Proving $f \in L^{2}[0, 1] = 0$ a.e. if integral of $x^{n}f(x)$ is 0 for each n

Question

Let $f \in L^{2}[0, 1]$ satisfy $\int_{0}^{1} x^{n} f(x) dx = 0$ for each $n = 0, 1, 2, ...$. Show that $f = 0$ a.e.

I know this will involve Holder's Inequality somewhere. So far it's clear that since $x^{n}$ is also in $L^{2}[0, 1]$ for each $n$, we have $0 = \int_{0}^{1} x^{n} f(x) dx \leq ||x^{n}||_{2} ||f||_{2} \leq ||f||_{2}$.

If I could show $||f||_{2} < 1/n$ for each $n$ that would tell me the norm is at most 0, implying $|f|^{2}$ = 0, implying $f$ = 0. How would I go about that?

Answer 1

First note that by Cauchy-Schwartz (or Holder) that $L^2[0,1] \subset L^1[0,1]$. Suppose first that $f$ is continuous. Then for every $\epsilon > 0$, there is a polynomial $p_n(x)$ such that $|f(x) - p_n(x)| < \epsilon$. Then by the criterion, $\int f^2(x) \le \int f(x)(f(x) - p(x)) < \epsilon \int f$. Thus $0 \le \int f^2 = 0$, and $f^2$ must be zero a.e.

Now use the density of continuous functions with compact support in the $L^2$ norm.

Answer 2

Let $\mathcal{H}=L^{2}[0,1]$ be a real Hilbert space with the usual inner-product $\langle f,g\rangle=\int_{0}^{1}f(x)g(x)dx$. Let $f\in\mathcal{H}$ be given such that $\int_{0}^{1}x^{n}f(x)dx=0$ for $n=0,1,2,\ldots$. By Riesz-Representation Theorem, $\theta:\mathcal{H}\rightarrow\mathbb{R}$, $\theta(g)=\langle f,g\rangle$ is a bounded linear functional. Note that if $g\in\mathcal{H}$ is a polynomial function, then $\theta(g)=0$. Recall that the linear subspace $\mathcal{P}$ of all polynomial functions is $||\cdot||_{2}$ dense in $\mathcal{H}$. (I elaborate a little bit: By Stone-Weierstrass Theorem, polynomial functions can uniformly approximate a continuous function on $[0,1]$ and hence polynomial functions can approximate a continuous function in $||\cdot||_2$-norm because the Lebesgue measure restricted on $[0,1]$ is finite. Lastly, it is well-known that the set of all continuous functions defined on $[0,1]$ is $||\cdot||_2$-dense in $L^2[0,1]$.) By continuity of $\theta$ and density of $\mathcal{P}$ in $\mathcal{H}$, we have that $\theta(g)=0$ for all $g\in\mathcal{H}$. That is, $\theta=0$. Therefore, $||f||_{2}=||\theta||=0$ and hence $f=0$ a.e.

Answer 3

Shifted Legendre polynomials $P_n(2x-1)$ provide a complete orthogonal base of $L^2(0,1)$. If the integral of $x^n f(x)$ over $(0,1)$ equals zero for any $n\in\mathbb{N}$, it follows that $\int_{0}^{1} f(x)P_n(2x-1)\,dx$ also equals zero, hence $\int_{0}^{1}f(x)^2\,dx = 0$ by Parseval's identity and $f(x)=0$ almost everywhere in $[0,1]$.