$f \in C([-1,1])$, given $\int_{-1}^{1}f(x)x^ndx = 0$ for $n = 0,1,2,...$ then $f(x)=0, \ \forall x \in [-1,1]$

Question

Let $f:[-1,1] \to \mathbb{R}$ be continuous on $[-1,1]$.

Assume $\displaystyle \int_{-1}^{1}f(x)x^ndx = 0$ for $n = 0,1,2,...$

Then show $f(x)=0, \ \forall x \in [-1,1]$

I would like to use Weierstrass Approximation Theorem (WAT) to prove this. Here is my incomplete attempt:

$f$ is continuous on $[-1,1]$ so by WAT $\exists$ a sequence polynomials $p_1, p_2, p_3,...,p_n,...$ s.t.

$\displaystyle \lim_{n \to \infty} |p_n - f| = 0$, so $\displaystyle \lim_{n \to \infty} \int_{-1}^{1}|f-p_n| = 0$.

Now, I would like to show $\int_{-1}^{1} f^2 = 0$ and from there I would like to conclude that $f=0$. So, \begin{align} \int_{-1}^{1} (f(x))^2 dx & = \int_{-1}^{1} f(x)f(x) \\ & = \int_{-1}^{1} \left[ f(x)f(x)-f(x)p_n(x)+f(x)p_n(x) \right] dx \\ & = \int_{-1}^{1} f(x) (f(x) - p_n(x)) dx + \int_{-1}^{1}f(x) p_n(x)dx \end{align}

$\int_{-1}^{1} f(x) p_n(x) dx = 0$ by the hypothesis since $\int_{-1}^{1} f(x) x^n dx = 0$ for all $n \in \mathbb{N}$.

So that leaves me to show,

$\int_{-1}^{1}f(x)(f(x)-p_n(x)) = 0$. I know $\lim_{n \to \infty}|f-p_n|=0$, but I am not clear how I can use this to show the integral is $0$. First of all, we are working with a fixed $n$, and second of all the limit of absolute difference is $0$.

Should I start the proof with limits and show the integral is arbitrarily small?

Thanks.

Answer 1

\begin{align*} \int_{-1}^{1}|f(x)||f(x)-p_{n}(x)|dx&\leq\sup_{x\in[-1,1]}|f(x)-p_{n}(x)|\int_{-1}^{1}|f(x)|dx\\ &=K\sup_{x\in[-1,1]}|f(x)-p_{n}(x)|, \end{align*} where $K$ is the integral, which is a finite number, now the supremum can be made arbitrarily small by Weierstrass theorem.

Answer 2

Since $$\limsup_{n\to\infty}|f(x)-p_n(x)|=0,$$ we know for any $\epsilon>0$, there is $N\in\mathbb{N}$ such that for $\forall n> N$ $$|f-p_n|<\frac{\epsilon}{2 M},$$ where $M:=\max_{[-1,1]} |f(x)|$.

Hence $$\left|\int_{-1}^1f(x) \left(f(x)-p_n(x)\right)\,dx \right|<\frac{\epsilon}{2 M}\int_{-1}^1 |f(x)|\,dx \leq\epsilon,\quad \forall n>N.$$