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!


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 '14 at 23:13
  • $\begingroup$ For example, all the functions $\{\exp(-a|x|^2) :a>0 \}\subset {\mathscr S}'(\mathbb R^n)$ are linearly independent. $\endgroup$ – Yiorgos S. Smyrlis Feb 16 '14 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 '14 at 11:11
  • $\begingroup$ @Vobo: In LCTVS. Corrected. $\endgroup$ – Yiorgos S. Smyrlis Feb 17 '14 at 12:36
  • $\begingroup$ You meant $y_n^\ast(y) = 1$ at the end, didn't you? $\endgroup$ – Daniel Fischer Feb 17 '14 at 12:41

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.