How to show negative entropy function $f(x)=\sum_{i=1}^nx_i\log(x_i)$ is strongly convex? Let $x \in \mathbb{R}^n$ belongs to $S$ where
$$
S= \{x \in \mathbb{R}^n \mid x \succ 0, \|x\|_{\infty} \leq M\}
$$
where $\succ$ is the generalized inequality which means all elements of $x$ are positive and $\log$ is natural logarithm. Use the following theorem to show that $f(x)=\sum_{i=1}^nx_i\log(x_i)$ is $\frac{1}{M}$-strongly convex over $S$.
Theorem: f is $\alpha$-strongly convex if and only if $\nabla^2f(x) \succeq \alpha I$ for all $x$.
Definition:$f$ is $\alpha$-strongly convex if there exist a constant $\alpha$ such that
$$ f(y) \geq f(x)+\left<f '(x),y-x\right>+\frac{\alpha}{2}\|y-x\|^2$$
or 
$$ \left<f'(y)-f '(x),y-x\right> \geq \alpha\|y-x\|^2$$
for all $x,y$.
 A: First of all, It seems to me that in the definition of $\alpha$-strongly convex function should have a coefficient of $\alpha$ but not $\frac{\alpha}{2}$.
Now here goes the proof
$$
\begin{aligned}
\frac{\partial^2 f}{\partial x_i \partial_j} & = &\frac{\partial}{\partial x_j}[\log(x_i) + 1]  \\
& = & \left \{
    \begin{aligned}
    1/x_i \quad \text{if } i =j \\
    0 \quad \text{if } i \neq j
    \end{aligned}
\right . 
\end{aligned}
$$
Since $0<x_i\leq M$, then $1/x_i \geq 1/M$
Thus $\nabla^2 f \geq \frac{1}{M} I$, which is the theorem for $1/M$-strongly convex function.
A: Perhaps you are interested the fact that the negative entropy is (1/M)-strongly convex w.r.t. $\left\|\cdot\right\|_{1}$ (and not only w.r.t. $\left\|\cdot\right\|_{2}$). Here is the proof: 
W.l.o.g. M=1. Let be $f^{*}$ the convex conjugate of $f$:
\begin{equation*}
f^{*}(y)=\max_{x\in S}\left<y,x\right>-f(x).
\end{equation*}
It holds:
\begin{equation*}
\partial_{y_{i}} f^{*}(y)=\exp(y_{i}-1)\quad \text{if}\quad\sum_{j}\exp(y_{j}-1)\leq 1,
\end{equation*}
and otherwise:
\begin{equation*}
\partial_{y_{i}} f^{*}(y)=\frac{\exp(y_{i}-1)}{\sum_{j}\exp(y_{j}-1)}.
\end{equation*}
Thus $\left\|\nabla f^{*}(y)\right\|_{\infty}\leq 1$ and consequently $f^{*}$ is 1-smooth w.r.t. $\left\|\cdot\right\|_{\infty}$. By e.g. Theorem 3 in https://arxiv.org/pdf/0910.0610.pdf it follows that $f$ is $1$-strongly convex (on $S$) w.r.t. $\left\|\cdot\right\|_{1}$.  
