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Given a set $S = \{x_1,\dotsc,x_n\} \subset \mathbb{R}$, is the function

\begin{align} f&: (0,\infty) \to \mathbb{R} \\ f&(p) = 2p^2 + \frac{1}{n}\sum_{i=1}^n \max(0, -p^2-x_i) \end{align}

strongly convex? (What if $f: \mathbb{R} \to \mathbb{R}$?)

I know $p^2$ is strongly convex. Let $P = \{x_i \in S : x_i + p^2 \geq 0\}$ and $N = S\setminus P$ then

\begin{align} f(p) = \begin{cases} 2p^2 & \text{ for all } x_i \in P \\ 2p^2 - \frac{1}{|N|}\sum_{i=1}^{|N|} -p^2 -\frac{1}{|N|}\sum_{i=1}^{|N|}x_i & \text{ for all } x_i \in N \end{cases} \end{align} so in both cases $f$ is strongly convex. Is this argumentation valid?

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  • $\begingroup$ Could you please specify the sign of the $x_i$'s and $p$? Is $p$ a continuous variable? $\endgroup$ – AndreaCassioli Jul 15 '14 at 16:37
  • $\begingroup$ The $x_i$ are $p$ real numbers and $p>0$ if necessary. Why could the sign of $x_i$ play a role? They are constants. $\endgroup$ – Manuel Schmidt Jul 15 '14 at 17:47
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First we can assume $x_i<0,\forall i$. We can write

$f(p) = \frac{1}{N} \sum_i^N ( p^2+\max\{0, -p^2-x_i\} )= \frac{1}{2}\sum_i^n \phi(p)_i,$

Then it is easy to see that $\phi(p)_i = -x_i$ for $p\in [-\sqrt x_i,\sqrt x_i]$, while $\phi(p)_i= p^2 $ otherwise. This is strictly convex. Each $\phi(p)_i$ is known as Huber penalty function (see http://web.stanford.edu/~boyd/cvxbook/ pag 299).

The $f(p)$ function is then the sum of strictly convex functions, and therefore strictly convex.

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  • $\begingroup$ I think in a special case I a can add a convex and a concave function and get a convex function. For example $p^2-p^2$ is constant $0$ and convex. $\endgroup$ – Manuel Schmidt Jul 15 '14 at 19:13
  • $\begingroup$ Even more $2p^2-p^2$ is the sum of a strongly convex and strongly concave function that is strongly convex. $\endgroup$ – Manuel Schmidt Jul 15 '14 at 19:38
  • $\begingroup$ You are right. I update the post. $\endgroup$ – AndreaCassioli Jul 15 '14 at 20:26
  • $\begingroup$ Thanks for your help. Please, note that I am talking about strong not strict convexity. As in my original question your argumentation has to consider two cases. $\endgroup$ – Manuel Schmidt Jul 15 '14 at 20:47

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