# Is $f(x)f(y)=f(x+y)$ enough to determin $f$? [duplicate]

I had a discussion with a friend and there it came up the question whether $f(x)f(y)=f(x+y)$, $f(0)=1$ and the existence of $f'(x)$ implies that $f(x)=\exp(a x)$. This seems very reasonable but I cannot figure this out. Can differentiability be pushed to smoothness in this case?

In a similar manner on would expect that $g(x)+g(y)=g(xy)$ and $g(1)=0$ would imply that $g(x)=\log(a x)$.

Any ideas?

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## marked as duplicate by user17762, ShreevatsaR, Pedro Tamaroff♦, Asaf Karagila, InquestApr 29 '13 at 2:02

$f(x)(fy) = f(x+y)$ gives you $f(x) = f(1)^x$ for rational $x$. Then continuity at any one point, or even just measurability, will tell you that it's true at all $x$. Else, there "exist" (assuming Axiom of Choice) discontinuous such functions, though one can't be written down explicitly. See e.g. math.mit.edu/~stevenj/exponential.pdf – ShreevatsaR Apr 29 '13 at 2:00
Oh, right. This was rather easy. – rom Apr 29 '13 at 2:05

Write $f(1) = a > 0$. Then $f(2) = a^2$ and so on. So for positive integers $n$, we have $f(n) = a^n$.

Note that $f(0) = f(1 + (-1)) = f(1)f(-1)$, so $f(-1) = a^{-1}$. Again by induction $f(-n) = a^{-n}$.

Now move on to rationals. Let $p/q \in \mathbb{Q}$ be in lowest terms. We have $f(p/q) = f(1/q + \cdots + 1/q) = f(1/q)^p$. Also, $f(1) = f(q/q) = f(1/q)^q = a$. So $f(1/q) = a^{1/q}$ and $f(p/q) = a^{p/q}$.

You demand that this function be differentiable and therefore continuous. To define $f$ on irrational numbers, pick any approximating sequence of rationals $\{c_i\}$ and examine the sequence $f(c_i) = a^{c_i}$. We already know that the exponential function is continuous, so $f(c) = a^{\lim c_i} = a^c$. (If you've never done this before, you might want to prove that any approximating sequence will do.)

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Let $f(1)=a$. We have $f(2)=f(1)^2=a^2$, $f(3)=f(2+1)=f(2)f(1)=a^3$, etc. But now $f(\frac 12)^2=f(1)$ so $f(\frac{1}{2})=a^{1/2}$. $f(\frac{1}{4})^2=f(\frac{1}{2})=a^{1/2}$, so $f(\frac{1}{4})=a^{1/4}$. It takes some induction-type argument, but we can get $f(x)=a^x$ for any rational $x$ this way. To get irrational $x$ we use continuity of $f(x)$ (we don't even need differentiability).

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