TITLE: In search of nontrivial solutions to $f'(t)=f(t/2)$
QUESTION [9 upvotes]: Just then my friend gave me a kinda interesting problem: find all non-trivial solutions to the DE $$f'(t)=f(\frac t2),\quad f:\Bbb R\to\Bbb R.$$
Regularity concerns are temporarily ignored. You may just assume sufficient smoothness if needed.
(Note also that this is NOT an ODE!)
So far the only solution I can find is $f\equiv 0$, the trivial one. And surprisingly, I can't find any useful results of this problem, simple as it looks.
Any full or non full answer or just attempt is welcome!

REPLY [2 votes]: Here is a complementary answer to Thomas Andrews':

First observe that your problem is linear in $f$, which I will assume is of class $C^1$.
We will prove now that if $f$ solves your equation, with the added condition that $f(0)=0$, then $f(t) \equiv 0$ for all $t \in \mathbb{R}$:

We have, by the fundamental theorem of calculus
$$f(t)=f(0)+\int_{0}^t f'(t_1) \mathrm d t_1=\int_0^t f\left( \frac{t_1}{2} \right) \mathrm{d} t_1. $$
Iterating this rule we get
$$f(t)=\int_0^t \int_0^{t_1/2}f\left(\frac{t_2}{2}\right) \mathrm{d} t_2 \mathrm{d} t_1=\int_0^t \int_0^{t_1/2} \int_0^{t_2/2}f\left(\frac{t_3}{2}\right) \mathrm{d} t_3 \mathrm{d} t_2 \mathrm{d} t_1, $$
and continuing in this fashion gives
$$f(t)=\int_0^t \int_0^{t_1/2} \int_{0}^{t_2/2} \cdots \int_0^{t_{n-1}/2} f \left( \frac{t_n}{2} \right) \mathrm{d} t_n  \cdots \mathrm{d} t_{3} \mathrm{d} t_2 \mathrm{d} t_1, $$
for all $n \in \mathbb{N}$. Let $M:=\max_{-1 \leq x \leq 1} |f(x)|$, for $n$ large enough (i.e. large enough so that $\frac{|t|}{2^{n-1}}<1$) we can estimate this integral as 
$$|f(t)| \leq MV, $$
where $V$ is the volume of the $n$-dimensional simplex on which the integration takes place. I will leave it to you to show that $V=\frac{1}{n!2^{n(n-1)/2}}$, which gives
$$|f(t)| \leq \frac{M}{n!2^{n(n-1)/2}} .$$
Letting $n \to \infty$ gives $f(t)=0$, and this applies to any $t \in \mathbb{R}$.

Now say $f(t)$ is a solution of your equation with a general initial value, $f(0)=a$. Let $g(t)$ be an analytic solution of your equation (as Thomas defined) with the same initial value. The difference $h:=f-g$ also satisfies your equation (by linearity), and $h(0)=0$. The above gives $h \equiv 0$, and hence all solutions are analytic.