For any two functions $f,g: \mathbb{R}^n\rightarrow \mathbb{R}$, we have $f\square g(x)=\underset{u\in\mathbb{R}^n}{\mathrm{inf}}\{f(u)+g(x-u)\}$. We need to show that for any two proper functions $f,g: \mathbb{R}^n\rightarrow \mathbb{R}\cup\{\infty\}$, the property $(f\square g)^*=f^*+g^*$. Then, assuming $f:\mathbb{R}^n\rightarrow (-\infty,\infty]$ is a convex function and $g:\mathbb{R}^n\rightarrow \mathbb{R}$ is a real-valued convex function, we need to show that $f^*\square g^*=(f+g)^*$.
I started using the definition of the conjugate, namely that $f^*(y)=\underset{x}{\mathrm{sup}}\: x^Ty-f(x)$. Using this, we have that $$(f\square g)^*(y) = \underset{x}{\mathrm{sup}}\:\{ x^Ty-\underset{u}{\mathrm{inf}}\:\{ f(u)+g(x-u)\}\} = \underset{x}{\mathrm{sup}}\:\underset{u}{\mathrm{inf}}\:\{ x^Ty-f(u)-g(x-u)\}.$$
However, my problem is: how can I set this equal to $f^*+g^*$ if this last equation does not involve $u$?
Here are some hints:
For $X\subset\mathbb{R}$, we have $-\textrm{inf}(X) =\mathbf{\sup}(-X)$ (not $\inf(-X)$, as currently stated).
Let's also note that $$x^\top y - f(u) - g(x-u) = \big[(x-u)^\top y - g(x-u)\big] + \big[u^\top y - f(u)\big].$$ Can we use a change-of-variables to have both bracketed terms look like conjugates?