Let $ A:L_2[0,\pi] \to L_2[0,\pi] $ is a linear operator such that $$(Ax)(t)=\sum_{n=1}^{\infty}[\int_{0}^{\pi}x(s)\sin(ns)ds]\cos(nt)$$
How do I work out that $A$ is a bounded linear operator for this specific example?
My Attempt:
I must show $ \|Ax\|\leq M.\|x\| $ there exists some $ M \geq 0 $ .
$$\|Ax\|^{2} =\int_{0}^{\pi}|\sum_{n=1}^{\infty}[\int_{0}^{\pi}x(s)\sin(ns)ds]\cos(nt)|^{2}dt\\ =\int_{0}^{\pi}\sum_{n=1}^{\infty}|\cos(nt)|^{2}|\int_{0}^{\pi}|x(s)\sin(ns)ds|^{2}dt\\ < \int_{0}^{\pi}\sum_{n=1}^{\infty}|\cos(nt)|^{2}[\underbrace{ \int_{0}^{\pi}|x(s)|^{2}ds}_{=\|x\|^{2}}\int_{0}^{\pi}|\sin(ns)|^{2}ds]dt$$
It's a bit easier to do the problem in abstract. Suppose that $\{e_n\}$ and $\{f_n\}$ are orthonormal sets in a Hilbert space $H$. Define $$\tag1 Ax=\sum_n \langle x,e_n\rangle\,f_n. $$ This is well-defined because for any finite sum we have, since $\{f_n\}$ is orthonormal, $$\tag2 \left\|\sum_{n=1}^N\langle x,e_n\rangle\,f_n\right\|^2=\sum_{n=1}^N|\langle x,e_n\rangle|^2\leq\|x\|^2, $$ where the inequality is Bessel's inequality. Similarly, $$ \left\|\sum_{n=M}^N\langle x,e_n\rangle\,f_n\right\|^2=\sum_{n=M}^N|\langle x,e_n\rangle|^2, $$ showing that the sequence of partial sums is Cauchy. So the series in $(1)$ exists in $H$, and by $(2)$ we have $\|Ax\|\leq\|x\|$.