Suppose $g \in L^{1}([0,1])$ and define:
$\hat{g}(n)=\int g(x)e^{-2\pi inx}dx$ for $n\in\mathbb{Z}$.
I want to show that if $g\in L^{1}([0,1])$
and $\{\hat{g}(n)\}\in\ell^{1}(\mathbb{Z})$
then $S_{N}g(x)=\underset{|n|\leq N}{\sum}\hat{g}(n)e^{2\pi inx}$ converges on $[0,1]$.
Attempt so far:
We know that $\sum|\hat{g_n}|<\infty$.
Being absolutely convergent, can define $g_0:\mathbb{R}\rightarrow C$ as: $$g_0(x)=\sum^{\infty}_{n=-\infty}\hat{g}e^{2\pi inx}=lim_{N \rightarrow \infty}S_Ng(x)$$ We can observe that the limit is uniform in $x$ because: $|g_0(x)-S_Ng| \leq \sum_{|n|>N}\hat{g}(n)$ for $x \in [0,1]$
Which leads to: $$\lVert g_0 - S_N g \rVert_{\infty}\leq \sum_{|n|>N}\hat{g}(n)\rightarrow 0$$ as $N \rightarrow \infty$
So,
$$lim_{N \rightarrow \infty} \lVert \frac{S_0g + S_1g+...+S_{N-1}g}{N} \rVert_{\infty}=0$$
Is the above attempt legitimate? Any clarifications and all criticisms are welcome.
You have proved the Fourier series converges uniformly to some function $g_0$. To show that $g=g_0$ a.e. it is enough to show that $\hat g(n)=\hat{g_0}(n)$ for all $n$. For this write down the definition of $\hat{g_0}(n)$ and interchange the integral and the limit (which is permissible by uniform convergence).