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I have some doubts on the interpretation and properties of the support function of a convex subset of $\mathbb{R}^d$.

(1)Let $K$ be a convex set in $\mathbb{R}^d$.

(2) The support function of $K$ is a function $h_{K}:\mathbb{R}^d\rightarrow \mathbb{R}$ such that $$ h_K(u):=\sup_{k\in K} <k,u> $$ where $<k,u>:=\sum_{i=1}^d k_i u_i$.

(3) $h_K$ is sub-linear:

-positive homogeneous: $h_K(au)=ah_K(u)$ $\forall a \geq 0$, $\forall u \in \mathbb{R}^d$.

-sub-additive: $h_K(u+y)\leq h_K(u)+h_K(y)$ $\forall u \in \mathbb{R}^d$, $\forall y \in \mathbb{R}^d$.

(4) Sub-addivity implies that $h_K$ is a convex function (from Wikipedia).

(5) We can show that $$ K=\cap_{u \in B^d}\{k \text{ s.t. } <k,u>\leq h_K(u)\}=\cap_{u\in S^{d-1}}\{k \text{ s.t. } <k,u>\leq h_K(u)\} $$ where $B^d$ is the unit ball in $\mathbb{R}^d$ centered at the origin and $S^{d-1}$ is the sphere in $\mathbb{R}^d$ centered at the origin.

Discussion on point (5)

(i) Point (5) shows that $K$ is uniquely determined by $h_K$. Knowing $h_K$ means knowing $K$.

(ii) Point (5) is strictly related with the notion of hyperplanes.

In fact, consider the hyperplane $H(u,b)$. The set of points lying on a side of $H(u,b)$ can be written as $\{k \in \mathbb{R}^d \text{ s.t. } <k,u>\leq b\}$. Since $h_{K}(u)$ is such that $\forall k \in K$ $<k,u>\leq h_K(u)$ this implies that $K$ lies on a side of $H(u,h_K(u))$. Hence, if we consider the half-spaces defined by $H(u,h_K(u))$ $\forall u\in \mathbb{R}^d$ and take their intersection, we get $K$.

Questions:

(A) According to point (5) it seems sufficient to consider all $u\in B^d$ or all $u\in S^{d-1}$. Why? From sub-linearity?

(B) Consider the example given here with $d=2$ and $K=[0,1]\times [0,1]$: is there any general reason for why it is sufficient to consider just $u=(1,0), u=(0,1), u=(-1,0), u=(0,-1)$?

(C) How do I know the side of the hyperplane to consider for each $u$ when I take the intersection?

(D) Moreover, is the support function continuous in $u$?

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  • $\begingroup$ Please use \lange \rangle for inner product. <> looks weird. $\endgroup$ – user251257 Apr 12 '16 at 17:17
  • $\begingroup$ @user251257 you have enough reputation to perform the edits yourself... $\endgroup$ – Michael Grant Apr 12 '16 at 17:55
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    $\begingroup$ @MichaelGrant: Yes, but OP should probably know there is a better alternative... $\endgroup$ – user251257 Apr 12 '16 at 17:58

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