I wish to extend Parseval's Equality (or disprove that the following doesn't hold):
Conjecture. Let $f,g\in L^1[0,1)$ such that $f\overline{g}\in L^1[0,1)$. Then $$ \int_0^1 f(x)\overline{g(x)}\,dx = \sum_{n\in \mathbf{Z}}\widehat{f}[n]\overline{\widehat{g}[n]}. $$
I've tried doing an extension argument which can be seen below, but unfortunately it does not quite work.
Ideas. Let $\{f_k\}\subseteq L^2[0,1)$ and $\{\overline{g_k}\}\subseteq L^2[0,1)$ be truncated approximating sequences for $f$ and $\overline{g}$ respectively. Specifically, $$ f_k = \begin{cases} f(x), & |f(x)|\leq k \\ 0, & \text{otherwise}. \end{cases} $$ Define $\overline{g_k}$ similarly. Then the following equalities hold: $$ \int_0^1 f(x)\overline{g(x)}\,dx = \lim_{k\to \infty}\int_{0}^1 f_k(x)\overline{g_k}(x)\,dx = \lim_{k\to\infty}\sum_{n\in \mathbf{Z}}\widehat{f_k}[n]\overline{\widehat{g_k}[n]}, $$ where the first equality holds as $|f_k| \leq |f|$ and $|\overline{g_k}|\leq |\overline{g}|$, hence we can apply DCT. The second equality holds as $f_k, \overline{g_k}\in L^2[0,1)$, therefore, we can directly apply Parseval's Equality. Unfortunately, with the way I defined $f_k, \overline{g_k}$ and the lack of strict assumptions I do not see any way to interchange the limit and summation in the last term.
Any advice or counterexample would be greatly appreciate. If you see a way to add more assumptions, without it implying $f,g\in L^2[0,1)$, such that the equality holds, I would also be interested.