# Normal cone

1. Oct 16, 2014

### moh salem

$$Let \text{ } H \text{ }be \text{ }a \text{ } Hilbert \text{ } space, \text{ }K \text{ }be \text{ }a \text{ }closed\text{ }convex\text{ }subset \text{ } of \text{ }H \text{ }and \text{ }x_{0}\in K. \text{ }Then \\N_{K}(x_{0}) =\{y\in K:\langle y,x-x_{0}\rangle \leq 0,\forall x\in K\} .\text{ }Hence, \text{ }if \text{ }K=\left\{ (x,y):x^{2}+y^{2}\leq 1\right\}\\ is \text{ }closed\text{ } and\text{ }convex, \text{ }find \text{ }N_{K}((0,0))?$$
Thanks.

2. Oct 17, 2014

### mathman

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.

3. Oct 17, 2014

### moh salem

$$\text{ }yes,\text{ } x_{0} = (0,0)$$

4. Oct 17, 2014

### moh salem

Yes, I mean $$\text{ } x_{0} = (0,0).$$
but, if $$\text{ } x_{0} = (0,1).$$ What is equal to $$N_{K}((0,1))?$$

5. Oct 18, 2014

### mathman

I haven't checked it throughly, but it looks like all y in K of the form (0,c) where c ≥ 0.

6. Oct 18, 2014

### moh salem

Thank u Mr. mathman.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook