Let $\Omega = [0, 1)$ with Lebesgue measure.
If $f\in L^2 (\Omega)$ and we have the following series
$$ f(x) = \sum_{n=-\infty}^{\infty} a_n \exp(2 \pi i n x),$$
where $i =\sqrt{-1}$, then can we use this series for $f\in L^1(\Omega)$?
Let $\Omega = [0, 1)$ with Lebesgue measure.
If $f\in L^2 (\Omega)$ and we have the following series
$$ f(x) = \sum_{n=-\infty}^{\infty} a_n \exp(2 \pi i n x),$$
where $i =\sqrt{-1}$, then can we use this series for $f\in L^1(\Omega)$?
On
If $f\in L^2[0,1]$, and if the series $\sum_{n=-\infty}^{\infty}c_n e^{2\pi int}$ converges in $L^2$ to $f$, then $f\in L^1$ and $$ \|f-\sum_{n=-N}^{N}c_ne^{2\pi nt}\|_{L^1}=\int_{-\pi}^{\pi}|f-\sum_{n=-N}^{N}c_ne^{2\pi int}|dt \\ \le \left(\int_{-\pi}^{\pi}1^2dt\right)^{1/2}\left(\int_{-\pi}^{\pi}|f-\sum_{n=-N}^{N}c_ne^{2\pi nt}|^2\right)^{1/2}\rightarrow 0 \mbox{ as } N\rightarrow\infty. $$
When $f \in L^{2}$ the infinite sum is interpreted as a sum in the $L^{2}$ norm sense. For $f \in L^{1}$ the series need not converge w.r.t either the $L^{1}$ norm or the $L^{2}$ norm.