I thought I needed help proving the above statement, but during typing I found a proof. Since I had already written it all down I will post it anyway, maybe in the future someone can benefit from it.

One question remains though: I'm not sure whether/where I have used the property that the space is complete in the proof. It was stated explicitly that the space has to be a Banach space, so it's likely that I've missed something. I would appreciate if somebody could point it out to me.

  • $\begingroup$ The statement also holds for noncomplete spaces as long as they are infinite dimensional, so you did not necessarily miss something. As to noted in your proof, the essential step is that weakly convergent sequences are bounded. But using the embedding in the (complete!) bidual, this holds also for noncomplete spaces. Your proof looks good btw. $\endgroup$ – PhoemueX Sep 1 '15 at 17:23
  • $\begingroup$ Isn't it true that on finite-dimensional spaces, norm and weak topology coincide. Does this mean then, that there are cases of non-complete finite-dimensional spaces in which the induced topology is not first-countable? Maybe you could provide me with an example, since I'm still quite new to this. And thank you. $\endgroup$ – azureai Sep 1 '15 at 17:42
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    $\begingroup$ I don't really get your question. Yes, on finite dimensional spaces, norm and weak topology coincide, so that the weak topology is first countable in this case. Also, every finite dimensional normed vector space is complete. Still, for every infinite dimensional normed vector space (complete or not), the weak topology is not first countable. Did I somehow answer your question? :) $\endgroup$ – PhoemueX Sep 1 '15 at 17:46
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    $\begingroup$ You may compare this with the famous theorem on topological groups. A topological group is first-countable if and only if it is metrisable. The weak topology on an ininite-dimensional Banach space is not metrisable. $\endgroup$ – Tomek Kania Sep 1 '15 at 20:06
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    $\begingroup$ The crucial point is the dimension of the dual. If $X$ is an infinite-dimensional normed space, its dual is an infinite-dimensional Banach space, and hence its dimension is uncountable. If the space you're starting with has a dual of countably infinite dimension, then the weak topology is first-countable. $\endgroup$ – Daniel Fischer Feb 4 '17 at 22:53

I'm going to prove the statement by contradiction.

The assumption that the weak topology $\tau_w$ on the $\infty$-dimensional linear space X is first-countable, is equivalent to saying that each point in $X$ has a countable neighborhood basis. In particular for $0_X \in X$ there exists a N.B. $(U_n)_n\subset\tau_w$. We define the sequence $(A_n)_n\subset\tau_w $ by$$A_n:=\bigcap_{i=1}^n U_i$$ obtaining another neighborhood basis of $0_X$, but this time decreasing.

It's easy to prove that in a normed linear space, each weakly open neighborhood of $0$ contains a non-trivial closed subspace. So for every $n$, $A_n$ contains a non-trivial subspace $Y_n$ of X, and we can pick a vector $y_n\in V_n\setminus\{0_x\}$.

We then consider $x_n:=n\frac{v_n}{\lvert\lvert v_n\rvert\rvert}\in V_n$. Because of $\lvert\lvert x_n\rvert\rvert=n$, $(x_n)_n$ is an unbounded sequence, so it is not weakly convergent (this is also easy to show, prove the contrapositive statement by using the canonical embedding and the uniform boundedness theorem).

On the other hand, for any $\varepsilon>0$ and $\phi \in X^*$ the set $$U_{\varepsilon,\phi}=\{x\in X \vert\ \lvert \phi(x) \rvert<\varepsilon\}$$ is a neighborhood of $0_X$, and because $(A_n)_n$ is a decreasing neighborhood basis, there exists an $N \in \mathbb{N}$ such that $$\forall n\ge N: U_{\varepsilon,\phi}\supseteq A_n\supseteq\ V_n$$ In particular $\lvert\phi(x_n)\rvert<\varepsilon,$ since $x_n\in V_n$. Because $\varepsilon$ was picked arbitrarily, we get $\lvert \phi(x)\rvert\rightarrow0$. We also considered an arbitrary $\phi \in X^*$, and hence $x\overset{w}\rightarrow 0$. This is a contradiction to what we showed in the above paragraph.

This proves the statement.

  • $\begingroup$ It is easy to prove that in an infinite dimensional normed vector space ... $\endgroup$ – PhoemueX Sep 1 '15 at 17:24

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