4
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Nathanael Schilling Feb 14 '19 at 22:37
  • $\begingroup$ @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. $\endgroup$ – paul garrett Feb 14 '19 at 22:44
  • 1
    $\begingroup$ 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? $\endgroup$ – Nathanael Schilling Feb 15 '19 at 14:19
  • 1
    $\begingroup$ 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. $\endgroup$ – paul garrett Feb 16 '19 at 18:26
  • $\begingroup$ @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). $\endgroup$ – Matthias Hübner Mar 28 '19 at 11:41
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $X$.

$\endgroup$

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.