Problem:
Prove that if $f:\left [ 0,1 \right ]\rightarrow \mathbb{R}$ is a continuous function such that: $\int_{0}^{1}f(x)e^{nx}dx=0$ for all $n=0,1,2,…$. Prove that $f(x)=0$ for all $0\leqslant x\leqslant 1$ using two methods:
1/ change of variables and then apply Weiestrasss Theorem.
2/ Apply Stone Weierstrass Theorem .
I already know how to prove that $f(x)=0$ for all $0\leqslant x\leqslant 1$ if $\int_{0}^{1}f(x)x^{n}dx=0$. I did the following change of variable: $e^{x}=y$ and then I got $\int_{1}^{e}f(y)y^{n-1}dy$. since $\int_{0}^{1}f(y)x^{n-1}dy=0$, we are left with $\int_{1}^{e}f(y)x^{n-1}dy$ which I want to prove equal to zero. Any help with this?
For the second part, I don't have any idea how to use the Stone Weierstrass Theorem to prove it. I have never used this theorem before in solving problems, so I appreciate if someone helps with details for this part.
Best Answer
I think you are on the wrong track with the substitution. Do try to prove:
if $A$ is a dense subset of $C([0,1])$ (wrt uniform convergence, of course) and, for $f\in C([0,1]),$ $\int_0^1 f(x) g(x) dx = 0 \quad\forall g\in A$ then $ f=0$.
Weierstrass/Stone-Weierstrass give you dense subsets of $C([0,1])$. The Weierstrass part should be more or less obvious with the statement I asked you to prove. For the Stone-Weierstrass part try to find an algebra satisfying the hypothesis of the theorem by looking at the functions which you are given, $e^{nx}$.