# 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.

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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 $y=0$.
How does $l_1$ comes and why? what does $L^1[0,2\pi]$ means, could you tell me please? – La Belle Noiseuse Jan 31 '13 at 7:09
The sequence of Fourier coefficients is summable, so the introduction of $\ell^1$ is natural. $L^1[0,2\pi]$ means the set of equivalence classes for equality almost everywhere of integrable functions on $[0,2\pi]$. – Davide Giraudo Jan 31 '13 at 10:18