Finding operator norm I have to solve the following problem: 
Find a norm of operator $$A:L^2[-\pi,\pi]\rightarrow L^2[-\pi,\pi]$$ given with $$Af(x)=\int_{-\pi}^{\pi} \cos^2{\left(\frac{x-t}{2}\right)}f(t) \,dt.$$
I proved that $\|A\|\leq 2\pi$, by using Holder's inequality, but I'm having difficulties to determine it exactly. Any help is welcome. 
 A: If you don't want to use the Fourier transform techniques suggested in the comments, here's something you can do. Recall that $L^2([-\pi, \pi])$ has an orthogonal basis consisting of the functions $1$, $\cos(mx)$ for $m\geq 1$ and $\sin(nx)$ for $n\geq 1$. You can explicitly compute what $A$ does to this orthogonal basis. For instance, if $f = \sin(nx)$, then $$(Af)(x) = \int_{-\pi}^\pi\cos^2\left(\frac{x-t}{2}\right)\sin(nt)\,dt = \begin{cases}\frac{1}{2}\pi\sin(x) & n = 1.\\
0 & n>1.\end{cases}$$ 
Similarly, if $f = \cos(nx)$, then  $$(Af)(x) = \int_{-\pi}^\pi\cos^2\left(\frac{x-t}{2}\right)\cos(nt)\,dt = \begin{cases}\frac{1}{2}\pi\cos(x) & n = 1.\\ 0 & n>1.\end{cases}$$
Finally, when $f \equiv 1$, you have $$(Af)(x) = \int_{-\pi}^\pi \cos^2\left(\frac{x-t}{2}\right)\,dt \equiv \pi.$$
(You should check these computations, as I could easily have made a mistake)
If these computations are indeed correct, this implies the norm of $A$ is $\|A\| = \pi$, and moreover $\|Af\| = \pi\|f\|$ when $f$ is a constant function. 
