Please, help me with a proof of this (apparently) known fact whose proof is out of my reach, even though I spent a considerable amount of time looking it up:

The weak$^*$ topology on the space of tempered distributions on $\mathbb{R}^n$ is not first-countable.

This very statement is highly counter-intuitive to me, since I know that the Schwartz test-functions space is metrizable, it has a metric derived from the countable family of semi-norms. Why would its topological dual with respect to pointwise convergence not be first countable?

Thank you!


1 Answer 1


In general, if $X$ is a locally convex topological vector space of uncountable dimension (as a linear space), then the weak$^*$ topology on $X^*$ is not first countable.

Proof. In the weak$^*$ topology a sub-base of the neighborhoods of $0$ is obtained by sets of the form $$ W_{x,\varepsilon}=\{x^*\in X^*: |x^*(x)|<\varepsilon\}, \quad \varepsilon>0,\, x\in X, $$ and hence a base 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 $$ W_{x_1,1/k_1}\cap\cdots\cap W_{x_n,1/k_n}\subset U. $$ In fact, if each $U$ in $\mathcal N$ is replaced by the intersection $W_{x_1,1/k_1}\cap\cdots\cap W_{x_n,1/k_n}$, then the new collection $\mathcal N'$ of these intersection is still a base of neighbourhoods of $0$.

One step further: The $x_n$'s, can be considered 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 $\ell>j(|c_1|+\cdots+|c_k|)$.

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 $$ y_n^*(y)=1 \quad\text{and}\quad y^*_n(x_j)=\frac{1}{n},\,\,\text{for $j=1,\ldots,n$.} $$ Clearly, every open set of the form $W_{x_1,1/k_1}\cap\cdots\cap W_{x_n,1/k_n}$ contains all but finitely many terms of the sequence $\{y_n^*\}_{n\in\mathbb N}$, while $W_{y,1}$ contains none of them. Hence $$ U\not\subseteq W_{y,1}, $$ for all $U\in\mathcal N$.

  • $\begingroup$ Hi, I totally understood your proof, except for the trivial fact in which the test functions space has an uncountable dimension (as a vector space I presume). How does that follow ? Thank you! $\endgroup$
    – DanielC
    Feb 16, 2014 at 23:13
  • $\begingroup$ For example, all the functions $\{\exp(-a|x|^2) :a>0 \}\subset {\mathscr S}'(\mathbb R^n)$ are linearly independent. $\endgroup$ Feb 16, 2014 at 23:16
  • $\begingroup$ @YiorgosS.Smyrlis How do you extend the functionals $y_n^\star$ in arbritrary topological vector spaces, where you may not use a Hahn-Banach theorem? And actually, why is $y_1\notin W_{y,1}$ as $y_1(y)=0$? $\endgroup$
    – Vobo
    Feb 17, 2014 at 11:11
  • $\begingroup$ @Vobo: In LCTVS. Corrected. $\endgroup$ Feb 17, 2014 at 12:36
  • $\begingroup$ You meant $y_n^\ast(y) = 1$ at the end, didn't you? $\endgroup$ Feb 17, 2014 at 12:41

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