Difference between revisions of "Reverse triangle inequality"

Latest revision as of 09:28, 13 March 2016

Statement

Given a normed space $(X,\Vert\cdot\Vert)$ other than the defining properties of a norm we also have:

• $$\Big\vert \Vert a\Vert - \Vert b\Vert\Big\vert\le \Vert a-b\Vert$$

Proof

Notice that $\Vert a\Vert=\Vert (a-b)+b\Vert$ and:

• $\Vert (a-b)+b\Vert\le \Vert a-b\Vert + \Vert b\Vert$ by the "triangle inequality" property of a norm
• Now we have $\Vert a\Vert = \Vert (a-b)+b\Vert \le \Vert a-b\Vert + \Vert b\Vert$ so:
  • $\Vert a\Vert \le \Vert a-b\Vert + \Vert b\Vert$
• Thus $\Vert a\Vert - \Vert b\Vert \le \Vert a-b\Vert$
• If we were to instead define $b$ as $a$ and $a$ as $b$ we would obtain:
  • $\Vert b\Vert - \Vert a\Vert \le \Vert b-a\Vert$,
    • But! $\Vert b-a\Vert = \Vert a-b\Vert$
  • Thus $\Vert b\Vert - \Vert a\Vert \le \Vert a-b\Vert$
    • Which is $-(\Vert a\Vert - \Vert b\Vert)\le\Vert a-b\Vert$
      • Or more usefully $\Vert a\Vert - \Vert b\Vert\ge-\Vert a-b\Vert$
• So we have:
  1. $\Vert a\Vert - \Vert b\Vert \le \Vert a-b\Vert$
  2. $\Vert a\Vert - \Vert b\Vert\ge-\Vert a-b\Vert$
• If $\vert x\vert \le y$ then we have:
  • $-y\le x\le y$ or the two statements $-y\le x$ and $x\le y$, so we see:
• $\Big\vert \Vert a\Vert - \Vert b\Vert\Big\vert\le \Vert a-b\Vert\iff - \Vert a-b\Vert \le \Vert a\Vert - \Vert b\Vert \le \Vert a-b\Vert$

Which is exactly what we have.

This completes the proof.

References