# “Support function of a set” and supremum question.

I have already learned about what a supremum means from wikipedia and from another answer here. However I am not quite sure what 'supremum over a set of functions' means exactly.

As an example, my book gives an example called "Support function of a set", where it states that the support function "associated with a set C" is:

$$S_c(x) = sup \{ x^Ty \ |\ y \in C \}$$

It goes on to say:

"Since $x^Ty$ is an affine function of $x$, so $S_c$ is the pointwise supremum of a family of affine functions, hence convex"

What I am not too clear about, is, what exactly is a 'supremum of a set of functions' mean? Does this mean that in this case, I am simply taking the supremum of all the dot product results fpr any $x$, across all the vectors $y$ in $C$?

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Yes. Note that for fixed $x$, the set is just a set of real numbers, so the supremum is just the ordinary one. – LutzL Feb 23 '14 at 2:22
@LutzL Thanks, so what's the relevance of the word 'support' in this example then? Why is he using it? ('Support function'). What is being 'supported' and by what? – Spacey Feb 23 '14 at 2:24
This comes from convex geometry. A supporting line or supporting hyperplane of a convex set $C$ is an affine codimension 1 subspace outside of $C$ that touches $C$ in some point or larger facette. $S_C(x)$ is how much you have to shift the plane $\{y:x^Ty=0\}$ to be a supporting hyperplane. – LutzL Feb 23 '14 at 2:31
@LutzL OOHHHH! Ok that makes sense. Then if we shift the plane by $S_c(x)$, we guarantee that the set $C$ is in the half space made by $x = S_c(x)$ right? – Spacey Feb 23 '14 at 2:35
Be careful with the dual of the dual, $C\subset \{y:x^Ty\le S_C(x)\}$, actually the intersection over all $x\ne 0$ is the closure of the convex hull of any $C$. – LutzL Feb 23 '14 at 2:38

In convex geometry, a supporting line or supporting hyperplane of a convex set $C$ is an affine co-dimension-$1$ subspace $\{y:a^Ty=b\}$ outside of $C$ that touches $C$ in some point or larger facet, i.e., so that $a^TC\le b$ with equality assumed at least once.

From there it is just a twist of thought to compute the constant as $b=\sup a^TC$ and formalize it by giving it a name as a function

$$S_C(x)=\sup\{x^Ty:y\in C\},$$

i.e., $S_C(x)$ is how much you have to shift the plane $\{y:x^Ty=0\}$ to be a supporting hyperplane, thus support function.

Then $C⊂\{y:x^Ty≤S_C(x)\}$. Starting from any even non-convex $C$, the intersection of these supporting half-spaces is the closure of the convex hull of $C$.

As to $S_C$ itself, convex functions are closed under suprema, and since the family of linear functions $x\mapsto x^Ty$, $y\in C$ is a family of convex functions, their pointwise supremum is again convex.

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Thanks LutzL, so just to be clear though, $S_c(x)$ is simply a scalar right? – Spacey Feb 23 '14 at 3:04
Yes, $S_C:\mathbb R^n\to\mathbb R\cup\{\infty\}$. – LutzL Feb 23 '14 at 3:06
Oh! We are cycling across all $x$'s, (that is, all possible vectors "$a$", and for EACH one, we get a $sup(x^Ty)$, (a scalar), and that defines how much to move the hyperplane by, such that it touches $C$, FOR that particular "$a$". Is that right? – Spacey Feb 23 '14 at 3:09
Yes. In the function $S_C$, it is $x$. Note that $S_C$ is positive homogenous, for $t>0$ one has $S_C(tx)=tS_C(x)$, so that one only needs one $x$ per ray from the origin. – LutzL Feb 23 '14 at 3:13
Right exactly. Thanks it is a great answer! :-) – Spacey Feb 23 '14 at 3:14