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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?

mowzorn
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1 Answers1

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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.

JackEight
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    I love this approach, but just to be sure, can you explain what you're using to justify the limit interchange? It's probably obvious but I'm not quite that advanced in this topic – Stephen Donovan Jan 04 '23 at 20:45
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    $ \left| \displaystyle\int_a^b p_k(t) (f(t) - g(t)) , dt - \displaystyle\int_a^b (f(t) - g(t))^2 , dt \right| = \left| \displaystyle\int_a^b (p_k(t) - (f(t) - g(t))) (f(t) - g(t)) , dt \right| \leq \underset{x \in [a, b]}{\sup} | p_k(x) - (f(x) - g(x)) | \times \displaystyle\int_a^b \left| f(t) - g(t) \right| , dt \underset{k \to +\infty}{\longrightarrow} 0$ – JackEight Jan 04 '23 at 20:48
  • That makes a lot of sense actually, thank you – Stephen Donovan Jan 04 '23 at 20:51