MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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!)

share|cite|improve this question

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:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, David White, Ricky Demer, j.c.
If this question can be reworded to fit the rules in the help center, please edit the question.

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
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
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!

share|cite|improve this answer
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.