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

Question

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 1

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 $:

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

Answer 2

The function $g$ defined by $g(x9=b^{-x/a}f(x)$ clearly satisfies $g(x+a)=g(x)$, hence the periodicity. The given differentiality is not used, except that you may use it to conclude that $g$ is differentiable.

Answer 3

$$f(x+a)=bf(x)$$

$$f'(x+a)=bf'(x)$$

Then

$$\frac{f'(x+a)}{f(x+a)}=\frac{f'(x)}{f(x)}$$

Let

$$\frac{f'(x)}{f(x)}=h(x)$$

$$f(x)=\exp\left(\int{h(x)}{dx}\right)$$

And we require :

$$h(x+a):=h(x)$$

Now from $f(x+a)=bf(x)$ :

$$\exp\left(\int^{x+a}{h(x)}{dx}\right)=b \,\, \exp\left(\int^x{h(x)}{dx}\right)$$

$$\exp\left(\int^{x+a}_x{h(x)}{dx}\right)=b$$

Now your book suggests $f(x)=b^\frac{x}{a}g(x)=e^{\frac{x \ln b}{a}}g(x)$

This gets :

$$f'(x)=\frac{\ln b}{a}e^{\frac{x \ln b}{a}}g(x)+e^{\frac{x \ln b}{a}}g'(x)$$

$$h(x)=\frac{f'(x)}{f(x)}=\frac{\ln b}{a}+\frac{g'(x)}{g(x)}$$

As $g(x)$ and $g'(x)$ are both periodic we can see that this makes $h(x)$ periodic as well. So the form we derived earlier is simply related to the form the book describes.

And equating $\frac{f'(x+a)}{f(x+a)}=\frac{f'(x)}{f(x)}$ we get

$$\frac{g'(x+a)}{g(x+a)}=\frac{g'(x)}{g(x)}$$

Which is correct if $g(x)$ has a period $a$.