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Science Forum Index » Mathematics Forum » Hilbert space related ...
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| Yokeire |
 Posted: Sun Dec 14, 2003 3:12 pm |
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H a Hilbert space, K a closed linear subspace. Let x0 be any pt in H.
w.t.s. Min{||x0-y|| : y in K} = Max{|<x0,y>| : y in K^perp, ||y||=1}
I'm getting stuck but should be pretty easy :(
know from given info that ||x0 -y|| >= ||x0|| for all y in K
Also that there exists a y in K, z in K^perp s.t. x0 = y + z
so ||x0 - y|| = ||z|| > = ||x0|| ... I think I'm going wrong way with this. |
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| Ronald Bruck |
| Posted: Fri Dec 19, 2003 6:40 pm |
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In article <200312142210.hBEMAnS10243@proapp.mathforum.org>, Yokeire
<yellowassedcow@hotmail.com> wrote:
Quote: H a Hilbert space, K a closed linear subspace. Let x0 be any pt in H.
w.t.s. Min{||x0-y|| : y in K} = Max{|<x0,y>| : y in K^perp, ||y||=1}
I'm getting stuck but should be pretty easy :(
know from given info that ||x0 -y|| >= ||x0|| for all y in K
Also that there exists a y in K, z in K^perp s.t. x0 = y + z
so ||x0 - y|| = ||z|| > = ||x0|| ... I think I'm going wrong way with this.
This result is false as stated. If K = H then the min is 0; but the
sup of the empty set is -infinity. (The max doesn't exist.)
Always watch the boundary cases...
--Ron Bruck |
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