Dual space of C 0(x) vs C b(x)

Introduction

In the case that the space $X$ is compact then all continuous functions belongs to $C_{0}(X)$ as we will show in the next section. On the other hand if the space $X$ is not compact while we always have the inclusion $C(X)\supseteq C_{0}(X)$ there may be some continuous function that do not belong to $C_{0}(X)$. Some of them may even even be bounded and still not belong to $C_{0}(X)$, this motivates us to consider the dual space of $C_{0}(X)$ and the dual space of $C_{b}(X)$.

Background and Statement

Let $C_{0}(X)=\{f\in C(X){\text{ and }}\forall \epsilon >0{\text{ }}\exists {\text{ a compact set}}K\subset X{\text{ s.t. }}\mid f(x)\mid <\epsilon {\text{ }}\forall x\in X\setminus K\}$ equipped with the sup norm. In other words this is the space of continuous function vanishing at infinity. Note that in the case $X$ is compact we can choose $K=X$ in the previous definition, since properties on the empty set are trivially true, we can conclude that $C(X)=C_{0}(X)$. Let $C_{b}(X)$ be the space of bounded continuous functions on $X$ together with the sup norm, Again in the case $X$ is compact we have not introduced a new space since every continuous function on a compact metric space is bounded, to see this assume on the contrary that there is a sequence $\{x_{n}\}_{n\in \mathbb {N} }$ such that $|f(x_{n})|\rightarrow \infty$ as $n\rightarrow \infty$. I can then extract by compactness a sub sequence $\{x_{n_{k}}\}_{i\in \mathbb {N} }$ converging to a point ${\hat {x}}\in X$. Therefore by continuity of $f$ we have $|f(x_{n_{k}})|\rightarrow |f({\hat {x}})|<\infty$, and this is our desired contradiction. We conclude that $C(X)=C_{b}(X)=C_{0}(X)$.

The rest of this discussion will consider the case where $X$ is not compact. We wont have the last chain of equalities but only the inequalities $C(X)\supset C_{b}(X)\supset C_{0}(X)$

The case of $C_{0}(X)'$

The representation of the dual space of $C_{0}(X)$ is a described by the following well known result in Functional Analysis (Riesz Representation Theorem 6.19 in Rudin ):

Let $X$ be a locally compact Hausdorff space, for any bounded linear function $\phi$, i.e. an element of the dual space $C_{0}(X)'$, there is a unique complex Borel measure $\mu$ such that the following holds:

$<\phi ,f>=\int _{X}fd\mu ,{\text{ for every }}f\in C_{0}(X)$.

This allows us to identify $C_{0}(X)'$ with ${\mathcal {M}}(X)$, space of complex Borel measures. Moreover we can endow $C_{0}(X)'$ with the total variation norm: $\|\phi \|=|\mu |(X)$.

The case of $C_{b}(X)'$

To descrive the dual space of $C_{b}(X)$, I will focus my attention to only one property of functions which is the behavior at infinity, Exercise 1.23 of . First we need a preliminary result: $C_{0}(X)$ is a closed (vector) subspace of $C_{b}(X)$. In other words $C_{0}(X)$ contains all its limit points. Let $f_{n}(x)$ be a convergent sequence in $C_{0}(X)'$, $f_{n}(x)$ are continuous functions vanishing at infinity and let $f(x)$ their limit. f is continuous since the uniform norm we are using provides uniform convergence. It remains to show that such limit $f(x)$ vanishes at infinity: let $n\in \mathbb {N}$ be such that $\|f-f_{n}\|_{\infty }<\epsilon /2$, now since each $f_{n}(x)$ vanishes at infinity, we can find $K_{n}\subset X$ such that $|f_{n}(x)|<\epsilon /2$ for any $x\in X\setminus K_{n}$. Then we can conclude by triangle inequality that

$|f(x)|\leq |f(x)-f_{n}(x)|+|f_{n}(x)|<\epsilon$,

That is, $f(x)\in C_{0}(X)$. I will now need to carefully specify the local property at infinity for $C_{b}(X)$.

