$\int_0^1 x^n f(x)\,\mathrm{d}x = 0$ for all $n$ implies $f=0$ almost everywhere

Question

It has been shown that given $f \in \mathcal C[0,1]$, we have that if $\int_0^1 x^nf(x)\,dx = 0$ for all $n \in \mathbb N$, then $f = 0$.

I was thinking of generalizing this statement to $L^p$ spaces. For instance, if we have that $f \in L^\infty[0,1]$ instead, and $\int_0^1 x^nf(x)\,dx = 0$ for all $n \in \mathbb N$. Can we say that $f = 0$ a.e on $[0,1]$?

That is, if $f$ is measurable on $[0,1]$, with $\|f\|_\infty = \inf\{a \geq 0 \mid \lambda(|f|^{-1}(a,\infty]) = 0\} < \infty$ given $|f|^{-1}(a,\infty] = \{x \in [0,1] \mid |f(x)| > a\}$, can we show that $f = 0$ for all points in $[0,1]$ except some set $B$ where $\lambda(B) = 0$? In this case, we use $\lambda$ to denote the Lebesgue measure.

Answer 1

Suppose $f\in L^1([0,1])$ (this is weaker than $f\in L^{\infty}([0,1])$) is such that there is an integer $n_0\geq 0$ such that for all $n\geq n_0$, we have $\int_0^1x^nf(x)\,dx=0$. Then, define $\phi(x)=x^{n_0}f(x)$. This lies in $L^1([0,1])$, and for each $n\geq 0$, we have $\int_0^1x^n\phi(x)\,dx=0$. By linearity, this implies that for all polynomials $P(x)$, $\int_0^1P(x)\phi(x)\,dx=0$, and so for all continuous functions $g\in C([0,1])$, and all polynomials $P$, we have \begin{align} \left|\int_0^1g(x)\phi(x)\,dx\right|&=\left|\int_0^1g(x)\phi(x)\,dx-\int_0^1P(x)\phi(x)\,dx\right|\\ &\leq\|\phi\|_1\cdot \|g-P\|_{\infty}. \end{align} By Weierstrass’ approximation, we can uniformly approximate $g$ by polynomials, so $\|g-P\|_{\infty}\to 0$ along a suitable sequence of polynomials, and hence $\int_0^1g(x)\phi(x)\,dx=0$. A basic variant of the fundamental lemma of calculus of variations now tells us that arbitrariness of $g$ implies that $\phi=0$ a.e on $[0,1]$, and so dividing by $x^{n_0}$ on $(0,1]$ implies that $f=0$ a.e on $[0,1]$.

Answer 2

If $\mu=\mu_+-\mu_-$ is the difference of two bounded positive measures on $[0,1]$ then from the F. Riesz theorem, the map $g\mapsto \int_0^1g(x)\mu(dx)$ is a continuous linear form on the space $C[0,1]$ of continuous real functions on $[0,1]$ endowed with the sup norm. Since from the Wierstrass theorem the set of polynomials is dense in $C[0,1]$ then $\int_0^1 x^n\mu(dx)$ for all $n$ implies $\mu=0.$

Answer 3

For fun. What about a solution that relies on Fourier analysis?

Let $f\in L^1((0,1))$ satisfying the condition in your question, and consider it as a function in $L^1(\mathbb R)$ by extending it as zero outside $(0,1)$. By Paley-Wiener’s theorem, the Fourier transform $\widehat f$ of $f$ is analytic on $\mathbb R$. Since the Fourier transform conjugates derivatives into multiplication by polynomials and viceversa, from the fact that $\int x^n f(x)dx=0$, it follows that $$ \frac{d^n\widehat f}{dx^n}(0)=0 $$ for all $n$. In partiular, its Taylor series at the origin is identically zero. Since $\widehat f$ is analytic, it must be $\widehat f\equiv 0$. By the injectivity of the Fourier transform on $L^1(\mathbb R)$, $f=0$ a.e..

(Edit: if the integral is zero only for $n$ large enough, the above shows that $\widehat f$ is a polynomial. But since $f\in L^1$, $\widehat f$ is continuous and goes to zero at infinity, so $\widehat f$ must still be identically zero.)

(Edit 2: this proof has the advantage that it works equally well for distributions $f\in\mathscr D’((0,1))$ with compact support in $(0,1)$.)