Is this operator compact and how do I prove it?

Question

I have a very big problem with the following question: Is the operator $T$ defined by $(Tx)t=tx(t)$, $(0<t<1)$ compact in $L_2(0,1)$?

My guess is no and I've tried 3 different approaches to show this. I've been considering the closed unit ball $\overline{B}_1(0)$ and I have assumed that ${T\overline{B}_1(0)}$ is compact in desire to derive at a contradiction.

1. I've considered the image $\{Tx_n\}$ of a sequence $\{x_n\}$ in $\overline{B}_1(0)$ which has no cauchy subsequences, using that $L_2(0,1)$ is a infinite dimensional complete NLS. I've tried to show that neighter the sequence $\{Tx_n\}$ can have any convergent subsequences, but this failed.

2. Then I tried to show that if there is no $\epsilon > 0$ and no $x\in {T\overline{B}_1(0)}$ such that the closed ball $\overline{B}_{\epsilon}(x)$ is fully contained in the image of $T$ then ${T\overline{B}_1(0)}$ can not be compact, but I can't seam to make the argument. Sadly, since othervise, the implication $\overline{B}_{\epsilon}(x)$ is a closed set of a compact set would result in a contradiction. (Since $\overline{B}_{\epsilon}(x)$ is not compact since $L_2(0,1)$ is not finite)

3. My last effort was considering the inverse of $T$ restricted to ${T\overline{B}_1(0)}$, I managed to show that $T$ is one to one and hence the inverse welldefined, but it seams to fail to be bounded and hence the desired contradiction failed. (The contradiction being $\overline{B}_1(0)$ is a continous image of a compact set.)

Now, I turn to you for help! Preferably, also with some comments on my different approaches... Can you find a proof in any of these lines or is it not possible?

I also must add that I may not use the theory of Hilbert spoaces because we have not yet reached those sections in my course! So for example there is no use proving that $T$ is not the limit of finite range bounded operators...

Answer 1

Do you know that if $T$ is compact and $f_n \rightharpoonup f$ then $Tf_n \to Tf$?

There exists a sequence such that $f_n \rightharpoonup 0$, $\|f_n\|_2 = 1$ and $\mbox{supp} f_n \subset (\frac{1}{2},1)$. To construct this sequence, you could just take an orthonormal basis of $L^2(\frac{1}{2},1)$, extend it by $0$ to $(0,1)$ and re-normalize (if you are not allowed to use Hilbert space concepts, you can still do this by constructing the functions explicitly). Then you have $$\|Tf_n\|^2 = \int_0^1 \! (tf_n(t))^2 \, dt= \int_{\frac{1}{2}}^1 \! (tf_n(t))^2 \, dt$$ $$ \geq \frac{1}{4}\int_{\frac{1}{2}}^1 \! f_n(t)^2 \, dt = \frac{1}{4}\int_{0}^1 \! f_n(t)^2 \, dt = \frac{1}{4}$$ so $\|Tf_n\|_2 \geq \frac{1}{2}$ and we cannot have $Tf_n \to 0$. Hence $T$ is not compact.

This is more or less a reformulation of your first approach to the problem.

Answer 2

Note that $T$ is a multiplication operator (multiplication by the identity function). It is not hard to check that the spectrum of a multiplication operator is the closure of the range of the corresponding function. That is, the spectrum of $T$ is $[0,1]$. So $T$ is not compact, as the spectrum of a compact operator can only accumulate at $0$.