Norm of linear operator $T(f) = \pi \int_0^{\pi} f(x)\,dx + i/2 \int_\pi^{2\pi} f(x)/x \,dx$

Question

I succeeded in finding the norm of operator $f \to \int_0^x f(t)\,dt$ in $C[0,1]$ with the infinite norm. (It was $1,$ correct me if I'm wrong).

I now need to find the norm of the linear operator $T(f) = \pi \int_{0}^{\pi} f(x)\,dx + \frac i 2 \int_\pi^{2\pi} f(x)/x \;dx$ I do:

$$\|T(f)\|_2^2 = \left| \pi \int_0^{\pi} f(x) \, dx + \frac i 2 \int_\pi^{2\pi} \frac{f(x)} x \;dx \right| \le \pi \left| \int_0^{\pi} f(x)\,dx \right| + \frac 1 2 \left| \int_\pi^{2\pi} \frac{f(x)} x \;dx \right| $$

Perhaps here I should be using Cauchy-Schwartz inequality but it seems too messy for determining I precise constant.

Is my procedure right or should I be looking for another strategy?

Answer 1

Note that

$$\pi\left|\int_0^{2\pi} f(x)\,dx\right| \le \pi \int_0^{2\pi} |f(x)|\,dx$$

and that

\begin{align} \left|\int_{\pi}^{2\pi} \frac{f(x)}{x}\,dx\right| & \le \int_{\pi}^{2\pi}\frac{|f(x)|}{x}\,dx \\ & \le \int_{\pi}^{2\pi} \frac{|f(x)|}{\pi}\,dx \\ &= \frac{1}{\pi}\int_{\pi}^{2\pi} |f(x)|\,dx. \end{align}

Can you take it from here?