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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?

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 $p_xcolonphimapsto|phi(x)|$ for $xin X$. If you view $X^ast$ as a subspace of $mathbbR^X$, 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 $xin X$ let $K_x$ be the closure of $Lambda(x)midLambdain K$. By assumption, $K_x$ is compact. Then $K$ is a subset of $prod_xin XK_x$, 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 $mathbbR^X$, 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 $X$.

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.