Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normal cone

  1. Oct 16, 2014 #1
    [tex]Let \text{ } H \text{ }be \text{ }a \text{ } Hilbert \text{ } space, \text{ }K \text{ }be \text{ }a \text{ }closed\text{ }convex\text{ }subset \text{ } of \text{ }H \text{ }and \text{ }x_{0}\in K. \text{ }Then \\N_{K}(x_{0}) =\{y\in K:\langle y,x-x_{0}\rangle \leq 0,\forall x\in K\} .\text{ }Hence, \text{ }if \text{ }K=\left\{ (x,y):x^{2}+y^{2}\leq 1\right\}\\ is \text{ }closed\text{ } and\text{ }convex, \text{ }find \text{ }N_{K}((0,0))? [/tex]
    Thanks.
     
  2. jcsd
  3. Oct 17, 2014 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    If x0 = (0,0), then you have a two dimensional Euclidean space. K is the unit disc. Nk((0,0)) has just one point (0,0).

    If you meant something else, I suggest you rewrite it.
     
  4. Oct 17, 2014 #3
    [tex]\text{ }yes,\text{ } x_{0} = (0,0)[/tex]
     
  5. Oct 17, 2014 #4
    Yes, I mean [tex]\text{ } x_{0} = (0,0).[/tex]
    but, if [tex]\text{ } x_{0} = (0,1).[/tex] What is equal to [tex]N_{K}((0,1))?[/tex]
     
  6. Oct 18, 2014 #5

    mathman

    User Avatar
    Science Advisor
    Gold Member

    I haven't checked it throughly, but it looks like all y in K of the form (0,c) where c ≥ 0.
     
  7. Oct 18, 2014 #6
    Thank u Mr. mathman.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook