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



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> **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.
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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.



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> Could you explain why $\Xi \equiv 0$ if $E=C\_{0}(X \times Y)$?
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Why is $\Xi \equiv 0$ if $E=C_{0}(X \times Y)$?
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.


