0

Let $2 \leq p < \infty$ and consider the function $f : \mathbb{R}^n \to \mathbb{R}$ defined by $$f(x) := |x|^p.$$ Then this function is mid-point convex (in fact strictly), i.e. we have that $$f\left(\frac{x + y}{2}\right) \leq \frac{f(x) + f(y)}{2}$$ holds for all $x,y \in \mathbb{R}^n$. Is there a nice way of showing this?

TheGeekGreek
  • 7,869
  • @JohnMa My answer uses an old geometric idea. It is different from all the answers in the post you have referred to. I hope it has some merit, so this question need not be closed. – Kavi Rama Murthy Jul 18 '18 at 09:11
  • Then you should post your answer to that question instead of this one. The goal is to put all good answers in a same place! @KaviRamaMurthy –  Jul 18 '18 at 09:12
  • I am doing that but that post is 1 year and 4 months old. It is unlikely that anyone would look at my answer. – Kavi Rama Murthy Jul 18 '18 at 09:14
  • That question was found when I put "|x|^p is convex" is google. That's the first post I find and has 1k views. I don't see why this question will be more popular in the long run? @KaviRamaMurthy –  Jul 18 '18 at 09:16

2 Answers2

0

Here is an elementary argument which uses only MVT: Claim: for $x \geq 0$ we have $x^{p}=\sup \{a^{p}+(x-a)pa^{p-1}: a\geq 0\}$. To prove that apply MVT to $x^{p}-a^{p}$ and consider the cases $a<x$ and $a \geq x$ separately. [ The idea for this comes from the geometric fact that a convex function is the upper envelope of the tangent lines]. Now define $f_a(x)=a^{p}+(x-a)pa^{p-1}$. [MVT gives $a^{p}+(x-a)pa^{p-1}\leq x^{p}$. To show that the supremum of the left side equals $x^{p}$ you just have to note that $a^{p}+(x-a)pa^{p-1}=x^{p}$ when $a=x$]. Then $f_a(\frac {x+y} {2})=\frac {f_a(x)+f_a(y)} 2$ (since $f_a$ is an affine function ). Just take supremum over $a \geq 0$ to get the desired inequality.

Kavi Rama Murthy
  • 311,013
0

After finishing my answer I realized I assumed $p$ to be an integer. I'll post my answer anyway for completeness:

The following inequality trivially holds for all $x, y \in \mathbb{R}$, $0 \leq n \leq p$:

$$0 \leq (|x|^n - |y|^n)(|x|^{p-n} - |y|^{p-n})$$

Multiplying all crossing terms and moving negative terms to the left, we get:

$$|x|^n|y|^{p-n} +|x|^{p-n}|y|^n \leq |x|^p + |y|^p$$

We sum over all $S \subseteq [1..p]$, substituting $n=|S|$:

$$2\sum_{S \subseteq [1..p]} |x|^{|S|}|y|^{p-|S|} \leq \sum_{S \subseteq [1..p]} |x|^p + |y|^p$$

Both sides can be simplified:

$$2(|x|+|y|)^p \leq 2^p(|x|^p+|y|^p)$$

By the triangle inequality the left hand side bounds $2|x+y|^p$ from above. We divide by $2^{p+1}$, getting:

$$\frac{|x+y|^p}{2^p} \leq \frac{|x|^p + |y|^p}{2}$$

Timon Knigge
  • 473