\geqslant t]=\int_0^\infty \mathbf 1_{x\geqslant t}f(x)\mathrm dx,
$$
hence Fubini again yields
$$
E[X]=\int_0^\infty\int_0^\infty \mathbf 1_{x\geqslant t}f(x)\mathrm dx\mathrm dt=\int_0^\infty\int_0^\infty \mathbf 1_{x\geqslant t}\mathrm dtf(x)\mathrm dx=\int_0^\infty xf(x)\mathrm dx.
$$
If $X$ is real valued with $P[X\gt0]\cdot P[X\lt0]\ne0$, use the identity $E[X]=aE[Y]-bE[Z]$ where $a=P[X\gt0]$, $b=P[X\lt0]$, $Y$ has the distribution of $X$ conditioned on $X\gt0$ and $Z$ has the distribution of $-X$ conditioned on $X\lt0$, that is, $a=1-F(0)$, $b=F(0)$,
$$
f_Y(y)=\frac{f(y)}{a}\mathbf 1_{y\gt0},\qquad f_Z(z)=\frac{f(-z)}{b}\mathbf 1_{z\gt0}.
$$
This yields
$$
E[X]=a\int_0^\infty y\frac{f(y)}{a}\mathrm dy-b\int_0^\infty z\frac{f(-z)}{b}\mathrm dz=\int_{-\infty}^{+\infty}xf(x)\mathrm dx.
$$</p>


问:
Why is $|\mathbf{a}|^{2T}\mathbf{b}=\mathbf{a}^{T}*diag(\mathbf{b})*\mathbf{a}$ true?
I haven't been able to work this problem out myself so I'm coming here for some inspiration about this simple problem (as stated in the title):

Why is the following true:

$|\mathbf{a}|^{2T}\mathbf{b}=\mathbf{a}^{T}*diag(\mathbf{b})*\mathbf{a}$

with $\mathbf{a}$ a vector of dimensions $N\times1$, $\mathbf{b}$ also a vector with dimensions $N\times1$, and $diag(\mathbf{b})$ a diagonal matrix with on its diagonal the elements of the vector $\mathbf{b}$. (The $*$-sign is just a sign for normal matrix multiplication.) Also,
$|\mathbf{a}|^{2T}$ is the transpose of the square of the absolute values of the vector $\mathbf{a}$.

答:
Edit: I'm not sure if I should leave my answer here or remove it.
I now think that the accepted answer is correct.
Origianlly I gave my answer, because I just got confused by the notation $|\mathbf|$ for the element wise absolute value. I thought it's refering to the absolute value of a vector. Apparently it's a common source of confusion: Is there a mathematical operator that will turn a matrix into its absolute values?

My comments

In the question I assume "absolute value" refers to the euclidean norm of the vector. If that's true $|\mathbf{a}|$ should be a scalar and thus $|\mathbf{a}|^{2}$. A transposed scalar is still a scalar, i.e. $|\mathbf{a}|^{2}\mathbf{b} = |\mathbf{a}|^{2T}\mathbf{b}$.

The accepted answer defines ${\bf v} = |{\bf a}|^2$ with ${\bf a} = [a_1, \ldots, a_N]^\top$.
He then concludes, that ${\bf v} = [a_1^2, \ldots, a_N^2]^\top.$ which I don't understand, because since the absolute value (aka euclidean norm) of a vector is a scaler so should be it's square.
Thus we should have
$\mathbf{v} = |\mathbf{a}|^2 = \left( \left( \sum_{i=1}^N a_i^2 \right)^{1/2} \right)^2 = \sum_{i=1}^N a_i^2$

My answer

What follows is my attempt at answering the original question.

As far as I understand the question is:

Given $\mathbf{a} \in \mathbb{R}^N$ and $\mathbf{b} \in \mathbb{R}^N$ we want to show, that
$$|\mathbf{a}|^{2}\mathbf{b} = \mathbf{a}^\top \text{diag}(\mathbf{b}) \mathbf{a}$$
where $\text{diag}(\mathbf{b})$ is the $N\mathsf{\times} N$ diagonal matrix with the elements of $\mathbf{b}$ on the diagonal and $|\mathbf{a}| = \left( \sum_{i=1}^N a_i^2 \right)^{1/2} = \sqrt{\sum_{i=1}^N a_i^2}$ is the euclidean norm of $\mathbf{a}$.

As far as I can tell the statement we're supposed to proof is not true, because the left side of the equation is a vector, i.e. the vector $\mathbf{b}$ scaled by the scalar $|\mathbf{a}|^2$:

$$|\mathbf{a}|^2 \mathbf{b} = \left( \left( \sum_{i=1}^N a_i^2 \right)^{1/2} \right)^2 \mathbf{b} = \left(\sum_{i=1}^N a_i^2 \right) \mathbf{b}$$
while the right side evaluates to a scalar:
$$
\underset{\text{scalar}}{|\mathbf{a}|^2} \underset{N\times 1}{\mathbf{b}} = \underset{1\times N}{\mathbf{a}^\top} \underset{N\times N}{\text{diag}(\mathbf{b})} \underset{N \times 1}{\mathbf{a}}
$$


问:
absolute value global/local Minima and Maxima (piecewise function)

given the function $
f(x)=
\begin{cases}
 |x-2|&\text {} \, 0<x<3\\
 |x-5|&\text{}\, 3≤x≤7\\
     \end{cases}$
which of the following statements is true?(there can be more than one true or all false)

In the open interval $(0,7)$ there isn't a global maxima

In the open interval $(4,5)$ there is a global minima

The point $x=3$ is a local maxima or minima point in the interval $(0,7)$

What I tried:

First thing was to write it in a more comfortable way without the absolute value

$
f(x)=
\begin{cases}
 -x+2&\text {} \, 0<x<2\\
 x-2 &\text{}\, 2<x<3\\
-x+5 &\text{}\, 3<x<5\\
x-5 &\text{}\, 5<x<7\\
 \end{cases}$

then I checked if there is continuity at these points $(x=2,x=3,x=5)$  $\lim \limits_{x \to 2^+}=0$ $\lim \limits_{x \to 2^-}=0$ ,  $\lim \limits_{x \to 3^+}=2$ $\lim \limits_{x \to 3^-}=1$ (not continuous at x=3) and $\lim \limits_{x \to 5^+}=0$ $\lim \limits_{x \to 5^-}=0$.

then I calculated the derivative $
f'(x)=
\begin{cases}
 -1&\text {} \, 0<x<2\\
 1 &\text{}\, 2<x<3\\
-1 &\text{}\, 3<x<5\\
1 &\text{}\, 5<x<7\\
 \end{cases}$
checked the limits to see if it is differentiable at the points mentioned above and found that they are all critical points (sorry if it is not the right word but what I mean is "suspicious" points).

Did a small table to check if the points are minima or maxima. found out that $x=2$ is a minima , $x=3$ is a maxima and $x=5$ is also a minima but I cannot tell if local or global so I cannot tell if statements 1 or 2 are right since I cannot calculate $f(2)$ or $f(5)$ $f(3)$

Can anyone give me hints on how to continue from here? and if my way was correct?
Thank you!

Edit: I just checked the limits of the interval but what can I learn from that?
$\lim \limits_{x \to 0^+}=2$ $\lim \limits_{x \to 7^-}=2$

from what I personally know I can conclude from that if the limits are infinite

答: Actually, you can calculate $f(2)$ and $f(5)$. To check which of the two minima is the global one (both of them are local minima), you can see a bit more carefully the domains on which $f$ is defined. By doing this you should also discover that the function has an additional maximum, other than the one you found.


问:
Finding $\lim_{x\to-\infty}x\cdot e^x$ without using L'Hospital rule
I have to evaluate
$$\lim_{x\to-\infty}x\cdot e^{x}$$
I know how I could do using L'Hospital rule, but I can't use it. I tried to rewrite it as:
$$\lim_{x\to-\infty}\frac{x}{e^{-x}}$$
but I still get $-\infty/\infty$ and I don't what to do next.

答:
For $x \gt 0$ you have $e^x \gt 1 + x + x^2/2 \gt x^2/2$ and therefore for $x \lt 0$ $e^{-x} \gt x^2/2$. Hence for $x \lt 0$

$$0 \lt \vert x \cdot e^x  \vert=\frac{\vert x \vert}{e^{-x}} \lt  \frac{2}{\vert x \vert}$$ which allows to conclude with the squeeze theorem.

Note (following comments discussion): in this kind of question, the definition (or properties) that you use for the exponential map is critical.

The inequality $e^x \gt 1 + x + x^2/2 \gt x^2/2$ can be derived in several ways:

Using $e^x= 1 + x + \frac{x^2}{2} + \dots$ Taylor series.

Using the fact that $\left(e^x\right)^\prime = e^x$ and  $e^0=1$.

Using growth properties like $e^x \gt x^n$ for any $n \in \mathbb N$ and $x$ large enough.


问:
Itō Integral has expectation zero
I have a question about the following property, which I didn't know so far:

Why does the Itō integral have zero expectation? Is this true for every integrator and integrand? Or is this restricted to special processes, i.e. is
  $$\mathbb{E}\left[\int f \, \mathrm{d}M\right]=0$$
  for all local Martingales $M$ and predictable $f$, such that the integral is well defined?

Thank you for clarification.

答:
Intuitive answer: for an Ito integral with respect to Brownian motion (and nice enough $f$), $\mathbb{E}\left[\int_0^t f(B_s) dB_s\right] = 0$ because each little $dB$ has mean zero - in fact, has distribution that is symmetric about zero (and, independent of where $B$ is!).  You can think of the integral, just like a normal integral, as a weighted sum of lots of little $dB$'s; and the fact that you multiply them by a factor doesn't change the fact that their mean is zero. The fact that's being used here is exactly the martingale property.

[EDIT:] But, as pointed out by others, this intuitive answer does not necessarily hold: the Ito integral might turn out to be only a local martingale, not a martingale.  However, being a local martingale means it has mean zero "locally": see the definition.

As for a more general statement: in Kallenberg (15.12) I find that if $M$ is a continuous local martingale with (finite) quadratic variation process $[M]$, and $V$ is a progressive process (implies predictable) with $\mathbb{E}[\int_0^t V^2_s d[M]_s] < \infty$ for all $t>0$, then $N_t = \int_0^t V_s dM_s$ is a continuous local martingale. If this is a martingale, then $\mathbb{E}[N_t]=0$ for all $t>0$.  Since in general it's only a local martingale, instead there is a sequence of increasing stopping times $\tau_k$ tending to $\infty$ such that $\mathbb{E}[N_{\min(t,\tau_k)}] = 0$ for each $k$.

Title: Itō Integral has expectation zero


问:
Set of diameter $\le 1$ contained in set of constant width $1$
I'm reading the paper Minimal universal covers in $E^n$ by H.G. Eggleston and they state that every set $A\subseteq{\bf R}^2$ of diameter at most $1$ (the diameter of $A$ is defined as $\sup_{x,y\in A}|x-y|$) is contained in a set which has width $1$ along any direction.  I can't see how the proof of this should work, though.

EDIT: Maybe the proof is something along the lines of constructing a thing that has width $1$ along any rational direction, then it should has width $1$ along all directions.  For this, it should suffice to show that for any $A$ of diameter $\le 1$ you can add a translate $I$ of the unit interval (embedded into ${\bf R}^2$) such that $A\cup I$ has diameter $1$.  Then you are able to add an infinite amount of intervals at different angles without the diameter exceeding $1$.

答:
WLOG we can assume that $X$ is convex. $p_1,\
p_2\in X$ s.t. $|p_1-p_2|=1$. Then there is a circular sector $V_1$ at $p_1$ whose angle is $\theta_1$.

And $V_1$ contains arc $p_2p_3$. Here $\theta_1$ is the largest when $V_1-X$ is empty.

Now we do the same thing at $p_2$. Hence we have $\sum_{i=1}^{2m+1}
\ \theta_i =\pi,\ \theta_i>0$ and $\bigcup_{i=1}^{2m+1} \ V_i$ contains $X$.


问:
Is it possible to have 2 random variables with the same distribution have different means.
I know that $E[X]=\int X dp$. If $X'$ is a independent copy of $X$. is it necessary that $E[X]=E[X']$?
I cant seem to come up with a proof.

