Separation and Decomposition into Pieces

Separation on Functions

Separation is considered one of those "troll" methods that if it is applicable in a problem then it will surely insta-kill it.

I'll give an example and then dissect the steps. Here is a made up problem:

Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for all real $x$ and $y$, $x,y \ne 0$ we have that

$$3x^2f(y) - 3y^2f(x)=3xy(x-y)$$

Solution:

We expand the right side to obtain that $3x^2f(y)-3y^2f(x) = 3x^2y-3xy^2$

Rearranging we get that $3x^2f(y) - 3x^2y= 3y^2f(x) - 3xy^2$

We factor each side to get $3x^2(f(y)-y) = 3y^2(f(x)-x)$

Thus $\dfrac{f(x)-x}{3x^2} =\dfrac{f(y)-y}{3y^2}$

Define a $g(x)$ such that $g(x) = \dfrac{f(x)-x}{3x^2}$. We have that $g(x) = g(y)$ for all real $x$ and $y$. Thus, $g(x)$ must be constant.

Let $g(x)=c$ where $c$ is a real constant. Then $\frac{f(x)-x}{3x^2}=c$

Thus we obtain that $f(x) = 3x^2c + x$ where $c$ is a real constant.

We plug this back into the original equation and get $3x^2(3y^2c+y) - 3y^2(3x^2c+x) = 9x^2y^2c+3x^2y-9x^2y^2c-3xy^2 = 3x^2y-3xy^2 = 3xy(x-y)$ as desired.

There are a few key things to note in this solution.

One of those things is the fact that if $g(x) = g(y)$ then $g(x)$ must be a constant function. This is very intuitive but there are many ways to explain this result. I shall leave finding one of these ways as an exercise for the reader.

Also some may question "what happens if it was $\dfrac{3x^2}{f(x)-x} = g(x)$ and we set $g(x)=c$?"

Well, I won't do it out, but you would get $f(x) = 3x^2*\frac{1}{c}+x$. But since $\frac{1}{c}$ is also a real constant, we can simply replace $\frac{1}{c}$ with another real constant such as $d$. We note that $d$ because if $d=0$ then $f(x) = 3x^2d + x = x$ thus making $f(x)-x = 0$ meaning that $\dfrac{3x^2}{f(x)-x}$ undefined.

And that is the process of separation. It's a lot like rearranging in the fact that you move like terms to one side. The Motivation to when you should use this technique is when you see some sort of symmetry among the terms of a functional equation such as the $3x^2$ and $3y^2$. To be honest most of the time this technique won't work but it's never a bad idea to give it a try.

Here is a practice problem that actually appeared on a Saudi Arabian test that uses this technique:

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real $x$ and $y$, we have

$$yf(2x)-xf(2y) = 8xy(x^2-y^2)$$

If you want an answer/solution then the best way to contact me is on AoPS where my username is efang.