Reverse triangle inequality
From Maths
Revision as of 09:28, 13 March 2016 by Alec (Talk | contribs) (Created page with "==Statement== Given a normed space {{M|(X,\Vert\cdot\Vert)}} other than the defining properties of a norm we also have: * {{MM|\Big\vert \Vert a\Vert - \Vert b\Vert\Bi...")
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]
- Which is [ilmath]-(\Vert a\Vert - \Vert b\Vert)\le\Vert a-b\Vert[/ilmath]
- [ilmath]\Vert b\Vert - \Vert a\Vert \le \Vert b-a\Vert[/ilmath],
- So we have:
- [ilmath]\Vert a\Vert - \Vert b\Vert \le \Vert a-b\Vert[/ilmath]
- [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.