# Compactness criterion for operator between reflexive Banach spaces

I found (without any proof) the following proposition:

Let $T \in \mathcal{L}(X,Y)$ be a linear continuous operator between two reflexive Banach spaces $X,Y$, then $T$ is compact if and only if for every sequence $\left(x_n\right)_{n\in\mathbb{N}} \subseteq X$ weakly converging to $0$ and for every sequence $\left(y^*_n\right)_{n\in\mathbb{N}} \subseteq Y^*$ weakly-$*$ converging to $0$ it turns out that $\left< y^*_n,T x_n \right> \to 0$

If $X$ is reflexive then $T$ is compact if and only if for every every sequence $\left(x_n\right)_{n\in\mathbb{N}} \subseteq X$ weakly converging to $0$ the sequence $\left(T x_n\right)_{n\in\mathbb{N}}$ converges strongly in $Y$.

I tried to prove that given $\left( y_n \right) \subseteq Y$ (with $Y$ reflexive) that converges weakly to $0$ if for every $\left( y^*_n \right) \subseteq Y^*$ that converges weakly-$*$ to zero it turns out that $\left< y^*_n, y_n \right> \to 0$ then $y_n \to 0$ strongly, but without success.

So I have a couple of questions:

• Q1: is my last proposition true? Can we infer strong convergence from the weak one and with this "dual tests"?
• Q2: how can I prove the original proposition?
• I'm guessing you mean, "given $(y_n) \subseteq Y$..." instead of "given $(y_n^*) \subseteq Y$..."? – Theo Bendit Jul 4 '18 at 12:14
• Yes, that's a typo. I'll edit the question. Thank you! – Enrico Polesel Jul 4 '18 at 12:21
• You only need the $\Leftarrow$ part of the original proposition, right? Otherwise note that in general if $x_n \rightarrow x$ strongly and $y_n' \to y$ weakly-$\ast$, then $\left<y_n',x_n\right> \rightarrow \left<y, x\right>$ strongly. – ComFreek Jul 4 '18 at 16:17
• Yes, I need only the $\Leftarrow$ since, as you said, I can use the second proposition to prove that. I know that in general with only weak convergence $\left< y^*_n, x_n \right> \not\to \left<y^*,x\right>$, so that's way I tried to prove this thing for reflexive spaces. (actually it would be enough to prove that exists a subsequence $n_k$ such that $Tx_{n_k}$ converges) – Enrico Polesel Jul 4 '18 at 16:49

I'll write down the proof of a friend of mine, we can prove both the second proposition (the "strong converging criterion" for sequences in a reflexive Banach space).

Let $$(y_n)\subseteq Y$$ (with $$Y$$ reflexive) be a sequence that converges weakly to $$0$$. If for every sequence $$(y^*_n)\subseteq Y^*$$ tha converges weakly-$$*$$ to zero it turns out that $$\left \rightarrow 0$$ then $$y_n \rightarrow 0$$ strongly.

It's easy to prove that $$y_n \rightarrow 0$$ if and only if for every subseqence $$(y_{n_k})$$ we can find a sub-subseqence $$(y_{n_{k_i}})$$ such that $$y_{n_{k_i}} \rightarrow 0$$. To keep the notation simple we will just prove that $$(y_n)$$ has a subsequence that converges to $$0$$, but the same argument can be applied to subsequences.

Let $$s_n := \left\lVert y_n \right\rVert$$, we know that $$\left\lVert y_n \right\rVert = \sup _{\left\lVert y^* \right\rVert =1} \left< y^*, y_n\right>$$ and so for every $$n$$ we can choose $$y^*_n$$ with norm $$1$$ such that $$\left< y^*_n, y_n\right> > s_n - \frac{1}{n}$$ and construct a sequence $$(y^*_n)\subseteq Y^*$$.

Since every $$y^*_n$$ as norm $$1$$ it is a sequence in the unitary ball and so, by compactness, exists $$(n_k)\subseteq \mathbb{N}$$ and $$y^*_\infty$$ such that $$y^*_{n_k} \overset{\ast}{\rightharpoonup} y^*_\infty$$, since $$Y$$ is reflexive we have also $$y^*_{n_k} \rightharpoonup y^*_\infty$$.

We have that $$\left \rightarrow 0$$ because by hypothesis $$y_n \rightharpoonup 0$$ (and so also $$\left \rightarrow 0$$).

Since $$y^*_{n_k} - y^*_\infty \rightharpoonup 0$$ we can write $$0 = \lim _{k\to \infty} \left = \lim _{k\to \infty} \left - \lim _{k\to \infty} \left = \lim _{k\to \infty} \left = \lim _{k\to \infty} \left( s_{n_k} - \frac{1}{n_k} \right)$$

And so $$\lim _{k \rightarrow \infty} s_{n_k} = 0$$ that means $$\left\lVert y_{n_k} \right\rVert \rightarrow 0$$