Difference between revisions of "Reverse triangle inequality"

Latest revision as of 09:28, 13 March 2016

Statement

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

• [math]\Big\vert \Vert a\Vert - \Vert b\Vert\Big\vert\le \Vert a-b\Vert[/math]

Proof

Notice that [ilmath]\Vert a\Vert=\Vert (a-b)+b\Vert[/ilmath] and:

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

Which is exactly what we have.

This completes the proof.

References