We say that a function $u\in C_{b}(X)$ admits a limit at infinity $u(\infty )$ if for any $\epsilon >0$ there exists a compact $K_{\epsilon }\subset X$ such that if $x\notin K_{\epsilon }$ implies $|u(x)-u(\infty )|\leq \epsilon$. We can see this operation as a linear function 'limit at infinity' thanks to Hahn-Banach we can build a continuous extension of it for all $C_{b}(X)$. This is another spectacular consequence of Axiom of Choice (Hahn-Banach theorem [1] in this case) since the extension assign a limit to any continuous bounded function! Intuitively we can partition the space $C_{b}(X)$ into equivalence classes of the equivalence relation of having the same limit at infinity. Then thanks to the axiom of choice choose a representative for each class. The problem with this argument that we don't know yet that every functions in $C_{b}(X)$ admits such limit. But this will not stop us from falling down the rabbit hole: Note that every function in $C_{0}(X)$ admit such limit, let $l:C_{0}(X)\rightarrow \mathbb {R}$

$u\rightarrow u(\infty )$,

Since the functions vanishes at infinity this operation of assigning limit at infinity is clearly a linear map. It's not hard to see that $l\in C_{0}(X)'$, i.e. a bounded linear operator on $C_{0}(X)$. We showed before that $C_{0}(X)$ is a closed (vector) subspace of $C_{b}(X)$ therefore we can extend $l$ to the whole $C_{b}(X)$ using the formulation of Hahn Banach Theorem for normed spaces. Let $L$ be such extension, $L\in C_{b}(X)'$ and $L=l$ on $C_{0}(X)$. Note that this functional is supported at infinity, in the sense that for any $u\in C_{0}(X)$ then $<L,u>=0$.

Kantorovich Duality for $C_{b}(X\times Y)$

As it can be found in Villani, Proposition 1.22 also [2], the following version of Kantorovich duality holds: let $X$ and $Y$ locally compact Polish spaces, let $c$ be a lower semi-continuous non negative function on $X\times Y$ and let $\mu$ and $\nu$ be two Borel probability measures on $X\times Y$ respectively then:

$\inf _{\pi \in \Pi (\mu ,\nu )}\int _{X\times Y}c(x,y)d\pi (x,y)=\sup _{(\phi ,\psi )\in \Phi _{c}}\int _{X}\phi d\mu +\int _{X}\psi d\nu$,

Here $\Pi (\mu ,\nu )$ is the set of all probability measures $\pi$ that satisfies $\pi (A\times Y)=\mu (A)$ and $\pi (X\times B)=\nu (B)$ for any measurable set $A\subset X$ and any measurable set $B\subset Y$; $\Phi _{c}$ is the set of all measurable functions $(\phi ,\psi )\in L^{1}(d\mu )\times L^{1}(d\nu )$ that satisfies $\phi (x)+\psi (x)\leq c(x,y)$ for $d\mu$ almost all $x\in X$ and for $d\nu$ almost all $y\in Y$.

If we try to extend the proof of the compact case we run into a problem since the dual of $C_{b}(X\times Y)$ strictly contains ${\mathcal {M}}(X\times Y)$. If we restrict to the closed subspace $C_{0}(X\times Y)\subset C_{b}(X\times Y)$ than any elements $C_{b}(X\times Y)'$ acts continuously, and mentioned before, can be represented by a unique $\pi \in {\mathcal {M}}(X\times Y)$ such that

$<l,f>=\int _{X\times Y}f(x,y)d\pi ,{\text{ for every }}f\in C_{0}(X\times Y)$.

We can then write $l=\pi +R$ where $R$ is a continuous linear functional supported at infinity, i.e. $\forall f\in C_{0}(X\times Y)$ implies $<R,f>=0$.

For what discussed in the previous section the behavior of some $R$ may not be clear at first glance as the following result implies:

let $\mu$ and $\nu$ be two Borel probability measures on $X\times Y$ respectively, there is a continuous linear functional $l$ on $C_{b}(X\times Y)$, supported at infinity, such that the following holds

$\forall (\phi ,\psi )\in C_{0}(X)\times C_{0}(Y),<l,\phi +\psi >=\int _{X}\phi d\mu +\int _{Y}\psi d\nu$.

Because