View source for Dual space of C 0(x) vs C b(x)


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

==Background and Statement==
Let <math> 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  \} </math> equipped with the sup norm. In other words this is the space of continuous function vanishing at infinity. Note that in the case <math>X</math> is compact we can choose <math>K=X</math> in the previous definition, since properties on the empty set are trivially true, we can conclude that <math> C(X) = C_{0}(X)  </math>. Let <math> C_{b}(X) </math> be the space of bounded continuous functions on <math> X </math> together with the sup norm, Again in the case <math>X</math> 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 <math> \{x_n\}_{n \in \N} </math> such that <math>  |f(x_n)| \rightarrow \infty</math> as <math> n \rightarrow \infty </math>. I can then extract by compactness a sub sequence <math> \{x_{n_k}\}_{i \in \N} </math> converging to a point <math> \hat{x} \in X </math>. Therefore by continuity of <math>f</math> we have <math>  |f(x_{n_k})| \rightarrow |f(\hat{x})| < \infty</math>, and this is our desired contradiction. We conclude that <math> C(X) = C_{b}(X) = C_{0}(X)  </math>.

The rest of this discussion will consider the case where <math>X</math> is not compact. We wont have the last chain of equalities but only the inequalities <math> C(X) \supset C_{b}(X) \supset C_{0}(X)  </math>
==The case of <math> C_{0}(X)' </math> ==
The representation of the dual space of <math> C_{0}(X) </math> is a described by the following well known result in Functional Analysis (Riesz Representation Theorem 6.19 in Rudin <ref>Rudin, Walter. Real and Complex Analysis, 1966.</ref>): 

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

:<math> <\phi, f > = \int_{X} f d \mu, \text{ for every }  f \in C_{0}(X) </math>.
This allows us to identify <math> C_{0}(X)' </math> with <math> \mathcal{M}(X) </math>, space of complex Borel measures. Moreover we can endow <math> C_{0}(X)' </math> with the total variation norm: <math> \| \phi \| = |\mu|(X)   </math>. 

==The case of <math> C_{b}(X)' </math> ==
To descrive the dual space of <math> C_{b}(X) </math>, I will focus my attention to only one property of functions which is the behavior at infinity, Exercise 1.23 of <ref>Villani, Cedric.  Topics In Optimal Transportation. American Mathematical Soc., 2003.</ref>. First we need a preliminary result: <math> C_{0}(X) </math> is a closed (vector) subspace of <math> C_{b}(X) </math>. In other words <math> C_{0}(X) </math> contains all its limit points. Let <math> f_n(x) </math> be a convergent sequence in <math> C_{0}(X)' </math>, <math> f_n(x) </math> are continuous functions vanishing at infinity and let <math> f(x) </math> their limit. f is continuous since the uniform norm we are using provides uniform convergence. It remains to show that such limit <math> f(x) </math> vanishes at infinity: let <math> n \in \N </math> be such that <math> \|f - f_n\|_\infty <\epsilon/2 </math>, now since each <math> f_n(x) </math> vanishes at infinity, we can find <math> K_n \subset X </math> such that <math> |f_n(x)|< \epsilon/2 </math> for any <math> x\in X \setminus K_n </math>. Then we can conclude by triangle inequality that

:<math>  |f(x)|  \leq |f(x) - f_n(x)| + |f_n(x)| < \epsilon </math>,
That is, <math> f(x) \in C_{0}(X) </math>. I will now need to carefully specify the local property at infinity for <math> C_{b}(X) </math>.

We say that a function <math> u \in C_{b}(X) </math> admits a limit at infinity <math> u(\infty) </math> if for any <math> \epsilon >0  </math> there exists a compact <math> K_{\epsilon} \sub X </math> such that if <math> x \notin K_{\epsilon} </math> implies <math> |u(x) - u(\infty)|\leq \epsilon </math>. 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 <math> C_{b}(X) </math>. This is another spectacular consequence of Axiom of Choice (Hahn-Banach theorem [https://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem] in this case) since the extension assign a limit to any continuous bounded function! Intuitively we can partition the space <math> C_{b}(X) </math> 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 <math> C_{b}(X) </math> admits such limit. But this will not stop us from falling down the rabbit hole: Note that every function in <math>  C_{0}(X) </math> admit such limit, let <math> l : C_{0}(X) \rightarrow \R </math>

:<math>  u \rightarrow u(\infty)  </math>,
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 <math> l\in C_{0}(X)' </math>, i.e. a bounded linear operator on <math> C_{0}(X) </math>. We showed before that <math> C_{0}(X) </math> is a closed (vector) subspace of <math> C_{b}(X) </math> therefore we can extend <math> l </math> to the whole <math> C_{b}(X) </math> using the formulation of Hahn Banach Theorem for normed spaces. Let <math> L </math> be such extension, <math> L \in C_{b}(X)' </math> and <math> L=l </math> on <math> C_{0}(X) </math>. Note that this functional is supported at infinity, in the sense that for any <math> u \in C_{0}(X) </math> then <math> <L,u>=0 </math>.

