Let $f:\mathbb{R} \rightarrow \mathbb{R}$.
Then consider the following delay equation : $$f'(x) = f(x-1)$$
Let $S$ be the set of solution ot this equation. Then I would like to prove that :
$\forall f \in S -\{ x \mapsto 0\}$, $f$ is not bounded.
What I've done so far is that :
The set of solutions is an infinite dimensional space, hence it's hard to get a general form of all solutions, hence I don't think that calculating the characteristic equation is a good idea.
I've noticed that this is true for all equations of the form : $f'(x) = f(x-a)$ where $-1 \leq a \leq 1$. Yet this is not true for $a = 3\pi/2$, because $x \mapsto \sin(x)$ is a trivial solution and clearly $x \mapsto \sin(x)$ is bounded.
Maybe using the mean value is a good idea, in order to get a contradiction if we suppose that $f$ is bounded. But the problem with this approach is that I don't get enough informations on $a$.
We also have the formula : $$f^{(n)}(x) = f(x-n)$$
Thank you !
AN IDEA
If $f$ is a tempered distributional solution, taking the Fourier transform, $$ i\xi\hat{f}(\xi)=e^{i a \xi}\hat{f}(\xi), $$ so $\hat{f}$ must be supported at $\xi=\pm 1$, and this is only possible if $\pm i=e^{ia}$. (Thanks LutzL for correcting the previous version of this).
Notice that the only non-zero distributions that are supported at a point are the Dirac delta and its derivatives. This corresponds to the $f=\sin$ solution you found above: notice that the Fourier transform of $\sin$ is $$ \mathcal F(\sin)(\xi)=\frac{\pi}{i}(\delta(\xi-1)-\delta(\xi+1)).$$
Therefore, either the solution is of this kind, or it is not a tempered distribution, and so, in particular, it is not a bounded function.