I need to show that: for $n=1,2,...$ if $f \in L^1(\mathbb{T}) $ with $\sum_{-\infty}^\infty |\hat{f}_n(k)| \leq 1$ for every $n$ and if $f_n$ converges uniformly to f on $\mathbb{T}$, then $\sum_{-\infty}^\infty |\hat{f}(k)|\leq 1.$
My attempt:
Since $f_n$ converges to f uniformly on the circle, we know $\forall \epsilon>0 \exists N \in \mathbb{N} s.t. \forall x \in \mathbb{T}$ and $\forall n \geq N, |f_n(x)-f(x)|\leq \epsilon.$
This implies $|\hat{f}(k)-\hat{f}_n(k)|\leq \epsilon $ $ \forall |n|\geq N$
Thus,
$\sum_{-\infty}^\infty |\hat{f}(k)|= \sum_{-\infty}^\infty |\hat{f}(k)- \hat{f}_n(k)+\hat{f}_n(k)|\leq 1+ \epsilon \sum_{-\infty}^\infty1$
This does not seem useful for my problem...
So $\widehat{f_{n}}(k)\rightarrow\widehat{f}(k)$ as $n\rightarrow\infty$, for each $k$, then Fatou's Lemma gives \begin{align*} \sum_{k=-\infty}^{\infty}|\widehat{f}(k)|&=\sum_{k=-\infty}^{\infty}\lim_{n\rightarrow\infty}|\widehat{f_{n}}(k)|\\ &\leq\liminf_{n\rightarrow\infty}\sum_{k=-\infty}^{\infty}|\widehat{f_{n}}(k)|\\ &\leq\liminf_{n\rightarrow\infty}1\\ &=1. \end{align*}