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Suppose $x \geq 0$, $y \geq 0$ and $0

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

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    The 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/1828/15062010-11-20

2 Answers 2

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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 $0 1$ 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 $0 (t-1)^{1-p}$ which is true since $p<1$ and $t>t-1>0$.

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    @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
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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.

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    @ AD: You should have $02010-11-19
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    @Sivaram: Of course, I add that. Thanks.2010-11-20
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    @Arturo Magidin: I know about the problem with <. I was just fixing that.2010-11-20
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    Sorry about that; saw the problem, so I tried to fix it. Sometimes it happens to me and I navigate away before realizing it.2010-11-20
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    @Arturo Magidin: Ok.2010-11-20
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    Very nice argument. For an alternative, we can prove the general by the concavity of $x\mapsto x^p$ and $0^p=0$.2016-01-05
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    @XiangYu Thanks, yes thanks.2016-01-07
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    @AD. Can we also say that $|x|^p-|y|^p\le p\max\{|x|^{p-1},|y|^{p-1}\}|x-y|$ ?2018-11-05
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    @Hirak No, that is false. What happens when $y=0$ and $x=1$?2018-11-05