答:
No.

$E(X) = \int x f_X(x) dx$
Nothing in that expression changes if you have the same distribution.  Expected value of the variable is a property of a distribution.


Title: How to solve an indefinite integral of this form?

Body: I have these two integrals. $$\int \frac{x^3}{(x^4+1)^{3/2}} dx$$$$\int x^2 (x^3+2)^{5/2} dx$$ I'm wondering how I can solve them. In the first one I place the $(x^4+1)$ in $dx$, so it becomes $d(x^4+1)$ and then I multiply everything by $1/4.$ But after that I get stuck and I don't know what to do with the $x^3$ and the $x^2$. Please help. Thank you.

Body: If you substitute $u = x^4+1$, then $du/dx = 4x^3$, so "multiplying by $dx$" gives $du = 4x^3dx$. Then $$\int \frac{x^3}{(x^4+1)^{5/2}} dx = \frac{1}{4} \int \frac{4x^3}{(x^4+1)^{5/2}} dx = \frac 1 4 \int \frac{1}{u^{5/2}}du = \frac 1 4 \cdot \frac{-2}{3} \frac{1}{u^{3/2}}$$ Substituting back we get $$\frac 1 4 \cdot \frac{-2}{3} \frac{1}{u^{3/2}} = -\frac{1}{6} \frac{1}{(x^4+1)^{3/2}}$$


答:
Let us denote
$$f(x)=x \mathrm{e}^x .$$

The derivative is $$f'(x)=(x+1)\mathrm{e}^x, $$
and therefore the function is decreasing in $(-\infty ,-1)$. In addition, $f(x)<0$ on that interval, so the monotone convergence theorem gives the existence of   $$L=\lim_{x \to -\infty} f(x) \in (-\infty , 0 ]. $$

On the other hand
$$\int_{-\infty}^0 f(x) \mathrm{d} x  = f(x)-\mathrm{e}^x \big|^0_{-\infty} = -1 - \lim_{x \to -\infty} f(x) + \lim_{x\to -\infty} \mathrm{e}^x = -1-L. $$
As the improper integral converges, we must have $L=0$.

Title: Finding $\lim_{x\to-\infty}x\cdot e^x$ without using L'Hospital rule

Body:
I have to evaluate
$$\lim_{x\to-\infty}x\cdot e^{x}$$
I know how I could do using L'Hospital rule, but I can't use it. I tried to rewrite it as:
$$\lim_{x\to-\infty}\frac{x}{e^{-x}}$$
but I still get $-\infty/\infty$ and I don't what to do next.


问:
Sequential Criteria of Limit Points in Hausdorff Spcaes
By the sequential criteria, I convey that :
A point p of set S is limit point of S if there exists a sequence of distinct points in S that converge to p.

I understand that if the space is metriziable, then this holds.
But for general spaces I haven't been able to find a counter example.
Any help would be appreciated.

答:
You have to look at big spaces. As in, really big. Consider the long line $L=\omega_1\times[0,1)$ with the order topology and let $X=L \cup \{\infty\}$ be its compactification (the extended long line), i.e. $\infty$ is greater than any other element of $L$. Let $S=X$. Then $\infty$ is a limit point of $S$ but no sequence (of distinct points) converges to $\infty$. It is just too far away. That's because cofinality of $\omega_1$ is uncountable (at least assuming ZFC).

Note that there is a proper generalization of that concept. In a topological space, $p$ is a limit point if and only if there is a net of distinct points convergent to $p$, not a sequence. The example above shows that some sequences, unlike nets, just don't reach far enough.

Typical topological space, where we can replace nets with sequences when talking about limit points, is a first countable space. In general such spaces are known as Fréchet–Urysohn spaces.


问:
What does meromorphic continuation mean?
Sorry if this question is slightly out of context.

I've learned that $\zeta(k)=\sum\limits^{\infty}_{z=1} \frac{1}{z^k}$ has meromorphic continuation to $Re(s)>0$.

I know that, from convergence test, $\zeta(s)$ converges when $Re(s)>1$. Also, it is clear that $\zeta(k)$ diverges at $k=0.5$. Meromorphic continuation of $\zeta(s)$ to $Re(s)>0$ means that we would have some meromorphic function $f(s)$ that has the same value with $\zeta(s)$ at $Re(s)>1$ but $f(s)$ can be defined at $Re(s)>0$, with the "compensation" of having some poles. Is this correct?

答: It's a generalization of analytic continuation; just as meromorphic is a generalization of analytic.</p>


问:
Can the exterior algebra be seen as a "model for area/volume"?
I am tying to understand the exterior product. Unfortunately I am not sure what it is supposed to describe. I found a wikipedia article, which I think says that the exterior algebra is sort of a formalisation/model of the notion of (signed) area. https://en.wikipedia.org/wiki/Exterior_algebra

I find this point of view really convincing. However what makes me wonder is that I haven't seen any such explanation before. Neither in the lecture nor in any book I looked at. As a matter of fact all I found regarding the exterior algebra are abstract definitions and not a single note to how one might come up with it or some other "intuitive" explanation.

Is this a valid way of looking at the exterior algebra?

The article then defines the exterior algebra in terms of a quotient.

From what I understand the above construction is/can be viewed as a realisation of a "model/formalisation of area". Is that correct?

Furthermore I would like to make sense of why we "mod out" by $I$.

If we only look at $v\otimes w$ it does not satisfy all of properties 1 to 5. In fact only the first and the 4th property is met only partially. In order to have property 4 we would need $w\otimes w$ to be zero. $w\otimes w=0$ is property 2. I take this as a hint to consider $(V\otimes V)/I$, since for the equivalence class $[v\otimes v]=\pi(v\otimes v)=[0]$, since $v\otimes v\in\ker\pi$. The other properties follow if we look at $\pi(v\otimes w)$.
Hence, it is reasonable to consider it a model of area/volume.

It is all a bit vague, but I hope it still makes some sense. Many thanks in advance.

答:
This point of view is adopted implicitly in multiple situations in linear algebra, analysis and differential geometry.

Linear algebra: The determinant of a linear transformation $T$ measures the signed volume of a unit parallel-epiped after applying $T$ to it. Also, the determinant is an alternating multilinear form on the columns of $T$ (identified with its matrix representation). Or more abstractly, a determinant map is a linear map $\det:\bigwedge^n V\to K$, and the determinant $\det T$ of a linear transformation $T:V\to V$ is defined as the unique number satisfying $\det(Tv_1\wedge\dots\wedge Tv_n)=(\det T)\det(v_1\wedge\dots\wedge v_n)$ for all exterior products $v_1\wedge\dots\wedge v_n\in\bigwedge^n V$. That's the linear algebraic connection between signed volume and the exterior product.

Analysis: The transformation theorem for volume integrals says that if $\Omega\subseteq\mathbb R^n$ is open, $\Phi:\Omega\to\Phi(\Omega)$ a diffeomorphism, and $f:\Phi(\Omega)\to\mathbb R$. Then $f$ is integrable if $\vert\det\mathrm D\Phi\vert f\circ\Phi$ is integrable, and it holds that
$$\int_{\Phi(\Omega)} f\mathrm dV=\int_\Omega \vert\det\mathrm D\Phi\vert f\circ\Phi\mathrm dV.$$

Here, $\vert\det\mathrm D\Phi\vert$ is a measure of how much $\Phi$ distorts volumes: $\mathrm D\Phi$ is a linear approximation of $\Phi$, and $\det\mathrm D\Phi$ measures how this linear approximation distorts signed volumes, and accordingly, $\vert\det\mathrm D\Phi\vert$ measures how much it distorts unsigned volumes. Since we already saw that determinants can be defined through exterior powers, this also gives way to understanding exterior powers in terms of signed volume.

Differential Geometry: On smooth manifolds, there is no intrinsic concept of volume. But we can give a manifold such a concept by defining a volume form: If $M$ is an $n$-dimensional smooth manifold, then a volume form is a nowhere vanishing differential $n$-form $\omega$. On tangent spaces, it acts exactly as the determinant map from the linear algebraic part at the beginning of my answer. And we can now define the volume of an open subset $\Omega\subseteq M$ as
$$V(\Omega):=\int_\Omega\omega.$$
Essentially, $\omega$ behaves locally like a signed volume does on the tangent space, but globally, it gives different portions of the manifold different scales. For instance, if we have the unit sphere $S^2$, then the standard volume form would yield $\int_{S^2}\omega=4\pi$, the surface area of the unit sphere. But if we want to model a sphere with radius $r$, we could instead take the volume form $\omega':=r^2\omega$, which would give our manifold the "volume" $\int\omega'=4\pi r^2$. Or we could multiply $\omega$ by some other nowhere vanishing smooth function $f:S^2\to\mathbb R$ to get a volume form $\omega'':=f\omega$, which assigns different parts of the sphere different "area densities", so it essentially distorts the sphere.

Why the exterior product? A bit more abstractly, you shouldn't view the exterior product/power/algebra as a quotient of the tensor product/power/algebra. You should instead consider what properties it has. The tensor power $V^{\otimes n}$, for instance, has the property that for all multilinear maps $m:V^n\to W$, there is a unique linear map $\tilde m:V^{\otimes n}\to W$ such that $m(v_1,\dots, v_n)=\tilde m(v_1\otimes\dots\otimes v_n)$. This is nice, since it makes multilinear things linear, and linear is easy. But for signed volumes, we are only interested in alternating multilinear maps, not in general multilinear ones. So it would be nice to have a similar constuct to the tensor power, but one which works with alternating maps instead. This is the exterior power: $\bigwedge^n V$ is a vector space with the property that for all alternating multilinear maps $a:V^n\to W$, there is a unique linear map $\tilde a:\bigwedge^n V\to W$ such that $a(v_1,\dots, v_n)=\tilde a(v_1\wedge\dots\wedge v_n)$. So the exterior power allows us to make alternating multilinear things linear. Signed volumes included! This is the reason why I defined the determinant map above as a linear map on the exterior power of $V$. It's also the reason why we define differential forms as smooth sections of an exterior power of the cotangent bundle: Differential forms are meant to be integrated, but to integrate, we need to give a certain scale to our volume elements.


答:
For the first, put
$$x^2=\sinh(t), 2xdx=\cosh(t)dt$$

$$(x^4+1)^\frac 32=\cosh^3(t)$$


问:
Analysis question from a past exam paper
Assume that $f \in C^1 [a,b].$ Prove that $$\forall \epsilon \  \exists \delta > 0 : \\
x, y \in [a,b] , |x-y| < \delta \implies \left\vert\frac{f(x)-f(y)}{x-y} - f'(x) \right\vert < \epsilon $$

I did he following proof and I wonder if its entirely correct as the question itself would have carried 10 % of a 2hr exam.

Edit: edited the proof to make $\delta$ independent of $x$

Let $x,y \in [a,b]$

Since $f'$ is continuous on a compact set, then its uniformly continuous.
Hence
$\forall x,y \in [a,b], |x-y| < \delta \implies |f'(y)-f'(x)| < \epsilon $

By MVT, since $f \in C^1 [a,b], \ \exists c \in (x,y) : f'(c) = \frac{f(x)-f(y)}{x-y}$ .

Thus in particular, $|x-c|<|x-y| < \delta \implies |f'(c)-f'(x)| < \epsilon $ and we're done.

答:
I'll give a slightly different proof:

Define the function $g\colon[a,b]\times[a,b]\to\mathbb{R}$ by
$$
g(x,y)=
\begin{cases}
\frac{f(x)-f(y)}{x-y} & x\neq y\\
f'(x) & x=y.
\end{cases}
$$
Now $f\in C^1$ gives $g$ is continuous.  Since $[a,b]\times[a,b]$ is compact, we have $g$ continuous implies $g$ uniformly continuous (Heine-Cantor), i.e., for every $\epsilon>0$ there exists $\delta>0$ such that
$$
\max\{\lvert x-x'\rvert,\lvert y-y'\rvert\}<\delta\implies\lvert g(x,y) - g(x',y')\rvert<\epsilon\quad\forall x,y,x',y'\in[a,b]
$$
and our result follows from the case $x'=y'=x$.


