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Suppose a Hilbert space W can be written as the direct sum (not necessarily orthogonal) of the closed subspaces H and V, where H is assumed to be of finite dimension. Define a new inner product via

||h+v||^2:=q(h)+|v|^2,

where |.| denotes the original norm on the Hilbert space and q is a positive definite quadratic form on H (one can assume w.l.o.g. q=|.|^2).

QUESTION: Are |.| and ||.|| equivalent?

||.||^2 is easily seen to be dominated by 2|.|^2, but I don't know about the other direction. (Also notice that the question is obviously true if V and H were orthogonal!)

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This might be overkill, but whenever you have two closed subspaces of a Hilbert space K, which form a direct sum decomposition of K, then there is a positive invertible operator R on K such that conjugation by R takes the two subspaces to orthogonal ones. So I think your problem reduces to the case where V and H are orthogonal –  Yemon Choi Sep 21 '10 at 10:46
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Is this a homework problem? It is a special case of the following standard exercise, which in textbooks usually comes in the section where the open mapping/closed graph theorems are proved: If the Banach space $(W,\|\cdot \|) $ is the direct sum of two closed subspaces $H$ and $V$, then $\|\cdot \|$ is equivalent to the norm defined by $\|h\| + \|v\|$ (where $h$ is in $H$ and $v$ is in $V$). –  Bill Johnson Sep 21 '10 at 10:52
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closed as off-topic by Ricardo Andrade, David White, Ricky Demer, j.c., Carlo Beenakker Oct 21 '13 at 13:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

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1 Answer

up vote 0 down vote accepted

Thanks to Bill Johnson!

My question is easily answered by a direct application of the closed graph theorem (one shows that the diagonal is closed in the mixed norms). Unfortunately, I did not have this one as an exercise in my functional analysis class!

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The question is much easier when it comes after a chapter on the closed graph theorem. I'm glad that my "hint" helped. –  Bill Johnson Sep 22 '10 at 1:27
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