Finding all $f\in L^1(\mathbb{R})\cap C^1(\mathbb{R})$ and $c\in\mathbb{R}$ so $f'(x)=f(x+c)$

Question

If we apply the Fourier Transform to our equation we get $$ix\hat{f}(x)=e^{icx}\hat{f}(x)$$ So, if $f\neq 0$ we conclude $ix=e^{icx}$. Plugging $x=1$ we get $c=-i\ln(i)=\frac{\pi}{2}$. But putting this value in the last equation we get $ix=e^{-ln(i)x}$, which only works for a specific values of $x$. I am at a loss, does this implies that the only solution is $f=0$?

Answer 1

From $ix\hat{f}(x)=e^{icx}\hat{f}(x)$ we conclude that the support of $\hat{f}$ is contained in the zero-set $Z := \{x \in \mathbb C\;|\; ix=e^{icx}\}$. This is a countable discrete set in $\mathbb C$. But note that for some functions $f$ it can turn out that the distribution $\hat{f}$ has discrete support. For example, if $f$ is the constant $1$, then $\hat{f} = \delta/(2\pi)$, the Dirac delta, with support $\{0\}$.

Here is one example of a solution: $$ f(x) = {{\rm e}^{ 0.3181315052\,x}}\cos \left( 1.337235701\,x \right) $$ satisfies $f'(x) = f(x+1)$.

The support of its Fourier transform is two solutions $z_1,z_2$ of $ix = e^{ix}$, given by $z_1 = -1.337235701 - i 0.3181315052, z_2 = +1.337235701 - i 0.3181315052$.

Of course a Fourier transform cannot have discrete support if $f \in L^1(\mathbb{R})\cap C^1(\mathbb{R})$.