1
$\begingroup$

[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?

$\endgroup$

0

You must log in to answer this question.