I will write a bit about the history of functional equations, their importance in mathematics and discuss the well-known Cauchy’s functional equation and some related solutions. We will also explore a few more functional equations that are somewhat related to Cauchy’s.

But, let us first discuss Cauchy’s functional equation.

$f(x+y) = f(x) + f(y). \quad (1)$

Our goal is to determine $f$. Now, notice that nothing is said about the domain or the range of $f$. So, equation $(1)$ is quite general. Its solution will vary depending on what domain/range we pick for $f$. So, let’s see how much information we can “extract” from $(1)$ without too many additional assumptions. (We will be using purely elementary methods here.)

Our first assumption is as follows.

$(i) \quad f: \mathbb{Z} \to \mathbb{R}$.

Here, we restrict our domain to the set of integers and the co-domain to the set of reals. Setting $y = 0$ yields $f(x) = f(x) + f(0)$, which implies $f(0) = 0.$ And, setting $y = -x$, we obtain $f(0) = f(x) + f(-x)$, or $f(-x) = -f(x)$, which, of course, means $f$ is an odd function. It also means that it is enough to determine $f$ for $x > 0.$

Now, notice that putting $y = x$, we get $f(2x) = 2f(x)$. And, putting $y=2x$, we get $f(3x) = f(x) + f(2x) = f(x) + 2f(x) = 3f(x).$ Similarly, $f(4x) = 4f(x)$. The pattern is now evidently clear. A simple induction shows that

$f(nx) = nf(x)$ for all $n \in \mathbb{N}$.

And, in particular, if $f(1)$ is some real, then we can write our solution as

$f(n) = nf(1)$ for all $n \in \mathbb{Z}$.

The above, therefore, is our solution to Cauchy’s functional equation under the stated assumption.

Now, what if we tweak our assumption a little bit? So, here’s our second assumption.

$(ii) \quad f: \mathbb{Q} \to \mathbb{R}$.

It turns out that the solution in this case is exactly the same as in $(i)$, i.e. the solution is

$f(x) = xf(1)$ for all $x \in \mathbb{Q}$.

Here’s why. Suppose $x = m/n$ is some non-zero rational number, where $m, n \in \mathbb{Z}$. Then, $nx = m$, which implies $f(nx) = f(m)$, or $nf(x) = mf(1)$, which yields $f(x) = (m/n) f(1) = xf(1)$. And, we are done.

So, we conclude that if the domain of $f$ is the set of rationals (or integers), then the solution is simply $f(x) = xf(1)$, where $f(1)$ is some real number. Here, we assume that the co-domain is $\mathbb{R}$. Now, it is natural to ask, “Can this be true if the domain is the set of reals?”. It turns out that the answer is NO. To extend the result to the set of reals, we need some additional assumption. In fact, $f(x) = xf(1)$ for all $x \in \mathbb{R}$ if we assume any one of the following:

• $f$ is continuous (which was weakened by Darboux who proved that continuity of $f$ at a single point is sufficient.)
• $f$ is monotonic on some interval.
• $f$ is bounded on some interval.

I will discuss the above in my next post.