I know this fact: "Suppose that $f\in L^{1}(\mathbb T)$ and $\hat{f}(n)=0$ for all $n\in \mathbb Z,$ then $f=0 $ all most everywhere on $\mathbb T$."
My Question is: Suppose that $f\in L^{1}(\mathbb T)$ and $\hat{f}(n)=0$ for all $n\in \mathbb Z \setminus \{ 0 \},$ and $\hat{f}(0) \in \mathbb C.$ Can we say $f=0$ all most every where on $\mathbb T$ ? If not (example?), what can we say about $f$?
Edit: In view of comments below, how to show $f$ is constant a. e. on $\mathbb T$?
Following David C. Ullrich's suggestion in the comments, let $g=f-\hat f(0)$. It is fairly straightforward to show $\hat g(n)=0$ for all $n\in\mathbb Z$, so by the usual uniqueness theorem, $g=0$ almost everywhere. This implies $f=\hat f(0)$ almost everywhere, i.e. $f$ is constant almost everywhere.