Let $f$ be a function on $[0,2\pi)$ satisfying $(\hat{f}(k))_{k\in \mathbb{Z}}\in l^1(\mathbb{Z}),$ where $$\hat{f}(k)=\int_0^{2\pi}f(t)e^{-ikt}dt.$$
Is it always true that $$f(x)=\sum_\mathbb{Z}\hat{f}(k)e^{ikx}, \text{ for almost every } x\in [0,2\pi)?$$
Which condition on $f$ guarantees above?
$(\hat{f}(k))_{k\in \mathbb{Z}} \in l^1(\mathbb{Z})$ implies that the Fourier series converges uniformly and absolutely to $f$ and $f$ is continuous. So am I required to assume that $f$ is continuous?
(I know there are some constant missing that normalises the transformation. Please ignore that)
True for any $f \in L^{1}$ with $\hat {f} \in \ell^{1}(\mathbb Z)$. Let $g(x)=\sum \hat {f} (k)e^{ikx}$. The series is uniformly convergent by M-test so $g$ is a continuous periodic function. By uniform convergence we get $\int e^{-int} g(t)dt=\sum \hat {f} (k) \int e^{i(k-n) t} dt $ which shows that $\hat {f}=\hat {g}$. Since $f$ and $g$ are integrable this implies $f=g$ a.e. and hence $f=\sum \hat {f} (k)e^{ikx}$ a.e.