Dual space of C 0(x) vs C b(x): Difference between revisions
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==Background and Statement== | ==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>. In other words this is the space of continuous function vanishing at infinity. | 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_i\}_{i \in \N} </math> such that <math> |f(x_i)| \rightarrow \infty</math>. With this norm <math> C_{0}(X) </math> is a closed subspace of <math> C_{b}(X) </math>. The representation of the dual space of <math> C_{0}(X) </math> is a well 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 locally compact Hausdorff space, <math> \phi | 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> | :<math> <\phi, f > = \int_{X} f d \mu, \text{ for every } f \in C_{0}(X) </math>. | ||
Moreover we can endow <math> C_{0}(X)' </math> with the total variation norm: <math> \| \phi \| = |\mu|(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>. This allows us to identify <math> C_{0}(X)' </math> with <math> \mathcal{M}(X) </math>, space of complex Borel measures. | ||
Revision as of 06:39, 6 June 2020
Introduction
In the case that the space is compact then all continuous functions belongs to as we will show in the next section. On the other hand if the space is not compact while we always have the inclusion there may be some continuous function that do not belong to . Some of them may even even be bounded and still not belong to , this motivates us to consider the dual space of and the dual space of .
Background and Statement
Let equipped with the sup norm. In other words this is the space of continuous function vanishing at infinity. Note that in the case is compact we can choose in the previous definition, since properties on the empty set are trivially true, we can conclude that . Let be the space of bounded continuous functions on together with the sup norm, Again in the case 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 such that . With this norm is a closed subspace of . The representation of the dual space of is a well described by the following well known result in Functional Analysis (Riesz Representation Theorem 6.19 in Rudin [1]):
Let be a locally compact Hausdorff space, for any bounded linear function , i.e. an element of the dual space , there is a unique complex Borel measure such that the following holds:
- .
Moreover we can endow with the total variation norm: . This allows us to identify with , space of complex Borel measures.
The case of
We want now to investigate the properties of elements of the dual space of , not by looking at local properties but rather a very property condition on the limit at infinity of our elements of , it turns out that this allow us to extract also global information on the whole .
We say that a function admits a limit at infinity if for any there exists a compact such that if implies . 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 . This is another spectacular consequence of Axiom of Choice (Hahn-Banach theorem in this case) since the extension assign a limit to any continuous bounded function!
Kantorovich Duality for
As it can be found in Villani [2], the following version of Kantorovich duality holds: let and locally compact Polish spaces, let be a lower semi-continuous non negative function on and let and be two Borel probability measures on respectively then:
- ,
Here is the set of all probability measures that satisfies and for any measurable set and any measurable set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B \subset Y } ; is the set of all measurable functions that satisfies for almost all and for almost all .
If we try to extend the proof of the compact case we run into a problem since the dual of strictly contains . If we restrict to the closed subspace than any elements acts continuously, and mentioned before, can be represented by a unique such that
- .
We can then write where is a continuous linear functional supported at infinity, i.e. implies .
For what discussed in the previous section the behavior of some may not be clear at first glance as the following result implies:
let and be two Borel probability measures on respectively, there is a continuous linear functional on , supported at infinity, such that the following holds
- .
Because