Reverse triangle inequality

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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]


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.