Is the metric obtained by altering the metric of a Hilbert space on a finite-dimensional subspace equivalent to the original one? [closed]

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!)

-

closed as off-topic by Ricardo Andrade, David White, Ricky Demer, j.c., Carlo BeenakkerOct 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