Proving that $f\equiv g$ if $\int_a^bt^nf(t)dt=\int_a^bt^ng(t)dt$

Question

I'm having some trouble with the following exercise:

Let $f,g:[a,b]\to\mathbb R$ be continuous functions such that:$$\int_a^bt^nf(t)dt=\int_a^bt^ng(t)dt$$ for all $n\in \mathbb N$. Prove that $f\equiv g$.

This exercise is at the end of the chapter about the Stone-Weierstrass theorem, so I tried to apply it but I got nowhere. I did the following:

We know that $$\int_a^bt^n(f(t)-g(t))dt = 0$$

for all $n\in \mathbb N$. According to the Stone-Weierstrass, there is a sequence of polynomials $p_k:[a,b]\to \mathbb R$ that converge uniformly to $f-g$. Using this and the integral, I tried to prove that $p_k\to0$, but I wasn't able to do so.

How can this be done?

Answer 1

We easily have for every polynomial $P$: $\displaystyle\int_a^b P(t) (f(t) - g(t)) \, dt = 0$.

Using your sequence $(p_k)$, we have for every $k$: $\displaystyle\int_a^b p_k(t) (f(t) - g(t)) \, dt = 0$.

Now all you basically have to do is take the limit $k \to +\infty$ to see that $\displaystyle\int_a^b (f(t) - g(t))^2 \, dt = 0$, and conclude.