# How do I show a linear operator is bounded?

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 $$\|Ax\|\leq\|x\|$$.