# how prove that $|f(x)|\ge2^n (n+1)$ under these conditions?

Assume $f:[0,1]\mapsto\mathbb{R}$ is continuous and satisfies

1. $\int_0^1x^kf(x) \, dx=0 \quad\forall k\in\{0,1,2,\ldots,n-1\}$,
2. $\int_0^1x^n f(x) \, dx=1$.

How do you prove that $\exists x\in[0 ,1]$ such that $|f(x)|\ge2^n (n+1)$?

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This a puzzler. :) If no one solves it here, but you, would appreciate if you post a solution as i failed fir an hour (except for the tremendous insight it doesn't hold for $n=0$...) – gnometorule Jan 28 '13 at 19:40
It doesn't hold for $n=0$? Why not? – user7530 Jan 28 '13 at 19:43
@User7530: my bad. – gnometorule Jan 28 '13 at 19:57

Let $\tilde P_n$ be the $n$-th shifted Legendre polynomial. Then $\tilde P_n$ has degree $n$, leading coefficient equal to $\binom{2n}n$ and

$$\int_0^1\tilde P_n(x)\tilde P_m(x)\,dx=\frac{\delta_{mn}}{2n+1}\ (\delta\ \style{font-family:inherit;}{\text{denotes the Kronecker delta}})\,.$$

The hypotheses on $f$ imply that $\int_0^1\tilde P_n(x)f(x)\,dx=\binom{2n}n$. If $a_n=(2n+1)\binom{2n}n$, then by the ortogonality of the polynomials $\tilde P_n$ we have

$$0\leq\int_0^1\bigl(f(x)-a_n\tilde P_n(x)\bigr)^2\,dx=\int_0^1f(x)^2\,dx-2a_n\binom{2n}n+\frac{a_n^2}{2n+1}$$

$$=\int_0^1f(x)^2\,dx-(2n+1)\binom{2n}n^2\,.$$

Since $f$ is continuous, then for some $x_0\in[0,1]$ we have $f(x_0)^2\geq(2n+1)\binom{2n}n^2\geq4^n(n+1)^2$ (this inequality can be easily proved by induction on $\boldsymbol{n\geq2}$) , and so $|f(x_0)|\geq2^n(n+1)$, as desired.

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You probably meant $f(x_0)^2\geq (2n+1)\binom{2n}{n}^2$. – 1015 Jan 28 '13 at 19:49
@julien I mean equality, by invoking mean value theorem for integrals, but yes, you can think it as an inequality. – Matemáticos Chibchas Jan 28 '13 at 19:53
Can you tell me how the mean value theorem for integrals gives you an equality here? – 1015 Jan 28 '13 at 19:58
@julien Apply mean value theorem to $F(x)=\int_0^x f(t)^2\,dt$, together with fundamental theorem of calculus. – Matemáticos Chibchas Jan 28 '13 at 19:59
If you mean $F(1)-F(0)=F'(x_0)(1-0)=f^2(x_0)$, I still only see $f(x_0)^2\geq (2n+1)\binom{2n}{n}^2$. But maybe you meant otherwise. – 1015 Jan 28 '13 at 20:06
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