Eigenvalues and norm of an operator

Question

Let $T:L^2[0,\pi]\to L^2[0,\pi]$ defined as $$Tu(x)=\sin(x)u(x)$$ I have to find its norm and determine if it has any eigenvalue. It is simple to see that $||T||_{op}\le 1$ but I don't know if it is the beast bound. Concerning the eigenvalues, since there isn't a non-zero (almost everywhere) function such that $$\lambda u(x)=\sin(x)u(x)$$ for constant $\lambda$, I don't think they exist but it seems too silly

Answer 1

For the norm:

The maximum of $\sin(x)$ on $[0,\pi]$ is $1$ attained at $x=\pi/2$.

Take $f_\epsilon(x)=1$ on $[\pi/2-\epsilon,\pi/2+\epsilon]$.

Then $\|f_\epsilon\|^2=2\epsilon$ and $\|T(f_\epsilon)\|^2=\int_{\pi/2-\epsilon}^{\pi/2+\epsilon}\sin^2(x)dx=\epsilon+\sin(\epsilon)\cos(\epsilon)$

Dividing $\frac{\|T(f_\epsilon)\|^2}{\|f_{\epsilon}\|^2}=\frac{1}{2}+\frac{\sin(\epsilon)}{2\epsilon}\cos(\epsilon)$ and tending $\epsilon\to0$ we get that the quotient $\frac{\|T(f_\epsilon)\|^2}{\|f_{\epsilon}\|^2}$ tends to $1$.

You can apply this idea to other multiplication operators by taking $f_\epsilon$ to be the indicator function of the set where the factor is not smaller than its maximum minus $\epsilon$. This allows you to prove that the norm is $\|\cdot\|_{\infty}$-norm of the factor.

For the eigenvalues:

Assume that $Tf=\lambda f$. Then $0=(\lambda-\sin(x))f$ on $[0,1]$.

If $\lambda\notin [0,1]$, then $\lambda-\sin(x)$ is never zero on $[0,\pi]$ and the equation implies that $f(x)=0$, for $x\in[0,\pi]$.

If $\lambda\in[0,1]$, then $f(x)=0$, for $x\in[0,\pi]\setminus A$, where $A$ is the set of solutions of $\sin(x)=\lambda$ in $[0,\pi]$. This set is finite. Therefore, $f(x)$ must be zero except possibly finitely many points. That belongs to the class of the zero function in $L^2[0,\pi]$.

Therefore, $T$ has no eigenvalues.