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I am fairly new with the concept of upper-hemicontinuity, i.e.

Let $X \subseteq E^{n}$, $Y \subseteq E^{m}$ and $\Psi: X \rightrightarrows Y$ be a set-valued map. $\Psi$ is upper hemicontinuous at $x_{0} \in X$ if, for every open set $V \supseteq \Psi (x_{0})$, there is an open set ${U}$ with $x_{0} \in {U}$ such that \begin{equation} \Psi (x) \subseteq {V} \text{ for every } x \in {U} \cap {X}. \end{equation} $\Psi$ is upper hemicontinuous if it is upper hemicontinuous at every $x \in {X}$.

I have been looking for a result of the type:

If $f$ be a proper convex function on $X$, then the subdifferential of $f$ denoted $\partial f$ is an upper hemicontinuous set-valued map from $X$ to $X$.

Does this result hold and if yes could you give me a reference? Theorem 3.1 of the paper of Gregory (1980) seems related but I do not believe this is exactly equivalent.

Thank you in advance for your help!

D. A. Gregory. Upper Semi-Continuity of Subdifferential Mappings. Canadian Mathematical Bulletin, 23 (1):11–19, 1980. doi:10.4153/CMB-1980-002-9.

Julien V
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1 Answers1

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The following is Proposition 6.1.1. from Convex Functions: Constructions, Characterizations and Counterexamples, by Jon Borwein and Jon Vanderwerff. Note here that $X$ is assumed to be a Banach space.

Suppose $f : X \to (-\infty, \infty]$ is a convex function that is continuous at $x_0 \in X$. Then the map $x \mapsto \partial f(x)$ is norm-to-weak$^*$ upper-semicontinuous at $x_0$.

The proof just uses the following ingredients:

  • Local boundedness of $\partial f$ at $x_0$ (continuity is important here)
  • Banach-Alaoglu's theorem, guaranteeing weak$^*$ cluster points of bounded sequences in $X^*$,
  • The fact that $\partial f$ has a norm-to-weak$^*$ closed graph.

In finite-dimensions, the weak$^*$ topology on $X^*$ is just the norm topology, which is identified with the norm topology on $X$ itself. So, you do get norm-to-norm upper-semicontinuity at points where $f$ is continuous, so long as $X$ is finite-dimensional.

That said, if $f$ is not continuous at $x_0$, then the subgradient map may not be upper-hemicontinuous at $x_0$ (even in finite-dimensions). For example, take $f$ to be the indicator function of the set $$C = B_{\Bbb{R}^2} = \{(x, y) : x^2 + y^2 \le 1\},$$ which is to say, $f(x) = 0$ if $x \in C$, and $f(x) = \infty$ otherwise. The subgradient of $f$ is therefore the normal cone operator, mapping a point $(x, y) \in C$ to the set of points that project onto $C$. In this case, if $(x_0, y_0)$ lie in the boundary circle of $C$, then $$\partial f(x_0, y_0) = \{\lambda(x_0, y_0) : \lambda \ge 1\}.$$ This subgradient map is not upper-semicontinuous. As we have seen, $$\partial f(1, 0) = \{\lambda(1, 0) : \lambda \ge 1\} = [1, \infty) \times \{0\}.$$ Consider the open set $$U = (0, \infty) \times (-1, 1) \supseteq \partial f(1, 0).$$ Note that $U$ contains none of the other normal cones; they are all rays, and the only ones that have bounded $y$ coordinates are the ones at $(\pm 1, 0)$. But, since the $x$ coordinates are bounded below by $0$, the only such ray begins at $(1, 0)$. This drastically contradicts $\partial f$ being UHC at $(1, 0)$.

Theo Bendit
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