Let $\{f_n\}_{n\in\mathbb{N}}$ be a Cauchy sequence of $L^2(\mathbb{T})$ functions, where $\mathbb{T}=[-\pi,\pi)$. Assume that, $\forall n\in\mathbb{N}$ and for $k\geq 0$, the following inequality holds:
$$\sum_{m=-\infty}^{+\infty} (1+m^2)^k |\hat{f_n}(m)|^2 \lt +\infty$$
where $\hat{f_n}$ is the Fourier transform of $f_n$.
I have to show that this sequence converges to a function $f\in L^2(\mathbb{T})$ which satisfies the same inequality.
This question is related to another problem I posted here a few days ago (the link is below), but I couldn't find a solution.
As another user said, probably the best way to approach this problem is to notice that, if $||\cdot||_2$ is the usual norm in $L^2(\mathbb{T})$, the sequence converges to a function $f$ in $L^2(\mathbb{T})$ and then to show that
$$\sum_{m=-\infty}^{+\infty} (1+m^2)^k |\hat{f_n}(m)-\hat{f}(m)|^2 \rightarrow 0$$
as $n \rightarrow+\infty$, but I really don't know how to show it. Any ideas?
Previous question: Real positive index Sobolev spaces are Hilbert spaces
The statement is not true for $k\ge 1$. ($k=0$ case is trivial by Parseval's identity.) Note that any trigonometric polynomial $f$ satisfies the condition $$ \sum_{j\in\mathbb{Z}} (1+|j|^2)^k |\widehat{f}(j)|^2 <\infty.\tag{*} $$
Let $F(x) = \sum_{n=1}^\infty \frac{\sin nx}{n}=\frac{\pi-x}{2}$, $0<x<2\pi$ and $f_n$ be the Fourier series of $F$. Then each $f_n$ converges to $F$ in $L^2$-norm and satisfies $(*)$. However, since $|\widehat{F}(j)| \ge \frac{c}{|j|}, j\ne 0$ for some $c>0$, $(*)$ cannot hold for $F$ for all $k\ge 1$. This can be a counterexample : $f_n$ is Cauchy in $L^2$ and satisfies $(*)$ but the limit $F$ does not.
EDIT: Define a norm as$$ \|f\|_s^2 = \sum_{j\in\Bbb Z}(1+|j|^2)^{s}|\widehat{f}(j)|^2. $$ Suppose $f_n$ is Cauchy in $\|\cdot\|_s$, i.e. $$ \sum_{j\in\Bbb Z}(1+|j|^2)^{s}|\widehat{f_n}(j)-\widehat{f_m}(j)|^2\to 0 $$ as $n,m\to\infty$. Then for each fixed $j$, $\widehat{f_n}(j)$ is a Cauchy sequence in $\Bbb C$ and $\lim_n \widehat{f_n}(j) =: s_j$ exists. Since $\|f_n-f_m\|^2_{L^2(\Bbb T)}\le \|f_n-f_m\|^2_s$, there exists $f\in L^2(\Bbb T)$ such that $f_n\to f$ in $L^2(\Bbb T)$ and $\widehat{f}(j) = s_j$. Now, for a given $\epsilon>0$, we can find $N\in\Bbb N$ such that $$ \sum_{j\in\Bbb Z}(1+|j|^2)^{s}|\widehat{f_n}(j)-\widehat{f_m}(j)|^2\le \epsilon^2 $$ for all $n,m\ge N$. Observe that for all finite $J\in\Bbb N$, $$ \sum_{j=-J}^J(1+|j|^2)^{s}|\widehat{f_n}(j)-\widehat{f_m}(j)|^2\le \epsilon^2 $$ for all $n,m\ge N$. Take $m\to\infty$. Then $$ \sum_{j=-J}^J(1+|j|^2)^{s}|\widehat{f_n}(j)-\widehat{f}(j)|^2\le \epsilon^2. $$ Take $J\to\infty$. Then $$ \sum_{j\in\Bbb Z}(1+|j|^2)^{s}|\widehat{f_n}(j)-\widehat{f}(j)|^2\le \epsilon^2 $$ for all $n\ge N$. This proves $$\lim_{n\to\infty}\|f_n -f\|_s^2 =0$$ and $$ \|f\|_s^2<\infty $$as desired.