Given a closed, convex, non-empty set K⊆Rn the support function hK:Rn→(−∞,∞] is defined as

hK(y):=supIt 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.

- 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.
- 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.

**Answer**

**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.

**Attribution***Source : Link , Question Author : Johannes Hahn , Answer Author : Community*