Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following inequality is from the proof that the $L^p$ norm is Gâteaux differentiable for $ 1 < p<\infty$ (from "Analysis" by Lieb and Loss).

Let $a$, $b\in\mathbb{C}$ and $-1\leq t\leq 1$, $t\not=0.$ Then $$|a|^p-|a-b|^p\leq\frac{1}{t}(|a+tb|^p-|a|^p) \leq |a+b|^p-|a|^p.$$

I managed to prove the second inequality for positive $t$ by writing $a+tb=(1-t)a+t(a+b)$and using the convexity of $\cdot^p$. From this the first inequality follows for negative $t$ by substituting $-b$ for $b$. The same trick would finish the proof, if I could prove either the second inequality for negative $t$ or the first inequality for positive $t$.

share|improve this question
add comment

1 Answer 1

up vote 4 down vote accepted

The same basic idea works for negative values of $t$:

What you want to prove is equivalent to $|a + tb|^p - |a|^p \geq t|a + b|^p - t|a|^p$. This in turn is the same as $(1 - t)|a|^p \leq |a + tb|^p + (-t)|a + b|^p$, which in turn is equivalent to $|a|^p \leq {1 \over 1 - t} |a + tb|^p + {-t \over 1 - t} |a + b|^p$. Note that $a = {1 \over 1 - t} (a + tb) + {-t \over 1 - t} (a + b)$, so you can use convexity as you used before.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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