I want to show that the inequality

$$2^{1-p}|x-y |^p \leq \left|\, x \vert x \vert^{p-1} - y \vert y \vert^{p-1} \,\right|$$

holds for every $x,y \in \mathbb{R}$ and every $p \geq 1$. I found this in my analysis paper but sadly I could not prove it. I tried to use the convexity of the function $x \mapsto \vert x \vert^p$ and also tried to use an integral representation. Can someone give me a hint or a link where this is shown? Thank you very much in advance.


There are basically two cases:

  • $x>0>y$, in which case replace $y$ by $-y$ and rewrite the inequality as $$\dfrac{x^p+y^p}{2} \geq \left (\dfrac{x+y}{2} \right )^p,$$ where $x,y>0$. This follows from the fact that the graph of $f(x)=x^p$ is concave up.

  • $x>y>0$. In this case rewrite the inequality as $$\dfrac{x^p-y^p}{2} \geq \left (\dfrac{x-y}{2} \right )^p,$$ but a stronger inequality holds in this case: $$x^p-y^p \geq (x-y)^p \geq 2(x-y)^p/2^p.$$

  • $\begingroup$ Can you suggest how to prove the first part of 3rd inequality? $\endgroup$ – Multigrid Aug 13 '18 at 15:09
  • $\begingroup$ Generally $(a+b)^p \geq a^p +b^p$ for $p\geq 1$. To prove this, for example divide by $b^p$ and reduce to $(t+1)^p \geq t^p+1$ which follows from showing that both expressions are 0 at $t=0$ but the derivative of the left hand side is greater than the derivative of the left hand side for $t>0$ and $p\geq 1$. $\endgroup$ – Marco Aug 13 '18 at 15:13
  • $\begingroup$ No the problem is $x^p -y^p \geq (x-y)^p$, if we take $y^p$ to right hand side, we have similar form as you said but then $x-y$ can be negative $\endgroup$ – Multigrid Aug 13 '18 at 15:20
  • 1
    $\begingroup$ We assumed $x>y$ for that part. The case $y>x$ follows similarly. $\endgroup$ – Marco Aug 13 '18 at 15:33
  • $\begingroup$ Ah, sorry for not observing properly! $\endgroup$ – Multigrid Aug 13 '18 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.