$p(t)=\begin{cases}
t\log{t}-t &\quad\text{if} \quad t>0\\
0 &\quad\text{if} \quad t=0 \\
+\infty &\quad\text{otherwise.} \\
\end{cases}$
Now we define the function on $\mathbb{R^n}$ , $f(x)= \sum_{i=1}^{n}p(x_i)$.
We want to prove that $f$ is strictly convex on $\mathbb{R^n}$.
Best Answer
Since the sum of strictly convex functions is strictly convex (see here), it suffices to prove that $p$ is convex. Clearly $p$ is convex on $(0,\infty)$, as the second derivative is strictly positive.
The definition of strict convexity is that $p(\lambda a + (1-\lambda)b) < \lambda p(a) + (1-\lambda)p(b)$ for all $a,b \in \mathbb{R}$ ($a \neq b$) and for all $0<\lambda<1$. Assume wlog that $a<b$. If $a<0$ or $a>0$, the condition is clearly satisfied.
What remains is proving: $$p(\lambda 0 + (1-\lambda)b) < \lambda p(0) + (1-\lambda)p(b)$$ for $b>0$ and $0<\lambda<1$. That is: $$p((1-\lambda)b) < (1-\lambda)p(b)$$ or $$(1-\lambda)b\log((1-\lambda)b) < (1-\lambda)b \log(b)$$ Since $(1-\lambda)b > 0$, this simplifies to $\log((1-\lambda)b) < \log(b)$, which is true since the logarithm is strictly increasing.