Cauchy's Functional Equation with Extra Conditions

The following equaton is called Cauchy's functional equation:

$$f(x+y) = f(x)+f(y)$$

Functions that follow this equation are also called additive functions.

Solving Cauchy's Functional Equation

Solving a functional equation means finding all the families of functions satisfying that equation. These functions are also called the solutions of that equation. While solving functional equations, often it's useful to guess the solutions beforehand.

For now suppose $f: \mathbb N \to \mathbb N$. So can you guess what the solutions can be? Well, an obvious solution is $f(x)=x$. And you can very easily see $f(x) = cx$, for any positive integer $c$, also satisfies this equation. Are there any other solutions? After a bit of thinking (do it yourself :p ) we find no more elementery functions satisfying this equation. So maybe $f(x)= cx$ is the only solution to this function.

Now it's time to take down this functional equation.

For brevity, suppose $P(x,y)\implies f(x+y)=f(x)+f(y)$

It's always helpful to find out small values like $f(1), f(2)$ or $f(0)$ (not present in this case). So we're gonna find out $f(2)$ from this equation.

$P(1,1)\implies f(1+1)=f(1)+f(1) = 2f(1)$

$$\begin{align*}P(2,1)\implies f(3) &= f(2)+f(1)\\ &= 2f(1)+f(1)\quad \text{(from the last equation)}\\ &= 3f(1)\end{align*}$$

Now can we find $f(4)$? Yeah, just like before

$P(3,1)\implies f(4) = f(3)+f(1) = 4f(1)$

So, if we know $f(n)=nf(1)$, we can find $f(n+1)$:

$P(n,1) \implies f(n+1)= f(n)+f(1)=(n+1)f(1)$

So, by induction, $f(m) = mf(1)$ for all positive integer $m$.

Exercise: Generalize this solution when the function is $f: \mathbb Z \to \mathbb Z$.

When $f:\mathbb Q\to \mathbb Q$

Now again we can deduce $f(x) = cx$ for any integer $x$ and rational $c$. Also note that, $$P\left (\frac 1 n, \frac 1 n \right) \implies f\left ( \frac 2 n\right ) = 2f\left ( \frac 1 n\right )$$ And following the inductve argument of the previous section, we see

$$f\left ( \frac m n\right ) = m f\left ( \frac 1 n\right ) \text { for all } m,n\in \mathbb Z, n\neq 0.$$

Setting $m=n$ yields $$f(1) = nf\left ( \frac 1 n\right )\Longleftrightarrow f\left ( \frac 1 n\right ) = \frac 1 n \cdot f(1)$$ Therefore we can conclude,

$$f\left ( \frac m n\right )=mf\left ( \frac 1 n\right )=\frac m n \cdot f(1)$$

When $f:\mathbb R\to \mathbb R$

Well, now we can't come down to $f(x) = cx\;\forall x\in \mathbb R$ as easily as before. In fact, assuming Axiom of Choice, it's not even possible. We need other restrictions for that. Some of them are:

1. $f$ is continuous at at least one point.

2. $f$ is monotone

3. $f$ is multiplicative

4. $f$ is bounded at either above or below.

Exercise: Can we say $f(x) = cx$ if-

1. $f:\mathbb R_+\to \mathbb R$?

2. $f:\mathbb R\to \mathbb R_+$?

Quick introduction: It is known that Cauchy's functional equation,

$$f(x+y)= f(x) + f(y)$$ from reals to reals has $f(x)=cx$ as a set of solutions for any real constant $c$.

Let's attempt (but as you may guess from the title, this attempt does not succeed) to prove this forms the set of all solutions:

First, let $f(1)=c$. Obviously $f(0) = f(0) + f(0) = 0$. Then $f(2) = f(1)+f(1)=2c$, and so on. Thus we conclude $f(n)=nc$ for nonnegative integers $n$.

We can see $c=f(1) = f(0.5) + f(0.5) =2f(0.5)$ so that $f(0.5) = \dfrac{c}{2}$.

