Suppose $x \geq 0$, $y \geq 0$ and $0 . Why is the following inequality true? $|x^{p}-y^{p}| \leq |x-y|^p$
Please explain inequality $|x^{p}-y^{p}| \leq |x-y|^p$
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0The inequality $\|x^p-y^p\|\leq\|x-y\|^p$ holds even when $x$ and $y$ are positive operators on Hilbert space. This follows from a more general result proved as Theorem 1.5 in the paper [1] by J. Phillips, which gives the inequality $\|f(x)-f(y)\|\leq f(\|x-y\|)$ whenever $f$ is an operator monotone function on $[0,\infty)$ such that $f(0)=0$. The fact that $t\mapsto t^p$ is operator monotone if $0\lt p\leq 1$ is proven as Proposition 1.3.8 in Pedersen's C*-algebras book. Your inequality is the $1$-dimensional case. [1]: https://dspace.library.uvic.ca:8443/dspace/handle/1828/1506 – 2010-11-20
2 Answers
Assume w.lo.g $x>y$. Then the statement becomes $x^p - y^p \leq (x-y)^p$. Since $y > 0$, divide through out by $y^p$. So we need to show $(\frac{x}{y})^p - 1 \leq (\frac{x}{y}-1)^p$ whenever $x > y >0$ and $0 . Let $t = \frac{x}{y}$. So we need to show $t^p - 1 \leq (t-1)^p$ whenever $t > 1$ and $0 . This is a calculus problem. Let $f(t) = (t-1)^p - (t^p -1) $ where $0 . Show that the function $f(t)$ is increasing for $t \geq 1$ and when $0 < p < 1$. So $f(t) \geq f(1)$ and $f(1) = 0$. So, we get the desired result. $\textbf{EDIT:}$ To show $f(t)$ is increasing for $t \geq 1$ and when $0 < p < 1$. We need to show 0 < f'(t) = p (t-1)^{p-1} - pt^{p-1} and since $p>0$, all we need to show is that $(t-1)^{p-1}>t^{p-1}$, $\forall t > 1$ and $0 . Since $0 , we need to show $t^{1-p} > (t-1)^{1-p}$ which is true since $p<1$ and $t>t-1>0$.
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11@user10: I see you're a new user to this site. Just so you know, it's considered polite on this site both to upvote the answers to your questions that you find helpful (click on the little up arrow next to the answer) and to formally accept the answer to each of your questions that you think is the best (click on the little check mark by the answer you want to accept). – 2010-11-20
This follows from the more general inequality \begin{equation}(1)\qquad\qquad|x+y|^p\le |x|^p+|y|^p\qquad\qquad\text{(for $x,y\in\mathbb{C}$ and $0\lt p\le1$)}\end{equation} Indeed, if we replace $x$ by $x-y$ in (1) we get $|x|^p\le |x-y|^p+|y|^p$ which imply $|x|^p-|y|^p\le |x-y|^p$ To prove (1), first note $|x+y|^p\le(|x|+|y|)^p$ Hence it is sufficient to prove (1) for $x,y\ge0$ in which case we may apply Sivaram's argument in the previous answer.