"× to ×" Maps: f(xy) = f(x)f(y)

(Quick note: If anyone can edit this and help me add spacing so this looks prettier I would appreciate it. Unfortunately I'm not too good with $LaTeX$...)

The most basic multiplicative functions, as you may guess from the title, are functions satisfying $f(xy)=f(x)f(y)$ for all $x,y$ in the domain of $f$.

In this post, I will focus on when the domain of $f$ is the set of real numbers (as this is categorized as an algebra topic...). The topic of multiplicative functions in number theory is quite interesting as well, and you can see this link for multiplicative functions in number theory.

First, a natural question to ask is

Well, we can try to prove $f(x)=x^c$ and $f(x)=0$ are the only solutions. First, we can attempt to plug in convenient values of $x,y$:

$x=0 \implies f(0)=f(0)f(y)$. So either $f(0)=0$, or $f(y)=0$ for all $y$. We may discard the second case. Now we know $f(0)=0$.

We can say something else about zeroes of $f$- Suppose $f(d)=0$ for any $d \neq 0$. Then we know $f(d)f(\frac{x}{d})=f(x)$ for all $x$, so $0=f(x)$, which means $f$ is identically zero. Thus we can assume that the only zero of $f$ is $x=0$.

Now $x=1 \implies f(y)=f(1)f(y)$. Since $f$ is not identically zero, we can choose a value $y$ with $f(y)\neq 0$ and then dividing by $f(y)$ yields $f(1)=1$.

We can try finding $f(2)$ now... but it's easy to see that we immediately run into problems. All we know are the values of $f(0), f(1)$, and no multiplication of zeroes and ones could possibly give a value of $f(2)$.

We can set $f(2)=d$ for any value of $d$ (but recall $d\neq 0$ from our earlier comments). Then $f(2^2)=f(2)f(2)=d^2, f(2^3)=f(2^2)f(2)=d^3$, and so on... So we conclude $f(2^n)=d^n$ for positive integers $n$.

Exercise: Show that this holds for negative integers $n$ as well.

Now, repeating an argument that may seem familiar, we can try to find $f(2^{\frac{p}{q}})$ for $p,q\neq 0$.

In fact, note that from the property $f(xy)=f(x)f(y)$, we can generate a more extended result: $f(x_1x_2...x_n)=f(x_1)f(x_2)...f(x_n)$ for all $x_1,x_2,...x_n \in \mathbb R$!

We can use this to our advantage- Let $n=q$ and $x_i = 2^{\frac{p}{q}}$ for each $i$. Then we easily obtain $f(2^{\frac{p}{q}} )^q = f(2^p) = d^p$, implying $f(2^{\frac{p}{q}} ) = d^{\frac{p}{q}}$.

Thus we have found all values $f$ when the input is $2^q$ for a rational $q$.

Unfortunately, this is where we start to get some problems- now we are actually out of things to plug in , because any combination of rational powers of two will yield another rational power of two! We can try setting $f(3)=e$ for a constant $e$, and get the same results, but then we'll be stuck again...

So we can try to find some similarities between multiplicative and additive functions- they are, after all, quite similar.

Well, we can first note that $f(x^2)=f(x)^2 >0$ for all values $x \neq 0$. Thus we can define a new function $g: \mathbb R^+ \to \mathbb R$ with $g(x) = \log f(x)$. Then $g(xy) = g(x)+g(y)$.

This still isn't quite additive, although we've made some progress. Since $x,y>0$, we can try the substitution $x=e^X, y=e^Y$ to obtain $g(e^{X+Y}) = g(e^X)+g(e^Y)$.

Ah-ha! Now we can define a new function in terms of $g$- it's staring us in the face!

The substitution $h(x)=g(e^x)$ brings us to $h(x+y)=h(x)+h(y)$. Now notice that $x,y$ can take on all real values, and $h$ can take on any real value- it's the log of another real number.

So, by this explanation on Solving Cauchy's Functional Equation with extra conditions, we can force $h$ to be linear if any of the following hold-

1. $h$ is continuous

2. $h$ is increasing

3. $h$ is positive on the positive reals.

(A proof may be found in the link I gave above. Since it's another expii topic, the proof will be omitted from here.)

Recall $h(x)=g(e^x)$. So what do these conditions tell us about $g$?

All three of those answers are sufficient conditions for $g$.

Noting $g(x)=\log f(x)$, we can obtain another set of conditions for our original $f$. This isn't exactly super exciting so I'll just list them below-

1'') $f$ is continuous

2'') $f$ is increasing

3'') $f$ is $> 1$ on the interval $(1, \infty)$.

Note that by our domain restriction on $g$, we only require conditions 1'', 2'' to hold on the positive reals. (In other words, $f$ does not have to be completely increasing).

Any of these three conditions holding will imply $h(x)=cx \implies g(x) = c \log x =\log x^c \implies \boxed{f(x) = e^{g(x)} = x^c}$ for a constant $c$, as desired.

(Note that choosing $c=\log_2 d$ from our earlier assignment $d=f(2)$ will give an agreeing function.)

In conclusion- in this post, we have shown that given a certain set of conditions, any multiplicative function must be either zero or of the form $x^c$. If these conditions are absent, strange things can happen... See exercise 3 below for more on this!

Now, with your knowledge of multiplicative functions on the reals, you can try solving some functional equations and exploring a bit more!

1. Find all functions $f: \mathbb R \to \mathbb R$ with $f(xy)=f(x)f(y)$, $f(xy) \ge \text{max} (x,y)f(\text{min} (x,y))$ for all $x,y \in \mathbb R$.

2. Find all functions $f:\mathbb R \to \mathbb R$ with $f(xy)=f(x)f(y), f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb R$.

3. Read about some non-linear "monster" solutions to Cauchy's Functional Equation in this blog post and see if you can construct a similar multiplicative function.