Ytukyg

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 $left(x_nright)_{ninmathbb{N}} subseteq X$ weakly converging to $0$ and for every sequence $left(y^*_nright)_{ninmathbb{N}} subseteq Y^*$ 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 $left(x_nright)_{ninmathbb{N}} subseteq X$ weakly converging to $0$ the sequence $left(T x_nright)_{ninmathbb{N}}$ 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'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
$left<y^*_n,y_nright> rightarrow 0$ 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 $s_n := leftlVert y_n rightrVert$, we know that $leftlVert y_n rightrVert = sup _{leftlVert y^* rightrVert =1} left< y^*, y_nright>$ and so for every $n$ we can choose $y^*_n$ with norm $1$ such that $left< y^*_n, y_nright> > s_n - frac{1}{n}$ 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 $left<y^*_infty, y_nright> rightarrow 0$ 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
$$ 0 = lim _{kto infty} left<y^*_n - y^*_infty, y_{n_k}right> = lim _{kto infty} left<y^*_{n_k}, y_{n_k}right> - lim _{kto infty} left<y^*_infty, y_{n_k}right> = lim _{kto infty} left<y^*_{n_k}, y_{n_k}right> = lim _{kto infty} left( s_{n_k} - frac{1}{n_k} right) $$

And so $lim _{k rightarrow infty} s_{n_k} = 0$ that means $leftlVert y_{n_k} rightrVert rightarrow 0$