Let $H = L_{2}([-\pi,\pi])$. For each $m \in \mathbb{N^{*}}$ and define $T_{m}:H \rightarrow H$ by $$T_{m}(f)(t) = \frac{\int_{-\pi}^{\pi} f(s) \sin(2m (s+t)) ds}{2^m},$$ Let $T = \sum_{m=1}^{\infty} T_{m}$.
Find $||T_{m}||$?
Could anyone help me finding this norm please?
The following is not a complete solution for the question.
Let $f(s)=u(s)+iv(s)$, then \begin{align*} f(s)\sin(2m(s+t))&=u(s)\sin(2m(s+t))+iv(s)\sin(2m(s+t))\\ &=(u(s)\sin 2ms)\cos 2mt+(u(s)\cos 2ms)\sin 2mt\\ &~~~~+i[(v(s)\sin 2ms)\cos 2mt+(v(s)\cos 2ms)\sin 2mt]. \end{align*} Let \begin{align*} & A=\int_{-\pi}^{\pi}u(s)\sin(2ms)ds\\ & B=\int_{-\pi}^{\pi}u(s)\cos(2ms)ds\\ & C=\int_{-\pi}^{\pi}v(s)\sin(2ms)ds\\ & D=\int_{-\pi}^{\pi}v(s)\cos(2ms)ds, \end{align*} then \begin{align*} &2^{m}\|T_{m}(f)\|_{L^{2}[-\pi,\pi]}\\ &=\left(\int_{-\pi}^{\pi}(A\cos 2mt+B\sin 2mt)^{2}+(C\cos 2mt+D\sin 2mt)^{2}dt\right)^{1/2}\\ &=\left(\int_{-\pi}^{\pi}(A^{2}+C^{2})\cos^{2}(2mt)+(B^{2}+D^{2})\sin^{2}(2mt)+2AB\sin(2mt)\cos(2mt)\\ +2CD\sin(2mt)\cos(2mt)dt\right)^{1/2}\\ &=\left(\dfrac{1}{2}(A^{2}+C^{2})\int_{-\pi}^{\pi}1+\cos(4mt)dt+\dfrac{1}{2}(B^{2}+D^{2})\int_{-\pi}^{\pi}1-\cos(4mt)dt\right)^{1/2}\\ &=\left(\dfrac{1}{2}(A^{2}+C^{2})2\pi+\dfrac{1}{2}(B^{2}+D^{2})2\pi\right)^{1/2}\\ &=\sqrt{\pi}(A^{2}+B^{2}+C^{2}+D^{2})^{1/2}. \end{align*} However, \begin{align*} A^{2}+B^{2}&=\left|\int_{-\pi}^{\pi}u(s)(\cos(2ms)+i\sin(2ms))ds\right|^{2}\\ &=\left|\int_{-\pi}^{\pi}u(s)e^{2mis}ds\right|^{2}\\ &\leq 2\pi\left(\int_{-\pi}^{\pi}|u(s)|^{2}ds\right)\\ &= 2\pi\|u\|_{L^{2}[-\pi,\pi]}^{2}, \end{align*} similarly, \begin{align*} C^{2}+D^{2}\leq 2\pi\|v\|_{L^{2}[-\pi,\pi]}, \end{align*} so \begin{align*} A^{2}+B^{2}+C^{2}+D^{2}\leq 2\pi(\|u\|_{L^{2}[-\pi,\pi]}+\|v\|_{L^{2}[-\pi,\pi]})=2\pi\|f\|_{L^{2}[-\pi,\pi]}^{2}, \end{align*} so \begin{align*} (A^{2}+B^{2}+C^{2}+D^{2})^{1/2}\leq\sqrt{2\pi}\|f\|_{L^{2}[-\pi,\pi]}. \end{align*} Summing up what we have, then \begin{align*} \|T_{m}\|=\sup_{f\ne 0}\dfrac{\|T_{m}(f)\|_{L^{2}[-\pi,\pi]}}{\|f\|_{L^{2}[-\pi,\pi]}}\leq\sup_{f\ne 0}\dfrac{\sqrt{2}\pi\cdot 2^{-m}\|f\|_{L^{2}[-\pi,\pi]}}{\|f\|_{L^{2}[-\pi,\pi]}}=\dfrac{\sqrt{2}\pi}{2^{m}}. \end{align*}
Guess: To obtain the bound $\sqrt{2}\pi/2^{m}$, it might be having something to do with the Fourier coefficients of $u$ and $v$, but I have no idea how to continue.