# 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.

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It's a convolution operator. Do you know Fourier transform of periodic functions? –  Michael Jun 22 '13 at 1:24
No, I don't know. –  alans Jun 22 '13 at 1:25
I just checked using Mathematica, it should be $2\pi$, if you can find an $f$ such that $\|Af \| = 2\pi \|f\|$ then you are done, for $2\pi \|f\|\leq \|A\|\,\|f\|$ and you already have $\|A\|\leq 2\pi$. –  Shuhao Cao Jun 22 '13 at 2:12

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.

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But shouldn't the answer be $2\pi$, as suggested in comment above? –  alans Jun 22 '13 at 8:30
@alans: Unless my computations are incorrect, I think $\|A\| = \pi$. Another way to see this is to use Young's inquality for convolutions. If $g(x) = \cos^2(x/2)$, then $(Af)(x) = (f*g)(x)$. By Young's inequality, $\|Af\|_2 \leq \|f\|_2\|g\|_1$. The $L^1$-norm of $g$ is $\|g\|_1 = \pi$, so $\|Af\|_2\leq \pi\|f\|_2$. Therefore $\|A\|\leq \pi$. Of course, we know it is exactly $\pi$ by looking at $A1$ as above. –  froggie Jun 22 '13 at 13:27