Given an arbitrary set $A ⊂ \mathbb{R}^n$ , the support function associated with the set $A$

$ σ_A : \mathbb{R}^n \to \mathbb{R} ∪ \{+\infty\}$ is defined as

$\sigma_A(x):= \sup_{z \in A} \langle x,z \rangle$

Prove that $\sigma_A$ is convex.

Can I confirm that the inner product $\langle x, z \rangle$ is just a line orthogonal to $x$

I am confused by the meaning of "support function" and also what $\sigma_A(x)$ is. If it's the supremum of the inner product of $x$ and $z$ (which tells us what exactly?) isn't it just going to be a single point/value?

How can I prove this, and where has my interpretation failed me?

  • 1
    $\begingroup$ Any supremum of convex functions is convex. For each $z$ the function $x \mapsto \langle z, x \rangle$ is convex. So $\sigma_A(x) = \sup_{z \in A} \langle z,x \rangle$ is convex. $\endgroup$ – littleO Jun 6 at 1:45

First of all, \begin{align*} \sigma_A(x)=\sup_A\langle x, z\rangle=\sup\{\langle x,z\rangle\ |\ z\in A\}. \end{align*} To prove that $\sigma_A$ is convex, let $x,y\in\mathbb R^n$ and $t\in[0,1]$. Then \begin{align*} \sigma_A(tx+(1-t)y)&=\sup_{z\in A}\langle tx+(1-t)y,z\rangle \\ &= \sup_{z\in A}\left(t\langle x,z\rangle+(1-t)\langle y,z\rangle\right) \\ &\leq t\sup_{z\in A}\langle x,z\rangle+(1-t)\sup_{z\in A}\langle y,z\rangle \\ &=t\sigma_A(x)+(1-t)\sigma_A(y), \end{align*} which shows that $\sigma_A$ is convex.

  • $\begingroup$ $\langle x, y \rangle$ does not measure the angle between $x$ and $y$, which is given by $\arccos((\langle x, y \rangle)/(||x||\cdot ||y||))$. What $\sigma_A(x)$ represents is the largest value $\alpha$ such that $A$ is contained in the hyperplane $\{z \in \mathbb{R}^n \mid \langle z,x \rangle \leq \alpha\}$. $\endgroup$ – Pedro M. Jan 20 '15 at 18:13
  • $\begingroup$ I know that this is not really the angle, but I thought that this might be easier to imagine than the hyperplane-thing. But you are right, I changed my answer. $\endgroup$ – sranthrop Jan 20 '15 at 18:18
  • $\begingroup$ The problem was not whether it is easier or not to imagine, but the conclusion that $\sigma_A$ has something to do with "the largest angle" was not correct: for instance, if $A$ is the unit sphere, then $\sigma_A(x) = ||x||$ (you could argue that this has to do with the smallest angle, but that also does not hold every time). $\endgroup$ – Pedro M. Jan 20 '15 at 18:27
  • $\begingroup$ So what exactly does $\langle x,z \rangle$ represent? What is the purpose of a 'support function' and why are we taking the supremum of $\langle x,z \rangle$ ? @sranthrop thank you for your answer $\endgroup$ – diabloescobar Jan 20 '15 at 19:30

When $f(x)=\sup_{a\in A}\ a\cdot x$, then for $x,\ y$ s.t. they are close, there are two affine hyperplane $P_x,\ P_y$ s.t. $$(1)\ P_x\perp x\ {\rm and}\ P_y\perp y$$ $(2)$ $A,\ x$ (resp $y$) are in different sides wrt $P_x$ (resp $ P_y$) and $(3)$ $P_x\bigcap A,\ P_y\bigcap A$ are non-empty.

Note that $f(x)= p\cdot x$ for any $p\in P_x$. When $z$ is a mid point in $[xy]$, then \begin{align*} f(z) &\leq \sup_{p\in P_x^+ \bigcup P_y^+}\ z\cdot p \\ &=\frac{1}{2}\sup_{p\in P_x^+ \bigcup P_y^+}\ x\cdot p + y\cdot p\\ &= \frac{1}{2} \sup_{p\in P_x^+\bigcup P_y^+}\ x\cdot p + \frac{1}{2}\sup_{p\in P_x^+ \bigcup P_y^+}\ y\cdot p\\&= \frac{1}{2}\{ f(x)+f(y)\} \end{align*}

where $P_x^+$ is a closed half plane intersecting $A$ whose boundary is $P_x\bigcap P_y$.


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