问:
How do we prove that the different definitions of the derivative are equivalent?
I was thinking about the different definitions of the derivative, and by using geometric arguments you can convince yourself that
\begin{align} f'(x) = \lim _{h \rightarrow 0} \frac{f(x+ah) - f(x-bh)}{(a+b)h}, \end{align}
for any choice of positive (or $0$) $a$ and $b$. However, is it possible to rigorously prove that these definitions of the derivative are the same, regardless of $a$ and $b$? I've tried for quite a while now, and seeing as \begin{align} f'(x) = \lim _{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \end{align}
and
\begin{align} f'(x) = \lim _{h \rightarrow 0} \frac{f(x+h) - f(x-h)}{2h} \end{align}
are both quite common definitions, it seems important that we be sure they are the same thing.

Any ideas on the proof of this?

答:
The two definitions are identical if you consider the domain of problems for which they are applied - problems where $$\lim_{x\to x_0^+} = \lim_{x\to x_0^-}.$$  However, in other situations, those two definitions are not exactly identical. Think about $y = |x|$ at $x = 0$.  The former will give you $1$ and the latter will give you $0$.

EDIT - commenter is correct, the former only gives $1$ if it is $\lim_{h\to 0^+}$.  It is undefined otherwise.


Title: How do we prove that the different definitions of the derivative are equivalent?

Body:
I was thinking about the different definitions of the derivative, and by using geometric arguments you can convince yourself that
\begin{align} f'(x) = \lim _{h \rightarrow 0} \frac{f(x+ah) - f(x-bh)}{(a+b)h}, \end{align}
for any choice of positive (or $0$) $a$ and $b$. However, is it possible to rigorously prove that these definitions of the derivative are the same, regardless of $a$ and $b$? I've tried for quite a while now, and seeing as \begin{align} f'(x) = \lim _{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \end{align}
and
\begin{align} f'(x) = \lim _{h \rightarrow 0} \frac{f(x+h) - f(x-h)}{2h} \end{align}
are both quite common definitions, it seems important that we be sure they are the same thing.

Any ideas on the proof of this?

Body:
hint

$$\Delta=f(x+ah)-f(x-bh)=(f(x+ah)-f(x))-(f(x-bh)-f(x))$$

$$\lim_{h\to 0}\frac{f(x+ah)-f(x)}{ah}=f'(x)$$

So,

$$f(x+ah)-f(x)=$$
$$ahf'(x)+h\epsilon_1(h)$$

and

$$f(x-bh)-f(x)=$$
$$-bhf'(x)+h\epsilon_2(h))$$

thus

$$\Delta=(a+b)hf'(x)+h\epsilon(h)$$
Done.


问:
Using Quantifiers to express "At Least Three"
I am having trouble interpreting this quantified statement:

All prime numbers at least 3 are even.

This is the pre-defined predicates given:

Square(x): x is a square number

Prime(x) : x is a prime number

Even(x)  : x is an even number

This is what I thought of:

$$\forall n \in \Z (Prime(n) \implies \exists x, y, z \in \Z (Prime(x) \land Prime(y) \land Prime(z) \land Even(x) \land Even(y) \land Even(z)))$$

答:
$\exists x,y,z \in P: [x\neq y ~\land x\neq z ~\land y\neq z ~\land x,y,z\in E ] $

where $P$  is the set of prime numbers, and $E$ is the set of even numbers.


问:
Conditional Expectation with independent sub-sigma fields
Let X and Y be bounded random variables on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Consider two independent sub-$\sigma$ fields $\mathcal{G}$ and $\mathcal{H}$ of $\mathcal{F}$. We assume X is $\mathcal{G}$-measurable and Y is $\mathcal{H}$-measurable. Prove the following equality holds:
$$\mathbb{E} (XY\,\, |\, \sigma (\mathcal{H},\mathcal{G})) =  \mathbb{E} (X\,\, |\mathcal{H}) \mathbb{E} (Y\,\, |\mathcal{G})$$ </p>

答:
It doesn't.

If $X$ is $\mathcal G$-measurable and
$Y$ is $\mathcal H$-measurable then $XY$ is $\sigma(\mathcal G, \mathcal H)$-measurable.  Therefore $\mathbb E(XY|\sigma(\mathcal G, \mathcal H))=XY$.

But if $\mathcal G$ and $\mathcal H$ are independent then
$\mathbb E(X|\mathcal H)= \mathbb E(X)$ and $\mathbb E(Y|\mathcal G)= \mathbb E(Y)$.

So you claim that $XY = \mathbb E(X)\mathbb E(Y)$, which holds only if $X$ and $Y$ are almost surely constant.


问:
How to give an Epsilon-Delta proof of this limit $ \lim_{x\to 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2} $?
How to give an Epsilon Delta proof of the following limit:

$$
\lim_{x\to 1}  \frac{1-\sqrt{x}}{1-x} = \frac{1}{2}
$$

Thanks for your help in advance!

答:
Subtract $1/2$ from the function to get

$\dfrac{1-\sqrt x}{1-x}-\dfrac12=\dfrac{1-2\sqrt x+x}{2(1-x)}$

Apply $(a-b)^2=a^2-2ab+b^2$ and $(c+d)(c-d)=c^2-d^2$ (do not be afraid to put in square roots for $b$ and $d$) to obtain

$\dfrac{1-2\sqrt x+x}{2(1-x)}=\dfrac{1-\sqrt x}{2(1+\sqrt x)}=\dfrac{1-x}{2(1+\sqrt x)^2}$

Argue that if $|1-x|=\delta$ is sufficiently small, then the denominator in the last expression given above will be greater than $2$.  Use this to find $\delta$ as a function of $\epsilon$ so that the absolute value of $\dfrac{1-\sqrt x}{1-x}-\dfrac12=\dfrac{1-x}{2(1+\sqrt x)^2}$ will be $<\epsilon$.

$\delta=2\epsilon$.  You may find a somewhat larger $\delta$ is allowed if you can sharpen the bound on the denominator in Step 3.


问:
Prove that infinitely many instances of ZF schemata are necessary in ZF
I want to prove that infinitely many instances of the ZF schemata (Separation, Replacement) are needed in ZF, i.e. that no finite subset $S$ of their instances is such that $\textbf{ZF} - \textbf{Separation} - \textbf{Replacement} + S \vdash \textbf{ZF}$.

My strategy: construct a "finitary" version of the Gödel constructible universe $L^-$ allowing only sets constructible by the finitely many axioms in $\textbf{ZF} - \textbf{Separation}- \textbf{Replacement} + S$. Then show there is a ZF set not in $L^-$.

But the second step is proving harder than I thought.
Surely there is a standard example of this that I am missing -- does anyone have any ideas?

答:
The key point is the reflection theorem, which says that for every finite subtheory $A\subset \mathsf{ZF}$ there is an ordinal $\alpha$ such that $V_\alpha\models A$. This lets us prove the non-finite-axiomatizability of $\mathsf{ZF}$ in two ways, one set-theoretic and the other proof-theoretic. (Incidentally, see here for another example of such a proof/set argument divide.)

(Recall that the reflection theorem is proved outside $\mathsf{ZF}$, so there's no contradiction with Godel here. There is an internal-to-$\mathsf{ZF}$ version, but it's weaker than one might expect: $\mathsf{ZF}$ proves that $\mathsf{ZF}$ proves $Con(A)$ for each finite $A\subseteq \mathsf{ZF}$, but this isn't saying much since - as far as $\mathsf{ZF}$ knows - $\mathsf{ZF}$ could be inconsistent and hence prove everything.)

Set-theoretic argument: Given a finite $A\subset\mathsf{ZF}$, let $\alpha_A=\min\{\alpha: V_\alpha\models A\}$. Inside $V_{\alpha_A}$ there is no level of the cumulative hierarchy which satisfies $A$ - so $V_{\alpha_A}$ can't satisfy $\mathsf{ZF}$, since $\mathsf{ZF}$ proves that there is such a level.

This tacitly used a bit of absoluteness, namely with respect to the cumulative hierarchy (for $\beta<\alpha$ we have $V_\beta^{V_\alpha}=V_\beta$) and to first-order satisfaction (for $M\in V_\alpha$ we have $V_\alpha\models(M\models A )$ iff $M\models A$ in reality). These absoluteness facts are true, important, and not hard to prove, but it's a good exercise to show that we can in fact avoid invoking them entirely:

 Suppose $A\subset\mathsf{ZF}$ is finite and axiomatizes $\mathsf{ZF}$. Let $\alpha_0$ be some ordinal such that $V_{\alpha_0}\models A$ and let $M_0=V_{\alpha_0}$. Now having defined $\alpha_i$ and $M_i$, let $\alpha_{i+1}$ be some ordinal $\beta<\alpha_i$ such that $M_i\models (V_{\beta}\models A)$ and let $M_{i+1}=V_{\alpha_{i+1}}^{M_i}$. This defines a descending sequence of ordinals, so it must stop at some point $n$; but then $M_n$ is a model of $A$, hence of $\mathsf{ZF}$, which thinks that no level of the cumulative hierarchy satisfies $A$, contradicting reflection.

Proof-theoretic argument: Suppose $T$ were a finite fragment of $\mathsf{ZF}$ which axiomatizes the whole. By the reflection principle we have $\mathsf{ZF}\vdash Con(T)$; since proofs are finite, there is some finite $S\subset\mathsf{ZF}$ such that $S\vdash Con(T)$. Since $S$ is finite and $T$ axiomatizes all of $\mathsf{ZF}$, we get $T\vdash S$ - the finiteness of $S$ being used to make sure this actually makes sense - and by composing proofs we get $T\vdash Con(T)$, contradicting Godel.

Note that there's a seemingly-simpler argument here: since $\mathsf{ZF}\vdash Con(T)$ we must have that $T$ doesn't axiomatize $\mathsf{ZF}$ since otherwise we'd get $\mathsf{ZF}\vdash Con(\mathsf{ZF})$, contradicting Godel. What's the issue?

 The above relies on $\mathsf{ZF}$ proving that $T$ axiomatizes $\mathsf{ZF}$, and this isn't trivial.

As a coda, note that no "$L$-flavored" argument will work since there is a finite $A\subset\mathsf{ZF}$ such that $L$ is the smallest transitive class containing the ordinals and satisfying $A$. This is basically a consequence of the construction of $L$ via the finitely many Godel operations. The reflection theorem is very un-$L$-flavored in that the models it yields are "short, wide, and (possibly) wild" rather than "tall, thin, and tame."


问:
If $A$ is injective and $\text{Im}(A) = Y$, then $A$ is invertible.
I am studying A Primer Nonlinear Analysis of A.Ambrosetti and there is something I do not understand in the second chapter. He said that the following result is a consequence of the Closed Graph Theorem.

Let $A \in L(X, Y)$ be a continuous linear map between two Banach spaces. If $A$ is injective and $\text{Im}(A) = Y$, then $A$ is invertible.

Why do we need the closed graph theorem to prove this result ? Injectivity and surjectivity is not enough to show the inversibility of $A$ ?

答: I think that the author meant that $A$ is invertible and continuous. This is why he leverages the closed graph theorem.


答:
You also have to show that $A^{-1}:Y \to X$ is a bounded linear operator.

This is a consequence of the open mapping theorem.

Body:
I am studying A Primer Nonlinear Analysis of A.Ambrosetti and there is something I do not understand in the second chapter. He said that the following result is a consequence of the Closed Graph Theorem.

Let $A \in L(X, Y)$ be a continuous linear map between two Banach spaces. If $A$ is injective and $\text{Im}(A) = Y$, then $A$ is invertible.

Why do we need the closed graph theorem to prove this result ? Injectivity and surjectivity is not enough to show the inversibility of $A$ ?


问:
Show that the sequence defined as $x_{n+1}=1+\frac{1}{x_n}$ converges and calculate it's limit (proof check)
Show that the sequence $(x_n)$ defined as $x_{n+1}=1+\frac{1}{x_n},x_0=1$ converges and calculate it's limit

Probably there are already posts concerning this sequence, but i would like to have a feedback on my proof, please. My attempt is to show that $(x_n)$ is Cauchy.

