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I am going back through a bunch of calculus I learned in high school and proving the stuff that they just told us was true. Along the way, I found I had to prove that if $f(x+y)=f(x)f(y)$ then $f$ is an exponential function.

I managed to do it by induction on the integers, then generalizing to the rationals by the fundamental theorem of arithmetic, then to the reals by continuity. It involved some rather ugly analysis for $f(1/2)$ because there are two possible values for the square root, which I didn't have to worry about for the other primes because, over $R$, odd powers are invertible. But this prevents my argument from being generalized to the complex numbers.

There has to be a more elegant way to get this result. I'm hoping you can point me in the right direction toward a better proof (preferably without giving the whole thing away -- the point, after all, is to develop my ingenuity).

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This is not an answer, but more a question of my own. Your function $f$ induces a morphism $\mathbf{R}\longrightarrow \mathbf{R}^\ast$ of groups. Assuming $f$ is smooth, this is a morphism of Lie groups. Maybe you can use this to get a proof that "looks" better. – Rayleigh Oct 9 '11 at 20:13
Hmmm...over the reals every nonzero number is invertible. – Mark Oct 9 '11 at 20:30
In the complex case, the (continuous nonzero) solutions to your equation have the form $f(z)=e^cz$ for some complex constant $c$, and this cannot always be written unambiguously as $a^z$ for any $a$. So you'd have to define "exponential function" to mean the former, if you want any hope of generalizing to the complex case. – Henning Makholm Oct 9 '11 at 21:14
@HenningMakholm, I suspect you mean $f(z) = e^{cz}$? – luqui Oct 9 '11 at 21:19
@luqui, yes, stupid typo (comments don't preview properly in Iceweasel 3). And too late to fix now. – Henning Makholm Oct 9 '11 at 21:21
up vote 3 down vote accepted

Here's a different proof under the stronger assumption that $f$ has a derivative at $0$, say $f'(0) = \lambda$. For any $x$, we have

$$\frac{f(x+h)-f(0+x)}{h} = \frac{f(h)-f(0)}{h} f(x)$$

Taking limit when $h$ goes to $0$, you find that $f'(x)$ exists and

$$f'(x) = \lambda f(x)$$

And setting $g(x) = f(x) e^{-\lambda x}$, one easly check that $g'(x) = 0$, so $g$ is the constant function. So

$$f(x) = f(0) e^{\lambda x}$$

To conclude, just notice that $f(0)$ is either $0$ or $1$. An interesting feature of this proof, is that is generalizes to complex variable or matrix-valued functions.

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Ooh I like that $g(x)$ trick. – luqui Oct 10 '11 at 2:36
Starting with the functional equation and deriving a differential equation also works for ln ($f(x)+f(y) = f(xy)$) and atan ($f(x+y) = (f(x) + f(y))/(1-f(x)f(y)$). – marty cohen Oct 10 '11 at 3:43

Note that either $f(x)=0$ for all $x$, or else $f(x)\gt 0$ for all $x$: indeed, since $x = \frac{1}{2}x + \frac{1}{2}x$, it follows that $f(x)$ is a square, so $f(x)\geq 0$. If $f(a)=0$ for some $a$, then $f(x) = f((x-a)+a) = f(x-a)f(a) = 0$ for all $x$.

So one possibility is the constant function $0$. I guess you can call that an exponential function...

Otherwise, composing $f$ with the natural logarithm yields an additive function $F\colon\mathbb{R}\to\mathbb{R}$ with $F(x) = \ln(f(x))$, and $$F(x+y) = \ln(f(x+y))= \ln(f(x)f(y)) = \ln(f(x)) + \ln(f(y)) = F(x)+F(y).$$

And additive function from $\mathbb{R}$ to itself must be $\mathbb{Q}$-linear (you may need to do some induction here if you are not familiar with this fact), so $F(qx) = qF(x)$ for all $q\in\mathbb{Q}$.

At this point, if $f$ is continuous, then you can conclude that $F(x) = F(1)x$ for all $x$, by approximating $x$ by rationals: if $q_n\to x$, then $F(x) = F(\lim q_n) = \lim F(q_n) = \lim q_nF(1) = xF(1)$.

Therefore, taking exponentials we have that $f(x) = \exp(F(x)) = \exp(F(1)x) = F(1)^x$, so $f(x)$ is an exponential function.

But there are other functions that satisfy the original property, just like there are non-continuous functions that satisfy Cauchy's functional equation (at least, assuming the Axiom of Choice).

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One way to approach this functional equation is through the Cauchy functional equation $$ T(x+y) = T(x)+T(y) . \tag{1} $$ Any continuous function $T: \mathbb R \to \mathbb R$ satisfying $(1)$ must be a linear function; that is, there exists $\lambda \in \mathbb R$ such that $T(x)=\lambda x$ for all $x$.

It is a bit simpler to work with $(1)$ directly (because it doesn't have the square root issues that you faced with $f$). So in this answer, I will assume that you know how to solve $(1)$. I will focus on transforming the given equation to $(1)$. Think about proving the following claims:

  1. For any $x$, $f(x) \geq 0$. Moreover, if $f(x_0)=0$ for some $x_0 \in \mathbb R$, then $f$ must be the zero function. So (discarding the trivial case where $f$ is zero), $f$ is strictly positive everywhere.

  2. Since $f(x) > 0$, the function $g$ defined by $g(x) = \ln f(x)$ is well-defined. Can you transform the functional equation that you have into an equation for $g$? (A big hint: Why did I mention $(1)$? :))

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