If mathematicians study an object they sometimes study the properties of the object. E.g. you can study the object “2” (that is the natural number 2) by saying that it denotes a certain cardinality, name the cardinality of two things. Well, that does not give you anything new. However, mathematicians often study objects by studying what either the object can do to other things or what other things can do the objects. In our (quite silly) example of the number two you see that you can use 2 to add it to other numbers (or, conversely, you can add other numbers to 2). One may feel that this does not add any insight but for objects which are more complicated (e.g. like sets of other objects or sets of these) this view can be particularly useful: If you consider a normed vector space , you can ask either for properties of the space itself (e.g. completeness) or you ask how one can map the objects of the space to other sets. Using the underlying ingredients to give more structure on the mappings under consideration this gives new constructions: In the case of a normed vector space you have the underlying field . This leads to mappings from to . You also have a linear structure on . This leads to linear mappings from to . Finally, there is a norm on and if there is an absolute value on this leads you to bounded linear mappings from to . Any guess what: These objects form the so-called dual space.

Dual spaces of (say) Banach spaces are telling you new things about the space itself and it seems to me that if you want to understand the nature of a Banach space you need to know the dual space (or a space of which your space is the dual, i.e. a pre-dual space).

This post shall not be about duality in general but on dual spaces of a particular type of spaces, namely of spaces which consist of continuous functions. While these spaces usually form nice Banach spaces, they are not as simple as Lebesgue spaces or Sobolev spaces in that there are no Hilbert spaces among them and even no reflexive spaces.

**1. What spaces of continuous functions? **

For sets and one can study continuous functions as soon as both and are topological spaces. For these functions to form a vector space we also want that is a vector space over a field . To get a normed space of continuous functions we endow with a norm and try to define a norm of as

But wait: The supremum may not exist. We better add the assumption that is bounded. Now this gives us a normed space and we denote:

Together with the norm this gives a Banach space as soon as is a Banach space. In the case that is *compact*, we can omit the condition of boundedness and obtain

and again get a Banach space as soon as is a Banach space.

Instead of assuming compactness of the whole space we could shift this property to all functions in this space. This leads to

This does not lead to a Banach space in the case that is not compact. Indeed, there is a last possibility by considering the closure of with respect to leading to the space

Informally one says that this are the continuous function “that vanish on the boundary of ”.

All these spaces admit dual spaces and all these spaces are described by some kind of “Riesz representation theorem”. These dual spaces consists of some kind of measures and the dual pairing is then defined as integrating the continuous functions with respect to these measures. Here I want to clarify (at least for myself) what these dual spaces are and how they are related.

The case in which (i.e. the real or complex numbers) is a little easier and we omit the argument in the spaces in this case: , ,…

**2. The dual of for normal **

We deal with real or complex valued functions on a set and assume that in a *normal topological space* (also called -space), that is the points in are all closed and two disjoint closed sets can be separated by open neighborhoods. Of course a metric space is a normal topological space.

** 2.1. Regular bounded finitely additive measures **

To see what the dual space of could be, we have to think of integrating a bounded continuous functions against a measure . Here the topological structure of comes into play because it naturally leads to a -algebra on , namely the one which is generated by the closed sets in (or equivalently be the open sets) which is also called the Borel algebra on . On a -algebra we can define a set function (which shall be a measure) by mapping the elements in to real (or complex) numbers. Such a set function is called *bounded* if for all and *finitely additive* if for any finite number of disjoint set it holds that

The set of all bounded and finitely additive set functions (nowadays often called bounded finitely additive measures) is denoted by . Equipped with the variational norm , becomes a Banach space. The space is a fairly large space of measures: Note that we did neither assume that the member are *countably* additive nor that the values shall be non-negative. However, we assumed the all values are finite which excludes, e.g. the Lebesgue-measure on (while *weighted* Lebesgue measures can be allowed if the weight is integrable). The space is slightly too large to be the dual of and we need another restriction. We call an element *regular*, if for any and any there exists a closed set and an open set such that for every it holds that . The space of *regular bounded finitely additive measures* is denoted by and is closed subspace of

** 2.2. Intgration with respect to **

Now we can define the integral of a bounded continuous function with respect to a regular bounded finitely additive measure as follows: Since is a bounded set in we can cover is with open sets with diameter less than a given . Define and . If is not empty, choose (otherwise choose ). Since is continuous, the sets are in and we can define

which is an -approximation to and indeed, is the uniform limit of the . For each , the integral is naturally defined as

and the limit of this expression is used as integral for .

** 2.3. Duality **

Since it holds that

every defines a continuous linear functional an . This shows that the dual space of contains . Indeed both spaces are equal:

Theorem 1For a normal topological space it holds that

The proof is lengthy and technical and the prime reference is “Linear Operators” by Dunford and Schwartz (Theorem IV.6.2).

