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Homework Help: Closed subset of R^n has an element of minimal norm

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data
    a) Let V be a normed vector space. Then show that (by the triangle inequality) the function f(x)=||x|| is a Lipschitz function from V into [0,∞). In particular, f is uniformly continuous on V.

    b) Show that a closed subset F of contains an element of minimal norm, that is, there is an x E F such that ||x||≤||y|| for all y E F. (here ||x|| refers to the usual Euclidean norm).
    (hint: F may not be compact, so work on a suitable compact subset of F.)


    2. Relevant equations
    3. The attempt at a solution
    I proved part a, but I really have no idea how to do part b. Why do we need compactness, and what is the suitable compact subset of F?

    I hope someone can help me out! Thank you!
     
  2. jcsd
  3. Apr 13, 2010 #2

    lanedance

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    just ideas that will need some working but hope they help

    you want to take a compact subset of F, is a closed an bounded subset enough to be compact (it is in Rn - Hiene Borel)?

    you could pick any element of F, then use it to set your bound, and show elements outside the bounded set have a larger norm

    now you have a compact subset of a metric space and i seem to remember something along the lines of a continuous real valued function always takes a maximal & minimal value on a compact set...
     
    Last edited: Apr 13, 2010
  4. Apr 13, 2010 #3
    If we intersect F with a compact set K, will F intersect K stilll be compact? Why or why not?
    Also, should we choose K to be very small? How can we formalize this choice of K? (i.e. what is the so-called suitable compact subset of F mentioned in the hint?)


    "...and show elements outside the bounded set have a larger norm"
    How??
     
  5. Apr 13, 2010 #4
    Consider the closed balls of radius r, centred at the origin:

    [tex]B\left(0,r\right)=\left\{x \in V:\left\|x\right\|\leq r\right\}[/tex]

    There must be one that intersects F. How do you go on from there?
     
  6. Apr 13, 2010 #5
    Why does it have to intersect F?

    And how can we show elements outside the bounded set have a larger norm? I don't get this idea at all. How is this possible? Can someone please explain this in greater detail?

    Thanks a lot!
     
  7. Apr 13, 2010 #6

    Office_Shredder

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    By definition, elements outside the ball of radius r have a larger norm than elements inside a ball of radius r. Look at how Suarez described the set
     
  8. Apr 14, 2010 #7
    For sufficiently large r, it must intersect F; otherwise, F would be empty.
     
  9. Apr 17, 2010 #8
    What would go wrong if F were empty??
     
  10. Apr 17, 2010 #9
    It wouldn't have an element with minimum norm; in fact, the original statement of your question should be "a closed nonempty subset F contains an element of minumum norm". As the empty set is closed, the statement is false without that condition.
     
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