About an inequality in Royden's Real Analysis

Question

I want to show for $p\geq1$

$|\text{sgn}(a)|a|^{1/p}-\text{sgn}(b)|b|^{1/p}|^{p}\leq 2^{p}|a-b|$.

But I am not getting minus sign on right hand side. Any hint will be appreciated.

Answer 1

I hope this will do.

Let $a>0,b>0$. Let $a^{1/p}-b^{1/p}>0$. Thus $a=(a^{1/p}-b^{1/p}+b^{1/p})^p\geq (a^{1/p}-b^{1/p})^p+b\implies (a^{1/p}-b^{1/p})^p\leq a-b\leq 2^p|a-b|$. Similarly, if $b^{1/p}-a^{1/p}>0$, then we can show that $(b^{1/p}-a^{1/p})^p\leq 2^p|a-b|$. Thus $|\text{sgn}(a)|a|^{1/p}-\text{sgn}(b)|b|^{1/p}|^p\leq 2^p|a-b|$.

Let $a>0,b<0$. Thus $|\text{sgn}(a)|a|^{1/p}-\text{sgn}(b)|b|^{1/p}|^p=(|a|^{1/p}+|b|^{1/p})^p\leq 2^{p-1}(|a|+|b|)\leq 2^p|a-b|$ as $x\mapsto x^p$ is convex.

Let $a<0,b>0$. Then $\text{sgn}(a)|a|^{1/p}-\text{sgn}(b)|b|^{1/p}=|b|^{1/p}-|a|^{1/q}$. Thus as we have done in the first case we can show that $|\text{sgn}(a)|a|^{1/p}-\text{sgn}(b)|b|^{1/p}|^p\leq 2^p|a-b|$.

The case $a<0,b<0$ will be similar to second case.