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$$f(s)=-\left(\min_{w\in C}(w^\top s+\frac12\|w\|_2^2)\right)$$ where $C$ is a compact and convex set in $\mathbb{R}^n$

I believe that the function is concave, but I need to show that it is convex (which could be false). Please give me any result. I am thinking of using this link which answers that the minimum of a parametric convex function is convex to prove that it is concave.

7337dtd
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  • Do you want to prove that the optimization problem is convex? Or else? – the_candyman Nov 20 '20 at 00:56
  • @the_candyman, I believe that the function is concave, but I need to show that it is convex (which could be false). Please give me any result. I am thinking of using https://math.stackexchange.com/questions/1407374/is-the-minimum-of-a-parametric-convex-function-convex-again to prove it as concave. – 7337dtd Nov 20 '20 at 02:17
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    The minimum of linear functions is concave, and the negative of a concave function is convex. – p.s. Nov 20 '20 at 02:51
  • @p.s. Sorry, I think I do not completely understand your statement, the minimum function includes two terms. – 7337dtd Nov 20 '20 at 18:27
  • I guess, to be more precise, it's an affine function of $s$ (though it's common to call affine functions linear). – p.s. Nov 20 '20 at 18:35
  • @p.s. Could you give a proof, thank you. – 7337dtd Nov 20 '20 at 18:37
  • Here's a proof for the maximum of convex functions being convex. Minimum of concave is similar: https://math.stackexchange.com/questions/373229/show-that-the-maximum-of-a-set-of-convex-functions-is-again-convex – p.s. Nov 20 '20 at 18:40
  • @p.s. So your conclusion is that $f(s)$ is convex? Could you see the answer below, is it correct? It says my $f(s)$ is concave. – 7337dtd Nov 20 '20 at 19:01

1 Answers1

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Given any family of convex functions, the pointwise supremum is convex. See the answer here: Prove the supremum of the set of affine functions is convex

(It assumes that the domain is compact, but the proof does not use this, and in any case one can always assume the domain is compact by restricting to a line segment.)

Multiplying by $-1$, we obtain that the pointwise infimum of concave functions is concave.

Affine functions are concave, hence so is the infimum in the definition of $f(s)$, so that $f(s)$ is convex. This argument does not require that $C$ is convex.

Note: it is not automatic that the minimum in your question exists (this can fail for certain $C$ if the term $\frac12 \lVert w \rVert^2$ is not there). But you can write it as $$f(s) = \frac12 \lVert s \rVert^2 - \inf_{w \in C} \frac12 \lVert s + w \rVert^2 $$

and this infimum is attained because $C$ is closed. Moreover, we see that $$f(s) = \frac12 \lVert s \rVert^2 - \frac12 d(s, -C)^2 \,.$$

Bart Michels
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  • Hi thank you for the solution, there is one question, so you are saying my $f(s)$ is concave, is that a typo? By adding a negative sign there does that mean the function is convex. – 7337dtd Nov 20 '20 at 19:19
  • I think there's a sign error here. The conclusion should be f is convex. For example, when $C=R^n$, $f(s)=\frac{1}{2}|s|^2$ which is clearly convex. – p.s. Nov 20 '20 at 19:26
  • @p.s. Does that mean any family (even the variable is not based on $s$) is also true that the infimum that family of concave functions is concave? – 7337dtd Nov 20 '20 at 19:35
  • @7337dtd Yes, the infimum of a family of concave functions is concave. – Bart Michels Nov 21 '20 at 17:50