First, we have $\forall p\ge1$ that:

$|x_p-x_{p-1}|=\Big|\dfrac{x_{p-2}-x_{p-1}}{\underbrace{x_{p-1}\cdot x_{p-2}}_{>2}}\Big| \le\frac{1}{2}|x_{p-1}-x_{p-2}|=...=(\frac{1}{2})^{p-1}|x_1-x_0|=(\frac{1}{2})^{p-1}$

Then we want to show that $(x_n)$ is Cauchy. By definition we must have: $\forall \varepsilon>0 \ \exists N \ \forall m,n\ge N \ (m>n)$:

$|x_m-x_n|<\varepsilon$.

But,

$|x_m-x_n|\le |x_m-x_{m-1}|+|x_{m-1}-x_{m-2}|+...+|x_{n+1}-x_n|\le (\frac{1}{2})^{m-1}+(\frac{1}{2})^{m-2}+...+(\frac{1}{2})^{n}=\sum_{k=0}^{m-1}(\frac{1}{2})^k-\sum_{k=0}^{n-1}(\frac{1}{2})^k=\frac{1}{2^{n-1}}-\frac{1}{2^{m-1}}<\frac{1}{2^{n-1}}$

So, one can chose $N$, such that $\frac{1}{2^{N-1}}<\varepsilon$ to satisfy Cauchy sequence definition.  So, $(x_n)$ is Cauchy and so convergent.

$\bullet$Limit calculation:

We can write that: $x_n=(x_n-x_{n-1})+(x_{n-1}+x_{n-2})+...+(x_1+x_0)+x_0$ and that $\forall n\ge1$: $x_n-x_{n-1}\le (-\frac{1}{2})^{n-1}$

Therefore:

$x_n\le 1+\sum_{k=0}^{n-1}\Big(-\frac{1}{2}\Big)^k=\frac{1+(\frac{1}{2})^n}{1+\frac{1}{2}}+1$

So, $\lim_{n\to \infty}x_n=\frac{5}{3}$. We conclude then that $(x_n)$ converges to $\frac{5}{3}$

Edit: I noticed that my limit calculation is wrong as i bound $(x_n)$ instead of "calculating" it directly.

So, to calculate the exact limit value i would pass by the fact that $\lim_{n \to \infty}x_n=\lim_{n\to \infty}x_{n+1}$ and then solve an equation $L^2-L-1=0$ (where $L$ denotes sequence's limit as it goes to infinity) and conclude that it's limit is equal to $\frac{1+\sqrt{5}}{2}$(golden ratio). (Of course by proving before that, that $(x_n)>0$ by induction)

答: Once you have established that the sequence converges (it's Cauchy), you know that the limit, say $L$, must satisfy $L = 1+\frac 1L$, simply because $\lim x_n = \lim x_{n+1}$. So, we narrow the possibilities to $L = \frac{1\pm \sqrt{5}}{2}$. Since you can easily prove that if $x_0=1$ the sequence remains positive, you conclude that the limit is $L = \frac{1+\sqrt{5}}{2}$.


问:
How I can write this in polar form?
$\cos x +i(1+\sin x)$

I already know that: $\bbox[yellow] {\sin x=2\sin(\frac{x}{2})\cos(\frac{x}{2})}$ and $\bbox[yellow]{\cos x=2\cos^2(\frac{x}{2})-1}$

But this doesn't help me in this case.

答:
Using your tips and $\cos(\pi/2-x)=\sin(x)$,

$$\cos x+i(1+\sin x)=\sin(\frac\pi2-x)+i(1+\cos(\frac\pi2-x))\\
=2\sin(\frac\pi4-\frac x2)\cos(\frac\pi4-\frac x2)+2i\cos^2(\frac\pi4-\frac x2)\\
=2\cos(\frac\pi4-\frac x2)\left[\sin(\frac\pi4-\frac x2)+i\cos(\frac\pi4-\frac x2)\right]\\
=2\cos(\frac\pi4-\frac x2)\left[\cos(\frac\pi4+\frac x2)+i\sin(\frac\pi4+\frac x2)\right]\\
=2\cos(\frac\pi 4-\frac x2)\exp(i(\frac\pi4+\frac x2))$$

Title: How I can write this in polar form?


答:
Let $$z=\cos x+(1+\sin x)i=\rho \,e^{i\phi}.$$
You can use the identity $e^{i\phi}=\cos\phi+i\sin\phi$ and demand that both hand sides of the first relation match. That is, the real part of the RHS should match with the real part of the LHS and the same for the imaginary parts. Upon doing so, you will find a $2\times2$ system of algebraic equations involving $\rho$ and $\phi$. Then you can solve it and obtain what you want.

Title: How I can write this in polar form?

Body:
$\cos x +i(1+\sin x)$

I already know that: $\bbox[yellow] {\sin x=2\sin(\frac{x}{2})\cos(\frac{x}{2})}$ and $\bbox[yellow]{\cos x=2\cos^2(\frac{x}{2})-1}$

But this doesn't help me in this case.


答:
observe $$ \dfrac{1-\sqrt{x}}{1-x} = \dfrac{1}{1+\sqrt{x}}  $$
and you can now guess that the limit is 0.5 to prove find
$$\begin{align*}
\left| \dfrac{1}{1+\sqrt{x}} - \dfrac{1}{2}  \right| &= \left| \dfrac{1-\sqrt{x}}{2(1+\sqrt{x})} \right|\\
&= \left| \dfrac{x-1}{2(1+\sqrt{x})^2} \right|\\
&< \dfrac{|x-1|}{2}  \end{align*} $$
Now just take $\delta = 2\varepsilon$

Title: How to give an Epsilon-Delta proof of this limit $ \lim_{x\to 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2} $?

Body:
How to give an Epsilon Delta proof of the following limit:

$$
\lim_{x\to 1}  \frac{1-\sqrt{x}}{1-x} = \frac{1}{2}
$$

Thanks for your help in advance!


问:
Find the $Ker(\theta)$ of the given ring homomorphism
$\theta :$ $K[x,y]$$\to$ $K[t]$ a ring homomorphism such that

$x$$\to$$t^a$  and

$y$$\to$$t^b$ given, $gcd(a,b)=1$

Prove that $Ker(\theta)=\langle y^a-x^b\rangle$ .

答:
Any element $p(x,y)$ $\epsilon$ $K[x,y]$ such that $p(x,y)$ $\epsilon$ $\langle y^a-x^b\rangle$ implies $p(x,y)=(y^a-x^b)q(x,y)$, where $q(x,y)$ $\epsilon$ $K[x,y]$, then we can easily say $\theta(p(x,y))=0$ ,

Hence $p(x,y)$ $\epsilon$ $Ker(\theta)$, so, $\langle y^a-x^b\rangle$ $\subseteq$ $Ker(\theta)$

Now conversely, say any element $p(x,y)$ $\epsilon$ $Ker(\theta)$.  So, $\theta(p(x,y))=0$

Say,  $p(x,y)=(y^a-x^b)q(x,y)+\sum_{i=1}^a P_i (x) y^{a-i}$ , where $P_i (x)$ $\epsilon$ $K[x]$. Since $\theta(p(x,y))=0$ , so, $\theta(\sum_{i=1}^a P_i (x) y^{a-i})=0$

$\to$ $\sum_{i=1}^a P_i (t^a) (t^b)^{a-i}=0$ .

Now suppose, $P_i(x)=a_{0_{i}}+a_{1_{i}}x+....+a_{n_{i_{i}}}x^{n_{i}}$, where $a_{n_{i}}\neq0$.

So, now $\theta(P_i (x) y^{a-i})=$ $P_i (t^a) (t^b)^{a-i}=[a_{0_{i}}+a_{1_{i}}(t^a)+....+a_{n_{i_{i}}}(t^a)^{n_{i}}]\times(t^b)^{a-i}$

$=a_{0_{i}}t^{(ab-ib)}+a_{1_{i}}t^{(a+ab-ib)}+....+a_{n_{i_{i}}}t^{(a{n_{i}+ab-ib)}}$

$=\sum_{j=0}^{n_i}a_{j_i}t^{(ja+ab-ib)}$, since we have $\sum_{i=1}^a P_i (t^a) (t^b)^{a-i}=0$

$\to$ $\sum_{i=0}^a\sum_{j=0}^{n_i}a_{j_i}t^{(ja+ab-ib)}=0$, now all these $\sum_{j=0}^{n_i}a_{j_i}t^{(ja+ab-ib)}$ are polynomials of $t$ having degree $(a{n_{i}+ab-ib})$ i.e  of different degree. So the sum of these polynomials is equals to $0$ implies all the polynomials are individually equals to $0$. So, $\sum_{j=0}^{n_i}a_{j_i}t^{(ja+ab-ib)}=0$ where, $1\leq i \leq a$ . This implies that $\sum_{j=0}^{n_i}a_{j_i}t^{ja}=0$, since $t^b=y$ is a variable then $t^{b(a-i)}$$\neq0$ for non zero $t$. So,$\sum_{j=0}^{n_i}a_{j_i}(t^{a})^j=0$

$\to$ $\sum_{j=0}^{n_i}a_{j_i}(x)^j=0$ , [here $t^a=x$ so replacing]

$\to$ $a_{0_{i}}+a_{1_{i}}x+....+a_{n_{i_{i}}}x^{n_{i}}=0$

$\to$ $P_i(x)=0$ , where $1\leq i \leq a$. So, we get, $p(x,y)=(y^a-x^b)q(x,y)$, hence $p(x,y)$ $\epsilon$ $\langle y^a-x^b\rangle$ and

So $Ker(\theta)$ $\subseteq$ $\langle y^a-x^b\rangle$

So, $Ker(\theta)=\langle y^a-x^b\rangle$ (Hence proved).


问:
Continuous extension of $f(x)=x\log(x)$: Show $\lvert f(a)-f(b)\rvert\leq\lvert f(b-a)\rvert$ for $0<a<b$ small.
I am considering the function
$$
f(x)=\begin{cases}x\log x, & x>0\\
0, & x=0\end{cases}
$$
on an interval $[0,\varepsilon]$ for small $\varepsilon>0$.

I would like to argue that for $x,y\in [0,\varepsilon]$ with $0<a<b$ one has
$$
\lvert f(a)-f(b)\rvert\leq\lvert f(b-a)\rvert.
$$

I think, in order to show this, one could use that near $x=0$ the function is convex and $f(0)=0$.

This immediately implies superadditivity, that is,
$$
f(a+b)\geq f(a)+f(b).
$$

Can one use this to prove the desired statement?

I do not see it.

答:
Your idea works. $f$ is convex with $f(0) = 0$ and therefore superadditive. $f$ is decreasing on $[0, 1/e]$, so that for $0 \le a < b \le 1/e$
$$
 |f(a) - f(b)| = f(a) - f(b) = f(a) - f(a+b-a) \\
\le f(a) - (f(a) + f(b-a)) = -f(b-a)
= |f(b-a)|\, .
$$


问:
What's the difference between an ergodic measure, ergodic sequence or ergodic mapping?
So I work a lot with ergodic theorems but it is a bit confusing.
We have a probability space $(\Omega,\mathcal{F},P\}$ and a measure preserving mapping $T:\Omega\rightarrow\Omega$.
We say that $T$ is ergodic (or $P$-ergodic) if any event in the $T$-invariant $\sigma$-algebra has probability $1$ or $0$. This is not entirely clear to me.
As example consider a simple random walk in a random environment, with $\{\omega_i\}_{i\in\mathbb{Z}}$ and $\omega_i$ the probability to jump to the right at position $i$.
Now how do I show that this sequence is ergodic? There is no mapping $T$ involved.

答: The notions of ergodic measure and ergodic mapping are more or less interchangeable. If you already have a probability space $(\Omega, \mathcal{F}, \mathbb{P}),$ you would say that a measure-preserving transformation $T: X \rightarrow X$ is ergodic if $\mathbb{P}(A) \in \{0, 1\}$ for all $A \in \mathcal{F}$ for which $T^{-1}(A) = A$ (this is one of many equivalent definitions). You could also say in this scenario that $\mathbb{P}$ is ergodic (with respect to $T$). However, there are situations in which you are only given a mapping $T$ and are asked if there exists an ergodic measure with respect to $T$ (a very well known theorem shows that if $X$ is a complete metric space and $T$ is continuous, then indeed there exists an ergodic measure). So it really depends on the context, really. As for ergodic sequences, I believe that the wikipedia page on ergodic sequences is a good reference. I hope this helps. :)


问:
Is $\mathbb{S}^3 \backslash \mathbb{S}^1$ homotopic equivalent to $\mathbb{S}^1$?
Is $\mathbb{S}^3 \backslash \mathbb{S}^1$ homotopic equivalent to $\mathbb{S}^1$?

