# Potential for Monotone Operator

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. Normally, when I have seen that results about the supremum of a family of affine functionals is convex, the family of functionals does not vary point to point, whereas here it seems to, as long as $$\textrm{dom}(M)$$ is not convex (which the authors are explicitly allowing for). For example, if I have two points $$x,y \in \textrm{dom}(M)$$ but for which the line segment connecting them is not, it is not clear to me that how the suprema of the set of affine functionals given by refining a path $$\pi_x$$ from $$x_0$$ to $$x$$ relates to the suprema over refinements for a given path $$\pi_y$$.

I'd be happy to provide my attempt to verify too and where I get stuck, but as this question is already fairly long I'll leave it for now, and I suspect the answer is probably something simple.

Question: Why is $$f$$ lower semicontinuous/convex?

• If the function $f$ is finite on $\text{dom}(M)$ and $\infty$ outside $\text{dom}(M)$, then $\text{dom}(M)$ needs to be convex for $f$ to be convex. Dec 5, 2018 at 2:58
• @LinAlg Correct. I had difficulty understanding the claim; my impression was that they are claiming there exists some lower semicontinous proper convex $g: X \to \mathbb{R} \cup \{+\infty\}$ which is finite on a convex set containing $\textrm{dom}(M)$, such that $g \vert_{\textrm{dom}(M)} = f$, with $f$ defined simply to be $+\infty$ elsewhere. However, even this I cannot figure out how to guarantee. Dec 5, 2018 at 3:23
• @LinAlg Alternatively, I can also see that taking the pointwise supremum of the collection of affine functionals given by every refinement of every polygonal path in $\textrm{dom}(M)$ yields a lower semicontinuous proper convex functional, but then I do not see at all how the values that functional takes on $\textrm{dom}(M)$ are related in any way to those determined by $f$. Dec 5, 2018 at 3:37
• Would you understand the proof if you had the additional assumption that $\textrm{dom}(M)$ is convex? Dec 6, 2018 at 17:48
• @supinf Yes I would. My confusion in fact mostly stemmed from the fact that the authors elsewhere carefully specify which results depend upon the assumption of convexity of the domain and which do not. I'm on board if the domain is convex. As of yet though I'm not if it isn't (and that distinction was precisely what prompted me to ask the question). In fact, I couldn't even make the argument worked for star-shaped domains, which seemed the natural next step. Dec 6, 2018 at 21:16