Convolution in $L^2(-\pi ,\pi )$ I'm reading about the Fourier transform and it's properties in The Spectral Analysis of Time Series - Koopmans, L. and I found a particular statement that makes me feel a little bit skeptical.
Lets assume that $x(t)$ and $y(t)$ belong to $L^2( -\pi ,\pi )$ and let $z=x*y$ where $*$ indicates the convolution. The author states that as a consequence of Schwartz inequality we have that $z(t)$ belongs to $L^2(-\pi ,\pi)$.
I know that Fubini's Theorem gives us a similar result but considering $L^1$, not $L^2$ and after a quick search in the site I found also some counterexamples of a similar statement (considering $L^2(R)$ ). So I would like to know if I am missing something, or the author is making further assumptions that I don't see. 
Thanks.
 A: Note that if $x(t)=\sum_k a_k\mathrm{e}^{ikx}$, then $\|x\|_{L^2}^2=\sum_k|a_k|^2$. 
If $x(t)=\sum_k a_k\mathrm{e}^{ikx}$ and $y(t)=\sum_k b_k\mathrm{e}^{ikx}$, then
$(x*y)(t)=\sum_k c_k\mathrm{e}^{ikx}$, where $c_k=\sum_j a_jb_{k-j}$. 
Hence
$$
\|x*y\|^2=\sum_k |c_k|^2=\sum_k\left|\sum_j a_jb_{k-j}\,\right|^2\le \sum_{k,j}|a_k|^2|b_j|^2=\|x\|^2\|y\|^2
$$
A: Holder's inequality shows immediately that the convolution is bounded and any $L^{\infty}$ function is an $L^{2}$ function on a finite interval. Some details: $|\int f(x-y)g(y)\, dy| \leq \sqrt {\int |f(x-y)|^{2} \, dy}\sqrt {\int |g(y)|^{2} \, dy}$. Since $\sqrt {\int |f(x-y)|^{2} \, dy}=\sqrt {\int |f(y)|^{2} \, dy} $ for all $x$ we get $|(f*g)(x)| \leq \|f\|_2 \|g|_2$ for all $x$. Hence $f*g$ is a bounded measurable function. This implies that it is also in every $L^{p}$, in particular $L^{2}$.
A: I'll use $f,g$ for ease. I'll also assume they are real so that I don't have to conjugate stuff.
$$||f*g||_2^2 =\int |\int f(y)g(x-y)dy|^2 dx = \int \int \int f(y)g(x-y)f(z)g(x-z)dxdydz \le \int \int |f(y)||f(z)| (\int |g(x-y)|^2 dx)^{1/2}(\int |g(x-z)|^2 dx)^{1/2} dydz = \int \int |f(y)||f(z)| ||g||_2^2 \le ||f||_1^2||g||_2^2$$
