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Given a closed, convex, non-empty set $K\subseteq\mathbb{R}^n$ the support function $h_K:\mathbb{R}^n\to (-\infty,\infty]$ is defined as $$h_K(y) := \sup_{x\in K} \langle x,y\rangle$$

It is easy to see that $h_K$ is convex, $\mathbb{R}_{\geq 0}$-homogenuous and that all sublevel sets $\{y \mid h_K(y)\leq C\}$ are closed (in other words: $h_K$ is lower semicontinuous). I have several questions which I haven't been able to answer satisfactorily through google. Note that I'm interested in the case of general $K$ not just compact ones.

  1. Do these properties characterize support functions? There is an obvious way to define a convex set given such a function $h$ by setting $K:=\{x \mid \forall y: \langle x,y\rangle \leq h(y)\}$ and obviously $h_K\leq h$, but I cannot prove equality.
  2. Is there an description of $h_{K_1\cap K_2}$ in terms of $h_{K_1}$ and $h_{K_2}$ similarly to the descriptions $h_{K_1+K_2} = h_{K_1}+h_{K_2}$, $h_{conv(K_1\cup K_2)} = \max\{h_{K_1},h_{K_2}\}$ ? I think that something like $h_{K_1\cap K_2} = \sup\{h \leq \min\{h_{K_1},h_{K_2}\} \mid h \,\text{convex}\}$ might be true but I cannot prove it.
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1. Yes, this is true. This can be derived from the involutive property of the Legendre-Fenchel transform: if $f:\mathbb R^n\to (-\infty,+\infty]$ is a closed convex function, then $f^{**}=f$ (Theorem 12.2 in Rockafellar's Convex Analysis). Recall that $$f^*(x) = \sup\{\langle x,y\rangle - f(y) : y\in\mathbb R^n\}$$ and a convex function is closed if its epigraph is.

With every convex closed set $K$ we can associate a closed convex function $\delta_K$ by letting $\delta_K(x)=0$ when $x\in K$ and $\delta_K(x)=+\infty$ otherwise. Observe that $\delta_K^*$ is exactly $h_K$.

Given $h$ as in your question, define $K= \{x : \langle x,y\rangle \le h(y) \ \forall y\}$. Observe that $h^*(x)=0$ when $x\in K$, because the supremum is attained by $y=0$. If $x\notin K$, then there is $y$ such that $\langle x,y\rangle > h(y)$. Considering large multiples of such $y$, we conclude that $h^*(x)=\infty$.

Thus, $h^* = \delta_K$. By the involutive property, $h=h^{**} = \delta_K^* = h_K$.

By the way, this result is Theorem 13.2 in Rockafellar's book.

2. Yes, this is correct. Another way to state this fact: the epigraph of $h_{K_1\cap K_2}$ is the convex hull of $\operatorname{epi} h_{K_1} \cup \operatorname{epi} h_{K_2}$. To see this, observe that the epigraph of $h_K$ is determined by its intersection with the horizontal plane $z=1$. This intersection is nothing but $K^\circ$, the polar of $K$. It remains to use the fact that $(K_1\cap K_2)^\circ = \operatorname{conv}(K_1^\circ \cup K_2^\circ)$. See also Corollary 16.5.1 in Rockafellar's book.

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  • $\begingroup$ There is something off in the second case of your proof of statement 1: You implicitly claim that $\{y \mid h(y)<\infty\}$ is closed when you assert that $x$ has a neighbourhood contained in $\{h=\infty\}$. That is not true: Consider the parabola $K:=\{(x_1,x_2)\in\mathbb{R}^2 \mid x_2\geq x_1^2\}$. Then $h_K$ is finite exactly on the set $\{(y_1,y_2) \mid y_2<0\}\cup\{0\}$ which is not closed. $\endgroup$ – Johannes Hahn May 14 '14 at 0:06
  • $\begingroup$ Theorem 2.29 in Hörmander's book: "Notions of convexity" treats the intersection, the support function is the lower semi-continuous regularization of $$x \mapsto \inf_{x_1 + x_2 = x \in \mathbb{R}^n} \{h_1(x_1) +h_2(x_2)\}$$. $\endgroup$ – Jorge E. Cardona Jan 30 '17 at 15:09

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