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$p(t)=\begin{cases} t\log{t}-t &\quad\text{if} \quad t>0\\ 0 &\quad\text{if} \quad t=0 \\ +\infty &\quad\text{otherwise.} \\ \end{cases}$

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 $\mathbb{R^n}$.

BrianTag
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  • $p(t)=+\infty$ has no meaning – Martigan Dec 19 '18 at 15:56
  • @Martigan What do you mean? – BigbearZzz Dec 19 '18 at 16:02
  • @BigbearZzz $\infty$ is not a number, and it is no value that a function can take. You should probably define your function so that it is only defined on $\mathbb{R}^+$. – Martigan Dec 19 '18 at 16:04
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    @Martigan It's common in convex analysis to let the range of a convex function be the extended real number system $\Bbb R\cup {\infty}$. – BigbearZzz Dec 19 '18 at 16:05
  • @Martigan Think of $p$ as defining the set ${(x,y) \in \mathbb{R}^2 : p(x) \le y}$ then $p$ is convex iff this set is convex. Having $p$ be $+ \infty$ at some points fits in perfectly. – Trevor Gunn Dec 19 '18 at 16:06
  • @Martigan you might find some searches on the topic "extended real valued functions" helpful here. – Michael Grant Dec 19 '18 at 17:43
  • @MichaelGrant I know what it is. It always shocks me to see $f(x)=+\infty$. To me there is a difference between a function mapping the extended real numbers and a function taking the VALUE of something which, even in that context, are not numbers... – Martigan Dec 20 '18 at 08:43
  • Fair enough. but it is an incredibly useful and widely used convention in this space (convex analysis and optimization). It's not going away, and I'm glad! – Michael Grant Dec 20 '18 at 13:11

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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.

LinAlg
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