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I'm reading the proof of Kantorovich duality from Villani's book Topics in Optimal Transportation. In page 28, the author said that

Exercise 1.11. Let us try to extend this proof to the non-compact case. Why would we like to replace $C_{b}(X \times Y)$ by $C_{0}(X \times Y) ?$ Show that if we do so in the definition of $\Xi$, then the latter turns out to be trivial: $\Xi \equiv 0$. This shows that the proof as such does not work in a non-compact setting. Still, a variation of it can be used, as we shall see later in Appendix 1.3.

Let $X$ and $Y$ be Polish spaces. Let $E:=C_{0}(X \times Y)$ be the space of all continuous vanishing at infinity functions on $X \times Y$, equipped with its usual supremum norm $\|\cdot\|_{\infty}$. Let $$ \Xi: u \in E \longmapsto\left\{\begin{array}{l} \int_{X} \varphi d \mu+\int_{Y} \psi d \nu &\begin{align*} &\text {if } u = \varphi \oplus \psi \text{ for some } (\varphi, \psi) \in C_b(X) \times C_b(Y) \end{align*}\\ +\infty &\text {else. } \end{array}\right. $$

I could not see how $\Xi \equiv 0$. For example, take some $(\varphi, \psi) \in C_0(X) \times C_0(Y)$. Then define $u$ by $u (x,y) := \varphi(x)+ \psi(y)$. Then $u \in E$. However, $\int_{X} \varphi d \mu+\int_{Y} \psi d \nu$ is possibly nonzero.

Could you explain why $\Xi \equiv 0$ if $E=C_{0}(X \times Y)$?

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  • $\begingroup$ Are you sure that $X$ and $Y$ are Polish only? Maybe they are locally compact too. $\endgroup$
    – Dieter Kadelka
    Commented May 13, 2022 at 11:57
  • $\begingroup$ @DieterKadelka I guess to prove $\Xi \equiv 0$, it's sufficient to assume that $X,Y$ are Polish. $\endgroup$
    – Akira
    Commented May 13, 2022 at 11:59
  • $\begingroup$ The problem only is: What is "vanishing at infinity", f.i. what if $X = Y = C([0,1])? $\endgroup$
    – Dieter Kadelka
    Commented May 13, 2022 at 12:12
  • $\begingroup$ @DieterKadelka in the book $f:X \to \mathbb R$ is vanishing at infinity if $\forall \varepsilon>0, \exists K \subset X$ compact such that $|f(x)| \le \varepsilon$ for all $x \in X \setminus K$. $\endgroup$
    – Akira
    Commented May 13, 2022 at 12:14

1 Answer 1

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A function $u(x,y) = \varphi(x) + \psi(y)$ with $\varphi\in C_0(X)$ and $\psi\in C_0(Y)$ will not be in $C_0(X,Y)$ if $\varphi$ or $\psi$ is not constant zero: If $x$ "tends to infinity", we have $\varphi(x)\to 0$, but $u(x,y)\to\psi(y)$ which is not zero in general.

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  • $\begingroup$ It seems the statement in the exercise holds only when $X,Y$ are both non-compact. Could you confirm if my observation is correct? [...] $\endgroup$
    – Akira
    Commented May 13, 2022 at 22:22
  • $\begingroup$ [...] Assume $X$ is not compact and $Y$ is compact. Let $(\varphi, \psi) \in C_0(X) \times C_0(Y)$ such that $\psi \equiv 0$. Define $u (x,y) := \varphi(x)+ \psi(y)$. Indeed, $u \in C_0(X \times Y)$. Fix $\varepsilon>0$. There is a compact set $K \subset X$ such that $\sup_{x \notin K} |\varphi(x)| \le \varepsilon$. Clearly, $K \times Y$ is compact. Also, $$ \sup \{ |u(x,y)| \mid (x,y) \notin K \times Y\} = \sup \{ |\varphi(x)| \mid x \notin K \} \le \varepsilon. $$ $\endgroup$
    – Akira
    Commented May 13, 2022 at 22:22
  • $\begingroup$ If you don't mind, please have a look at another closely related question here. $\endgroup$
    – Akira
    Commented May 14, 2022 at 8:09
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    $\begingroup$ Phew, I am not sure if I find the time and energy to chase down that detail… $\endgroup$
    – Dirk
    Commented May 14, 2022 at 10:40
  • $\begingroup$ Thank you so much for your help :v $\endgroup$
    – Akira
    Commented May 14, 2022 at 10:51

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