Obviously every exponential function has a constant time $\tau$ such that $f(x+\tau) = 2f(x)$ for all $x$. I am interested in the reverse problem: Let $f: \mathbb{R} \to \mathbb{R}$ be a function which has a constant doubling time (WLOG this can be $\tau=1$), i.e. the property that:
$$\forall x \quad f(x+1) = 2f(x)$$
Must $f$ be an exponential function? i.e. in this case, must $f(x) \propto 2^x$? With no further conditions, the answer is clearly no: If $h: [0,1] \to \mathbb{R}$ is any function at all, then the function $f(x) = 2^{\lfloor x \rfloor}h(\{x\})$ has the desired property. If $h(0) = h(1) = 0$, then the function is even continuous. I guess by using an appropriate bump function, we can even ensure that $f$ is smooth.
I wonder whether either of the following assumptions ensures that $f$ must be an exponential function?
- $f$ is continuous but $f(x)$ is never 0.
- $f$ is analytic.
I didn't get anywhere considering either of these assumptions.
Best Answer
No, neither assumption is enough. Taking logs, we get
$$\log(f(x+1)) - \log(f(x)) = \log(2)$$
Letting $L(x) = \log(f(x)) + \log(2) x$, we have
$$\forall x\quad L(x+1) = L(x)$$
So if $L$ is any function with period 1, then $f(x) = e^{L(x)}2^{x}$ solves the equation. Moreover, if $L$ is analytic then so is $f$. In particular, say, if $L(x) = \sin(2\pi x)$ then
$$f(x) = e^{\sin(2\pi x)}2^{x}$$
is an analytic function with constant doubling time.
Edit Even simpler argument: Just let $f'(x) = f(x)/2^{x}$. Then $f$ has doubling time 1 if and only if $f'$ is periodic with period 1. So this characterises all examples.