Norm of $Tf(x)=\int_{-\pi}^{\pi} \cos(x-y)f(y)\,dy$

Question

I'm dealing with this operator $T\in \mathcal{L}(H)$ where $H=L^2([-\pi,\pi])$ $$ Tf(x) = \int_{-\pi}^{\pi} \cos(x-y)f(y)\,\mathrm{d}y $$ One of the questions request to compute the norm of the operator. I tried but only show that $$ \|Tf\|\leq \sqrt{2}\pi \|f\| $$ with Cauchy-Schwarz. I already show that this operator is compact and self-adjoint hence $$ \|T\| = \rho(T) = \sup_{\lambda\in\sigma_p(T)}|\lambda| $$ Since $\sigma(T)=\{0,\pi\}$, then $\|T\|=\pi$. I want to get the same result by definition of norm but I can't find another bound. Does anyone have any suggestion for compute the norm?

Answer 1

You can write $$ Tf = \langle f,\cos\rangle\cos +\langle f\sin\rangle\sin $$ where $\langle g,h\rangle=\int_{-\pi}^{\pi}g(x)h(x)dx$ if your space is a real space, and where $\langle g,h\rangle=\int_{-\pi}^{\pi}g(x)\overline{h(x)}dx$ if your space is complex. Whether the space is complex or real, $\|\sin\|=\sqrt{\pi}$ and $\|\cos\|=\sqrt{\pi}$, and $\{ \frac{\sin}{\sqrt{\pi}},\frac{\cos}{\sqrt{\pi}}\}$ is an orthonormal subject of $L^2[-\pi,\pi]$; and $Tf$ is a constant multiple (the multiple is $\sqrt{\pi}$) of the orthonormal projection of $f$ onto the subspace spanned by $\{\cos(x),\sin(x)\}$. So the spectrum of $T$ is $\{ 0, \sqrt{\pi}\}$. The norm of $T$ is $\sqrt{\pi}$.