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The function $f(x)$ is continuous on the interval $[0,1]$ and we have that $$ \int_0^1f(x)x^ndx = 0, \quad\, n=0,1,2,\dots $$ Prove that $f(x)= 0$ on the interval $[0,1]$.

So I am thinking like this, I can expand $f(x)$ to an even function and for even $n$ I got this

$$\frac{1}{2}\int_{-1}^1f(x)x^ndx$$ and then $f(x)$ must be zero for all $x$ in this interval, is this right? What about odd $n$? Not even sure if this is the right way to attack this.

This is in my Fourier Analysis course and its in the chapter that contains orthogonal set of functions. So I need to prove this with tools within this area.

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    Do you know that any continuous function (on a bounded closed interval) can be approximated uniformly by polynomials? – Thomas Jan 01 '16 at 15:49
  • Can't say that it is, I need to prove that it is in fact zero. And its in my fourier analysis course. –  Jan 01 '16 at 16:31
  • Have you studied Legendre polynomials yet in your chapter on orthogonal functions? – John Dawkins Jan 01 '16 at 17:15
  • Yes I have! But those use the interval [-1,1] right? –  Jan 01 '16 at 17:27
  • The "shifted" Legendre polynomials ($Q_n(x) := P_n(2x-1)$ where $P_n$ are Legendre polynomials) form a complete orthonormal system in $L^2[0,1]$. – John Dawkins Jan 01 '16 at 17:39
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    The suggested "duplicate" in no duplication at all, as the OP has requested a solution by the methods of Fourier analysis/orthogonal polynomials. (Not Weierstrass or variants thereof.) – John Dawkins Jan 01 '16 at 17:59
  • @JohnDawkins Indeed (although that was not part of the original question) – Clement C. Jan 01 '16 at 18:11
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    To me, it wasn't a duplicate even from the beginning. I tagged Fourier analysis and everything and the question isn't even the same but I guess people had their reasons. Anyway, anyone got any answer to this? –  Jan 01 '16 at 18:19
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    Compute the "Fourier coefficients" $c_n$, $n=0,1,2,\ldots$ of $f$ with respect to the basis ${Q_n}$ for $L^2[0,1]$ consisting of the shifted Legendre polynomials (from my earlier comment). Because $\int_0^1 f(x)x^n,dx=0$ for $n=0,1,\ldots$, you will have $c_n:=\int_0^1 f(x)Q_n(x),dx = 0$ for all $n$. – John Dawkins Jan 01 '16 at 18:24
  • Thank you sir, now I think I will handle this :) –  Jan 01 '16 at 20:35

1 Answers1

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Note: this answer was written before the editing of the OP's question, when was added the constraint that tools from Fourier analysis be used.

Hint:

  • get uniform convergence of some sequence of polynomials $(P_n)_n$ to $f$ on $[0,1]$. (How?)

    Invoke Stone—Weierstrass's theorem (see this).

  • From there, get $$0 = \int_0^1 fP_n \xrightarrow[n\to\infty]{} \int_0^1 f^2$$

  • conclude (since $f$ is continuous) that $f^2=0$.

Clement C.
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  • I'm not sure that I understand this, I need to prove this with tools from Fourier analysis. I tagged the question with it but someone took that tag away. I have another idea now tho, is it possible to show that the only function that is orthogonal to all polynomials in the stated space is in fact t $f(x)=0$? –  Jan 01 '16 at 16:38
  • If the Weierstrass approx theorem is not in your course then you can easily prove it from what is there. You know that you can uniformly approximate $2\pi$-periodic continuous functions with trigonometric polynomials. A change of variables (take $f(|\cos t|)$) will show fairly easily that continuous functions on $[0,1]$ can be approximated by ordinary polynomials. I point this out only so you appreciate that the answer and hints are not completely foreign to your concerns. The answer by Mr Dawkins is more in the spirit of the question since it is an "orthogonality" answer as requested. – B. S. Thomson Jan 01 '16 at 18:33