Functions that satisfy $f(x+y)=f(x)f(y)$ and $f(1)=e$

Solution 1:

First let's make things slightly simpler. Note that if $f(x+y)=f(x)f(y)$, then $\log f(x+y)=\log f(x)+\log f(y)$. So it sufficient to talk about $g(x+y)=g(x)+g(y)$. Moreover, $f$ is continuous if and only if $g$ is continuous.

Why is this simpler? Because now it's not difficult to show that $g$ is a linear function from $\Bbb R$ to itself, as a vector space over $\Bbb Q$.

We can show, in fact, that if $g$ is continuous, or measurable, or Baire measurable, then $g(x)=ax$, which means that $f(1)=e^a$, and for $a=1$ we simply get $e^x$.

But what about a discontinuous solution? That's harder to come up by. The reason is that it is consistent with the failure of the axiom of choice that all solutions are continuous. In which case, there are no other functions satisfying the first two conditions except $e^x$.

But with the axiom of choice we can show that $\Bbb R$ has a basis, as a vector space, over $\Bbb Q$. This basis is called a Hamel basis. Moreover, this basis, as any basis would, has the property that any function from it to $\Bbb R$ has a unique extension to a linear function on $\Bbb R$. And there's even more that we can do with the axiom of choice. We can extend any set of $\Bbb Q$-linearly independent real numbers into a basis.

Combining all these facts together, we extend $\{1,e\}$ to a Hamel basis, and consider any function such that $g(1)=1, g(e)=0$. Extend this $g$ to a linear function, and finally take $f(x)=e^{g(x)}$. And this indeed satisfying the first two conditions, but $f(e)=1\neq e^e$.

Let me also add somewhat of a historical perspective on this.

How can we prove something exists without writing it down explicitly? The axiom of choice allows us to conclude that from certain assumptions there exists a function with certain properties. That is what the axiom is for, letting us make this inference. And so we can make a lot of similar inferences as a consequence.

But that's fine. You know that there is no largest natural number. But can you write one which has more than $10^{100000000}$ digits explicitly? Probably not. How do you know it exists, then? Because you have a list of axioms, and you used them to conclude that there is no largest natural number, and that if $x,y$ are natural numbers then $x^y$ is a natural number, and therefore there is a natural number with so many digits, and more.

Historically, however, the axiom of choice caused some controversy with its consequences, but nowadays it is generally considered part of mathematics, and as someone whose research is models where the axiom of choice fails, I get more strange looks than supportive looks, which should indicate how far we've come from the early 20th century.