I am reading this from some notes on algebraic topology.

I am not even sure what does $\mathbb{S}^3 \backslash \mathbb{S}^1$ mean?

what is $\mathbb{S}^1 \in \mathbb{R}^3$? Is it any one particular circle? Can anyone please define this as a set?

How will $\mathbb{S}^2 \backslash \mathbb{S}^1$ look like? (homeomorphic to 2 open discs?)

答:
Another way to think about the case when $\mathbb{S}^1$ is embedded in $\mathbb{S}^3$ as an unknot (so for example as an equator of an equator) is the following. Treat $\mathbb{S}^3$ as formed by gluing two solid tori (see Complement of the Solid Torus in $S^3$ is Again a Solid Torus ). Then homotopically $\mathbb{S}^1$ is the same as the interior of one of that tori and hence complement has the same homotopy type as a solid torus. $\mathbb{S}^1$ is a deformation retract of solid torus, therefore in that case complement is homotopy equivalent to the circle.

As it was already noted, for different embeddings situation is far more complicated.


问:
Finding $\lim_{x\to-\infty}x\cdot e^x$ without using L'Hospital rule
I have to evaluate
$$\lim_{x\to-\infty}x\cdot e^{x}$$
I know how I could do using L'Hospital rule, but I can't use it. I tried to rewrite it as:
$$\lim_{x\to-\infty}\frac{x}{e^{-x}}$$
but I still get $-\infty/\infty$ and I don't what to do next.

答:
Even though there are already several fine answers, I think the instructor probably just wanted you to use some "notable limit" out of a list. A probable candidate would be
$$
\lim_{x\to +\infty} \frac{e^x}{x^p} = +\infty.
$$

in this case,

$$
\lim_{x\to -\infty}x e^x = \lim_{y\to +\infty} -y e^{-y} = \dfrac{-1}{\displaystyle \lim_{y\to +\infty} \frac{e^y}{y}} = \frac{-1}{+\infty} = 0.
$$


问:
Sequential Criteria of Limit Points in Hausdorff Spcaes
By the sequential criteria, I convey that :
A point p of set S is limit point of S if there exists a sequence of distinct points in S that converge to p.

I understand that if the space is metriziable, then this holds.
But for general spaces I haven't been able to find a counter example.
Any help would be appreciated.

答:
As an alternative to the ordinal example (which is a very standard one, and I usually reduce it to just the ordinal $\omega_1 + 1$ in the order topology):

Let $X=[0,1]^I$ where $I$ is an uncountable set, in the product topology.
For $f \in X$ let the support $s(f)$ of $f$ be defined by $s(f)=\{i \in I\mid f(i) \neq 0\}$, and define the subspace $S$ of $X$ by:

$$S = \{f \in X\mid s(f) \text{ is at most countable }\}$$

Then $S$ is dense in $X$ (so all functions are limit points of $S$), but no sequence of elements from $S$ can converge to a point $g \notin S$. So this is quite a strong example of even a compact Hausdorff space where the sequential criterion fails. BTW $S$ is also a sequentially compact subspace of $X$ which is not closed in $S$ and moreover $X$ itself is not sequentially compact despite being compact.


问:
Higher order factorials.
The function $y=x!$ can be drawn using the gamma functions, what could function would give higher order factorials like $y=x!!$ where $x!! = x\cdot(x-2)\cdot(x-4)\cdot\ldots\cdot5\cdot3\cdot1$. I have this:
$$\prod\limits_{n=0}^{\lfloor x/k\rfloor-1}(x-kn)$$
Where k is the amount of !'s but obviously this is only valid for integer k, but I'm looking for something which would make would make sense and be continuous for non-integers.

答:
For the odd case you could take

$$\Gamma_2(z) := 2^{(z-1)/2} \int_0^{\infty} t^{(z-1) /2} e^{-t} dt $$

For $z=1$ you have 1, and différentiation by parts you get

$$ \Gamma_2(z) = 2^{(z-1) /2} \frac{z-1}{2} \int_0^{\infty} t^{(z-3) /2} e^{-t}dt = (z-1) \Gamma_2(z-2) $$

So that by induction you get the right thing. In the even case you just get the semifactorial with a constant in front coming front $z=2$, i.e $\Gamma(1/2) $ which i think is something related to $\pi$ (maybe $\pi/4$). You can go around this problem by multiplying by an oscillating function :

$$ \left ( 1+ \cos( \pi z/2)^2  \alpha \right ) 2^{(z-1) /2} \Gamma((z-1) /2) $$

Where $ \alpha = \Gamma(1/2) -1$

Title: Higher order factorials.


问:
differential equations- area problem
I'm having some trouble with solving this problem.
Show there is one f(x) that its graph is in the first quadrant, and for every a>0 the area between the the graph above, x- axis down, y-axis left and x=a on the right is exactly $f(a)^3$.

I started by this; $\int_{0}^a f(x) dx=f(a)^3$ but I got stuck because I can't derive it since it is a number and not variable, so I don't know how to continue it. appreciate some help.

Title: differential equations- area problem

Body:
hint

$$\int_0^af(x)dx=(f(a))^3=\int_0^a3f^2(x)f'(x)dx$$


Title: differential equations- area problem

Body:
I'm having some trouble with solving this problem.
Show there is one f(x) that its graph is in the first quadrant, and for every a>0 the area between the the graph above, x- axis down, y-axis left and x=a on the right is exactly $f(a)^3$.

I started by this; $\int_{0}^a f(x) dx=f(a)^3$ but I got stuck because I can't derive it since it is a number and not variable, so I don't know how to continue it. appreciate some help.

Body:
Hint

$$f(a)=\frac{d}{da}\int_{0}^a f(x) dx=\frac{d}{da}(f(a))^3=3(f(a))^2f'(a).$$


问:
How many sides does a circle have?
My son is in 2nd grade.  His math teacher gave the class a quiz, and one question was this:

If a triangle has 3 sides, and a rectangle has 4 sides,
  how many sides does a circle have?

My first reaction was "0" or "undefined".  But my son wrote "$\infty$" which I think is a reasonable answer.  However, it was marked wrong with the comment, "the answer is 1".

Is there an accepted correct answer in geometry?

edit: I ran into this teacher recently and mentioned this quiz problem.  She said she thought my son had written "8." She didn't know that a sideways "8" means infinity.

答:
My third-grade son came home a few weeks ago with similar homework questions:

How many faces, edges and vertices do the following
   have?

cube

cylinder

cone

sphere

Like most mathematicians, my first reaction was that for
the latter objects the question would need a precise
definition of face, edge and vertex, and isn't really
sensible without such definitions.

But after talking about the problem with numerous people, conducting a kind of social/mathematical experiment, I observed something intriguing. What I observed was that
none of my non-mathematical friends and acquaintances had
any problem with using an intuitive geometric concept here,
and they all agreed completely that the answers should be

cube: 6 faces, 12 edges, 8 vertices

cylinder: 3 faces, 2 edges, 0 vertices

cone: 2 faces, 1 edge, 1 vertex

sphere: 1 face, 0 edges, 0 vertices

Indeed, these were also the answers desired by my
son's teacher (who is a truly outstanding teacher). Meanwhile, all of my mathematical
colleagues hemmed and hawed about how we can't really
answer, and what does "face" mean in this context anyway,
and so on; most of them wanted ultimately to say that a
sphere has infinitely many faces and infinitely many
vertices and so on. For the homework, my son wrote an explanation giving the answers above, but also explaining that there was a sense in which some of the answers were infinite, depending on what was meant.

At a party this past weekend full of
mathematicians and philosophers, it was a fun game to first
ask a mathematician the question, who invariably made various objections and refusals and and said it made no sense and so on, and then the
non-mathematical spouse would forthrightly give a completely clear
account. There were many friendly disputes about it that evening.

So it seems, evidently, that our extensive mathematical training has
interfered with our ability to grasp easily what children and
non-mathematicians find to be a clear and distinct
geometrical concept.

(My actual view, however, is that it is our training that has taught us that the concepts are not so clear and distinct, as witnessed by numerous borderline and counterexample cases in the historical struggle to find the right definitions for the $V-E+F$ and other theorems.)

Title: How many sides does a circle have?


问:
Let $f$ be a bounded uniformly continuous function in $R^1$. Then $X_n\to 0$ in pr. implies $E(f(X_n)) \to f(0)$.
The following is problem 4 from Section 4.2 of "A Course in Probability Theory" by Kai Lai Chung.

Let $f$ be a bounded uniformly continuous function in $R^1$. Then $X_n\to 0$ in pr. implies $E(f(X_n)) \to f(0)$.

I don't understand where the condition uniformly continuous is needed. Where is the flaw in the following argument that uses just $f$ is continuous at 0?

Consider an $\epsilon >0$.

Continuity of $f$ says that there exists $\delta > 0$ such that if $|x| \le \delta$ then $|f(x)-f(0)|<\epsilon/3$

That $f$ is bounded says that there exists $B$ such that $|f| N$ then $P(|X_n| > \delta) < \epsilon/3B$.

Then for each $n > N$ we have that
$$|E(f(X_n))-f(0)|=|\int f(X_n)dP - f(0)| = \left|\int_{|X_n|>\delta} f(X_n) + \int_{|X_n|\le \delta}(f(X_n)-f(0)) - \int_{|X_n|>\delta}f(0)\right|\le \left|\int_{|X_n|>\delta} f(X_n)\right| + \left|\int_{|X_n|\le \delta}(f(X_n)-f(0))\right| + \left|\int_{|X_n|>\delta}f(0)\right|\le \int_{|X_n|>\delta} |f(X_n)| + \int_{|X_n|\le \delta}|f(X_n)-f(0)| + \int_{|X_n|>\delta}|f(0)| \le \int_{|X_n|>\delta} B + \int_{|X_n|\le \delta}\epsilon/3 + \int_{|X_n|>\delta}B \le B*(\epsilon/3B)+\epsilon/3+B*(\epsilon/3B) = \epsilon$$

答:
You're correct.  The reason the author has chosen the uniform continuity condition is that this is a step in the proof of a stronger statement

Let $f$ be a bounded uniformly continuous function in $R^1$. Then $X_n\to X$ in pr. implies $E(f(X_n)) \to E(f(X))$.

For this statement you need the uniform continuity condition. So I'd guess that the author is trying to keep the conditions consistent to avoid confusing the reader.


问:
Can $R$ be isomorphic to $R[X]/\langle X^2-a\rangle$ where $a\in R$ and $R$ is an integral domain?
I have proved that if $a=0$, then $R[X]/\langle X^2\rangle\not\cong R$

If they are isomorphic then suppose $f:R[X]/\langle X^2\rangle\to R$ is an isomorphism. Then let $c\in R,c\ne 0$ such that $f(\overline{X})=c$. Again we have $\overline{X}\ \overline{X}=\overline{X^2}=0\implies 0=f(0)=c^2$. As $R$ is integral domain, we have $c=0$, contradiction.

Hence $R[X]/\langle X^2\rangle\not\cong R$.

But I am not sure about if $a\ne 0$. By divison algorithm, we can say that $R[X]/\langle X^2-a\rangle=\{\overline{\alpha X+\beta}|\alpha,\beta\in R\}$.

Can anyone help me in this regard?

答:
If there exists $b$ such that $b^2=a$, $\bar X-b$ has a divisor of zero.

If there does not exists $b$ such that $b^2=a$, $X^2-a$ is irreducible in $R$ but not in $R[X]/(X^2-a)$.


