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Let $f$ be a smooth function on a closed interval $[a,b]$. Assume that there is $M>0$ such that $\left|f^{(n)}\right|<M$ for all $n\geq1$ and all $x\in[a,b]$. Prove that if

$$\int _ {a}^{b}x^n f(x)dx=0$$

for all $n\geq0$ then $f=0$ on $[a,b]$

Hint: Prove that $\int _ {a}^{b}(f(x))^2dx=0$ by using the Taylor theorem.

I found that the given condition in the first line is simply the Remainder Estimation Theorem of the Taylor series. However, I don't know how to prove the hint by using the Taylor's theorem. Even I can prove the hint, I'm still blank and I don't know how the hint is related to the problem.

Could someone help me on this problem? Any help will be greatly appreciated.

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    $\begingroup$ This isn't true, take $\sin(x)$ on $[0,2\pi]$. $\endgroup$ – Wojowu Nov 4 '15 at 16:15
  • $\begingroup$ Generally if $\forall x \in [a,b] \space f(x)\geq 0$ and $\int_a^bf(x)dx=0$ then $f\equiv 0$ on $[a,b]$. $(f(x))^2$ is certainly non negative function thus $\int_a^b(f(x))^2dx=0$ implies $(f(x))^2\equiv 0$ and that happens iff $f\equiv 0$. $\endgroup$ – Alex Fish Nov 4 '15 at 16:22
  • $\begingroup$ Hmm that's an interesting that $\sin(x)$ makes the statement false. I think there might be sth wrong. My question is that what if $f(x)$ is always positive, maybe that's what the problem is trying to ask. $\endgroup$ – IgNite Nov 4 '15 at 16:29
  • $\begingroup$ If $f(x)$ is always non negative you don't need all the derivatives. $\endgroup$ – Alex Fish Nov 4 '15 at 16:38
  • $\begingroup$ Oh sorry I have made a mistake. It should be $$\int _ {a}^{b}x^n f^{(n)}(x)dx=0$$ $\endgroup$ – IgNite Nov 4 '15 at 17:43
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Since the derivatives $f^{(n)}$ are bounded, $f$ is analytic on $X = [a, b]$. (More generally, $f$ is analytic on the compact space $X$ iff $|f^{(n)}| \leq C^{n+1} n!$ for some constant $C$.) Furthermore, we have $f(x) = \sum a_n(x - a)^n$ with $|a_n| = |f^{(n)}(a)|/n! < M/n!$. It follows by the Weierstrass $M$-test that the sum $\sum a_n (x - a)^n$ converges uniformly. Hence \begin{align*} \int_X f^2 = \sum_{n=0}^\infty a_n \int_X (x - a)^n f = \sum_{n=0}^\infty a_n \left[ \sum_{m=0}^n \binom{n}{m} (-a)^{n-m} \int_X x^m f\right] = 0, \end{align*} forcing $f = 0$.

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Hint: Using Taylor, $f(x) = P_n(x) + R_n(x)$ with $R_n(x) \to 0$ fast. Use

$$\int_a^b f(x)^2\,dx = \int_a^b f(x)(P_n(x) + R_n(x))\,dx.$$

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