Extended reading: functional equations

An equation whose unknown is a function

In algebra, you solved equations such as $2x + 3 = 11$ for a number. A functional equation asks for a whole function. For example:

We do not look for one number. We look for every function $f$ that makes this true for all $x$ and $y$.

Four tools

1. Try special values. Put $x = 0$, $y = 0$, $x = 1$, or $y = -x$. The equation becomes simpler, and it often gives a fact such as $f(0) = 0$.
2. Replace a variable. If the equation has $f(1 - x)$, replace $x$ by $1 - x$. You get a second equation, and you solve the two like simultaneous equations.
3. Use one-one and onto. If $f(f(x)) = x$, then $f$ is one-one and onto, and it is its own inverse.
4. Guess, then check. A guess is not a proof. Put the final answer back in the original equation and see that it works for all $x$.

Example 1: the additive equation

Let $f(x + y) = f(x) + f(y)$ for all real $x$ and $y$, and let $f(1) = c$.

• Put $x = y = 0$: $f(0) = 2f(0)$, so $f(0) = 0$.
• Put $y = -x$: $f(0) = f(x) + f(-x)$, so $f(-x) = -f(x)$. The function is odd.
• Put $y = x$: $f(2x) = 2f(x)$. By repeating, $f(nx) = n\,f(x)$ for every positive integer $n$.
• So $f(n) = nc$. Also $n\,f\!\left(\dfrac{1}{n}\right) = f(1) = c$, so $f\!\left(\dfrac{1}{n}\right) = \dfrac{c}{n}$.
• Then for every rational number $\dfrac{p}{q}$: $f\!\left(\dfrac{p}{q}\right) = p\,f\!\left(\dfrac{1}{q}\right) = \dfrac{p}{q}\,c$.

On the rational numbers, then, $f(x) = cx$. For all the real numbers, mathematicians know that the equation alone is not enough. Strange solutions exist that are not lines. If we also say that $f$ is increasing (or continuous), then $f(x) = cx$ for every real $x$. In problems, the condition is given.

Example 2: a second equation from a replacement

Solve $f(x) + 2f(1 - x) = x^2$ for all real $x$.

Replace $x$ by $1 - x$. We get a second equation:

Now there are two equations with two unknowns, $f(x)$ and $f(1 - x)$. Multiply the second by $2$ and subtract the first:

Check. $f(1 - x) = \dfrac{(1 - x)^2 - 4(1 - x) + 2}{3} = \dfrac{x^2 + 2x - 1}{3}$. Then $f(x) + 2f(1 - x) = \dfrac{x^2 - 4x + 2 + 2x^2 + 4x - 2}{3} = x^2$. It works.

Example 3: a product equation

Let $f(xy) = f(x)\,f(y)$ for all real $x$ and $y$.

• Put $x = y = 1$: $f(1) = f(1)^2$, so $f(1) = 0$ or $f(1) = 1$.
• If $f(1) = 0$, then $f(x) = f(x \cdot 1) = f(x)\,f(1) = 0$ for every $x$. This is the zero function.
• If $f(1) = 1$, there are many solutions: $f(x) = x$, $f(x) = x^2$, $f(x) = |x|$, $f(x) = x^3$ and the constant function $1$ all work.

A functional equation can have no solution, one solution or many. The work is to find all of them, and to be sure that nothing else is possible.

Keep $c = 0$ and change $k$: every such line passes the test. Any $c \neq 0$ breaks it by exactly $c$.

Practise with new numbers

Try it in Python

Python cannot prove that a function solves an equation for all $x$ and $y$, but it can test many pairs. One pair that fails is enough to reject a function. This program tests five functions for two equations.

Remember these

• Try $0$, $1$ and $-x$ first. Replace a variable to get a second equation.
• For $f(x + y) = f(x) + f(y)$: $f(0) = 0$, $f$ is odd, and $f(p/q) = (p/q)\,f(1)$.
• A guess must be checked in the original equation.
• An equation may have no solution, one, or many.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. $f(x + y) = f(x) + f(y)$, $f$ is increasing, and $f(1) = 5$. Find $f\!\left(-\dfrac{3}{2}\right)$.

2. With $f(x) + 2f(1 - x) = x^2$, find $f(1)$.

3. Does $f(x) = x + 1$ satisfy $f(x + y) = f(x) + f(y)$?

4. $f(xy) = f(x) + f(y)$ for all positive $x$ and $y$. What is $f(1)$?