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Let S be a nonempty, bounded convex set in $\mathbb{R}^n$, and let $f: \mathbb{R}^n \to \mathbb{R} $ be defined as follows: $$ f(y) = \sup \{y^T x: x\in S\}. $$ The function $f$ is called the support function of S. Prove that $f$ is convex. Also, show that if $f(y) =y^T x^-,$ where $x^- \in S, \ x^-$ is a subgradient of $f$ at y.

This problem comes from “Nonlinear Programming Theory and Algorithms”.

Roy Sht
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751495642
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1 Answers1

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You need neither boundedness nor convexity for the claims to hold!

  • Step 1: Convexity. Let $x,y \in \mathbb R^n$ and $t \in [0, 1]$. Then

$$ \begin{split} f(tx+(1-t)y) &:= \sup_{z \in S}(tx+(1-t)y)^Tz\\ &= \sup_{z \in S}tx^Tz+(1-t)y^Tz\\ &\le \sup_{z \in S}tx^Tz + \sup_{z \in S}(1-t)y^Tz\\ &\le t\sup_{z \in S}x^Tz + (1-t)\sup_{z \in S}y^Tz\\ &=: tf(x) + (1-t)f(y) \end{split} $$

Thus $f$, is convex (and this doesn't depend on $S$ is convex / bounded or not!).

  • Step 2: Subgradients. So, suppose $f(x) = x^Tz$ for some $z \in S$. Lets show that $z$ is a subgradient of $f$ at $x$, i.e that $$ f(x') \ge f(x) + z^T(x'-x),\;\forall x' \in \mathbb R^n. $$

Indeed, for any $x' \in \mathbb R^n$, one computes

$$ f(x) + z^T(x'-x) = f(x) + - z^Tx + z^Tx' = 0 + z^Tx' = z^Tx' \le f(x') := \sup_{s \in S}s^Tx'. $$

Thus $z$ is a subgradient of $f$ at $x$, and once again, this doesn't depend on whether $S$ is convex / bounded or not!

dohmatob
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  • I think the definition of subgradient need S is convex? – 751495642 Jun 13 '20 at 02:09
  • No, definitely not. You can always write down ${v \in \mathbb R^n \mid f(x') \ge f(x) + v^T(x'-x)}$ and call it the "subdifferential" (i.e the set of all "subgradients") of $f$ at $x$. – dohmatob Jun 13 '20 at 06:42