Finding explicitly the operator norm of $T u = \sin (t) u $ I have the operator 
$$T: L^2(0,\pi)\to L^2(0,\pi), \ \ \ Tu=\sin(t)u.$$ 
I can show that the operator norm is bounded by $1$ but I can't find any function to show the norm is exactly that. I am asked to find It explicitly. Can anybody give me some technique? Thanks.
 A: It might actually be easier to prove in general that if $Tu=fu$ for some continuous function $f$ on $[0,\pi]$ then $\Vert T\Vert=\Vert f\Vert_\infty=\sup_{0\leq x\leq\pi}|f(x)|$.
Choose $x_0$ such that $|f(x_0)|=\Vert f\Vert_\infty$. On a neighbourhood $(x_0-\delta,x_0+\delta)$ of $x_0$ we have $|f(x)|\geq \Vert f\Vert_\infty-\epsilon$. Considering the function $u=\sqrt{1/2\delta}\chi_{(x_0-\delta,x_0+\delta)}$, which has $L^2$-norm $1$, we have
$$\Vert Tu\Vert_2^2=\int_{x_0-\delta}^{x_0+\delta} \frac{1}{2\delta}|f(x)|^2dx\geq (\Vert f\Vert_\infty-\epsilon)^2$$
so $\Vert Tf\Vert_2\geq\sup_{\epsilon>0}\Vert f\Vert_\infty-\epsilon=\Vert f\Vert_\infty$.
A: If you take 
$$
u=\sqrt{n}\,1_{[\pi/2,\pi/2+1/n]},
$$
then, using that $\sin(\pi/2+1n)=\cos1/n\geq1-1/n^2$,
$$
\|u\,\sin t\|_2^2=  n\,\int_{\pi/2}^{\pi/2+1/n}\sin^2t\,dt\geq n\left(1-\frac1{n^2}\right)\,
\int_{\pi/2}^{\pi/2+1/n}1\,dt=\left(1-\frac1{n^2}\right),
$$
you show that $\|T\|\geq\sqrt{1-1/n^2}$ for all $n$; thus $\|T\|\geq1$. 
A: For $n \in \mathbb{N}$ consider $$f_n \equiv 1_{\left[\frac\pi2 - \frac1n, \frac\pi2 + \frac1n\right]}$$
We have
$$\|f_n\|_2^2 = \int_{[0,\pi]} |f_n|^2 = \int_{\left[\frac\pi2 - \frac1n, \frac\pi2 + \frac1n\right]} 1 = \frac2n$$
$$\|Tf_n\|_2^2 = \int_{[0,\pi]} |f(t)|^2\sin^2 t\,dt = \int_{\left[\frac\pi2 - \frac1n, \frac\pi2 + \frac1n\right]} \sin^2t\,dt = \frac1n + \frac12 \sin\frac2n$$
Therefore
$$\|T\|^2 \ge \frac{\|Tf_n\|_2^2}{\|f_n\|_2^2} = \frac{\frac1n + \frac12 \sin\frac2n}{\frac2n} = \frac12 + \frac12 \frac{\sin\frac2n}{\frac2n} \xrightarrow{n\to\infty} 1$$
We conclude $\|T\| \ge 1$.
