5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.