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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.

Clayton
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Rupert R
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1 Answers1

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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: $$x^p-y^p \geq (x-y)^p \geq 2(x-y)^p/2^p.$$

Marco
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  • Can you suggest how to prove the first part of 3rd inequality? – Multigrid Aug 13 '18 at 15:09
  • 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 $p\geq 1$. – Marco Aug 13 '18 at 15:13
  • 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 – Multigrid Aug 13 '18 at 15:20
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    We assumed $x>y$ for that part. The case $y>x$ follows similarly. – Marco Aug 13 '18 at 15:33
  • Ah, sorry for not observing properly! – Multigrid Aug 13 '18 at 15:36