答: We will assume the regularity condition $f \in C^1([0, \infty)).$ Well, what you could do is define the function $F(x) := \int_0^x f(t) dt$ and you know that $F(x) = (f(x))^3$ for all $x > 0.$ Then you have that $F$ is differentiable and $F'(x) = f(x)$ (this is just the Fundamental Theorem of Calculus). It follows that $F'(x) = f(x) = 3 (f(x))^2 f'(x).$ You get the stationary solution $f(x) = 0.$ Otherwise, assuming $f(x) \neq 0$ for all $x > 0,$ we get that $3 f(x) f'(x) = 1,$ i.e. $\frac32 ((f(x))^2)' = 1,$ so that $(f(x))^2 = \frac23 x + c,$ where $c \in \mathbb{R}.$ Note that you also have that $(f(0))^3 = \int_0^0 f(t) dt = 0$ (by continuity of $F,$ for example), so that $c = 0.$ It follows that $f(x) = \frac{\sqrt{6 x} }{3},$ which is a solution to your equation. I hope this helps. :)

Title: differential equations- area problem


问:
Algebraic inequality $\sum \frac{x^3}{(x+y)(x+z)(x+t)}\geq \frac{1}{2}$
The inequality is
$$\frac{x^3}{(x+y)(x+z)(x+t)}+\frac{y^3}{(y+x)(y+z)(y+t)}+\frac{z^3}{(z+x)(z+y)(z+t)}+\frac{t^3}{(t+x)(t+y)(t+z)}\geq \frac{1}{2},$$
for $x,y,z,t>0$.

It originates from a 3-D geometry problem involving volumes of tetrahedra etc. Actually, it is equivalent with that problem (see Let ABCD be a tetrahedron of volume 1 and M,N,P,Q,R,S on AB,BC,CD,DA,AC,BD s.t. MP,NQ,RS are concurrent. Then the volume of MNRSPQ is less than 1/2.).

The three variables simpler case
$$\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+x)(y+z)}+\frac{z^2}{(z+x)(z+y)}\geq \frac{3}{4}$$
can be proved using the Cauchy-Schwarz inequality in "Engel's form".

I have tried variants of Holder type inequalities, until now unsuccessfully.

Body:
As inequality homogenous WLOG $x+y+z+t=4$

By AM-GM$$\sum \frac{x^3}{(x+y)(x+t)(x+z)} \ge\sum \frac{27x^3}{{(3x+y+z+t)}^3}= \sum \frac{27x^3}{{(2x+4)}^3}$$  By tangent line method $$\frac{x^{3}}{\left(2x+4\right)^{3}}\ge \frac{1}{6^{3}}+\frac{1}{108}\left(x-1\right)$$ $$\iff -\frac{\left(x-1\right)^{2}\left(x^{2}-6x-4\right)}{108\left(x+2\right)^{3}}\ge 0$$ which is true as $x^2-6x-4<0,\forall x\in[0,4]$ So $$  \sum \frac{27x^3}{{(2x+4)}^3}\ge \frac{27 \cdot 4}{6^3}+\frac{0}{108}=\frac{1}{2}$$ done!!


问:
Killing form of complexification of a Lie algebra
I am studying Helgason's Differential Geometry and Symmetric Spaces, trying to understand real forms of Lie Algebras.

My problem is related to Lemma 6.1 (1st edition):

Let $K_0$ be the Killing form of a Lie algebra $\mathfrak{g}$ over $\mathbb{R}$, and $K$ the Killing form of the complexification Lie algebra $\mathfrak{g}_{\mathbb{C}}$. Then
$$K_0(X,Y) = K(X,Y)~~\forall X,Y \in \mathfrak{g}$$

My problem:

Understanding the previous equation as $K(X,Y)\equiv K(X+i0,Y+i0)$ I would get twice the result stated in the book. This is due to the following facts.

$\text{ad}_{\mathbb{C}}(X)\in \text{End}_{\mathbb{C}}(\mathfrak{g}_{\mathbb{C}})$, acting as
$$\text{ad}_{\mathbb{C}}(X) [A+iB] = [X+i0,A+iB]_{\mathbb{C}}= [X,A]+ i [X,B] $$
therefore I can write it as a direct sum of the real adjoint map
$$\text{ad}_{\mathbb{C}}(X) = \text{ad}(X)\oplus \text{ad}(X)$$

$\text{Trace}[\text{ad}_{\mathbb{C}}(X)\circ \text{ad}_{\mathbb{C}}(Y)] = \text{Trace}[\text{ad}(X)\circ \text{ad}(Y)\oplus \text{ad}(X)\circ \text{ad}(Y)] = 2\text{Trace}[\text{ad}(X)\circ \text{ad}(Y)]$

Background:

Helgason's definition of complexification: $\mathfrak{g}_{\mathbb{C}} = \mathfrak{g}\times \mathfrak{g} \simeq \mathfrak{g}\oplus \mathfrak{g}$ with the complex structure
$$J:(X,Y)\equiv X+iY\mapsto (-Y,X)\equiv -Y + i X$$
extending the Lie bracket by $\mathbb{C}$-linearity:
$$[X+iY,Z+iT]_{\mathbb{C}} = [X,Z]-[Y,T] + i [X,T]+i[Y,Z] $$

Killing form of any Lie algebra over an arbitrary field $\mathbb{K}$:
$$B(X,Y) = \text{Trace}[\text{ad}(X) \circ \text{ad}(Y)] $$
where $\text{ad}(X)$ is the $\mathbb{K}$-linear map $[X,\cdot]$.

答: $K(a,b)=Tr(ad_a\circ ad_b)$ a basis of $g$ induces a basis of $g_{\mathbb{C}}$ and $ad_a\circ ad_b$ have the same matrix in both basis.


问:
closed ε-ball around x
Let (X,d) be a metric space. Let $\varepsilon > 0$ and let $x \in X$.

We view X as a topological space w.r.t. the topology induced by d and write
$$\overline{B}_{\varepsilon}(x) = \{x' \in X \mid d(x,x') \leq \varepsilon\}$$
for the closed $\varepsilon$-ball around x.

I need to show that: $\overline{B_{\varepsilon}(x)} \subset \overline{B}_{\varepsilon}(x)$

I know that $B_{\varepsilon}(x) \subset \overline{B}_{\varepsilon}(x)$

So if I show that $\overline{B}_{\varepsilon}(x) \subset X$ is closed, then I am ready I think,
since the closure of a set is the smallest closed subset of X that contains this set.

How do I show that $\overline{B}_{\varepsilon}(x) \subset X$ is closed?
I think I can show this to show that the complement is open. But I don't know how.

答:
Hint

Consider $y\notin \bar{B}_{\epsilon}(x).$ Thus $d(y,x)>\epsilon.$ Say $d(y,x)=\epsilon+r$ for some $r>0.$ Show that $B_r(y)\cap \bar{B}_{\epsilon}(x)=\emptyset.$

Title: closed ε-ball around x


问:
Help with functions
How many functions  $f\ne0$, $f:\mathbb{Z}\to\mathbb{C}$

Periodic in an integer $a$ are there so that

$$f(x+y)=f(x)f(y)$$

What I have sofar is that
$$f(a)=f(2a)=f(a)f(a)$$
So that $f(a)=1$,

Also if I multiply $f(x)$ together with itself $a$ times I get that $$f(x)^a=f(x)f(x)f(x)....f(x)=f(ax)=1$$

So that $f(x)^a=1$

I would appreciate any help

答:
By your argument, we can conclude that $f(1)$ is an $a$th root of unity. Once you know $f(1)$, then $f(n)$ is determined for positive $n\in \mathbb{N}$, as $f(n)=f(1)^n$.

We can also figure out $f(0)=1$ as $1=f(a)=f(0+a)=f(0)\cdot f(a)=f(0)$. That means that $1=f(0)=f(1+(-1))=f(1)f(-1)$, so we also know $f(-1)=\frac{1}{f(1)}$ and from there we know $f(-n)=f(1)^{-n}$ for positive $n\in \mathbb{N}$.

Therefore the function $f$ is determined by a choice of an $a$th root of unity for $f(1)$. Conversely, for any $a$th root of unity $\omega$, $n\mapsto \omega^n$ satisfies the requirement for $f$. However, to guarantee the period is exactly $a$ and no less, we must choose an $a$th root of unity $\omega$ with order $a$. There are $\phi(a)$ such choices.

Title: Help with functions


问:
Every compact subset of Baire Space (ω^ω) has empty interior
Here is the Well known proposition.
I assumed that K is the compact subset of ω^ω and it contains an open set Ns, so I thought if K is compact so Ns is because Ns is clopen set, If I can not find a finitely many subcover of Ns, solution is done. But I could not construct and don’t know how to start</p>

答:
Let $K$ be compact in $\omega^\omega$. Suppose $U$ is open and non-empty (so pick $x \in U$) where $U \subseteq K$.

Then there is a basic open subset $\prod_{n \in \omega} U_n$ (so we have a finite subset $F \subseteq \omega$ such that $n \notin F \to U_n = \omega$) such that $x \in \prod_{n \in \omega} U_n \subseteq U$.

If $m \notin F$ we then have $U_m = \omega \subseteq \pi_m[U] \subseteq \pi_m[K]\subseteq \omega$ so that $\pi_m[K] = \omega$ which is not compact (infinite and discrete), while the projection $\pi_m$ is continuous and $K$ is compact. Contradiction, so no such non-empty open subset of $K$ exists.

OwnerDisplayName: Etü


问:
Cardinality of the torsion subgroup of a field.
In an exercise I encountered the following question:

Let $k$ be a field, what is the cardinality of the torsion subgroup of $k$?

Let distinguish the following cases:

If $\textrm{char}(k)=0$, then $k_{\textrm{tor}}=\{0\}$. Indeed, if $x\in k^\times$, then for all $n\in\mathbb{N}$, $nx=0\implies n=0$. Whence: $$\#k_{\textrm{tor}}=1.$$

If $\textrm{char}(k)=p$, then if $k$ is a finite field using Lagrange's theorem, one has $k_{\textrm{tor}}=k$. Whence: $$\#k_{\textrm{tor}}=\#k.$$
When $k$ is a infinite field of characteristic $p$, e.g. $k=\mathbb{F}_p(X)$, I have no clue on how to proceed.

Any help would be greatly appreciated.

答:
To sum up your work and the remainder in the comments:

Let $k$ be a field of characteristic $n$.

If $n=0$ then for all $x\in k^{\times}$ we have $mx=0$ if and only if $m=0$, so no nonzero element of $k$ is torsion, i.e. $k_{\text{tor}}=\{0\}$

If $n>0$ then for all $x\in k$ we have $nx=0$, so every element of $k$ is torsion, i.e. $k_{\text{tor}}=k$.


答: Let $\omega=e^{2\pi i/a}$. You have $f(0)=1$, and $f(1)=\omega^k$ for some integer $k$. This completely determines $f$: $f(n)=\omega^{nk}$ for all $n\in\Bbb Z$. However, $f$ has period $a$ iff $k$ is relatively prime to $a$, so there are $\varphi(a)$ such functions, where $\varphi$ is the Euler totient function.</p>

Title: Help with functions

Body:
How many functions  $f\ne0$, $f:\mathbb{Z}\to\mathbb{C}$

Periodic in an integer $a$ are there so that

$$f(x+y)=f(x)f(y)$$

What I have sofar is that
$$f(a)=f(2a)=f(a)f(a)$$
So that $f(a)=1$,

Also if I multiply $f(x)$ together with itself $a$ times I get that $$f(x)^a=f(x)f(x)f(x)....f(x)=f(ax)=1$$

So that $f(x)^a=1$

I would appreciate any help


问:
Basis vectors question
Suppose the columns of a $5\times 5$ matrix $A$ are a basis for $\mathbb R^5$

If $b$ is in $\mathbb R^5$ then $Ax = b$ is solvable because the basis vectors span $\mathbb R^5$
.

Why is this the case? I don't get how the basis vectors spanning $\mathbb R^5$ says anything about it being solvable.

