As one knows, we have the duality theorem for $W_2^2$:
\begin{align} W_2^2(P,Q) = \inf_{\pi} \mathbb{E}_{\pi, X \sim P, Y\sim Q} |X-Y|^2 = \sup_{\phi,\psi \in C_b, \phi(x) + \psi(y)\leq (x-y)^2} \int \phi dP + \int \psi dQ. \end{align}
My question is, why in this case we cannot restrict the search of $\phi,\psi$ to $\psi(y) = -\phi(y)$? It seems to be reasonable since $\phi(x) + \psi(x)\leq 0$, which implies that $\psi(y) \leq -\phi(y)$.
Indeed, if we were able to do that and consider function $\phi$ that satisfies $\phi(x) - \phi(y) \leq (x-y)^2$, it immediately implies that $\phi'(x)=0$, which further implies that $\phi$ is a constant. What is wrong with this argument?
Let's put it another way: you are trying to find $$\sup_{(x,y)\in A}f(x,y),\quad\text{where $A\subset X\times Y$}$$ when you suddenly realize that, for some function $g$, $$f(x,y)\le f(x,g(x))$$ You then wish to prove $$\sup_{(x,y)\in A}f(x,y)=\sup_{(x,g(x))\in A}f(x,g(x))$$ But there is no such a logic. You can either derive $$\sup_{(x,y)\in A}f(x,y)\le\sup_{(x,y)\in A}f(x,g(x))=\sup_{x\in\pi_x(A)}f(x,g(x))$$ (here $\pi_x(A)$ denotes the projection of $A$ onto $X$), or $$\sup_{(x,g(x))\in A}f(x,g(x))\le\sup_{(x,y)\in A}f(x,y)$$ None of these two inequalities lead to the result we desired. The key fact here is that $$\{x:(x,g(x))\in A\}\not\supset\pi_x(A)$$