Recording of a talk I gave on this topic

The main idea that if you want to show a property is preserved under multiplication, try showing that the property is preserved under linear combinations and squaring. We could then use the identity to establish preservation under multiplication.

The main tool to achieve is the polarisation identity $$ab = \frac{1}{4}[(a+b)^2 - (a-b)^2]$$ which could be derived from $$a^2 - b^2 = (a+b)(a-b)$$

## Algebra of Limits

Suppose you were asked to show that product of two convergent sequences are convergent. One could first show that square of convergent sequences are convergent.

Proposition: If $$(a_n)$$ is convergent to L, then $$(a_n^2)$$ is convergent to $$L^2$$.

Proof: Take $$\epsilon > 0$$. We may assume that $$\epsilon < 1$$.

Suppose $$a_n \rightarrow L$$, then by definition $$\exists \, N \in \mathbb{Z}$$ such that if $$n \geq N$$ then $$\|a_n - L\| < \epsilon$$. We have,

\begin{align*} \|a_n^2 - L^2\| &= \|a_n + L\| \cdot \|a_n - L\| \\ &\leq (\|a_n\| + \|L\|) \cdot \|a_n - L\| \\ &\leq (2\|L\| + \epsilon) \cdot \epsilon \\ &\leq (2\|L\| + 1) \cdot \epsilon \end{align*}

As $$(2\|L\|+1)$$ is constant, this is enough to show that $$(a_n^2)$$ is convergent.

$\blacksquare$

We then use the identity $$a_n b_n = \frac{1}{4}(a_n+b_n)^2 - \frac{1}{4}(a_n-b_n)^2$$ and that sum of convergent sequences are convergent to the sum of their limits, to deduce that product of convergent sequences converges to the product of their limits.

## Product Rule

To prove the product rule, one could first show that squares of differentiable functions are differentiable.

Proposition: If $$f$$ if differentiable, then $$f^2$$ is differentiable and its derivative equal to $$2f \cdot f'$$.

Proof:

\begin{align*}\lim_{x \rightarrow x_0} \dfrac{f(x)^2 - f(x_0)^2}{x-x_0} &=\lim_{x \rightarrow x_0} \bigg[\big(f(x) +f(x_0)\big) \dfrac{f(x) - f(x_0)}{x-x_0} \bigg]\\ &= 2 f(x_0) \cdot f'(x_0) \end{align*} $\blacksquare$

Using the identity $$fg = \frac{1}{4}(f+g)^2 - \frac{1}{4}(f-g)^2$$ and that sum of differentiable functions are differentiable with derivative equal to the sum of the derivatives, we deduce

\begin{align*}(fg)' &= \dfrac{1}{4}\bigg\{\big[(f+g)^2\big]^{'} - \big[(f-g)^2\big]^{'} \bigg\} \\ &= \dfrac{1}{2}\bigg[(f+g)(f'+g') - (f-g)(f'-g')\bigg] \\ &= f'g + fg' \;\end{align*}

We could use similar methods to deduce $$\nabla (fg) = f \nabla g + g \nabla f$$ from $$\nabla f^2 = 2 f \nabla f$$. I find the later to be much easier to grasp geometrically.

## Integration

To show that the product of two Riemann integrable functions is integrable, one could

1. Show that if $$f$$ is integrable, then $$f^2$$ is integrable.

2. Use the identity to show that if $$f, g$$ is integrable then $$fg = \frac{1}{4}(f+g)^2 - \frac{1}{4}(f-g)^2$$ is integrable

For the full argument, you could refer to the answer by Philipp at StackExchange. This argument similarly works for measurable functions.

## Abstraction

For all of the examples above, we could generalise the idea as follows. We would use a variation of the identity instead, namely $$ab = \frac{1}{2}((a+b)^2 - a^2 - b^2 )$$

Proposition: Let $$V$$ be a vector space over some field $$F$$ with characteristic greater than 2. Let there be a commutative bilinear product $$V \times V \rightarrow V$$ denoted by $$\times \,$$. Let $$P$$ be a subspace of $$V$$. Then $$v \times v \in P \; \forall v \in P$$ is equivalent to $$v \times w \in P \; \forall v, w \in P$$

Proof: $$\impliedby:$$ Suppose $v, w \in P$, then $v+w , v-w \in P$ so $(v+w)\times(v+w)- v\times v - w\times w \in P$ by linearity and assumption of $P$. By commutativity of $\times$ we deduce that $(1_F + 1_F) v \times w \in P$. As the characteristic of $F$ is greater than 2, $(1_F + 1_F)^{-1}$ exists and is in $F$ so $v \times w \in P$.

$$\implies:$$ Immediate

$\blacksquare$

We could let $$P$$ be the set of vectors that satisfy the desired property. For example, in the case of algebra of limits, $$V$$ is the vector space of real sequences and $$P$$ is the subspace of convergent sequences.

This means that under sufficient conditions, preservation of properties under multiplication is equivalent to preservation of properties under squaring.

We know that bilinear forms can be solely determined by the associated quadratic forms using the identity. I.e.

$B(v,w) = \frac{1}{4} (B(v+w, v+w) - B(v-w, v-w))$

Hence, if $$V$$ is the space of differentiable functions from $$R$$ to $$R$$, for any real number $$c$$ we could define the bilinear form

$B_c(f,g) = \dfrac{d}{dx}\bigg\vert_{x=c}(fg)$

I believe much of the theory above could be viewed from this perspective, although that’s for another time.

## Geometry

The abstraction doesn’t all possible cases the idea could be used. Consider an example in geometry:

A map $$T$$ from $$\mathbb{R}^n$$ to $$\mathbb{R}^n$$ is called an isometry if $$\|T(v) -T(w) \| = \|v-w\|\, \forall v, w \in \mathbb{R}^n$$. Consider isometries $$T$$ with an additional condition that they map the zero vector to the zero vector. I.e. $$T(0) = 0$$

To show $$T(u) \cdot T(v) = u \cdot v$$, one could

1. Show that $$T(u) \cdot T(u) = u \cdot u \, \forall u \in \mathbb{R}^n$$

2. Use $$u \cdot v = \frac{1}{4}(u+v) \cdot (u+v) - \frac{1}{4}(u-v) \cdot (u-v)$$

## Multi-variable calculus

One can prove this rather complicated identity

$\nabla(\mathbf{u} \cdot \mathbf{v}) = (\mathbf{u} \cdot \nabla)\mathbf{v} + \mathbf{u} \times (\nabla \times \mathbf{v}) + (\mathbf{v} \cdot \nabla)\mathbf{u} + \mathbf{v} \times (\nabla \times \mathbf{u})$

from proving

$\nabla (\mathbf{u} \cdot \mathbf{u}) = 2[(\mathbf{u} \cdot \nabla)\mathbf{u} + \mathbf{u} \times (\nabla \times \mathbf{u})]$

alone. As an aside, you could prove these identities nicely using Levi-Civita symbols which I’ve written about here.

## Takeaway

The takeaway is that we can often transform facts about powers (squares) into facts about products. We could see this more generally through the Cauchy-Schwarz Inequality: $$\vert\braket{\mathbf{u},\mathbf{v}}\vert^2 \leq \braket{\mathbf{u},\mathbf{u}} \braket{\mathbf{v},\mathbf{v}}$$
which gives you a bound on products by “squares”. As a side note, Terry Tao and Timothy Gowers have written very nice articles on the inequality.

One could also view this as “spending the symmetry” of $$P(x), P(y) \implies P(x+y)$$ to prove $$P(x), P(y) \implies P(xy)$$ where $$P$$ is some property for any object $$x,y$$ in some class $$X$$.