By similar methods, we can easily obtain $$f \left( \dfrac{p}{q} \right) = \dfrac{cp}{q}$$

However- this only works when $p,q$ are positive integers. Essentially, we have shown $f$ is linear on the rational domain. For example, we can't say

$$f(\sqrt{2} ) = f(1) + f(1) + ... + f(1) = \sqrt{2} f(1)$$ since $\sqrt{2}$ is irrational.

Through some advanced techniques, however, we can find some messy and not-so-obvious solutions to the equation. (If you want more on this, see Cauchy's Functional Equation )

So now the question is, can we impose conditions on $f$ which force $f(x)=cx$ to be the only solutions?

Indeed, we can! Let's list some common conditions for functions- and I'll assume $c>0$ because $c<0$ uses the exact same arguments and conditions, only flipped.

In fact, all of these conditions independently force $f$ to be linear. Let's see why:

Constraint $(1)$ is obvious.

For constraint $(2)$, since $f$ is a polynomial, suppose it has degree $d$. We know that $f(1)=c,f(2)=2c,...f(d)=cd, f(d+1) = c(d+1).$. Thus, since each polynomial of degree $d$ is determined by $d+1$ points, $f(x)$ must be identically $cx$.

Now we move on to the not-so-obvious conditions... But before we do this, let's prove a lemma regarding rational numbers. Since we know for all rational $r>0$, $f(r) = cr$, it makes sense for the set of rationals to be our "starting point" to prove $f(x) =c x$ for all $x$.

Lemma Suppose $a,b$ are real numbers with $a>b$. Then there is a rational number $r$ with $a>r>b$.

Essentially, take $r \approx \dfrac{a+b}{2}$. This clearly satisfies the conditions. All we need to do is consider the decimal representation of $\frac{a+b}{2}$ and terminate it after a very very long time. (If you want to rigorify this, let $d$ be the position at which the decimal representations of $a,b$ differ, and choose the $d+5$th to cut off $\frac{a+b}{2}$. )

Now that we're done with that... we want to show that $f(x)=cx$ if $f$ is continuous.

Essentially, the lemma tells us that we can find rational numbers arbitrarily close to any given real number on either side. For example, take $a = b + 10^{-1000}$ and the lemma guarantees a rational number very very close to $a,b$.

Now we take advantage of the continuity- let $r_1$ be a rational approaching $x$. By continuity, essentially we have

$f(x)$ is close to $f(r_1) = cr_1$. Taking the limit as $r_1 \to x$, we know $f(x) = cx$. Intuitively, we are squishing the value of $f(x)$ by using inputs whose values we already know which are close to $x$.

Let $r_1>x>r_2$ where $r_1,r_2$ are rational numbers near $x$. Then by the increasing condition, this implies

$$cr_1 = f(r_1) > f(x) > f(r_2) = cr_2$$

Once again, take the limit as $r_1, r_2$ approach $x$ to see that $f(x) = cx$.

The final condition, luckily, reduces to the one we just did. Indeed, since $f(x+y) = f(x) + f(y)$ and $f(y)>0$ we know that $f(x+y) > f(x)$, so $f$ is increasing. Thus we are done.

Conclusion Although not all the solutions to the functional equation $f(x+y) = f(x) + f(y)$ are given by $f(x)=cx$, we may impose certain constraints on $f$ to force $f$ to be linear. These include, but are not limited to, $f$ being a polynomial, increasing, continuous, or positive on certain domains. Indeed, you may wish to try to find an original condition which implies linearity!

Note 1: For the proofs of constraints 3,4,5 we have essentially used the fact that the rational numbers are dense. You can read about (mathematical) density elsewhere for more information.

Note 2: If you want a fun problem to try to solve with this information, try the following: Let $f$ be a function from the reals to the reals with $f(x)f(y)=f(xy), f(x)+f(y) = f(x+y)$. Find all $f$.

Show that $f$ is positive on the positive reals.

In particular, $f(x)^2 = f(x^2)$. What can you conclude?