# Weak-$*$ topology on algebraic dual

I was looking at

Izzo, Alexander J., A functional analysis proof of the existence of Haar measure on locally compact Abelian groups, Proc. Am. Math. Soc. 115, No. 2, 581-583 (1992). ZBL0777.28006.

which proves existence of the Haar-measure for locally compact abelian groups using the Markov-Kakutani theorem.

What I find strange is that the Haar measure is constructed as an element of the dual of $$C_c(X)$$. But for noncompact $$X$$ (such as $$X$$ being the real numbers $$\Bbb R$$) this must be an unbounded functional (as the Lebesgue-measure on $$\Bbb R$$ is not finite). It seems like the author has no problem with this, and (without mentioning it further) goes on to define a weak-* topology for this case and even uses Banach-Alaoglu.

I have not seen this being done this way before, am I misunderstanding something or can one define a weak-* topology on the algebraic dual of a TVS without any problems?

Probably the topology on $$C^o_c(\mathbb R)$$ (or for other non-compact topological group in place of $$\mathbb R$$) is not what you anticipated. It is not sup norm, for example. It is an ascending union, categorically a "strict colimit" (=strict inductive limit=...) of spaces of the form $$C^o(K)$$ as $$K$$ ranges over compact subsets of $$\mathbb R$$. (These are Banach spaces.)

In particular, from the characterizing mapping property of "colimit", a linear map or functional from $$C^o_c(\mathbb R)$$ is continuous if and only if its restriction to each $$C^o(K)$$ is continuous.

So the dual of $$C^o_c(\mathbb R)$$ does not include "unbounded" functionals in a true sense, because even on such spaces "continuous" is still equivalent to "bounded"... but "bounded" has a more complicated sense.

• Sure, if I don't take C_c( R ) with the norm topology but with some different topology then I can get more continuous functionals. But this doesn't answer my question: in Izzo's proof (if I understand it correctly) he explicitly takes the space of all linear functionals on C_c and gives it a topology and claims that something like Banach-Alaouglo holds. – Nathanael Schilling Feb 14 '19 at 22:37
• @NathanaelSchilling, I would strongly doubt that the intent is to take all linear functionals on $C^o_c(\mathbb R)$, since things would run amok. Ah... Just realized: when functional analysts say "dual" they almost-surely mean "continuous dual". Rarely, if ever, "purely-algebraic dual". I'd bet a dollar that that's the intent. – paul garrett Feb 14 '19 at 22:44
• Yes usually I would agree. I've never seen the algebraic dual being used in functional analysis. Except (implicitly) in the Riesz-Markov theorem, which takes an arbitrary positive functional on C_c and shows it can be represented by a measure. But since the author emphasises the word "all" here and the context is very similar, it got me wondering if this perhaps is possible... I don't see any obvious obstructions to defining a weak^* topology on this space, and even if the result is not a topological vector space it could be that some sort of compactness results hold even in this setting? – Nathanael Schilling Feb 15 '19 at 14:19
• Also, btw, the "positive functional" version of Riesz-Markov-Kakutani is just a veiled form of a continuity assumption (Banach-Steinhaus...), but/and has a certain popularity because it's easier to say "positive" than to describe the topology on compactly-supported continuous functions. – paul garrett Feb 16 '19 at 18:26
• @paul garrett, I fully agree with you. A. Izzo formulates and proves in his Lemma 2 weak* compactness of closed and locally bounded subsets of the space of all linear functionals on $X$, and this is a nice piece of information (my opinion). But then he moves on with positive linear functionals on $C_c(G)$ and is thus back in the continuous functionals as a subset of the algebraic dual. Hence the full generality of Lemma 2 is neither used nor necessary to prove what is ultimately proved (the existence of the Haar measure). – Matthias Hübner Mar 28 '19 at 11:41

I agree with Paul Garrett that it seems more reasonable to endow $$C_c(G)$$ with the strict inductive limit topology and consider the topological dual, which is isomorphic to the space of all Radon measures on $$G$$ (in particular, infinite measures also give rise to continuous functionals).

However, the linked lemma also holds for the algebraic dual $$X^\ast$$. What the author calls the weak$$^\ast$$ topology on $$X^\ast$$ is the $$\sigma(X^\ast,X)$$ topology, i.e, the locally convex topology generated by the seminorms $$p_x\colon\phi\mapsto|\phi(x)|$$ for $$x\in X$$. If you view $$X^\ast$$ as a subspace of $$\mathbb{R}^X$$, then this is just the (induced topology of) the product topology. This already gives you a hint how to prove this generalized Banach-Alaoglu theorem.

For $$x\in X$$ let $$K_x$$ be the closure of $$\{\Lambda(x)\mid\Lambda\in K\}$$. By assumption, $$K_x$$ is compact. Then $$K$$ is a subset of $$\prod_{x\in X}K_x$$, which is compact by Tychonoff's theorem. Since $$K$$ is also assumed to be closed in $$X^\ast$$, it suffices to show that $$X^\ast$$ is a closed subset of $$\mathbb{R}^X$$, which is pretty straightforward.

The intent of the quoted lemma in Izzo's paper is to show the compactness of a certain subset of linear functionals in the weak* topology, and I agree with @MaoWao that the linked lemma also holds for the algebraic dual $$X^*$$. The definition of the weak* topology on a space $$X^*$$ of functionals or generally functions on a domain space $$X$$ does not depend on any topology on $$X$$, but rather only on the topology of the value field (e.g. $$R$$ or $$C$$), and if a theorem like Banach-Alaoglu is desired, then on the local compactness of the value field.

It is not the 1st time I see the weak* topology on the space of $$all$$ linear functionals, i.e. on the algebraic dual: on page 108 of John L. Kelley's famous Topology book there is exercise W on "Functionals on real linear spaces" which starts like follows:

"Let $$(X, +,.)$$ be a real linear space. A real-valued linear function on $$X$$ is called a $$linear functional$$. The set $$Z$$ of all linear functionals on $$X$$ is, with the natural definition of addition and scalar multiplication, a real linear space. It is clear that $$Z$$ is a subset of the product $$R^X$$, where $$R$$ is the set of real numbers. The relativized product topology for $$Z$$ is called the $$weak^*$$ or $$w^*$$ topology (the $$simple$$ topology)."

Kelley then moved on to formulate a $$Density$$ $$lemma$$ and an $$Evaluation$$ $$Theorem$$ as exercises with respect to the weak* topology on the Algebraic dual $$Z$$ of $$X$$.