there is $M<\infty$ such that $\sum_{n} |\hat{f}(n)|\le M\int_{0}^{2\pi}|f(t)|dt$ for each $f\in X$

for $f\in L^1[0,2\pi]$ define $$\hat{f}(n)=\int_{0}^{2\pi} f(t)e^{-int} dt$$ for $n\in\mathbb{Z}$, $X$ is a closed linear subspace of $L^1[0,2\pi]$ such that $\sum_{n} |\hat{f}(n)|<\infty$ for each $f\in X$, we need to show that there is $M<\infty$ such that $\sum_{n} |\hat{f}(n)|\le M\int_{0}^{2\pi}|f(t)|dt$ for each $f\in X$

please tell me how to solve this one, I have no clue.

Solutions Collecting From Web of "there is $M<\infty$ such that $\sum_{n} |\hat{f}(n)|\le M\int_{0}^{2\pi}|f(t)|dt$ for each $f\in X$"

Let $T\colon X\to \ell^1$ defined by $T(f)=\{\widehat f(n)\}_{n=-\infty}^{+\infty}$. As $X$ and $\ell^1$ are complete and $T$ is linear, we just need to check that the graph of $T$ is closed.

Let $\{f_k\}\subset X$ such that $f_k\to 0$ in $L^1$ and $T(f_k)\to y$ in $\ell^1$. We have for all $n$ that $\widehat{f_k}(n)\to 0$ and $y_n=\lim_{k\to +\infty}\widehat{f_k}(n)$ (as convergence in $\ell^1$ implies coordinatewise convergence. We conclude that $y=0$.