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What is the best way to prove the polarization identity in the following form: for any $u, v \in V$ $$4 \langle T(u), v \rangle = \langle T(u+v), u+v \rangle - \langle T(u-v), u-v \rangle + i\langle T(u+iv), u+iv \rangle - i\langle T(u-iv), u-iv \rangle$$

Here $T$ is a linear operator.

Context

Polarization identity is a useful result in Hilbert spaces. It expresses the inner product in terms of norm:
$$4 \langle u, v \rangle = \langle u+v, u+v \rangle - \langle u-v, u-v \rangle + i\langle u+iv, u+iv \rangle - i\langle u-iv, u-iv \rangle$$ where all terms on the right are just norms squared.

More generally, it gives a way to estimate a bilinear form $B(u,v) = \langle T(u), v \rangle $ in terms of quadratic form $Q(u)=B(u,u)$. This is the generalized form with $T$ above.

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What did you try? –  AD. Nov 4 '10 at 13:57
    
What is the property of the $\langle . , . \rangle$ operator? $\langle u, v + w \rangle = ...$ (Hint: in the real case, we speak about bilinear forms). –  Djaian Nov 4 '10 at 13:59
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As a side comment: you've posted 11 questions on this forum so far, most of them are homework type questions with initially incorrectly applied tags (this one also, which I'll fix in a minute). You may get better attention if you at least tag questions in the correct field. You may also garner more good will if you actually accept answers. At least 3 or 4 of the questions you've asked in the past month have received very nice answers which you've ignored. –  Willie Wong Nov 4 '10 at 14:02
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@AD. : I think you can delete the "What" in your question. –  Djaian Nov 4 '10 at 14:03
    
@Djaian: nice joke! :D –  J. M. Nov 4 '10 at 14:06

1 Answer 1

Expand RHS, using distributive property and the property that $\langle u,cv\rangle=\bar{c}\langle u,v\rangle$. Simplify and cancel many terms to get the LHS.

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