# About the linear functional equations: f(x+a) = bf(x) and f(ax) = bf(x).

About the linear functional equations: $$f(x + a) = bf(x)$$ and $$f(ax) = bf(x)$$, Marek Kuczma e Polyanin A.D. they got the respective solutions (http://eqworld.ipmnet.ru):

$$f(x) = g(x)b^{x/a}$$, where $$g(x) = g(x + a)$$ is an arbitrary periodic function with period $$a$$.

And

$$f(x) = g(\log x)x^{\log b/\log a}$$, where $$g(x) = g(x +\log a)$$ is an arbitrary periodic function with period loga.

By the induction method I got the particular solutions: $$f(x) = Cb^{x/a}$$ and $$f(x) = Cx^{\log b/\log a}$$, where $$C$$ is an arbitrary constant. But I did not understand how they arrived at the generic solutions with the arbitrary periodic function "$$g(x)$$"?

Would anyone have a demonstration of how they arrived at the generic solutions including the arbitrary periodic functions?

I searched all over the net and found no proof and no book accessible. Thank You.

I'll explain with a very simple example.

From the equation

$$f(x+1)=f(x)$$ you will conclude $$f(x)=c.$$

But as $$x$$ is a continuous variable it can take fractional values. Given that e.g. $$f(0)$$ and $$f(0.3)$$ are unrelated by the equation, you might very well have $$f(0)=f(1)=f(2)=\cdots=4$$ while $$f(0.3)=f(1.3)=f(2.3)=\cdots=-5$$.

In fact, $$c$$ is not a constant but a function of the fractional part of $$x$$, or if you prefer, $$f$$ is an arbitrary periodic function of period $$1$$.

For example,

$$f(x)=e^{\sin(2\pi x)}$$ or

$$f(x)=(x-\lfloor x\rfloor)^2$$ are solutions.

• Yes, I understood how the periodic function in question works in isolation. But what I did not understand is how from the equation f(x + a) = bf(x) the solution was obtained with the associated arbitrary periodic function g(x)? Note that the arbitrary periodic function multiplies a particular solution. How to demonstrate this? – JaberMac May 17 at 16:36
• @JaberMac: apply my method to your equation. – Yves Daoust May 17 at 16:51
• Sorry, I still do not understand how I can apply the method in the original equation. There are constants a and b that I can not linearize. I got something like this: f(x + na) = (b^n)f(x). Replacing t:= x + na, we have f(t) = [b^(t-x)/a] f(t-na) – JaberMac May 18 at 0:09
• But in equation f(t) = [b^(t-x)/a] f(t-na) I can not see or extract the periodic function. Help. – JaberMac May 18 at 0:20