I need to prove or disprove
$f_n\in R(\displaystyle \mathbb{T})$ sequence of integrable functions in $[0,2\pi]$. $$\forall g\in R(\displaystyle \mathbb{T}), \langle f_n,g \rangle \xrightarrow[]{n\rightarrow\infty} 0 \Rightarrow \|f_n\|_{L_2}\rightarrow0$$
What I've been thinking is this:
$\hat{f_n}(m)= \langle f_n,e_m \rangle, \hat{f_n}(m) \to_{n\to\infty }0$.
$S_Mf_n=\sum_\limits{m=-M}^{M}\hat f_n(m)(t)e^{imt} \xrightarrow[]{M\to\infty} f_n(t)$ in $L_2$.
$\lim\limits_{n \to \infty}(\lim\limits_{m \to \infty}\sum_{-M}^{M}\hat{f_n}(m)(t)e^{imt})=\lim\limits_{m \to \infty}(\lim\limits_{n \to \infty}\sum_{-M}^{M}\hat{f_n}(m)(t)e^{imt})=0$ if $S_Mf_n$ converges uniformly ($M\to\infty$), but not always. If we can change the order of the limits , we get
$\lim\limits_{n \to \infty}(\lim\limits_{m \to \infty}\|S_Mf_n-f_n\|_{L_2})= \lim\limits_{m \to \infty}(\lim\limits_{n \to \infty}\|S_Mf_n-f_n\|_{L_2})= \lim\limits_{n \to \infty}\|f_n\|_{L_2}=0$
So a counter example could be a sequence of functions such that $S_Mf_n$ doesn`t converge uniformly, and $\hat{f_n}(m) \to 0$ for all m.
Can anyone provide a counter-example or point out my mistake?
Best Answer
It's not true.
Consider the orthonormal basis $\{e_n : n \in \mathbb{Z}\}$ for $L^2[0, 2\pi]$, where $e_n(t) = \frac1{\sqrt{2\pi}}e^{int}$.
Then for any $g \in L^2[0,2\pi]$ we have $\langle e_n, g\rangle \xrightarrow{n\to\infty} 0$ by the Riemann-Lebesgue lemma, but $\|e_n\|_{L^2} = 1, \forall n\in\mathbb{N}$.