# If $X$ is reflexive and $A \in \mathcal{L}(c_0,X)$ then show that $A$ is a compact operator.

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.

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

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.

• @KaviRamaMurthy - See my edited post which answers your question and explains why we can say that in $\ell_1$ weak convergence implies norm convergence. – uniquesolution Mar 3 at 9:31
• @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. – uniquesolution Mar 3 at 9:38
• @Rebellos I understand your academic limitations. It is my modest opinion that problems mentioning the space $c_0$, and "reflexive Banach space" do not require the tools of Operator Theory, but rather, the tools of classical Banach spaces theory, some samples of which appear in my post. I doubt that the exercises you brought recently in several posts are part of an Operator Theory course. I strongly advise you to read an introductory text on Banach space theory. A friendly text is "A short course in Banach space theory", by N.L. Carothers (2004). – uniquesolution Mar 3 at 10:03
• @Rebellos I see. Then I would be extremely interested to see what kind of solutions to these exercises are proposed by your instructors, that use Spectral theory etc. – uniquesolution Mar 3 at 10:05
• @Rebellos - Could you kindly mention here the recommended textbook(s) for your Operator Theory course? – uniquesolution Mar 3 at 10:07