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

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 ?

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 $$L^2$$.

• I found dificulties in proving that $Tf_n=f_n$ – Anas BOUALII Dec 22 '18 at 9:08
• @AnasBOUALII I have added details – mathworker21 Dec 22 '18 at 9:18
• I see, thank you. – Anas BOUALII Dec 22 '18 at 9:24

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.

• By the theorem it follows that the spectrum is $\mathbb{C}$, which is a contradiction with the spectrum is compact. Right ? – Anas BOUALII Dec 22 '18 at 8:56
• I'll elaborate on this. In fact, $\ker(T-\lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $\sigma_p(T)\setminus\{0\} = \sigma(T)\setminus\{0\}$, where $\sigma_p(T) := \{\lambda\;|\;\ker(T-\lambda) \neq (0)\}$ is the point spectrum of $T$. For the given $T$, we have $\sigma_p(T) =\varnothing$ and hence it follows $\sigma(T) \subset \{0\}$. However, we can show that $\sigma(T) = [0,1]$ holds, leading to a contradiction! – Song Dec 22 '18 at 9:03
• I agree with you. so since $T$ is self-adjoint operator then $\sigma(T) \subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$? – Anas BOUALII Dec 22 '18 at 9:22
• Well, $\langle f,Tf\rangle\subset [0,1]$ is almost direct, and $f_n = \sqrt{n}1_{[0,\frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =\sqrt{n}1_{[1-\frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $\sigma(T) =[0,1]$ can be seen in a more direct way. Since for $\lambda \in [0,1]$, $x\mapsto \frac{1}{x-\lambda}$ is not bounded on $[0,1]$, it follows that $(T-\lambda)^{-1}f(x) =\frac{f(x)}{x-\lambda}$ is not a bounded inverse in $L^2$ . – Song Dec 22 '18 at 9:34