Multiplication Operator on $L^2( \mathbb{R})$ is not Compact Let $H=L^2(\mathbb{R})$, we define the multiplication operator with a function $h\in H$, ($0\leq h)$ and $||h||_1>0$
.
We have $$T_h:  H \rightarrow H, T_h(f)=h.f$$
Note that $T$ is bounded linear operator.
I want to prove that $T_h$ is not a compact operator.
So If we suppose that $T_h$ is compact, I need to prove for example that the spectral set is non countable. But I can't find a good start.
 A: The idea from MaoWao's answer can also be used to give a direct proof, without any spectral theory.
Since $h$ is not a.e. zero, there is a set $E$ of positive measure on which $h$ is bounded away from zero; say $|h| \ge \epsilon$ on $E$.  As in MaoWao's approach, find a sequence of disjoint sets $E_n \subset E$, each with finite positive measure.  Set $f_n = 1_{E_n} / \sqrt{m(E_n)}$; then the $f_n$ are orthonormal, and in particular they are all in the closed unit ball.  Now verify that the functions $T_h f_n$ are also orthogonal, and $\|T_h f_n\|_{2} \ge \epsilon$.  Conclude from this that $\{T_h f_n\}$ has no convergent subsequence.
A: If $T$ is compact, then every $\lambda\in \sigma(T)\setminus\{0\}$ is an eigenvalue of finite multiplicity. So it suffices to show that $T_h$ cannot have eigenvalues of finite multiplicity (if $\sigma(T_h)=\{0\}$, then it's easy to see that $h=0$ a.e.).
Look at the eigenvalue equation
$$
hf=T_h f=\lambda f.
$$
For it to have a non-trivial solution, $h$ must equal $\lambda$ on a set $E$ of positive measure. But if this is the case, then one can decompose $E$ into a disjoint union of sets $E_n$, $n\in\mathbb{N}$, that all have positive measure.  Then $f_n=1_{E_n}$ are orthogonal eigenfunctions to the eigenvalue $\lambda$. Thus $\lambda$ has infinite multiplicity.
Note that this decomposition property of the Lebesgue measure is crucial here. There are (non-trivial) compact multiplication operators on $\ell^2(\mathbb{N})$.
A: So If we suppse that $||h||_1>0$ then we can find the inverse operator for $(T_h-\lambda I) $ for selected $\lambda \in \mathbb{R}$.
So since, $||h||_1>0$, there exists $\alpha>0$ such that $||h||_1>\alpha$.
Now If we take $\lambda \in [0,\alpha]$ the operator $(T_h-\lambda I)$ will be invertible, and it's inverse will be: $\psi=\frac{I}{h-\lambda}$
And then, we come to the conclusion that $[0,\alpha]\subset \sigma(T_h)$ then $T_h$ is not compact.
