<s>[INST] How to show that T is a projection operator For $x ∈ [0, 2π]$ let $G(x) = π^{−1}\cos x$, and define an operator $T$ on $L^2([0, 2π])$ as follows:
$$(Tf)(x) = ∫_0^{2π}G(x − x')f(x') \,dx'. $$
Show that $T$ is a projection operator.
I guess I must show that $T$ is selfadjoint and idempotent, in other words that:
$T=T^*$ and $T^2=T$
I am quite new to this and have a little difficulty to getting started 
Sincerely Ingvar
 [/INST] To show $T^2 = T$, just compute $T^2$. We have for $f \in L^2([0,2\pi])$:
\begin{align*}
  (T^2 f)(x) &= T(Tf)(x)\\
             &= \int_0^{2\pi} G(x-x')(Tf)(x')\, dx'\\
             &= \int_0^{2\pi} G(x-x')\int_0^{2\pi} G(x'-x'')f(x'')\, dx''\, dx'\\
             &= \int_0^{2\pi} \int_0^{2\pi} G(x-x')G(x'-x'')\,dx'\, f(x'')\,dx''
\end{align*}
So we have to compute $\int_0^{2\pi} G(x-x')G(x'-x'')\, dx'$, plugin in the given $G$, we have
\begin{align*}
 \int_0^{2\pi} G(x-x')G(x'-x'')\, dx'
   &= \frac 1{\pi^2}\int_0^{2\pi} \cos(x-x')\cos(x'-x'')\, dx'\\
   &= \frac 1{2\pi^2} \int_0^{2\pi} \bigl(\cos(x-x'') + \cos(x+x''-2x')\bigl)\, dx'\\
   &= \frac 1{2\pi^2} \cdot \bigl( 2\pi \cos(x-x'') + 0\bigr)\\
   &= \frac 1{\pi}\cos(x-x'')\\
   &= G(x-x'').
\end{align*}
Continuing our calculation above, we have
\begin{align*}
  (T^2 f)(x)&= \int_0^{2\pi} \int_0^{2\pi} G(x-x')G(x'-x'')\,dx'\, f(x'')\,dx''\\
 &= \int_0^{2\pi} G(x-x'')f(x'')\, dx''\\
   &= (Tf)(x).
\end{align*}
As $f$ and $x$ were arbitrary, we have $T^2 = T$.
For the adjoint, we will do the same (i. e. computing $T^*$ from its definition), we have for $f,g \in L^2([0,2\pi])$:
\begin{align*}
  (T^*f, g) &= (f, Tg)\\
            &= \int_0^{2\pi} f(x)(Tg)(x)\, dx\\
            &=  \int_0^{2\pi} f(x)\int_0^{2\pi} G(x-x')g(x')\, dx'\, dx\\
            &= \int_0^{2\pi}\int_0^{2\pi} f(x)G(x-x')g(x')\,dx'\,dx
\end{align*}
Note now, that $G$ is symmetric, that is $G(-y) = G(y)$ for all $y$, hence $G(x-x') = G(x'-x)$, continuing we have
\begin{align*}
  (T^*f, g) &= \int_0^{2\pi} \int_0^{2\pi} f(x)G(x-x')g(x')\,dx'\,dx\\
            &= \int_0^{2\pi} \int_0^{2\pi} G(x'-x)f(x)\, dx\, dx'\\
            &= \int_0^{2\pi} (Tf)(x')g(x')\, dx'\\
            &= (Tf, g)
\end{align*}
Hence $T^* = T$.
</s>