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 ${X}$, 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 ${K}$. This leads to mappings from ${X}$ to ${K}$. You also have a linear structure on ${X}$. This leads to linear mappings from ${X}$ to ${K}$. Finally, there is a norm on ${X}$ and if there is an absolute value on ${K}$ this leads you to bounded linear mappings from ${X}$ to ${K}$. 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 ${\Omega}$ and ${V}$ one can study continuous functions ${f:\Omega\rightarrow V}$ as soon as both ${\Omega}$ and ${V}$ are topological spaces. For these functions to form a vector space we also want that ${V}$ is a vector space over a field ${K}$. To get a normed space of continuous functions we endow ${V}$ with a norm ${\|\cdot\|}$ and try to define a norm of ${f:\Omega \rightarrow V}$ as

$\displaystyle \|f\|_\infty = \sup_{x\in\Omega}\|f(x)\|.$

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

$\displaystyle C_b(\Omega,V) = \{f:\Omega\rightarrow V\ |\ f\ \text{continuous and bounded}\}.$

Together with the norm ${\|\cdot\|_\infty}$ this gives a Banach space as soon as ${V}$ is a Banach space. In the case that ${\Omega}$ is compact, we can omit the condition of boundedness and obtain

$\displaystyle C(\Omega,V) = \{f:\Omega\rightarrow V\ |\ f\ \text{continuous}\}$

and again get a Banach space as soon as ${V}$ is a Banach space.

Instead of assuming compactness of the whole space ${\Omega}$ we could shift this property to all functions in this space. This leads to

$\displaystyle C_c(\Omega,V) = \{f:\Omega\rightarrow V\ |\ f\ \text{continuous with compact support}\}.$

This does not lead to a Banach space in the case that ${\Omega}$ is not compact. Indeed, there is a last possibility by considering the closure of ${C_c(\Omega,V)}$ with respect to ${\|\cdot\|_\infty}$ leading to the space

$\displaystyle C_0(\Omega,V) = \overline{C_c(\Omega,V)}^{\|\cdot\|_\infty}.$

Informally one says that this are the continuous function “that vanish on the boundary of ${\Omega}$”.

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 ${V=\mathbb{K}}$ (i.e. the real or complex numbers) is a little easier and we omit the argument ${V}$ in the spaces in this case: ${C_b(\Omega) = C_b(\Omega,\mathbb{K})}$, ${C(\Omega) = C(\Omega,\mathbb{K})}$,…

2. The dual of ${C_b(\Omega)}$ for normal ${\Omega}$

We deal with real or complex valued functions on a set ${\Omega}$ and assume that ${\Omega}$ in a normal topological space (also called ${T_4}$-space), that is the points in ${\Omega}$ are all closed and two disjoint closed sets can be separated by open neighborhoods. Of course a metric space ${\Omega}$ is a normal topological space.

2.1. Regular bounded finitely additive measures

To see what the dual space of ${C_b(\Omega)}$ could be, we have to think of integrating a bounded continuous functions against a measure ${\mu}$. Here the topological structure of ${\Omega}$ comes into play because it naturally leads to a ${\sigma}$-algebra on ${\Omega}$, namely the one which is generated by the closed sets in ${\Omega}$ (or equivalently be the open sets) which is also called the Borel algebra ${B(\Omega)}$ on ${\Omega}$. On a ${\sigma}$-algebra we can define a set function (which shall be a measure) by mapping the elements in ${B(\Omega)}$ to real (or complex) numbers. Such a set function ${\mu}$ is called bounded if ${|\mu(E)|<\infty}$ for all ${E\in B(\Omega)}$ and finitely additive if for any finite number of disjoint set ${E_1,\dots,E_n\in B(\Omega)}$ it holds that

$\displaystyle \mu(\cup E_i) = \sum \mu(E_i). \ \ \ \ \ (1)$

The set of all bounded and finitely additive set functions (nowadays often called bounded finitely additive measures) is denoted by ${ba(\Omega)}$. Equipped with the variational norm ${|\mu| = |\mu|(\Omega)}$, ${ba(\Omega)}$ becomes a Banach space. The space ${ba(\Omega)}$ is a fairly large space of measures: Note that we did neither assume that the member are countably additive nor that the values ${\mu(E)}$ shall be non-negative. However, we assumed the all values ${|\mu(E)|}$ are finite which excludes, e.g. the Lebesgue-measure on ${\Omega=\mathbb{R}}$ (while weighted Lebesgue measures can be allowed if the weight is integrable). The space ${ba(\Omega)}$ is slightly too large to be the dual of ${C_b(\Omega)}$ and we need another restriction. We call an element ${\mu\in ba(\Omega)}$ regular, if for any ${E \in B(\Omega)}$ and any ${\epsilon>0}$ there exists a closed set ${F\subset E}$ and an open set ${G\supset E}$ such that for every ${C\subset G\setminus F}$ it holds that ${|\mu(C)|<\epsilon}$. The space of regular bounded finitely additive measures is denoted by ${rba(\Omega)}$ and is closed subspace of ${ba(\Omega)}$

