1
$\begingroup$

Exercise :

Let $X$ be a reflexive Banach space and $A \in \mathcal{L}(c_0,X)$. Show that $A$ is a compact operator.

Thoughts :

First of all, I know that if $X,Y$ are Banach then $A \in \mathcal{L}_c(X,Y)$ iff $A^* \in \mathcal{L}_c(X,Y)$. In our case, $c_0$ is a banach space (when equipped with the sup-norm). So essentialy it may come down to proving such a statement.

If the adjoint operator $A^* : X^* \to c^*_0$ is considered, then I know that $c_0^* \simeq \ell_1$ and that $X^*$ should be reflexive too, but I seem to struggle using sequential statements to work a proof.

Also, this is part of exercises on our notes after being introduced to Spectral Theory, so there may be a solution involving that, despite not having found anything that may help (as most Lemmas involve a Bounded Linear Functional from a space to itself).

Any hints or elaborations will be highly appreciated.

$\endgroup$
  • $\begingroup$ what is this, your homework? $\endgroup$ – uniquesolution Mar 2 at 21:40
  • $\begingroup$ @uniquesolution Self study exercises on our topics from notes. $\endgroup$ – Rebellos Mar 2 at 21:41
2
$\begingroup$

An operator $T$ from a Banach space $X$ to a Banach space $Y$ is compact if and only if for every bounded sequence $x_n$ in $X$, there is a subsequence $x_{n_k}$ such that $Tx_{n_k}$ converges in $Y$ (in the norm).

Consider the adjoint operator $A^*:X^*\to (c_0)^*\simeq\ell_1$. Let $(x_n^*)$ be a bounded sequence in $X^*$. Since $X$ is reflexive, so is $X^*$. Every bounded sequence in a reflexive Banach space has a weakly convergent subsequence, so there exists a subsequence $(x_{n_k}^*)$ converging weakly. Since $A^*$ is continuous, it is also weakly continuous, that is, $A^*x_{n_k}^*$ converges weakly to some element $y\in\ell_1$. But in $\ell_1$ weak convergence is equivalent to norm-convergence. Therefore $A^*x_{n_k}^*$ converges in norm. We have thus verified that the condition above holds, so $A^*$ is compact, and so is $A$.

Added in response to some requests for clarification.

The argument above relies on some classical theorems in Banach space theory, which apparently are not sufficiently well known.

The first theorem is that in a reflexive Banach space every bounded sequence has a weakly convergent subsequence. For a proof, see here.

The second theorem is that in $\ell_1$ the weak and norm-topologies are equivalent. In particular, weak convergence of sequences is equivalent to norm convergence of sequences. Most textbooks on Banach spaces contain this theorem. For several references, see here.

$\endgroup$
  • $\begingroup$ I am afraid to say that I fail to follow the proof given + the wealy convergent subsequence of $x^*_n$ as @KaviRamaMurthy mentioned. $\endgroup$ – Rebellos Mar 3 at 5:51
  • $\begingroup$ @KaviRamaMurthy - See my edited post which answers your question and explains why we can say that in $\ell_1$ weak convergence implies norm convergence. $\endgroup$ – uniquesolution Mar 3 at 9:31
  • $\begingroup$ @Rebellos. I do not wish to offend you but judging by your posts on Banach spaces you need to do a lot of reading before you can follow standard arguments in Banach space theory. $\endgroup$ – uniquesolution Mar 3 at 9:38
  • $\begingroup$ @uniquesolution A standard argument for someone may not be standard for someone else, note that I am an undergraduate as well. Also, we haven't been using any of these statements, so there could be another way around, more fitted to what we do. That doesn't mean I am not well studied or I do not know enough things. Thanks for your time though. $\endgroup$ – Rebellos Mar 3 at 9:42
  • $\begingroup$ Also to further clarify my knowledge standing, I have only been taught one semester of Functional Analysis I and the course Functional Analysis II isn't taught this year, due to some issues. Moreover, this is just the second week on a very demanding Operator Theory course (post-grad difficulty) which we are trying to cover up things we do not know, due to absence of Functinal Analysis II and Measure Theory classes (due to issues at uni). So, if my knowledge seems sub-par to you, I apologise, but all I do is trying to learn and work on my own to keep up and have a very respectable standing. $\endgroup$ – Rebellos Mar 3 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.