答: Set $A=(C_1, \cdots, C_5)$ where $C_i$ is the $i$-th column of $A.$ Now, $Ax=b$ is solvable if and only if there exists $x_1,\cdots, x_5\in \mathbb{R}$ such that $$x_1C_1+\cdots+x_5C_5=b.$$ Given $b\in \mathbb{R}^5$ the existence of $x_1,\cdots,x_5\in \mathbb{R}$ is guaranteed because $\{C_1, \cdots, C_5\}$ is a basis of $\mathbb{R}^5.$

Title: Basis vectors question


问:
Eigenvalues and semigroup generated by the Laplacian
Let $d\in\mathbb N$, $\Omega\subseteq\mathbb R^d$ be bounded and open and $$\mathfrak a(u,v):=\langle\nabla u,\nabla v\rangle_{L^2(\Omega)}\;\;\;\text{for }u,v\in H_0^1(\Omega).$$ By the Lax-Milgram theorem, $$\forall f\in L^2(\Omega):\exists! v_f\in H_0^1(\Omega):\forall u\in H_0^1(\Omega):\mathfrak a(u,v_f)=\langle u,f\rangle_{L^2(\Omega)}\tag1.$$ The map $$S:L^2(\Omega)\to H_0^1(\Omega)\;,\;\;\;f\mapsto v_f$$ is

a bounded linear operator from $L^2(\Omega)$ to $H_0^1(\Omega)$; and

a compact self-adjoint linear operator on $L^2(\Omega)$.

By (2.) and the spectral theorem, we know that there is an orthonormal basis $(e_i)_{i\in I}\subseteq H_0^1(\Omega)$ of the closed subspace $\overline{\mathcal R(S)}$ of $L^2(\Omega)$, where $I:=\mathbb N\cap[0,\operatorname{rank}S]$, such that $$S=\sum_{i\in I}\mu_ie_i\otimes e_i,\tag2$$ where the sum converges in $\mathfrak L(L^2(\Omega))$, for some nondecreasing $(\mu_i)_{i\in I}$.

As usual, $(e_i)_{i\in I}$ can be supplemented to an orthonormal basis $(e_n)_{n\in\mathbb N}$ of $L^2(\Omega)$ by an orthonormal basis $(e_n)_{n\in\mathbb N\setminus I}$ of $\mathcal N(S)$ and we may set $\mu_n:=0$ for $n\in\mathbb N\setminus I$.

Question 1: Is it actually possible that $I\ne\mathbb N$?

Now let $$P:=-\Delta:C_c^\infty(\Omega)\to C_c^\infty(\Omega)$$ and set $\lambda_n:=\frac1{\mu_n}$ for $n\in\mathbb N$ with the convention that $\frac10:=\infty$. We easily see that it holds $$Pe_i=\lambda_i\varphi\tag3$$ in the weak sense, i.e. $$\langle e_i,P\varphi\rangle_{L^2(\Omega)}=\lambda_i\langle e_i,\varphi\rangle_{L^2(\Omega)}\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega)\tag4,$$ for all $i\in I$ (even for all $i\in\mathbb N\setminus I$?).

Now, from Riesz' representation theorem, we see that there is a unique bounded linear operator $A$ on $H_0^1(\Omega)$ such that $$\mathfrak a(\;\cdot\;,v)=\langle\;\cdot\;,Av\rangle_{H^1(\Omega)}\;\;\;\text{for all }v\in H_0^1(\Omega)\tag5$$ and it is easy to see that $A$ is self-adjoint.

Question 2: Are we able to show that $\lambda_i$ is an eigenvalue of $A$ for all $i\in I$? And if we actually can, does $A$ admit any eigenvalue not contained in $\{0\}\cup\{\lambda_i:i\in I\}$?$^1$

From $(4)$ and the definition of $A$ we see that $$\langle\lambda_ie_i,\varphi\rangle_{L^2(\Omega)}=\langle Ae_i,\varphi\rangle_{H^1(\Omega)}\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega)\tag6,$$ but that doesn't allow us to conclude, since different inner products are taken on the left- and right-hand side.

So, maybe we need to consider the linear operator $$B:=\mathfrak a(u,\;\cdot\;)\;\;\;\text{for }u\in H_0^1(\Omega)$$ instead?

In any case, what I finally would like to understand is why it makes sense to say that $$e^{t\Delta}f:=\sum_{n\in\mathbb N}e^{-\lambda_nt}\langle f,e_n\rangle_{L^2(\Omega)}e_n\;\;\;\text{for }f\in L^2(\Omega)\tag7$$ is "the semigroup generated by the Dirichlet Laplacian on $\Omega$.

Please note that I'm familiar with semigroups and their generators. But I actually only know the definition of the operator exponential $e^{tT}$ for a bounded linear operator $T$. I guess the notation of the left-hand side in $(7)$ has to be understood with the "functional calculus" sense used in the spectral theorem for bounded linear self-adjoint operators ... But which bounded linear self-adjoint operator, corresponding to the Laplacian, are we considering here (if it's not $A$)?

$^1$ $A$ is certainly not compact and hence its spectrum $\sigma(A)$ should not be equal to the point spectrum $\sigma_p(A)$ (unless I'm terribly wrong). But that doesn't exclude the possibility that $\sigma_p(A)$ is countable ... So, maybe we can actually show that $\sigma_p(A)\setminus\{0\}=\{\lambda_i:i\in I\}$.

答:
Q1: the range of $S$ is $H^1_0$, which is dense in $L^2$, so the sum (2) has to be an infinite sum.

(3) makes no sense as $e_i\not\in C_c^\infty$ in general

Q2: The difference between (5) and the eigenvalues of $S$ is: (5) means
$$
\int_\Omega \nabla u\nabla v = \lambda \int_\Omega \nabla u\nabla v + uv \quad \forall v\in H^1_0(\Omega)
$$
while the eigenvalues of $S$ are defined by
$$
\int_\Omega \nabla u\nabla v = \lambda \int_\Omega uv\quad \forall v\in H^1_0(\Omega).
$$
The difference is in the inner products to form the right-hand side. $S$ is compact, but $A:H^1_0\to H^1_0$ is bijective.


问:
Define a function for which $f'(3)$ exists and $f''(3)$ doesn't exist
The task is to define a function for which $f'(3)$ exists and $f''(3)$ doesn't. My idea is: a function doesn't have a derivative at a point if $\lim\limits_{x\rightarrow x_0-}{f'(x)\neq \lim\limits_{x\rightarrow x_0+}{f'(x)}}$. Therefore if $f'(x)=|x-3|$, $f''(3)$ won't exist. I'll make $f'(x)=2|x-3|$, so that the original function looks better. If I integrate $2(x-3)$, I'll get $x^2-6x=x(x-6)$, so in order to preserve the module, I will define $f(x)=x|x-6|$. This solution is based on intutition and I'm not even sure it works. What is the formal way to solve the problem?

答:
Your idea is great. But the integration is not. We could rather do the following:

$$f(x)=\int_3^x 2|t-3|\mathrm{d}t$$
If $x > 3$, we have that:
$$f(x)=\int_3^x 2(t-3)\mathrm{d}t=(x-3)^2$$
And if $x < 3$, we have that:
$$f(x)=\int_{3}^x -2(t-3)\mathrm{d}x=-(x-3)^2$$
So:
$$f(x)=\text{sgn}(x-3) (x-3)^2 = |x-3|(x-3)$$


答:
Write the matrix $A$ as 5 columns vectors $A = (A_1 \vert A_2 \vert A_3 \vert A_4 \vert A_5)$ and the vector $x = (x_1, \dots, x_5)^T$.

The product $Ax$ is equal to the vector
$$x_1 A_1 + x_2 A_2+x_3 A_3+x_4 A_4+x_5 A_5.$$

Saying that the columns of $A$ span $\mathbb R^5$ means that any vector can be written as a linear combination of the columns $A_1, \dots, A_5$. This should be the case in particular of the vector $b$. This is exactly saying that $Ax=b$ has a solution.

Title: Basis vectors question

Body:
Suppose the columns of a $5\times 5$ matrix $A$ are a basis for $\mathbb R^5$

If $b$ is in $\mathbb R^5$ then $Ax = b$ is solvable because the basis vectors span $\mathbb R^5$
.

Why is this the case? I don't get how the basis vectors spanning $\mathbb R^5$ says anything about it being solvable.


问:
Inverse of an integral function
$$f(x) =\int_0^x  \sin(t)\,dt.$$

I have to compute $(f^{-1})'(0)$ and I know it involves the formula for the derivative of an inverse function $1/f'(f^{-1}(x))$ and that $f' = \sin(x)$ but I do not know how to find $f^{-1}$ of an inverse function.

答:
You need to find $(f^{-1})'(0)$, and you know from the formula you mentioned that $$(f^{-1})'(0) = \frac{1}{f'(f^{-1}(0))}.$$  Now, given that $$f(x) =\int_0^x  \sin(t)\,dt,$$ you have that $f(0) = 0$, and so $f^{-1}(0) = 0$ as well.  That simplifies the question to finding $$(f^{-1})'(0) = \frac{1}{f'(0)}.$$  You have also said that $f'(x) = \sin(x)$, and since $\sin(0) = 0$ you have that $(f^{-1})'(0)$ is undefined, and in fact $(f^{-1})'(x) \to \infty$ as $x\to 0^{+}$.

Alternatively, you can compute directly that $$f(x) = \int_{0}^{x}\sin(t)\,dt = -\cos(t)\bigg|_{0}^{x} = 1 - \cos(x)$$ and so $$f^{-1}(x) = \arccos(1-x).$$  Differentiating, then, gives us $$(f^{-1})'(x) = \frac{1}{\sqrt{1-(1-x)^{2}}} = \frac{1}{\sqrt{2x-x^{2}}}.$$  Again, we have that $(f^{-1})'(0)$ is undefined, and $(f^{-1})'(x) \to \infty$ as $x\to 0^{+}$.

Title: Inverse of an integral function


问:
Prove that $R$ does not have an identity
Let $R$ be the collection of sequences $(a_1,a_2,...)$ of integers $a_1$, $a_2$,..., where all but finitely many of the $a_i$ are $0$.

$1$.Prove that $R$ is a ring under componentwise addition and multiplication

$2.$ Prove that $R$ does not have an identity

My attempt : I can easily show the properties  of inverse ,associative and distributive  in $R$

I have no any problem with question $1$.

My doubt in question $2$ that $R$ does not have identity

I think $R$  have identity take  $e= (0,0,0,..0,1,1,1..)$

Note :where all but finitely many of the $a_i$ are $0$ mean that $R$ have infinitely many $0$

答: The issue is that an identity element of $R$ has to belong to $R$. Which is not the case of $e= (0,0,0,..0,1,1,1..)$ that only has a finite number of zeros.


问:
PDE: solve IVP where $u_{t} +3u u _{x} = 0$, $u(0,x) = \left \{ \begin{matrix} 2, x < 1 \\ 0, x>1 \end{matrix} \right.$
I just started learning PDEs and I am at the very beginning of the book by Peter J. Olver. Now when I search in the stackexchange I found similar problems, all relating to Cauchy or Burger equations. The book so far has not spoken of these, so I guess I should be able to solve this without knowledge of them.

I am asked to solve the IVP where
$$u_{t} +3u u _{x} = 0, \quad \quad u(0,x) = \left \{ \begin{matrix} 2, x < 1 \\  0, x>1 \end{matrix} \right.$$

My reasoning (likely to be wrong) so far is the following: we have a non-linear transport equation, where the speed of the wave is dictated by $3u$ (i.e. the wave is moving faster the bigger it is). We also know the solution is constant along the characteristic curve. We can see the wave being constant at $x>1$ but moving to the right when $x<1$. This leads me to the following:

$$
\begin{align}
\frac{ \partial x}{\partial t} &= 3u \\
\frac{\partial u }{\partial t} &= 0 \\
u(0,x) &= f(x),
\end{align}
$$

Then $u(t,x) = f(x_{0})$ and now I am stuck. I do not know how to proceed.

Any suggestions or comments on my train of thought or on how to proceed would be much appreciated.