2.2. Intgration with respect to ${\mu\in rba(\Omega)}$

Now we can define the integral of a bounded continuous function ${f\in C_b(\Omega)}$ with respect to a regular bounded finitely additive measure ${\mu\in rba(\Omega)}$ as follows: Since ${f(\Omega)}$ is a bounded set in ${\mathbb{K}}$ we can cover is with open sets ${G_1,\dots, G_n}$ with diameter less than a given ${\epsilon>0}$. Define ${A_1=G_1}$ and ${A_j = G_j\setminus \bigcup_{i=1}^{j-1} G_i}$. If ${A_i}$ is not empty, choose ${\alpha_j\in A_j}$ (otherwise choose ${\alpha_j=0}$). Since ${f}$ is continuous, the sets ${B_j = f^{-1}(A_j)\subset\Omega}$ are in ${B(\Omega)}$ and we can define

$\displaystyle f_\epsilon = \sum_{j=1}^n \alpha_j \chi_{B_j}$

which is an ${\epsilon}$-approximation to ${f}$ and indeed, ${f}$ is the uniform limit of the ${f_\epsilon}$. For each ${f_\epsilon}$, the integral is naturally defined as

$\displaystyle \int f_\epsilon d\mu = \sum \alpha_j \mu(B_j)$

and the limit of this expression is used as integral for ${f}$.

2.3. Duality

Since it holds that

$\displaystyle \Big| \int f d\mu\Big| \leq \|f\|_\infty |\mu|$

every ${\mu\in rba(\Omega)}$ defines a continuous linear functional an ${C_b(\Omega)}$. This shows that the dual space of ${C_b(\Omega)}$ contains ${rba(\Omega)}$. Indeed both spaces are equal:

Theorem 1 For a normal topological space ${\Omega}$ it holds that

$\displaystyle C_b(\Omega)^* = rba(\Omega).$

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 ${C(\Omega)}$ for compact Hausdorff ${\Omega}$

Now we assume that ${\Omega}$ is a compact Hausdorff space (think of closed and bounded subsets of ${\mathbb{R}^d}$). 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 ${E_i}$. Together with finiteness and regularity we arrive at the space ${rca(\Omega)}$ of regular bounded countably additive measures. In our case, where ${\Omega}$ is also a topological space and the ${\sigma}$-algebra is the Borel algebra this space is also called the space of regular Borel measures} or Radon measure} and often denoted by ${\mathfrak{M}(\Omega)}$.

3.2. Duality

Here we have the following representation theorem:

Theorem 2 If ${\Omega}$ is compact and Hausdorff it holds that

$\displaystyle C(\Omega)^* = \mathfrak{M}(\Omega)\ (=rca(\Omega)).$

By

$\displaystyle f\mapsto \int f d\mu$

every ${\mu\in\mathfrak{M}(\Omega)}$ determines an element in ${C(\Omega)^*}$. Moreover, it is clear that ${rba(\Omega)\supset C(\Omega)^*}$. It remains to show that every ${\lambda\in rba(\Omega)}$ determines a ${\mu\in rca(\Omega) = \mathfrak{M}(\Omega)}$ such that for all ${f\in C(\Omega)}$ it holds that ${\int fd\lambda = \int fd\mu}$.

4. The dual of ${C_0(\Omega)}$ for locally compact Hausdorff ${\Omega}$

This case is basically the same as for ${C(\Omega)}$ with compact Hausdorff ${\Omega}$. 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 ${\infty}$-norm) and hence, the result do not differ from the previous one:

Theorem 3 If ${\Omega}$ is locally compact and Hausdorff it holds that

$\displaystyle C_0(\Omega)^* = \mathfrak{M}(\Omega).$

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 ${\Omega}$. 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 ${\Omega\subset{\mathbb R}^d}$ the distinction is:
1. ${\Omega}$ arbitrary and bounded continuous functions give ${rba(\Omega)}$.
2. ${\Omega}$ compact and continuous functions give ${\mathfrak{M}(\Omega)}$.
3. ${\Omega}$ arbitrary and continuous functions vanishing on ${\partial\Omega}$ also give ${\mathfrak{M}(\Omega)}$.

• 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.