A difficulty in understanding an example of an infinite dimensional generalization of the spectral theorem for compact self adjoint operators.

Question

The first part of the example is given below:

Why this operator is bounded and why we are sure that its range is in $L_{2}([-\pi,\pi])$, could anyone explain this for me?

Thanks!

Answer 1

\begin{align*} \|Kf\|_{L^{2}[-\pi,\pi]}&=\left(\int_{-\pi}^{\pi}\left|\int_{-\pi}^{\pi}h(t-s)f(s)ds\right|^{2}dt\right)^{1/2}\\ &=\left(\int_{-\pi}^{\pi}\left|\int_{t-\pi}^{t+\pi}h(s)f(t-s)ds\right|^{2}dt\right)^{1/2}\\ &\leq\left(\int_{-\pi}^{\pi}\left(\int_{t-\pi}^{t+\pi}|h(s)|\cdot|f(t-s)|ds\right)^{2}dt\right)^{1/2}\\ &\leq M\left(\int_{-\pi}^{\pi}\left(\int_{t-\pi}^{t+\pi}|f(t-s)|ds\right)^{2}dt\right)^{1/2}\\ &=M\left(\int_{-\pi}^{\pi}\left(\int_{-\pi}^{\pi}|f(s)|ds\right)^{2}dt\right)^{1/2}\\ &\leq M\left(\int_{-\pi}^{\pi}2\pi\|f\|_{L^{2}[-\pi,\pi]}^{2}dt\right)^{1/2}\\ &=2M\pi\|f\|_{L^{2}[-\pi,\pi]}, \end{align*} where $M=\sup_{s\in[-\pi,\pi]}|h(s)|$ and note that $h$ is $2\pi$-periodic, then $\sup_{s\in[t-\pi,t+\pi]}|h(s)|=M$.
And the last inequality is by Holder's inequality.