2. Since $\psi$ is smooth all its derivatives on the compact space $\TT$ are bounded. Thus, by a result on parameter integrals: $$ (PA_f)\psi(x) =\frac{-i}{2\pi}\pa_x\int_{-\pi}^\pi\psi(x-y)f(y)\,dy =\frac{-i}{2\pi}\int_{-\pi}^\pi\psi^\prime(x-y)f(y)\,dy =A_f(P\psi)(x)~. $$ 3. $P\psi_n(x)=-i(in)e^{inx}=n\psi_n(x)$ and since $\psi_n(x-y)=\psi_n(x)\psi_n(-y)=\psi_n(x)e^{-iny}$, we get $$ A_f\psi_n(x) =\frac1{2\pi}\int_{-\pi}^\pi\psi_n(x-y)f(y)\,dy =\psi_n(x)\frac1{2\pi}\int_{-\pi}^\pi f(y)e^{-iny}\,dy =\wh f(n)\psi_n(x)~. $$ 4. By integration by parts and $2\pi$-periodicity of $\psi\bar\vp$ we get \begin{eqnarray*} \la P\psi,\vp\ra &=&\frac{-i}{2\pi}\int_{-\pi}^\pi\psi^\prime(x)\bar\vp(x)\,dx\\ &=&\frac{-i}{2\pi}\Big(\psi(x)\bar\vp(x)|_{-\pi}^\pi-\int_{-\pi}^\pi\psi(x)\bar\vp^\prime(x)\,dx\Big)\\ &=&\frac{i}{2\pi}\int_{-\pi}^\pi\psi(x)\bar\vp^\prime(x)\,dx =\la\psi,P\vp\ra~. \end{eqnarray*}