A non compact operator on $L^2[0,1]$

Question

Let $H$ be the Hilbert space $L^2[0,1]$.

and the operator $T : H\rightarrow H$, such as $T(f)(x)=x.f(x)$

Why $T$ isn't compact ?

Answer 1

Let $f_n(x) = \sqrt{n}1_{[1-\frac{1}{n},1]}(x)$. Then $||f_n||_2 = 1$ for each $n$ but $(Tf_n)_n$ has no convergent subsequence.

Indeed, suppose $Tf_{n_k} \to g$ in $L^2$ for some $g \in L^2$ and subsequence $(f_{n_k})_k$. Since $||f_{n_k}||_2 = 1$ for each $k$, $g$ cannot be $0$, so $\int_0^1 |g(x)|^2dx > 0$. Then $\int_0^a |g(x)|^2dx =: \epsilon > 0$ for some $a \in (0,1)$. But then for large $k$, $$\int_0^1 |g(x)-xf_{n_k}(x)|^2dx = \int_0^a |g(x)|^2 +\int_a^1 |g(x)-xf_{n_k}(x)|^2dx \ge \int_0^a |g(x)|^2dx \ge \epsilon,$$ which contradicts $f_{n_k} \to g$ in $L^2$.

Answer 2

Observe that $T$ is a self-adjoint operator. And also $$ \ker (T-\lambda)=(0) $$ holds for all $\lambda\in \mathbb{C}$. If $T$ is compact, then by the spectral theorem of compact operators, it follows that $$ T=0. $$ This leads to an obvious contradiction.