calculus - Differentiable function satisfying $f(x+a) = bf(x)$ for all $x$

This is an exercise from Apostol Calculus, (Exercise 10 on page 269).

What can you conclude about a function which has derivative everywhere and satisfies an equation of the form

$$f(x+a) = bf(x)$$
for all $x$, where $a$ and $b$ are positive constants?

The answer in the back of the book suggests that we should conclude $f(x) = b^{x/a} g(x)$ where $g(x)$ is a periodic function with period $a$. I'm not sure how to arrive at this.

One initial step is to say, by induction,

$$f(x+a) = bf(x) \implies f(x+na) = b^n f(x)$$
for all $x$. I'm not sure what to do with this though. I'm also not clear how to use the differentiability of $f$. (If I write down the limit definition of the derivative then I end up with a term $f(x+h)$, but I cannot use the functional equation on that since the functional equation is for a fixed constant $a$.)

Answer

One trivial solution that doesn't use the differentiability of $f(x)$:

From $f(x+na)=b^nf(x)$, letting $y=x+na$, and requiring that $x \in [0,a)$ and $n = \left\lfloor \frac{y}{a} \right\rfloor$ we get the following equivalent definition of $f$:

$$f(y)=b^{\frac{y-\left(y-\left\lfloor \frac{y}{a} \right\rfloor a\right)}{a}}f\left (y- \left\lfloor \frac{y}{a} \right\rfloor a \right)$$

By letting $g(y)=b^{-\frac{\left( y-\left\lfloor \frac{y}{a} \right\rfloor a\right)}{a}}f\left(y-\left\lfloor \frac{y}{a} \right\rfloor a \right)$ noting that $g$ is periodic with a period of $a$:

$$f(y)=b^{\frac{y}{a}}g(y)$$