Proof of $|x|^{q-2}x$ is strongly monotone

Question

Let $g\colon \mathbb{R}^{n} \to \mathbb{R}^{n}$ with $g(x)=|x|^{q-2}x$ and $q\geq 2$. Since I know what it should look like, I already got

$$ (|x|^{q-2}x-|y|^{q-2}y,x-y)=\frac{1}{2}(|x|^{q-2}+|y|^{q-2})|x-y|^2+\frac{1}{2}(|x|^{q-2}-|y|^{q-2})(|x|^2-|y|^2)\\ \geq 2^{-1}(|x|^{q-2}+|y|^{q-2})|x-y|^2\phantom{dddd}\\ \geq 2^{2-q}|x-y|^q.\phantom{ddddddddddddddd|} $$

I totally understand the first equality and ofc the first inequality but i don't get the last step. Maybe you can help me out.

Answer 1

This is not an answer as I am missing a factor two. But maybe the OP is still interested in this is a weaker bound that can be obtained by more elementary means than convexity.

Let $a,b\geq 0$. Then obviously $$\max\{a,b\}\leq a+b.$$ Now one has $$(a+b)^n=\sum_{k=0}^n{n \choose k}\underbrace{a^kb^{n-k}}_{\leq \max\{a^n,b^n\}}\leq \underbrace{\sum_{k=0}^n{n \choose k}}_{=(1+1)^n=2^n}\cdot \max\{a^n,b^n\}\leq 2^n(a^n+b^n),$$ Where in the last step I have used the previous bound.

Now apply this to your situation, after using the triangle inequality $$\|a-b\|\leq \|a\|+\|b\|$$

Another way to get this bad bound, maybe even more easily is to see that $$a+b\leq 2\cdot \max\{a,b\},$$ and so by monotonicity of power raising for positive numbers: $$(a+b)^n\leq (2\cdot \max\{a,b\})^n=2^n \max\{a^n,b^n\})\leq 2^n(a^n+b^n),$$ again by the same step used before.

Answer 2

the last step use the fact that the function $$ f(x)=|x|^{q-2} $$ is convex in $\mathbb{R}^n$ and then $$ \frac{|x+y|^{q-2}}{2^{q-2}}=f((x+y)/2)\leq \frac{1}{2}(f(x)+f(y))= \frac{|x|^{q-2}+|y|^{q-2}}{2} $$