Let $\mathbb{T}=[-\pi,\pi]$ and let $\hat{f}(n)$ be the $n$-th Fourier coefficient of a function $f$.
I am asked to give an example of a sequence $\{a_n\}_{n\in\mathbb{Z}}$ such that $a_n\to 0$ and such that $\hat{f}(n)\neq a_n$ for any function $f\in L^1({\mathbb{T}})$. Let $\alpha$ be a real number such that $0<\alpha\leq 1/2 $. I think the sequence defined by
\begin{align} a_n=\begin{cases} \frac{1}{n^\alpha} &\text{ if } n\geq1,\\ \frac{-1}{|n|^{\alpha}} &\text{ if } n\leq-1, \end{cases} \end{align} may work, but I'm not sure if the following argument is correct:
I can see that this sequence verifies $a_n\to0$ as $n\to\pm\infty$. Assume there exists a function $f\in L^1(\mathbb T)$ such that $\hat{f}(n)=a_n$ for all $n\in\mathbb{Z}$. Then, by Parseval identity,
$$ ||f||_1^2 = \sum_{n\in\mathbb{Z}} |\hat{f}(n)|^2=2\sum_{n\geq1}\frac{1}{n^{2\alpha}}.$$ This is a contradiction because RHS diverges but $||f||_1<\infty$ since $f\in L^{1}(\mathbb T)$.
My main doubt is whether Parseval identity holds in $L^1(\mathbb{T})$, or if I'm missing some detail I'm not aware of. Any help will be appreciated, thank you.
If $a_n$ decreases to $0$ then $\sum a_n \sin nx$ converges for all $x$. This series is the F.S. of a function in $L^{1}(T)$ if and only if $\sum\frac{a_n} n <\infty$. So a counter-example is provided by $a_n=\frac 1{\ln (n+1)}$
Ref: Theorem 7.3.3, p.115 of Fourier Series by Edwards.