I learnt without proof that if $X$ is a normed space of uncountable dimension, then the weak* topology on $X^*$ is not first countable. Can anyone point out how I should go about proving it?

I tried the following:

Suppose that the weak* topology on $X^*$ is first countable. Then we may take a sequence $(x_n)$ such that $$ V_n=\big\{x^*\in X^*: \lvert x^*(x_i)\rvert<1, \,\,\text{for all}\,\, i=1, 2, \dots, n\big\} $$ gives a (descending sequence of) countable base at $0\in X^*$. Because I want to find a contradiction, I guess I should try to show that $(x_n)$ span $X$.

Is the above attempt getting anywhere?

  • $\begingroup$ You mean this sequence in $X^*$? I see where this is going. The sequence is w*-bounded, but not norm-bounded. This is not quite a contradiction because w*-boundedness and norm-boundedness are not quite the same. $\endgroup$ – Chuwei Zhang Dec 31 '13 at 23:56
  • $\begingroup$ Thank you. But why are weak* convergent sequences norm-bounded? $\endgroup$ – Chuwei Zhang Jan 1 '14 at 0:10
  • $\begingroup$ But Uniform Boundedness Theorem requires the normed space in question to be complete, right? $\endgroup$ – Chuwei Zhang Jan 1 '14 at 0:13
  • $\begingroup$ @DavidMitra But to apply the uniform boundedness principle to a sequence in $X^\ast$, you need the completeness of $X$. $\endgroup$ – Daniel Fischer Jan 1 '14 at 0:18
  • $\begingroup$ @DanielFischer Ah, yes. Chuwei Zhang: Sorry, this argument will fail for general normed spaces $X$ (deleting previous comments). But I think you can show the $x_i$ span $X$. Any $x$ contains the intersection of the kernals of finitely many $x_i$ ($x$ generates a weak* nhood of $0$). This implies any $x$ is a linear combination of (finitely many) the $x_i$. $\endgroup$ – David Mitra Jan 1 '14 at 0:33

Let $V_n$, $x_n$ be as in the posted question, and let $\iota : X \to X^{**}$ be the natural embedding. The idea of the proof comes from the discussion in the comment field above.

Take $x\in X$. $$A:=\{x^*\in X^*: |x^*(x)|<1\}$$ is a weak* neighbourhood of $0$ in $X^*$. Therefore, for some $n$, $\bigcap_{1\le i\le n}\ker \iota(x_i)\subset V_n\subset A$. So if $x^*\in \bigcap_{1\le i\le n}\ker \iota(x_i)$, then for all scalar $\lambda$, $\lambda x^*\in \bigcap_{1\le i\le n}\ker \iota(x_i) \subset A$. So $|(\lambda x^*)(x)|<1$. This forces $x^*(x)=0$. So $\bigcap_{1\le i\le n}\ker \iota(x_i)\subset \ker \iota(x)$. So $x\in\langle x_1, \dots, x_n\rangle$. So $(x_n)_{n\in\mathbb{N}}$ is a Hamel basis for $X$, which is a contradiction.


In the weak$^*$ topology a sub-base of the neighborhoods of $0$ is obtained by sets of the form $$ W_{x,\varepsilon}=\big\{x^*\!\in X^*: \lvert x^*(x)\rvert<\varepsilon\big\}, \quad \varepsilon>0,\, x\in X, $$ and a local base of 0 (in the weak$^*$ topology) is obtained by finite intersections of the above sets. In particular, if ${\mathcal N}$ is a base of the neighbourhoods of $0\in X^*$, then for every $U\in\mathcal N$, there exist $n\in\mathbb N$, $x_1, \ldots x_n\in X$ and $k_1,\ldots k_n\in\mathbb N$, such that \begin{equation} W_{x_1,1/k_1}\cap\cdots\cap W_{x_n,1/k_n}\subset U. \tag{1} \end{equation} In fact, if each $U$ in $\mathcal N$ is replaced by an intersection $W_{x_1,1/k_1}\cap\cdots\cap W_{x_n,1/k_n}$ satisfying (1), then the new collection $\mathcal N'$ of these intersection is still a local base of $0$.

One step further: The $x_n$'s, can be assumed to be linearly independent, for if $$ x_m=c_1x_1+\cdot+c_kx_k, $$ then $W_{x_1,1/\ell}\cap\cdots\cap W_{x_k,1/\ell}\subset W_{x_m,1/j}$, if $\ell>j\big(\lvert c_1\rvert+\cdots+\lvert c_k\rvert\big)$.

From the above we derive that, if the weak$^*$ topology of $X$ is first countable, then there exists a linearly independent set $\{x_n\}_{n\in\mathbb N}\subset X$, such that the finite intersection of the open sets $$ W_{x_n,1/k}, \quad k,n\in\mathbb N, $$ form a local basis of $0$.

Now as $\dim X>\aleph_0$ we can find $y\in X\smallsetminus\mathrm{span}\,\{x_n :n\in\mathbb N\}$. Using Hahn-Banach we can construct a sequence $\{y_n^*\}_{n\in\mathbb N}\subset X^*$ satisfying $$ y_n^*(y)=0 \quad\text{and}\quad y^*_n(x_j)=\frac{1}{n},\,\,\text{for $j=1,\ldots,n$.} $$ Clearly, every open set of the form $W_{x_{i_1},1/k_1}\cap\cdots\cap W_{x_{i_n},1/k_n}$ contains all but finitely many terms of the sequence $\{y_n^*\}_{n\in\mathbb N}$, while $W_{y,1}$ contains none of them. Hence $$ W_{x_{i_1},1/k_1}\cap\cdots\cap W_{x_{i_n},1/k_n}\not\subseteq W_{y,1}. $$ Thus, the weak$^*$ topology on $X^*$ is not first countable if the dimension of $X$ is uncountable.

Corollary. Let $X$ be a normed space. The weak$^*$ topology on $X^*$ is not first countable if the dimension of $X$, as a linear space, is infinite.

Proof. If the dimension if $X$ is infinite, then the dimension of its completion $\hat{X}$ is at least $\mathfrak{c}$ (the cardinal of the continuum), while the weak$^*$ topologies of $X^*$ and $\hat{X}^*$ are identical.


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