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Find all the continuous functions on $[-1,1]$ such that $\int_{-1}^{1}f(x)x^ndx=0$ fof all the even integers $n$.

Clearly, if $f$ is an odd function, then it satisfies this condition. What else?

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Can you find a special case? – Mhenni Benghorbal Jan 6 '13 at 2:55
Follow-up exercise: Drop the continuity requirement and replace it with $f \in L_1([-1,1])$. Show that the answer doesn't change except for the qualifier that it only hold almost everywhere. – cardinal Jan 6 '13 at 21:34
up vote 10 down vote accepted

You observed that every odd continuous function $f$ satisfies all of these equations. Try to prove the converse also: If $f$ is a solution, then it is odd.

Hint: Every function $f$ can be split into its even and odd parts. It therefore suffices to prove the following: If $f$ is even and solves your equations, then it is identically equal to zero. Think about approximation by polynomials. Show that an even function $f\in C([-1,1])$ can be uniformly approximated by even polynomials. There are multiple ways of doing this.

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Since any $f(x)$ can be divided into a sum of odd and even functions, say $f(x) = g(h) + h(x)$, where $g(x)$ is odd and $h(x)$ is even, we may concentrate on asking whether or not even functions such that: $$\int_{-1}^{1}{h(x)x^ndx}=0$$ exist. Moreover if such a function does exist, it is also true that by construction for every even polynomial $P(x)$: $$\int_{-1}^{1}{h(x)P(x)dx}=0$$ We construct $P_m$, the $m$-polynomial expansion of: $$\frac{\sin(m (x-\alpha))}{m(x-\alpha)} + \frac{\sin(m (x+\alpha))}{m(x+\alpha)}$$ At the limit $m\rightarrow\infty$: $$\lim_{m\rightarrow\infty}\int_{-1}^{1}{h(x)P_m(x)dx}=2h(\alpha)$$ Since we can construct such a series for each $\alpha$, the function $h(x)$ must be arbitrarily small at each point, hence showing that no such even non-zero functions exist and that only odd functions fulfill the requirement.

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I am probably misreading, but you seem to be assuming $a_m \geq 0$. – cardinal Jan 5 '13 at 23:47
Not all continuous functions are analytic. Some are not even first differentiable. – Baby Dragon Jan 6 '13 at 0:54
@BabyDragon - I think it should be OK now. – nbubis Jan 6 '13 at 2:06
@cardinal - you were right, the proof now does not assume a series representation. – nbubis Jan 6 '13 at 6:41

Rewrite the integral as $$\int_0^1 [f(x)+f(-x)]x^n dx = 0,$$ which holds for all even $n$, and do a change of variables $y=x^2$ so that for all $m$, even or odd, we have $$ \int_0^1 \left[\frac{f(\sqrt{y})+f(-\sqrt{y})}{2 \sqrt{y}}\right] y^{m} dy = 0. $$

By the Stone Weierstrass Theorem, polynomials are uniformly dense in $C([0,1])$, and therefore, if the above is satisfied, we must have $$\frac{f(\sqrt{y})+f(-\sqrt{y})}{2 \sqrt{y}}=0$$ provided it is in $C([0,1])$. [EDIT: I just realized the singularity can be a problem for continuity--so perhaps jump to the density of polynomials in $L^1([0,1])$.] Substituting $x=\sqrt{y}$ and doing algebra, it follows $f$ is an odd function.

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