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F(x)=||x|| is Lipschitz function

  1. Apr 12, 2010 #1
    "Let (V,||.||) be a normed vector space. Then by the triangle inequality, the function f(x)=||x|| is a Lipschitz function from V into [0,∞)."

    I don't understand how we this follows from the triangle inequality. How does the proof look like?

    Any help is appreciated!
    Last edited: Apr 12, 2010
  2. jcsd
  3. Apr 12, 2010 #2
    [tex]\Big|\|x\| - \|y\| \Big|\le \|x-y\|[/tex]
  4. Apr 12, 2010 #3
    OK, I think that's the "variant triangle inequality". I've seen the proof in R (by the regular triangle inequality), but how can we prove it for normed vector spaces?
    Last edited: Apr 12, 2010
  5. Apr 12, 2010 #4

    [tex]\Big|\|x\| - \|y\| \Big|\le \|x-y\|[/tex] <=> -||x-y||[tex]\leq||x||-||y||\leq[/tex] ||x-y|| <=> -||x-y|| +||y||[tex]\leq ||x||\leq[/tex] ||x-y|| +||y|| which is correct since

    ||x|| =||x-y+y||[tex]\leq[/tex] ||y|| +||x-y|| and

    ||y|| = ||y-x+x||[tex]\leq[/tex] ||x-y||+ ||x||
  6. Apr 12, 2010 #5

    Just wondering...is there a analogue of the "variant triangle inequality" in general METRIC SPACES? If so, how does it look like and how does the proof change?
  7. Apr 13, 2010 #6

    [tex]\left|d(x,0)-d(y,0)\right|\leq d(x,y)[/tex]

    The same proof
    Last edited: Apr 13, 2010
  8. Apr 13, 2010 #7
    So in the metric space context, does it mean that the function "d" is always a Lipchitz function as well?
  9. Apr 14, 2010 #8
    The Lipschitz function is f(x) =d(x,0) and not d(x,y)
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