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 $L^2$.

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

  • $\begingroup$ By the theorem it follows that the spectrum is $\mathbb{C}$, which is a contradiction with the spectrum is compact. Right ? $\endgroup$ – Anas BOUALII Dec 22 '18 at 8:56
  • $\begingroup$ 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! $\endgroup$ – Song Dec 22 '18 at 9:03
  • $\begingroup$ 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 $M=1$? $\endgroup$ – Anas BOUALII Dec 22 '18 at 9:22
  • $\begingroup$ 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 $L^2$ . $\endgroup$ – Song Dec 22 '18 at 9:34

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