# The Unapologetic Mathematician

## The Parallelogram Law

There’s an interesting little identity that holds for norms — translation-invariant metrics on vector spaces over $\mathbb{R}$ or ${C}$ — that come from inner products. Even more interestingly, it actually characterizes such norms.

Geometrically, if we have a parallelogram whose two sides from the same point are given by the vectors $v$ and $w$, then we can construct the two diagonals $v+w$ and $v-w$. It then turns out that the sum of the squares on all four sides is equal to the sum of the squares on the diagonals. We write this formally by saying

$\displaystyle\lVert v+w\rVert^2+\lVert v-w\rVert^2=2\lVert v\rVert^2+2\lVert w\rVert^2$

where we’ve used the fact that opposite sides of a parallelogram have the same length. Verifying this identity is straightforward, using the definition of the norm-squared:

\displaystyle\begin{aligned}\lVert v+w\rVert^2+\lVert v-w\rVert^2&=\langle v+w,v+w\rangle+\langle v-w,v-w\rangle\\&=\langle v,v\rangle+\langle v,w\rangle+\langle w,v\rangle+\langle w,w\rangle\\&+\langle v,v\rangle-\langle v,w\rangle-\langle w,v\rangle+\langle w,w\rangle\\&=2\langle v,v\rangle+2\langle w,w\rangle\\&=2\lVert v\rVert^2+2\lVert w\rVert^2\end{aligned}

On the other hand, what if we have a norm that satisfies this parallelogram law? Then we can use the polarization identities to define a unique inner product.

$\displaystyle\langle v,w\rangle=\frac{\lVert v+w\rVert^2-\lVert v-w\rVert^2}{4}+i\frac{\lVert v-iw\rVert^2-\lVert v+iw\rVert^2}{4}$

where we ignore the second term when working over real vector spaces.

However, if we have a norm that does not satisfy the parallelogram law and try to use it in these formulas, then the resulting form must fail to be an inner product. If we did get an inner product, then the norm would satisfy the parallelogram law, which it doesn’t.

Now, I haven’t given any examples of norms on vector spaces which don’t satisfy the parallelogram law, but they show up all the time in functional analysis. For now I just want to point out that such things do, in fact, exist.

April 24, 2009 - Posted by | Algebra, Linear Algebra

1. i always thought one starts with an inner product and then the norms are “induced” by the inner product. never thought we start with a norm and derive an inner product from it …

Comment by awhan | April 26, 2009 | Reply

2. An inner product is basically a positive definite quadratic form (on finite dimensional spaces) and a positive definite quadratic form gives you a norm.

Comment by Zygmund | April 28, 2009 | Reply

3. Yes, Zygmund, that’s basically what I’ve said. awhan was commenting that he hadn’t seen before the idea of starting from a norm instead of ending with one.

Comment by John Armstrong | April 28, 2009 | Reply

4. There are discontinuous additive real functions on the real line. That is, there are everywhere discontinuous functions f:R->R such that f(x+y)=f(x)+f(y). The existence of such functions is intimately tied to the axiom of choice, and to the existence of non-measurable sets. One can arrange for f(x)=0 iff x=0. Using any such discontinuous f, define q(x) = f(x)f(x). The function q is non-negative, and obeys the parallelogram law. However, q(x)^{1/2} is not a norm on R because q(ax) != a^{2}q(x) for all scalars ‘a’.

If you start with a function q : X x X -> C satisfying the parallelogram law, then you can show that b(x,y) = 1/4\sum_{n=0}^{3} i^{n} q(x+i^{n}y) is additive in both coordinates. Therefore, for fixed x,y, the function t -> b(tx,y) is additive. The only remaining issue of linearity–the issue of whether or not b(ax,y)=ab(x,y) for all scalars ‘a’–reduces to the real case, provided q(ix)=q(x), where ‘i’ is the imaginary constant. This is not a straightforward matter, even for positive forms, as you can see. However, if q(x) >= 0, and q(ax)=|a|^{2}q(x) for all scalars ‘a’, then that is enough to guarantee the continuity of t->b(tx,y) and, hence, also the sesquilinear nature of b(x,y).

An additive real function f(x) is linear iff it is Lebesgue measurable. It is not hard to show that f(qx) = q f(x) for all rational numbers q; this follows directly from the additivity of f. Therefore, f is linear over the rational numbers Q. We end up with a natural decomposition of R into equivalence classes [x] containing all y in R such that x-y is rational. This is, in turn, related to the classical construction of a non-measurable set, where one picks a single element from each distinct equivalence class [x].

Very interesting stuff which is not easily summarized. The upshot: a positive definite function q on a complex vector space satisfying q(ix)=q(x) and q(x+y)+q(x-y)=2q(x)+2q(y) does not necessarily produce an inner product over X. However, if q(x)^{1/2} is a norm (which comes down to q(ax)=|a|^{2}q(x)), then a positive-definite q is generated by an inner-product. Or, equivalently, if t -> q(x+ty) is measurable for all x,y in X, then q is generated by an inner product.

Comment by Trent | June 15, 2012 | Reply