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?
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.