I have an equation $$xf(x)=f(x+1)$$ which is true for all real $x$.
Denote the Fourier transform of $f(x)$ as $F(z)$. Now perform Fourier transform on both sides, we get $$\frac{i}{2\pi}F’(z)=e^{2\pi iz}F(z)$$
Solving the differential equation, I obtain $$F(z)=C\cdot\text{exp}(-e^{2\pi iz})$$ Thus, $$f(x)=C\int^\infty_{-\infty} \text{exp}(-e^{2\pi iz}+2\pi ixz)dz$$
However, when I plug in $f(x)$ into the functional equation, it doesn’t seem right.
What is the mistake?
Thanks in advance.
Let $\mathcal{C} = \{ \varphi \in C^{\infty}(\mathbb{R}/\mathbb{Z}) : \varphi(0) = 0\}$.
Fix any $\varphi \in \mathcal{C}$ and let $f$ by $f(x) = \varphi(x)\Gamma(x)$. Since $\Gamma(x)$ has only simple poles and they are cancelled by zeros of $\varphi$, $f$ extends to a well-defined smooth function on $\mathbb{R}$ satisfying
$$ xf(x) = x\varphi(x)\Gamma(x) = \varphi(x+1)\Gamma(x+1) = f(x+1). $$
Conversely, for any smooth solution $f$ of the functional equation, define $\varphi(x) = f(x)/\Gamma(x)$. Since $1/\Gamma(x)$ is entire, $\varphi$ is well-defined. Moreover, this solves
$$\varphi(x+1) = \frac{xf(x)}{x\Gamma(x)} = \varphi(x). $$
Together with $\varphi(0) = f(0)/\Gamma(0) = 0$, we have $\varphi \in \mathcal{C}$.
This provides a 1-1 correspondence between the set of smooth solutions of the functional equation and the family $\mathcal{C}$.