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 $||h||_1&gt;0$
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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.
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.
So If we suppse that $||h||_1&gt;0$ then we can find the inverse operator for $(T_h-\lambda I) $ for selected $\lambda \in \mathbb{R}$.
So since, $||h||_1&gt;0$, there exists $\alpha&gt;0$ such that $||h||_1&gt;\alpha$.

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.