Proving that the Fourier coefficients of a functional determine it

Question

I have the following exercise, taken from old homework of a functional analysis course:

Let $\mu\in C(\mathbb{T})^{*}$. Define the Fourier coefficients of $\mu$ by $$ \hat{\mu_{n}}=\mu(e^{2\pi int}) $$

Prove that if $\mu_{1},\mu_{2}$ have the same Fourier coefficients then $\mu_{1}=\mu_{2}$

I have tried to prove this by calculating $\mu_{1}(f)$ for $f\in C(\mathbb{T})$ and tried to prove that it is the same as $\mu_{2}(f)$.

My problem is that I have used $$ f=\sum_{-\infty}^{\infty}\hat{f}(n)e^{2\pi int} $$

which I was told that is not necessarily true in $C(\mathbb{T})$, but rather in $L^{2}$.

I was also told that I should of used the Stone-Weierstrass theorem, but I didn't understand how.

Can someone please explain how can I use S-W in order to solve the question ?

Coefficients are not furry. I assume you talk about Fourier coefficients? — Hagen von Eitzen, Oct 04 '13 at 15:45
@HagenvonEitzen - Thanks for the correction. Spellcheck didn't catch it for obviousreasons so I assumed I typed it right :P — Belgi, Oct 04 '13 at 15:49
You could use Fejér's theorem. Or Stone-Weierstraß. And Hahn-Banach. — Daniel Fischer, Oct 04 '13 at 15:54

score 3 · Accepted Answer · answered Oct 04 '13 at 16:55

Letting $\mu = \mu_1-\mu_2$, the question boils down to showing that if $\hat{\mu}_n = 0$ for all $n$, then $\mu = 0$.

Let me use the notation $e_n(t) = e^{i 2 \pi n t}$.

Suppose $\mu(f) > 0 $ for some $f \in C(\mathbb{T})$. Let $p_n = \sum_i \alpha_n^i e_i$ be a sequence of trigonometric polynomials converging uniformly to $f$. Since $\mu(p_n) = \sum_i \alpha_n^i \mu(e_i)$, we see that for sufficiently large $n$, we have $\mu(p_n) > 0$. In particular, we must have $\hat{\mu}_k=\mu(e_k) \neq 0$ for some $k$.

Hence if $\mu$ is non zero, the $\hat{\mu}_k$ cannot all be zero.

Proving that the Fourier coefficients of a functional determine it

1 Answers1