# Weakly convergent subsequences in separable Fréchet spaces

The exercise is to prove that if $K\subseteq X$ is weakly compact, and if $x_0$ is a weak limit point of a countable set $E\subseteq K$, then there is a sequence of points in $E$ that converge to $x_0$. Here is my attempt:

a). $V^*_n=\frac{1}{n}\left \{\Lambda :\vert \Lambda x\vert \le 1;x\in X \right \}$ are weak$^*$ compact and $\bigcup V^*_n=X^*$.

b). Since $X$ is separable, it has a countable dense subset $\left \{ x_n \right \}$, from which we can define $\phi_n (\Lambda)=\Lambda x_n$ for $\Lambda \in X^*$. The $\phi_n$ are evidently continuous in the weak$^*$ topology. Moreover, if $\Lambda_1\neq \Lambda_2$ then there is an $x\in X$ and an $\epsilon >0$ such that $\vert(\Lambda_1-\Lambda_2)x\vert>\epsilon$. There is also an $x_n\in\left \{ x_n \right \}$ such that $d(x,x_n)$ is small enough so that $\vert (\Lambda_1-\Lambda_2)x-(\Lambda_1-\Lambda_2)x_n\vert<\epsilon /2$. Then, $\vert (\Lambda_1-\Lambda_2)x_n\vert>\epsilon /2\Rightarrow \Lambda_1 x_n\neq \Lambda_2 x_n \Rightarrow \phi_n (\Lambda_1)\neq \phi_n (\Lambda_2)$ so $\left \{ \phi_n \right \}$ separates the points of $X^*$.

c). Now we have that $V^*_n$ is compact and metrizable, hence separable, so that $X^*$ itself is separable in the weak$^*$ topology. Take a countable dense subset $\left \{ \Lambda_n \right \}$ of $X^*$ and observe that it separates points on $X$:

By Hahn-Banach ($X$ is Fréchet), $X^*$ separates points on $X$. If $x\neq y\in X$, we can find a $\Lambda\in X^*$ such that $\Lambda x\neq \Lambda y\Rightarrow \vert \Lambda(x-y)\vert >\epsilon$ for some $\epsilon>0$. Weak$^*$ density of $\left \{ \Lambda_n \right \}$ implies that there is a $\Lambda_n$ such that $\vert \Lambda_n(x-y)-\Lambda (x-y)\vert <\epsilon /2\Rightarrow \vert \Lambda_n(x-y)\vert>\epsilon /2.$

Now, (using the hint of the exercise), let $Y$ be the smallest closed subspace of $X$ containing $E$. Then, $Y\cap K$ is compact in the weak topology, and by c)., with $Y$ in place of $X$, we have that $Y\cap K$ is metrizable.

But there seems to be a problem with this last part of the argument. Why did we need $Y$ to be the $smallest$ closed subspace of $X$?

Your attempt builds a good argument (though that can be streamlined), but unfortunately it builds upon a wrong base. We can fix that, though. Let's go from top to bottom. You start in a) with

$V^*_n=\frac{1}{n}\left \{\Lambda :\vert \Lambda x\vert \le 1;x\in X \right \}$ are weak$^{\ast}$ compact and $\bigcup V_n^{\ast} = X^{\ast}$.

But that's wrong (unless $X = \{0\}$). As written, we have $V_n^{\ast} = \{0\}$ for all $n$, since $\lvert\Lambda x\rvert \le 1$ for all $x\in X$ implies $\Lambda = 0$. Probably you wanted to adapt the representation of the dual of a normed space as a countable union of weak$^{\ast}$ compact sets, $E^{\ast} = \bigcup n\cdot B^{\ast}$, where $B^{\ast}$ is the norm-closed unit ball in $E^{\ast}$. We can adapt it, but it works slightly differently. By the Alaoglu-Bourbaki theorem, for any neighbourhood $U$ of $0$ in $X$, its polar $U^0 = \{ \Lambda : \lvert \Lambda x\rvert \le 1\text{ for all } x \in U\}$ is weak$^{\ast}$ compact. Now one may be tempted to believe that $X^{\ast} = \bigcup n\cdot U^0$, but unless $X$ is normable, that is not the case. What one needs is a countable neighbourhood basis of $0$ in $X$. That exists if and only if $X$ is metrisable, which a Fréchet space of course is. So if we take a neighbourhood basis $\{ U_n : n \in \mathbb{N}\}$ of $0$ in $X$ and let $V_n^{\ast} = U_n^0 = \{\Lambda : \lvert\Lambda x\rvert \le 1\text{ for all } x \in U_n\}$, we have a countable family of weak$^{\ast}$ compact sets with $X^{\ast} = \bigcup V_n^{\ast}$. Let's verify that this last property does indeed hold: for every $\Lambda \in X^{\ast}$, by continuity there is a neighbourhood $U_\Lambda$ of $0$ in $X$ such that $x \in U_\Lambda \implies \lvert \Lambda x\rvert \le 1$. Since $\{ U_n : n \in \mathbb{N}\}$ is a neighbourhood basis of $0$, there is a $k \in \mathbb{N}$ with $U_k \subseteq U_\Lambda$. But then $\Lambda \in U_k^0 = V_k^{\ast}$.

