# Norm Satisfying the Parallelogram Law

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Let V be a vector space over the complex field.

If V has an inner product $$<\cdot,\cdot>$$, and $$||\cdot||$$ is the induced norm, then it's easy to show that the norm must satisfy the parallelogram law, to wit:

$$||x+y||^2 + ||x-y||^2 = 2||x||^2 + 2||y||^2$$

Much more interestingly, given an arbitrary norm on V, there exists an inner product that induces that norm IF AND ONLY IF the norm satisfies the parallelogram law.

Horn and Johnson's "Matrix Analysis" contains a proof of the "IF" part, which is trickier than one might expect. I'm trying to produce a simpler proof.

It's not too hard to show that if there exists an inner product $$<\cdot,\cdot>$$on V which induces a norm $$||\cdot||$$, then it must satisfy the polarization identity:

$$<u,v> = \frac{1}{4}\left(||u+v||^2 + ||u-v||^2 + i||u+iv||^2 - i||u-iv||^2\right)$$

So, to solve my problem, I assume that V is equipped with a norm that satisfies the parallelogram law, and then I define

$$p(u,v) = \frac{1}{4}\left(||u+v||^2 + ||u-v||^2 + i||u+iv||^2 - i||u-iv||^2\right)$$

and my goal is to show that $$p$$ satisfies all the requirements of an inner product:

(1) $$p(u,u) \geq 0$$ for all $$u \in V$$, with equality iff $$u = 0$$
(2) $$p(u+v,w) = p(u,w) + p(v,w)$$ for all $$u,v,w \in V$$
(3) $$p(\alpha v, w) = \alpha p(v,w)$$ for all $$\alpha \in \mathbb{C}$$, and all $$v,w \in V$$
(4) $$p(v,w) = \overline{p(w,v)}$$

It's easy to show that (1) and (4) are true. With a bit of crafty manipulation, (2) is also not too hard.

Horn and Johnson use (2) to establish (3) first for integer values of $$\alpha$$ and then for rational values. Then they resort to a limiting argument to show that (3) must also hold for all real (and then complex) $$\alpha$$.

I suspect that there is a more elementary, direct way to obtain (3) directly from the polarization identity, the parallelogram, and the norm axioms. I suspect this in part because this problem shows up as an exercise in Axler's "Linear Algebra Done Right," and if the Horn and Johnson solution really is the simplest, than that means this exercise is at least an order of magnitude harder than most of Axler's other exercises.

But I've been banging my head against the problem of trying to find such an elementary proof of (3), without success so far.

I have a hunch that there's a clever change of variables that will convert the parallelogram law

$$||x+y||^2 + ||x-y||^2 = 2||x||^2 + 2||y||^2$$

into a form that expresses

$$||x+y||^2 - ||x-y||^2$$ (note the sign change)

in a nice way that will allow me to "pull out" the $$\alpha$$ from the various terms of the polarization identity:

$$p(\alpha x, y) = \frac{1}{4}\left(||\alpha x + y||^2 - ||\alpha x - y||^2 + i ||\alpha x + iy||^2 - i ||\alpha x - iy||^2\right)$$

but I haven't been able to find a trick that works. Any ideas?

P.S. Using Google Books, I found another book, "Elements of Operator Theory," by Carlos S. Kubrusly, which proves it the same way as Horn and Johnson, so maybe that really IS the simplest way?

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