For $a,b \in \mathbb R$, $p\geq2$ I try to show $$\left|\frac{a+b}{2}\right|^p+\left|\frac{a-b}{2}\right|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p.$$

Is this a popular inequality (At least I could not find it in the list of popular inequalities from wikipedia)? It seems to be related to convexity but I did not succeed to show it. A related inequality seems to be for $p \geq 1,a,b\geq0$

$$\left(\frac{a+b}{2}\right)^p\leq \frac{1}{2}a^p+\frac{1}{2}b^p,$$ which directly follows from the convexity of $x^p$ for positive numbers.

  • 4
    $\begingroup$ It's a special case of Clarkson's inequalities (when your measure space is a single point with counting measure). $\endgroup$ – Jose27 Dec 31 '11 at 5:08

We have for all $x_1,x_2\geq 0$: $x_1^p+x_2^p\leqslant (x_1^2+x_2^2)^{p/2}$. Indeed, it suffice to show it when $x_2=1$, otherwise apply it to $\frac{x_1}{x_2}$. $f(t):=(t^2+1)^{p/2}-t^p-1$ is non-negative, since its derivative is $p(t^2+1)^{p/2-1}t-pt^{p-1}\geqslant 0$ and $f(0)=0$. Applying it to $x_1=\frac{a+b}2$ and $x_2=\frac{a-b}2$, we get \begin{align*} \left|\frac{a+b}2\right|^p+\left|\frac{a-b}2\right|^p&\leqslant \left(\left|\frac{a+b}2\right|^2+\left|\frac{a-b}2\right|^2\right)^{p/2}\\\ &=\left(\frac {2a^2+2b^2}4\right)^{p/2}\\\ &=\left(\frac {a^2}2+\frac{b^2}2\right)^{p/2}\\\ &\leqslant \frac 12|a|^p+\frac 12|b|^p, \end{align*} since the map $t\mapsto |t|^{p/2}$ is convex ($p\geqslant 2$).

  • 1
    $\begingroup$ Why is it sufficient to show it for $x_2=1$? $\endgroup$ – Listing Dec 31 '11 at 0:10
  • 1
    $\begingroup$ It's clear if $x_2=0$. Otherwise, you apply it to $\frac{x_1}{x_2}$. $\endgroup$ – Davide Giraudo Dec 31 '11 at 0:12
  • $\begingroup$ Does the last step even follow from the convexity of $t\mapsto |t|^{p/2}$? I don't see it :/ $\endgroup$ – Listing Dec 31 '11 at 14:42
  • $\begingroup$ I have added the details. $\endgroup$ – Davide Giraudo Dec 31 '11 at 14:58
  • 1
    $\begingroup$ Thank you, I did not see that one could remove the abs inside because of the squares. $\endgroup$ – Listing Dec 31 '11 at 15:07

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.