Hello,

I need help with this question..

Thank you .

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- Jan 13th 2009, 04:01 AMBlackWarriorHilbert Spaces
Hello,

I need help with this question..

Thank you . - Jan 13th 2009, 10:27 AMOpalg
Here's a sketch of the proof. Essentially, w_0 is defined as the closest point to u in W.

Let d be the distance from u to W, $\displaystyle d:= \inf\{\|u-w\|:w\in W\}$. We don't yet know that the infimum is attained, but we can choose a sequence (w_n) in W such that $\displaystyle \|u-w_n\|\to d$ as n→∞.

Apply the parallelogram identity $\displaystyle \|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2$, with $\displaystyle x = u-w_m,\ y=u-w_n$, to get $\displaystyle 4\|u-\tfrac12(w_m+w_n)\|^2 + \|w_m-w_n\|^2 = 2\|u-w_m\|^2 + 2\|u-w_n\|^2$. Since$\displaystyle \tfrac12(w_m+w_n)\in W$, the first term on the left side is ≥d^2, and you should be able to deduce that $\displaystyle \|w_m-w_n\|\to0$. Thus (w_n) is a Cauchy sequence and (by the completeness of V) it converges to a point $\displaystyle w_0\in W$, which is the orthogonal projection of u on W.