答:
Not quite, but part of it is correct. In fact, the method of characteristics giving
$$ [{\bf wrong!}]\qquad u(t,x) = \left\lbrace
\begin{aligned}
&2, & &x<1+6t \\
&0, & &x>1
\end{aligned}\right.
$$
is only valid until the characteristic curves meet the shock wave (which already develops at time $t=0$ here). Since shock wave solutions are discontinuous, they cannot be differentiated in space and time. Thus, solutions must be seeked in the weak sense. The procedure for solving PDE problems of this type, i.e. Riemann problems, is described here:
$$ u(t,x) = \left\lbrace
\begin{aligned}
&2, & &x<1+3t \\
&0, & &x>1+3t
\end{aligned}\right.
$$


问:
ordered pairs of distinct primes
This is an olympiad question, I don't know how to solve. Please tell me the logic/algorithms/steps I should follow to solve the question.

Find the number of ordered pairs of distinct positive primes $p, q$ ($p$ not equal to $q$) such  that  $$p^2 + 7pq + q^2$$ is the square of an integer.

答:
The first hint is that

$$ p^2 + 7pq + q^2 \equiv N^2 $$

and see if you can rearrange the expression by separating the squares; does that tell you anything about the nature of $p$ and $q$?

Title: ordered pairs of distinct primes


答: Let us assume that $R$ admits an identity element $1 \in R.$ Then we should have that $e_i \cdot 1 = 1 \cdot e_i = e_i$ for all $1 \leq i,$ where $e_i \in R$ is the element with $1$ on position $i$ and $0$ on all the others. Write $1 = (x_1, x_2, \dots, x_n, \dots) \in R.$ By the definition of $R$, there is $M \geq 1$ such that $x_j = 0$ for all $j \geq M.$ But then $e_M \cdot 1 = (0, 0, \dots, 0, \dots) \neq e_M,$ hence a contradiction. Thus, $R$ is not a unital ring.

Body:
Let $R$ be the collection of sequences $(a_1,a_2,...)$ of integers $a_1$, $a_2$,..., where all but finitely many of the $a_i$ are $0$.

$1$.Prove that $R$ is a ring under componentwise addition and multiplication

$2.$ Prove that $R$ does not have an identity

My attempt : I can easily show the properties  of inverse ,associative and distributive  in $R$

I have no any problem with question $1$.

My doubt in question $2$ that $R$ does not have identity

I think $R$  have identity take  $e= (0,0,0,..0,1,1,1..)$

Note :where all but finitely many of the $a_i$ are $0$ mean that $R$ have infinitely many $0$


问:
Sum of two positive elements is positive in a $C^*$ algebra is positive
I think this is easy to see, using Gelfand Transform. Using the transform we can see $spec(a)=\{\phi(a)| \phi\in Spec(A)\}$ thus $spec(a+b)=\{\phi(a)+\phi(b)|\phi\in Spec(A)\}$ Now if both $a$ and $b$ are positive then clearly $\phi(a)$ and $\phi(b)$ sperately are non-negative and thus their sum is also.

I think this is a nice proof but I was wondering if there is a way of proving this without appealing to the Gelfand transform.

答:
The following characterization of positivity will do the trick.

Let $A$ be unital (otherwise consider the minimal unitization $\tilde{A}$ instead of $A$), let $x \in A$ be self-adjoint and $\lambda \geq \|x\|$. Then $x$ is positive if and only if $\|\lambda 1_A -x\| \leq \lambda$.

Proof. Consider the constant function $1$ and the identity function $id$ in $C(spec(x))$. Since $x$ is self-adjoint and $\lambda \geq \|x\|$ we have $spec(x) \subseteq \left[-\lambda,\lambda\right]$; moreover observe that $spec(x) \geq 0$ if and only if $\|\lambda 1 - id\|_{\infty} \leq \lambda$. Then by functional calculus we infer that $x$ is positive if and only if  $\|\lambda 1_A -x\| = \|\lambda 1 - id\|_{\infty}\leq \lambda$. QED

Now let's go back to the core discussion:

Let $x,y$ be positive elements of $A$, then $x+y$ is positive.

Proof. By the previous characterization we have in particular that $\| \|x\| -x \| \leq \|x\|$ and the same for $y$. Now $ \| \|x\| + \|y\| -(x+y) \| \leq \| \|x\| -x \| + \| \|y\| -y \| \leq \|x\| + \|y\|$. Again by the characterization above (with $\lambda = \|x\|+\|y\|$, where the triangle inequality guarantees that $\|x+y\| \leq \lambda$) we conclude that $x+y$ is positive. QED


问:
maximal interval of existence of $\dot{x} = f(t,x)$
I was doing some exercises, and i met this one, that i really don't know how to solve. Does anybody know how to solve it? Also, hints are very (if not more) appreciated! Thanks very much

答:
The general idea is that according or similar to the Grönwall lemma, you get an upper bound $u$ with $|x(\pm t)|\le u(t)$ that is the solution of
$$
\dot u=Cu^M, ~~ u(0)=|x(0)|
$$
This is separable and further symbolically solvable
$$
u(t)^{1-M}-u(0)^{1-M}=-C(M-1)t
$$
This has a singularity where $u(0)^{1-M}=C(M-1)t$, inside that range the solution of the original DE is finite and thus well-defined and extendable. Thus
$$T\ge \frac{|x(0)|^{-(M-1)}}{C(M-1)}$$


问:
A basic probability doubt on independence
Let $X_1$ and $X_2$ be two i.i.d continuous random variable. I need to find the probability that $P(X_1 < X_2)$. I know how to formally find the probability by integrating over appropriate region and how we use the "identically distributed" and "independent" property here.

Now,consider the intuitive argument for this problem. $X_1$ and $X_2$ both have the same distribution. So, any one is equally likely to be the smallest. So, is n't identically distributed sufficient condition ? Where is "independence" needed here ?

答: Suppose $X_1$ is a continuous uniform variable on the unit interval $[0,1]$, and let $X_2=X_1+0.01-\lfloor X_1+0.01\rfloor$.</p>


问:
Prove a Vector set is linear independent?

problem:Let $V$ be a linear space,$\mathscr A $ is a linear translation on $V$.choose $0≠α∈V$.let $S=\{\mathscr A^k(α):k≥0\}$ and $U=span(S)$, prove that:
(i) $U$ is stable under $\mathscr A$.
(ii) assume $dim(U)=r$,then ${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$ is a basis of $U$ and find out the metric of  $\mathscr A|_U$ under basis ${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$.

My attempt

For(i) i know we only check $\mathscr A$ map the element of $S$ to $U$.

For(ii) i don't know how to do it.I try yes the definition of linear independent to prove it,but i failed. Then i want to use a proposition : if there is k such
that  $\mathscr A^{k-1}(α)≠0,\mathscr A^{k}(α)=0$  then ${α,\mathscr A(α),\cdots,\mathscr A^{k-1}(α)}$ is linear independent,i want to use $dim(U)=r$ to prove that k is exist, but i also failed.

答:
lemma: if ${α_1,α_2,\cdots,α_n}$ is linear independent,${α_1,α_2,\cdots,α_n,β}$ is linear dependent,then $β$ can be linear expression by ${α_1,α_2,\cdots,α_n}$,but $α_i$ can't be linear expression by the other.

From the comment:we know${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$ is linear dependent and ${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α),\mathscr A^r(α)}$ is linear dependent.

We write $\mathscr A^r(α)=x_0α+\cdots+x_{r-1}\mathscr A^{r-1}(α)$.

We have$(\mathscr A(α),\cdots,\mathscr A^r(α))⊆ (α,\cdots,\mathscr A^{r-1}(α))$. If ${\mathscr A(α),\cdots,\mathscr A^r(α)}$ is independent,then $α$ can be linear expression by ${\mathscr A(α),\cdots,\mathscr A^r(α)}$,by lemma,it is impossible, so ${\mathscr A(α),\cdots,\mathscr A^r(α)}$ is linear dependent,thus $\mathscr A^r(α)$ can be expression by ${\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$,so $x_0=0$.

we know${\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$ is linear dependent and ${\mathscr A(α),\cdots,\mathscr A^{r-1}(α),\mathscr A^r(α)}$ is linear dependent. Use similarly way with above,we can get $x_1=0$,this  process can continues. So $x_0=\cdots=x_{r-1}=0$.
So $\mathscr A^r(α)=0$.

Remark:$(\mathscr A(α),\cdots,\mathscr A^r(α))$ is subspace generated by${\mathscr A(α),\cdots,\mathscr A^r(α)}$.


问:
Rotate one plane to another in $\mathbb{R}^n$
I have some points that lie on the unit sphere in $\mathbb{R}^{n - 1}$, and I want to rotate them in $\mathbb{R}^n$ so that they lie on the plane defined by $\sum_{i=1}^n x_i = 0$, where $x_i$ are the coordinates of the points in $\mathbb{R}^n$.

My main question is how do I do this in $\mathbb{R}^n$, but at the moment, I can't even get it working in $\mathbb{R}^3$.  Here is my $\mathbb{R}^3$ attempt:

$u_1 = (0, 0, 1)^T$ is unit normal to the $x-y$ plane.
$u_2 = (1, 1, 1)^T$ is normal to $x + y + z = 0$.  Define the angle of rotation $\theta$ by $\cos(\theta) = (u_1 \cdot u_2)/|u_2|$ and the axis of rotation $a$ by $a = (u_1 \times u_2)/|u_2|$. Then apply the rotation matrix $R$ given on wikipedia.  This is essentially the answer given here.

I was expecting this rotation matrix to rotate one plane to another, then if I got the planes the wrong way around I would negate the angle.  But I don't get either plane rotating to the other:
$$
R\cdot(1, 1, 0)^T = (0.58, 0.58, -0.94)^T\\
R\cdot(1, -0.5, -0.5)^T = (0.56, -0.74, -0.52)^T.
$$

Any help in 3 dimensions or $n$ dimensions very much appreciated.

答:
Let's go a bit more general and consider two hyperplanes $H,H'\subset \mathbb R^n$, where by hyperplane I mean a linear subspace of dimension $n-1$. Assuming that $H\ne H'$, we're going to rotate $H$ into $H'$.
For that purpose, we choose unit normal vectors $\nu,\nu'\in S^{n-1}$ for $H$ and $H'$. Since rotations are orthogonal, it is enough to find one that takes $\nu$ to $\nu'$.

Since $H\ne H'$, the vectors $\nu$ and $\nu'$ are linearly independent and span a plane $E$ in $\mathbb R^n$. The vectors $f_1=\nu$ and $f_2= \frac{\nu'-\langle\nu,\nu'\rangle \nu}{|\nu'-\langle\nu,\nu'\rangle \nu|}$ are an orthonormal basis for $E$. This gives us a linear isometry $$\mathbb R^n= E\oplus E^\perp \cong \mathbb R^2\times E^\perp$$ with $\nu$ and $\nu'$ corresponding to $(1,0;0)$ and $(a,b;0)$, respectively, where $\nu'=a f_1+bf_2$. (I'll leave it to you to work out formulas for $a$ and $b$). The problem is now reduced to finding a rotation of $\mathbb R^2$ that takes $(1,0)$ to $(a,b)$. But that's easy. Since $1=|\nu'|^2=a^2+b^2$ we have $$\begin{pmatrix}a & -b\\b&a\end{pmatrix}\in SO(2)$$ Working backwards, note that $(-b,a;0)$ corresponds to $-bf_1+af_2\in\mathbb R^n$.  We now obtain our desired rotation of $\mathbb R^n$ by mapping $$\nu=f_1\mapsto af_1+bf_2=\nu', \quad f_2\mapsto -bf_1+af_2$$ and leaving all vectors in $E^\perp$ unchanged.

If you insist on a matrix representation of this rotation, you'd have to extend $f_1,f_2$ to an orthonormal basis of $\mathbb R^n$ by choosing an orthonormal basis $f_3,\dots,f_n\in E^\perp$. Then the matrix $F$ with columns $f_1,\dots,f_n$ is orthogonal and the desired roation matrix is $$R=F \begin{pmatrix}\begin{smallmatrix}a & -b \\ b&a \end{smallmatrix} & 0 \\ 0 & E_{n-2} \end{pmatrix}F^t\in SO(n)$$
where $E_{n-2}$ is the unit matrix in dimension $n-2$.

In concrete cases such as $\nu = (0,\dots,0,1)$ and $\nu'=\tfrac{1}{\sqrt n}(1,1,\dots,1)$ it is easy to write down a concrete matrix $F$ as above. Again, I'll leave this to you.