Question

$a \in \mathbb{R}$

$f:\mathbb{R} \rightarrow \mathbb{R}$

$g:\mathbb{R} \rightarrow \mathbb{R}$

For generic functions $f$ and $g$, how isolate $f(x)$ in the equation below?

$f(x+a)=f(x)+a\times g(x)$

I tried to use Fourier Transform and Inverse Fourier Transform but looks like this don't work very well.

$f(x - a)=$ $e^{-2\pi i a \xi} \hat{f}(\xi)$

$\hat{f}(\xi)=$ $\int_{-\infty}^{\infty}f(x) e^{-2\pi i x\xi}\, dx \quad$ (Fourier Transform)

I tried ZTransform too, but again, didn't worked very well.

Answer 1

If you're not concerned with continuity, integrability, or any of the niceties of real analysis (and there's nothing in the problem that says you are), then, given $a\ne 0$ and $g$, you can let $f$ be any function on the interval $[0,|a|)$ and simply extend it to all real numbers by repeated applications of the given functional equation (written in the form $f(x) = f(x+a) - ag(x)$ to extend it in the direction opposite to the sign of $a$). If $a=0$, it's clear $f$ can be anything.

Addition: Let me be somewhat more explicit. If $a=1$, we can let $f$ be identically zero on $[0,1)$ and equal to $\sum_{k=1}^{[x]} g(x-k)$ for $x\ge1$, with a similar formula for $x<0$.

Answer 2

The answer is not unique - you can add any function of period $a$ to f. This is what the singularities are trying to tell you. In a distributional sense, the Fourier transform of a function of period a is supported exactly on the zeros of $e^{2 \pi i a \xi}-1$.

Answer 3

Just as you said:$e^{2\pi i a \xi} \hat{f}(\xi)= \hat{f}(\xi) +a \hat{g}(\xi)$. In this way you get $\hat{f}(\xi)$ and you can use it to find $f(x)$.