==Kantorovich Duality for <math> C_{b}(X \times Y) </math> ==
As it can be found in Villani, Proposition 1.22 <ref>Villani, Cedric.  Topics In Optimal Transportation. American Mathematical Soc., 2003.</ref> also  [http://34.106.105.83/wiki/Kantorovich_Problem], the following version of Kantorovich duality holds: let <math> X </math> and <math> Y </math> locally compact Polish spaces, let <math> c </math> be a lower semi-continuous non negative function on <math> X \times Y </math> and let <math> \mu </math> and <math> \nu </math> be two Borel probability measures on <math> X \times Y </math> respectively then:

:<math> \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 </math>,

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

As mentioned in Villani Section 1.3 pag. 39<ref>Villani, Cedric.  Topics In Optimal Transportation. American Mathematical Soc., 2003.</ref>, if we try to extend the proof of the compact case we run into a problem since the dual of <math> C_{b}(X\times Y)</math> strictly contains <math> \mathcal{M}(X\times Y)</math>. If we restrict to the closed subspace <math> C_{0}(X\times Y) \subset C_{b}(X\times Y)</math> than any elements <math> C_{b}(X\times Y)'</math> acts continuously, as mentioned before, can be represented by a unique <math> \pi \in \mathcal{M}(X\times Y)</math> such that 

:<math> <l,f> = \int_{X\times Y} f(x,y) d \pi, \text{ for every }  f \in C_{0}(X \times Y) </math>.

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

For what discussed in the previous section the behavior of some <math> R </math> may not be clear at first glance as the following result showed in exercise 1.23 of <ref>Villani, Cedric.  Topics In Optimal Transportation. American Mathematical Soc., 2003.</ref>.

Let <math> \mu </math> and <math> \nu </math> be two Borel probability measures on <math> X \times Y </math> respectively, there is a continuous linear functional <math> L </math> on <math> C_{b}(X\times Y)</math>, supported at infinity, such that the following holds

:<math> \forall (\phi,\psi) \in C_{0}(X) \times C_{0}(Y), <L,\phi +\psi>= \int_{X}\phi d\mu + \int_{Y}\psi d\nu.  \text{        } \bigstar  </math>.
To prove this we want to apply what we have seen in the previous section, lets consider the function <math> u(x,y) \in C_0(X \times Y) </math>, for fixed <math> x </math> we can see this function as a function of <math> y </math> i,e, let <math> \hat{u}(y)=u(x,y) </math>. Noticing that <math>\hat{u} \in  C_0(Y) </math>, we can assign a limit at infinity <math> l </math> to <math>\hat{u} </math> and then extend it to all <math>\phi \in  C_b(Y) </math> following the construction of <math>L</math> in the precious section. Similarly we will do the same by considering for any fixed <math> y </math> the function <math> u(x,y) \in C_0(X \times Y) </math> as a function of <math> x </math>. Note that such extension is supported at infinity! This will allow me to first write the two functions:

:<math> u_1(x,\infty)= l(\hat{u}(y)), u_2(\infty,y)=l(\hat{u}(x)) </math>.
Since we are only considering functions that vanishing at infinity we can conclude that our <math> u_1 \in C_b(X) </math>  and <math> u_2 \in C_b(Y) </math> satisfy the following: 
:<math>  <l, u_1+u_2 >= \int_{X}u_1(x,\infty) d\mu (x) + \int_{Y}u_2(\infty,y) d\nu(y)     </math>.
Where <math> l </math>, with a slightly abuse of notation, is  the simultaneously assignment of limit at infinity in <math>  C_b(Y) </math> and <math>  C_b(X) </math>. The simultaneously extension <math> L </math>, again with a little abuse of notation, will be a bounded linear functional on <math> C_{b}(X\times Y)</math> will satisfy <math> \bigstar</math>