# Understanding the weak* topology on the space of distributions

I am studying distributions from Folland's book on real analysis. In this text he says

A distribution on $$U$$ is a continuous linear functional on $$C_C^\infty(U)$$ (where $$U$$ is an open subset of $$\mathbb{R}^n$$). The space of all distributions on $$U$$ is denoted by $$\mathcal{D}$$'(U). We impose the weak* topology on $$\mathcal{D}'(U)$$, that is, the topology of pointwise convergence on $$C_C^\infty(U)$$.

I have been trying to better understand this topology and why we would impose this in the first place. Part of my troubles could be that I have not used the weak* topology much and so I am not too comfortable/experienced with it. To my knowledge, for a topological vector space $$X$$, we look at the maps $$\mathscr{X}_x \in X^{**}$$ that are associated to some $$x \in X$$ defined as $$\mathscr{X}_x(f) = f(x)$$ for all $$f \in X^*$$, and require all these maps to be continuous. This is the weak* topology.

If this picture is correct, what interest is this topology in the context of distributions? Is it simply because it is weak enough to allow certain convergence results? Is there some better picture to have in mind?

• take a look here Sep 1, 2022 at 23:45
• Suppose you know that some sequence of functions $u_n$ converges to some function $w$ in some stronger topology. Then if you show that $u_n$ converges to some $u$ in a weaker topology, then by Hausdorffness of the weaker topology, $w = u$. Thus the weaker convergence allows you to identify the limit. This kind of argument can be used to prove that boundedness of difference quotients implies inclusion in a Sobolev space. Sep 1, 2022 at 23:53
• The weak$^*$-topology on $\mathscr D'(U)$ has some severe drawbacks. For example, it is neither complete (but only sequentially complete), nor bornological nor barrelled. If you want to apply, e.g., Banach-Steinhaus or closed graph theorems for operators on $\mathscr D'(U)$ you have to use the strong topology of uniform convergence on all bouded subsets of $C^\infty_c(U)$. Sep 2, 2022 at 6:56

Let $$X$$ be a topological vector space. Then the weak$$^\ast$$-topology on $$X'$$ is the initial topology on $$X'$$ with respect to the mappings $$\kappa_x \colon X' \to \mathbb C, \quad \kappa_x(x') := \langle x, x'\rangle.$$ Equivalently, the weak$$^\ast$$-topology is the weakest (or coarsest) topology on $$X'$$ such that all the mappings $$\kappa_x$$ are continuous. In particular, a net $$(x_i')_i$$ in $$X'$$ converges to $$x' \in X'$$ with respect to the weak$$^\ast$$-topology if and only if $$\kappa_x(x_i') = \langle x, x_i' \rangle \to \langle x, x'\rangle = \kappa_x(x').$$ In your case, one has $$X = C_c^\infty(U)$$ and $$\mathcal D'(U) = X' = \{u \colon X \to \mathbb C : u \text{ is linear and continuous} \}.$$ But one has to explain what continuity means in this context. Recall that $$X$$ is a locally convex vector space with respect to the family of seminorms given by $$\lVert \, \cdot \, \rVert_{\alpha} \colon X \to [0, \infty), \quad \lVert \varphi \rVert_{\alpha} := \sup_{x \in U} \lVert \partial^\alpha \varphi \rVert_\infty.$$ Thus, net $$(\varphi_i)_i$$ in $$X$$ converges to some $$\varphi \in X$$ if and only if $$\lVert \varphi_i - \varphi \rVert_\alpha \to 0 \qquad (\alpha \in \mathbb N_0^n)$$ and from standard topology one knows that for $$u \colon X \to \mathbb C$$ linear the following assertions are equivalent:

• $$u \in \mathcal D'(U)$$.
• $$u(\varphi_i) \to u(\varphi)$$ for each net $$(\varphi_i)_i$$ in $$X$$ such that $$\varphi_i \to \varphi \in X$$.

Now there are several different ways to define a topology on $$X'$$. In particular, if $$\mathcal B$$ is any collection of bounded subsets of $$X$$, then the seminorms $$p_B(x') := \sup_{x \in B} \langle x, x' \rangle \qquad (B \in \mathcal B)$$ induce a locally convex topology on $$X'$$ and this topology coincides with the weak$$^\ast$$-topology if you choose $$\mathcal B$$ to be the set of all finite subsets of $$X$$.

However, there is a reason why the weak$$^\ast$$-topology on $$X'$$ is so useful in the context of distributions: A net $$(u_i)_i$$ in $$\mathcal D'(U)$$ converges to $$u \in \mathcal D'(U)$$ with respect to the weak$$^\ast$$-topology if $$u_i(\varphi) \to u(\varphi)$$ for all $$\varphi \in X$$ (i.e., $$u_i \to u$$ pointwise).

But if you have a sequence $$(u_n)_{n \in \mathbb N}$$ in $$\mathcal D'(U)$$ that converges to some linear map $$u \colon X \to \mathbb C$$ with respect to the weak$$^\ast$$-topology, then it follows that $$u \in \mathcal D'(U)$$ due to the uniform boundedness principle. This shows that the weak$$^\ast$$-topology is pretty well-behaved with respect to sequence limits (in the sense that it yields that the limit is again a distribution) while being still quite weak in nature. Therefore, it is often easy to check in applications that a sequence of distributions converges to a linear map (which is at least a priori not a distribution) with respect to the weak$$^\ast$$-topology. I hope things got a little bit clearer :-)

• $X=C^\infty_c(U)$ isn't a Fréchet spaces (with respect to any reasonnable topology -- there are unreasonnable Fréchet topologies on $X$ which come from considerations about Hamel dimensions, but then not even the evaluations $f\mapsto f(x)$ for points $x\in U$ are continuous). The natural topology on $C_c^\infty(X)$ is an inductive limit of Fréchet spaces. Sep 2, 2022 at 6:51
• This answer is incredible, thank you! Do you have any references for building such an intuition on these topics? Sep 2, 2022 at 21:18
• @Jochen You are right. The Schwartz functions form a Fréchet space while the smooth functions with compact support do not. I corrected that part of my answer! Sep 2, 2022 at 22:26
• @CBBAM Do you speak german? If so I got a nice script covering basic distribution theory. Sep 2, 2022 at 22:29
• The continuity condition that you describe does not yield the space of all distributions! A linear map $:u:C^\infty_c\to\mathbb C$ is a distribution if, for every compact set $K\subseteq U$, there are $n\in\mathbb N$ and $c>0$ such that $|u(f)|\le c\sup\{|\partial^\alpha f(x)|: x\in U, |\alpha|\le n\}$ for all smooth functions $f$ with support in $K$. Sep 3, 2022 at 9:33