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 $\left< y^*_n,T x_n \right> \to 0$

I already know that:

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?
  • $\begingroup$ I'm guessing you mean, "given $(y_n) \subseteq Y$..." instead of "given $(y_n^*) \subseteq Y$..."? $\endgroup$ – Theo Bendit Jul 4 '18 at 12:14
  • $\begingroup$ Yes, that's a typo. I'll edit the question. Thank you! $\endgroup$ – Enrico Polesel Jul 4 '18 at 12:21
  • $\begingroup$ 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. $\endgroup$ – ComFreek Jul 4 '18 at 16:17
  • $\begingroup$ 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) $\endgroup$ – 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 $\left<y^*_n,y_n\right> \rightarrow 0$ 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 $\left<y^*_\infty, y_n\right> \rightarrow 0$ because by hypothesis $y_n \rightharpoonup 0$ (and so also $\left<y^*_\infty, y_{n_k}\right> \rightarrow 0$).

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

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


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