**3. The dual of for compact Hausdorff **

Now we assume that is a compact Hausdorff space (think of closed and bounded subsets of ). Due to compactness we can omit the boundedness assumption on our continuous functions (they are bounded anyway).

** 3.1. Regular bounded countably additive measures **

We move from finitely additive measures to countably additive measure that is, (1) holds for countably many disjoint sets . Together with finiteness and regularity we arrive at the space of regular bounded countably additive measures. In our case, where is also a topological space and the -algebra is the Borel algebra this space is also called the space of regular Borel measures} or Radon measure} and often denoted by .

** 3.2. Duality **

Here we have the following representation theorem:

Theorem 2If is compact and Hausdorff it holds that

By

every determines an element in . Moreover, it is clear that . It remains to show that every determines a such that for all it holds that .

**4. The dual of for locally compact Hausdorff **

This case is basically the same as for with compact Hausdorff . The absence of compactness is compensated by the fact that the function “vanish on the boundary”. Well, “vanishing on the boundary” really means, that the function can be approximated by continuous function with compact support (in the -norm) and hence, the result do not differ from the previous one:

Theorem 3If is locally compact and Hausdorff it holds that

Final remarks:

- It turned out that the situation is easier than I expected. Basically there are two cases: Bounded continuous functions on a normal set . Here we get the regular bounded and finitely additive measures as the dual space. The case are continuous functions on a compact space or continuous function which “vanish at the boundary” on a locally compact space. In both we arrive at cases regular bounded countably additive measure aka Radon measures.

- For the distinction is:
- arbitrary and bounded continuous functions give .
- compact and continuous functions give .
- arbitrary and continuous functions vanishing on also give .

- There are a lot of excellent references for the Riesz representation theorem on the web and you’ll easily find several proofs. However, the proof is always technical and lengthy. One attempt of a short proof in the AMM-article “The Riesz Representation Theorem Revisited”was successful, however uses several heavy tools: Stone-Cech compactification, Caratheodory extension and a result about clopen sets in extremally disconnected spaces.

December 27, 2011 at 11:35 am

Dear Prof. Lorenz, thank you for keeping this blog. I was searching for the duality that you present in section 2.3 and I’ve stumbled upon this page which proved to be exactly what I was looking for.

Please allow me to ask one technical question: do you know whether an analogue of Lebesgue’s dominated convergence theorem holds for r.b.a. measures? Usually, Lebesgue’s theorem is given for measure spaces, which implies countable-additivity and positivity of the measure, and I do not know whether the proof (or at least the result) can be carried over to regular bounded finitely-additive complex measures.

January 5, 2012 at 10:36 am

Sorry for the delay (due to holiday break). I am not aware of a result in this direction. I suggest to look at the proof of Fatou’s lemma (e.g. on Wikipedia) and check if it works for an rba measure (then dominated convergence should follow as usual). Probably you need some additional assumption (non-negativity of the measure?).

March 18, 2012 at 7:42 pm

I have a question concerning 2.1 / 2.3: In the literature (e.g. Dunford-Schwartz) rba(X) is defined as the set of all bounded finitely additive set functions on the “algebra” generated by the closed subsets of X. In your definition, “sigma-algebra” is used instead of “algebra”. Doesn’t this distinction matter ?

March 18, 2012 at 9:15 pm

In this case we are only deal with the Borel algebra which is indeed a sigma algebra. In general, only an algebra (no countable unions and intersections needed) is enough to define ba and rba.

March 19, 2012 at 12:07 am

The heart of my question was: If v is an rba set function on the “algebra”, then, does it have a unique extension to an rba set function on the Borel algebra ?

March 19, 2012 at 8:20 am

This is not true in general but under not too complicated sufficient conditions, see section 2 in this post: http://terrytao.wordpress.com/2010/10/30/245a-notes-6-outer-measures-pre-measures-and-product-measures/

May 4, 2012 at 11:00 am

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September 20, 2013 at 4:37 pm

Thanks for this overview which I just stumbled upon. One question: Since the continuous functions vanishing at infinity form a subspace of the bounded continuous functions, the dual space of the former, rca, should contain the dual space of the latter, rba. But countable additivity seems to be more restrictive than finite additivity. How does this fit together?

September 24, 2013 at 10:46 am

Good point. Actually I could dig up the answer yet. Should be related to the fact that this “vanishing at the boundary” in defined only for locally compact spaces . There may be something special to this case…

August 6, 2014 at 12:53 pm

You can restrict functionals on the bigger space to functionals on the smaller space, but this restriction is not injective (unless the subspace is dense). Therefore there is not an inversed inclusion of the dual spaces, i.e. the dual of the bigger space is not included in the dual of the smaller (just look at finite dimenions to get the idea of this). Indeed, in this case, since C_0 is closed in C_b, the dual of the C_0 is a quotient of C_b, hence smaller (see any functional analysis reference, e.g. Rudin Chapter 4)