Square-summable Fourier coefficients implies square integrable? Let $f\in L^1([0,1])$, the $n$-th Fouier coefficient of $f$ is $\widehat{f}(n)=\int_{0}^1 f(x)e^{-2\pi inx}\,dx$. Show that $\sum_{n=-\infty}^\infty|\widehat{f}(n)|^2<\infty$ implies $f\in L^2([0,1])$. I want to use the Parseval identity, but it need the assumption $f\in L^2([0,1])$. I have no idea how to start.
 A: Note: You have the wrong Fourier coefficents on $[0,1]$. The coefficients should be
$$
                     \hat{f}(n)=\int_{0}^{1}f(t)e^{-2\pi in t}dt.
$$
$\{ e^{2\pi in t}\}_{n=-\infty}^{\infty}$ is an orthonormal basis of $L^2[0,1]$. 
Every function $f \in L^2[0,1]$ is also in $L^1[0,1]$ because
$$
                    2|f| \le |f|^{2}+1.
$$
If you have $\sum_n |\hat{f}(n)|^{2}<\infty$ for some $f \in L^1$, then $g=\sum_n a_ne^{2\pi int}$ defines a function $g \in L^2\subset L^1$ whose Fourier coefficients are the same as those of $f$, i.e. $\hat{f}(n)=\hat{g}(n)$ for all $n$. So you have to be able to show that if $f,g\in L^1$ and $\hat{f}(n)=\hat{g}(n)$ for all $n$, then $f=g$ a.e..
The function $h=f-g$ is in $L^1$ and has all $0$ Fourier coefficients. The function $k(x)=\int_{0}^{x}h(t)dt$ is absolutely continuous with $k'=h$ a.e. by the Lebesgue differentiation theorem. Furthermore, $k$ is periodic on $[0,1]$ because $\hat{h}(0)=0$. Hence,
\begin{align}
            \hat{k}(n) & = \int_{0}^{1}e^{-2\pi int}k(t)dt \\
                & = \left.\frac{e^{-2\pi int}}{-2\pi in}k(t)\right|_{t=0}^{1}+\int_{0}^{1}\frac{e^{-2\pi int}}{-2\pi in}k'(t)dt \\
    & = \frac{\hat{h}(n)}{2\pi in} = 0,\;\;\; n=\pm 1,\pm 2,\pm 3,\cdots .
\end{align}
Because $k \in L^2[0,1]$, then uniqueness of Fourier coefficients implies that $k$ is equal a.e. to a constant function. Therefore, $0=k'=h$ a.e..
A: We will prove that if the sequence $(c_n)_{n \in \mathbb{Z}}$ is in $\ell^2(\mathbb{Z})$, then the series 
\begin{equation}
f(x) := \sum_{n \in \mathbb{Z}} c_n e^{2 \pi i n x}
\end{equation}
converges to a function $f(x)$ that is in $L^2(0,1)$. 
First recall that $(e^{2\pi i n x})_{n \in \mathbb{Z}}$ is a complete orthonormal system here in $L^2(0,1)$. Then
\begin{equation}
\left|\left| \sum_{ N \leq |n| \leq M} c_n e^{2\pi i n x} \right|\right|^2_{L^2(0,1)} = \sum_{ N \leq |n| \leq M} |c_n|^2.
\end{equation} 
Since the right-hand side can be made arbitrarily small choosing $M,N$ large, we see that 
\begin{equation}
f_N(x) = \sum_{|n| \leq N} c_n e^{2\pi i n x}
\end{equation}
is such that $(f_N(x))_{N\geq 1}$ is a Cauchy sequence in $L^2(0,1)$. Consequently, it converges to some $f \in L^2(0,1)$. One obtains $\hat{f}(n) = \langle f, e^{2\pi i n x} \rangle = c_n$, by orthogonality: 
\begin{align}
\langle f(x), e^{2\pi i n x} \rangle &=  \langle \lim_N f_N(x), e^{2\pi i n x} \rangle \\
&= \lim_N \langle f_N(x), e^{2\pi i n x} \rangle \\
&= \lim_N \sum_{|n'|\leq N} c_{n'} \langle e^{2\pi i n' x}, e^{2\pi i n x} \rangle \\
&= c_{n},
\end{align}
where we used continuity of the inner product, and at the last step, orthogonality of the basis functions.