The next part, b), is correct, but it is shorter to argue that every continuous function, in particular every $\Lambda \in X^{\ast}$, that vanishes on all $x_n$ must also vanish on $\overline{\{ x_n : n \in \mathbb{N}\}} = X$. Thus if $\Lambda_1 \neq \Lambda_2$, there must be an $n$ with $(\Lambda_1 - \Lambda_2)(x_n) \neq 0$.

In part c), you give no argument for the metrisability of $V_n^{\ast}$ (in the weak${\ast}$ topology). It seems you know the argument, but it is unlikely that it can be tacitly assumed that everybody concerned knows the argument, thus it should be made explicit. The topology induced by the family of seminorms $\{ \Lambda \mapsto \lvert\Lambda(x_n)\rvert : n \in \mathbb{N}\}$ induces a metrisable topology $\tau$ on $X^{\ast}$ that is coarser than the weak$^{\ast}$ topology. Therefore every weak$^{\ast}$ compact set $C$ is also $\tau$-compact, and the identity $(C,\sigma(X^{\ast},X)\lvert_C) \to (C,\tau\lvert_C)$ is a closed (since both topologies are Hausdorff) continuous bijection, hence a homeomorphism, i.e. $\tau\lvert_C = \sigma(X^{\ast},X)\lvert_C$. The argument that a weak$^{\ast}$-dense subset $\{ \Lambda_n : n \in \mathbb{N}\}$ separates points on $X$ can also be shortened, like in b): a weak$^{\ast}$ continuous function vanishing on all $\Lambda_n$ must vanish on $\operatorname{cl}_{w^{\ast}}\{\Lambda_n : n \in \mathbb{N}\} = X^{\ast}$. So $\Lambda_n x = 0$ for all $n$ implies $\Lambda x = 0$ for all $\Lambda \in X^{\ast}$, and hence $x = 0$.

Then we come to the end,

Now, (using the hint of the exercise), let $Y$ be the smallest closed subspace of $X$ containing $E$. Then, $Y\cap K$ is compact in the weak topology, and by c)., with $Y$ in place of $X$, we have that $Y\cap K$ is metrizable.

But there seems to be a problem with this last part of the argument. Why did we need $Y$ to be the smallest closed subspace of $X$?

Your confusion is comprehensible. One doesn't need to consider any subspace $Y$ if $X$ is a separable Fréchet space. The topology induced on $X$ by the family of seminorms $\{ x \mapsto \lvert\Lambda_n x\rvert : n \in \mathbb{N}\}$ is metrisable and coarser than the weak topology on $X$, and like above, we conclude that every weakly compact subset of $X$ is metrisable in the weak topology.

The hint to consider the smallest closed subspace $Y$ containing $E$ is needed when one drops the separability assumption on $X$, since the assertion holds in every Fréchet space, separable or not. Since $E$ is a countable set, the smallest closed subspace $Y$ containing $E$ is a separable Fréchet space. Then the above arguments applied to $Y$ and $Y^{\ast}$ show that $K\cap Y$ is metrisable in $\sigma(Y,Y^{\ast})$ (the weak topology on $Y$). Finally, one uses that [as a consequence of the Hahn-banach theorems] the weak topology on $Y$ coincides with the subspace topology induced by the weak topology on $X$, $\sigma(Y,Y^{\ast}) = \sigma(X,X^{\ast})\lvert_Y$ to conclude.

• Yes this is what I was after: a set of functions that separates the points of $X$,using Banach–Alaoglu to get there. Then, compactness of $K$ gives metrizability. As you surmised, I know the proof of this from Rudin's Functional Analysis. Thanks for the thorough (and clear) answer. – Matematleta Nov 15 '16 at 1:39