# Potential for a Monotone Operator

[Cross-posted from math.stackexchange]

I have a question about understanding the proof of Theorem 4.11 in the paper A Potential Theory for Monotone Multivalued Operators (accessible here). The authors claim to construct a convex functional and I'm not sure I follow their argument.

My specific question is at the end, but I provide some background from the paper below before.

Background:

The paper shows, how for a pair of dual locally convex topological vector spaces $$(X,X')$$ and a monotone set-valued operator $$M:X \to X'$$, one can define a notion of path integral along polygonal paths (as the restriction of $$M$$ to any straight line in one's is monotone and hence Riemann-integrable).

The authors call $$M$$ conservative if its path integral around any closed polygonal path in its domain (the set of points of $$X$$ where it is non-empty valued) is zero. The authors define the integral of $$M$$ along any line segment $$[x,y] \subseteq \textrm{dom}(M)$$ via:

$$\int_{0}^1 \langle M(x + t(y-x)), y-x \rangle \, dt = \sup \bigg\{\sum_{i=0}^{n-1}\langle x_i^*, x_{i+1} - x_i\rangle\bigg\} = \inf\bigg\{\sum_{i=0}^{n-1}\langle x_{i+1}^*, x_{i+1}- x_i\rangle \bigg\}$$ where $$x_i^* \in M(x_i)$$, and the sup/inf are over all refinements of the line segment, and just follow from their respective arguments being the left/right Riemann sums of monotone increasing functions.

The authors then state the following theorem (4.11, p. 623), which I reproduce below.

Theorem 4.11: To any conservative monotone multivalued map $$M:X \to X'$$ with a polygonally path connected domain, there corresponds, to within an arbitrary additive constant, a convex potential $$f: X \to \mathbb{R} \cup \{+\infty\}$$, which is the restriction on $$\textrm{dom}(M)$$ of a lower semicontinuous proper convex functional. The potential $$f$$ is assumed to be $$+\infty$$ outside $$\textrm{dom}(M)$$ and is defined on $$\textrm{dom}(M)$$ by: \begin{aligned} f(x) - f(x_0) & = \int_\pi \langle M(z), dz\rangle = \\ & =\sup\bigg\{\sum_{i=0}^{n-1}\langle x_i^*, x_{i+1}- x_i\rangle + \langle x_n^*, x- x_n\rangle \bigg\}\\ & = \inf\bigg\{\sum_{i=0}^{n-1} \langle x_{i+1}^*, x_{i+1} -x_i\rangle + \langle x^*, x-x_n\rangle\bigg\} \end{aligned} where the sup/inf are again over all refinements of the poylgonal path $$\pi$$.

Source of confusion:

The proof argues that, by definition, on $$\textrm{dom}(M)$$, $$f$$ is equal to the lower semicontinuous proper convex function defined as the pointwise supremum of a family of continuous affine functions, the Riemann sums. I don't follow this step without the additional assumption that $$\textrm{dom}(M)$$ is convex. The authors are careful elsewhere, however, to specify when that is an assumption that is made.

Question: Why is $$f$$ convex?