Let $X=L^2[a,b]$ and $ u\in C[a,b]$. $$A:X\rightarrow X$$ defind by $(Ax)(t)=u(t)x(t)$ for almost all $t\in [a,b]$. Then prove that:
1) $A$ is Bounded& normal operator,
2) $A$ is self-adjoint iff $u$ is real valued, and
3) $A$ is unitary iff $|u(t)|=1, \quad \forall t\in[a,b]$ .
Attempt: $||Ax||^2=||ux||^2=\int_{a}^{b}|u(t)x(t)|^2dt$ Now am not able to proceed further. Any help is appreciated. Thanx in advance
Note that since $u\in C[a,b]$ $u$ is a bounded function. It follows $|u(t)\,x(t)|≤\|u\|_{\infty}|x(t)|$ and for that reason $$\int_a^b|u(t)|^2|x(t)|^2\,dt≤\|u\|_\infty^2 \int_a^b|x(t)|^2\,dt=\|u\|_\infty^2\,\|x\|_2^2$$ Also: $$\langle x,Ay\rangle = \int_a^b\overline{y(t)}u(t)x(t)\,dt=\int_a^b\overline{\overline{u(t)}y(t)}x(t)\,dt=\langle A^*y,x\rangle$$ and it follows $A^*$ is the multiplication with the complex conjugate of $u$, so $A^*=A$ iff $\overline{u(t)}=u(t)$ almost everywhere. From continuity of $u$ it follows $\overline{u(t)}=u(t)$ everywhere.
Finally $A^*A(x)=\overline{u(t)}u(t)x(t)=|u(t)|^2x(t)=u(t)\overline{u(t)}x(t)=AA^*(x)$ for all $x$ and $A$ is normal. Also it is clear from that equation that $A^*A(x)=x$ for all $x$ iff $|u(t)|^2=1$ almost everywhere. But since $u$ is continuous it must follow $|u(t)|^2=1$ everywhere.