Lengendre-Fenchel transform of infimal convolution Wikipedia states the following interesting fact about the Lengendre-Fenchel transform:
Define the infimal convolution of two functions as
$$
(f\square g)(x) = \inf_y\big\{f(x-y) + g(y)\big\}
$$
Then, with certain caveats listed below, the Legendre-Fenchel transform ${}^*$ has the following property:
$$
(f_1\square\dots\square f_n)^*(x^*) = f_1^*(x^*) + \dots + f_n^*(x^*).
$$
I worked my way through this and it's not hard to see why it's true. I'll give my sketch proof below.
However, Wikipedia states that it's true only if the functions $f_1, \dots, f_n$ are proper convex functions and lower semicontinuous, and states also that the right hand side might not be a proper convex function.
What I want to understand is:


*

*Why these are the exact conditions for the theorem, and what happens if the functions fail to meet them?

*What conditions are needed to go the other way? That is, what conditions do functions $f_1, \dots, f_n$ need to satisfy in order to guarantee that $(f_1 + \dots + f_n)^* = f_1^* \square \dots \square f_n^*$?
I don't have access to the book Wikipedia cites, so I'm wondering if
    someone can post (or point me toward) a more detailed proof than my
    sketch below, which shows exactly where those assumptions are used.
Here's my sketch of the proof, for two functions. We have
$$
(f\square g)^*(x) = \sup_x \Big\{\langle 
x^*,x \rangle - \inf_y\big\{ f(x-y) + g(y) \big\} \Big\} \\
= \sup_{x,y}\big\{ \langle 
x^*,x \rangle - f(x-y) - g(y) \big\}.
$$
Let $k=x-y$. Then
$$
(f\square g)^*(x) = \sup_{k,y}\big\{ \langle 
x^*,k+y \rangle - f(k) - g(y) \big\} \\
= \sup_{k}\big\{ \langle 
x^*,k \rangle - f(k) \big\} + \sup_{y}\big\{ \langle 
x^*,y \rangle - f(y) \big\}\\
 = f^*(x^*) + g^*(x^*).
$$
 A: This theorem has been well proved in Rockafaller's book ``Convex Analysis'' (Theorem 16.4). In his theorem, it does not need $f_1,\dots,f_m$ to be closed (lower-semicontinuous). 
And the infimal convolution is defined on proper convex functions, thus we need $f_1,\dots,f_m$ to be proper.
In the other way, i.e., to prove $(f_1+\dots+f_m)^*=f_1^* \square \dots \square f_m^*$. 
We need one more condition: the relative interior of effective domain, $\operatorname{ri}(\operatorname{dom}f_i),i=1,\dots,m$, should have a point in common.
I am very glad to copy his theorem (Theorem 16.4 in his book) here.

Let $f_1,\dots,f_m$ be proper convex functions on $\mathbb{R}^n$, Then
  \begin{equation}
\begin{aligned}
\left(f_{1} \square \cdots \square f_{m}\right)^{*}, &=f_{1}^{*}+\cdots+f_{m}^{*} \\\left(\mathrm{cl} f_{1}+\cdots+\mathrm{cl} f_{m}\right)^{*} &=\mathrm{cl}\left(f_{1}^{*} \square \cdots \square f_{m}^{*}\right).
\end{aligned}
\end{equation}
  If the sets $\operatorname{ri}(\operatorname{dom}f_i),i=1,\dots,m$, have a point in common, the closure operation can be omitted from the second formula, and 
  \begin{equation*}
\left(f_{1}+\cdots+f_{m}\right)^{*}(\left.x^{*}\right) =\inf \left\{f_{1}^{*}\left(x_{1}^{*}\right)+\cdots+f_{m}^{*}\left(x_{m}^{*}\right) | x_{1}^{*}+\cdots+x_{m}^{*}=x^{*}\right\} ,
\end{equation*}
  where for each $x^*$ the infimum is attained.

