let $C$ be a closed convex subset of a Hilbert space $\mathscr{H}$. Show that for each $x\in\mathscr{H}$ there is a unique point $y_0\in C$ such that $||x-y_0||=\rho(x,C)=\inf\{||x-y|| \ : \ y\in C\}$

Question

let $C$ be a closed convex subset of a Hilbert space $\mathscr{H}$. Show that for each $x\in\mathscr{H}$ there is a unique point $y_0\in C$  such that  $||x-y_0||=\rho(x,C)=\inf\{||x-y|| \ : \ y\in C\}$

zzr0ck3r · Answer

@eliassaab I am thinking we can have some sequence that converges to $y_0$ since C is closed. s.t. we get the desired result. But as of now I have no idea how to go about it.

zzr0ck3r · Answer

Actually I think I found a good proof for this.

zzr0ck3r · Answer

https://www.math.lsu.edu/~sengupta/7330f02/7330f02projops.pdf

zzr0ck3r · Answer

ok the only other one that I have been having trouble with is 

A semi inner product on a linear space $X$ is a function $\langle ,\rangle :X \times X\rightarrow F$ satisfying the following conditions conditions 
a) $\langle ax+by,z\rangle =a\langle x,z\rangle +b\langle y,z\rangle $
b) $\langle x,y\rangle =\langle y,z\rangle $ if $F=\mathbb{R}$ or $\langle x,y\rangle =\overline{\langle y,x\rangle }$ if $F=\mathbb{C}$.
c) $\langle x,x\rangle  \ \ge  \ 0$
d) $\langle x,x\rangle =0 $ if $x=0$    (note that non iff statement for this last property)
I need to show that the following propositions hold, and I know that they hold for the case of an inner product.

1) $\langle x+y, x+y\rangle =\langle x,x\rangle +2\mathscr{R}\langle x,y\rangle +\langle y,y\rangle $
2) $\lvert\langle x,y\rangle \rvert^2\ \le \ \langle x,x\rangle \langle y,y\rangle $

anonymous · Answer

See this for 2) http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality

anonymous · Answer

$$ =++ +=\ ++\overline {}+= \langle x,x\rangle +2\mathscr{R}\langle x,y\rangle +\langle y,y\rangle $$ Since $$ a+ i b + a-i b = 2a $$

zzr0ck3r · Answer

I think I might be figuring it out as we speak

zzr0ck3r · Answer

sweet:) tnx. I will keep trying without looking to hard at what you did, and know that I can come back to this if it gets to late:)

zzr0ck3r · Answer

@eliassaab that proof of the Cauchy inequality uses the non constrained definition of inner product, so they can claim that since $v\ne0$ they can divide by $\langle v, v \rangle $ but we do not have this property for the semi inner product

anonymous · Answer

See http://www.ams.org/journals/tran/1967-129-03/S0002-9947-1967-0217574-1/S0002-9947-1967-0217574-1.pdf

zzr0ck3r · Answer

sorry but what part of that? Its getting a little over my head.

zzr0ck3r · Answer

@eliassaab  what about this one

Let $S$ be a subset of an inner product space $X$. Show that $S^{\perp}$ is a closed linear subspace of $X$.

I know how to show it is a linear subspace, that was easy enough, but what about the closed part?

anonymous · Answer

@zzr0ck3r . It is better to ask each question as a new problem, so new viewers would know which part we are talking about. Post it and tag me.