In search of nontrivial solutions to $f'(t)=f(t/2)$

For analytic solutions, if $f(t)=\sum a_n t^n$ then: $$f'(t)=\sum_{n=0}^{\infty} (n+1)a_{n+1}t^{n} = \sum_{n=0}^{\infty} a_n(t/2)^n,$$ so $$a_{n+1}=\frac{a_n}{(n+1)2^n}.$$ Therefore $$a_n = \frac{a_0}{n!2^{n(n-1)/2}}$$

That's unlikely to have a nice closed form.

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Now, if $g$ is any function with $g'(t)=g(t/2)$ then let $f(t)=g(0)\sum \frac{t^n}{n!2^{n(n-1)/2}}$. Then $h(t)=g(t)-f(t)$ satisfies $h'(t)=h(t/2)$ and $h(0)=0$. So $h^{(n)}(0)=0$. By the mean value theorem, for any $t_0$, there is a $t_1$ between $0$ and $t_0$ so that $h(t_0)=t_0h'(t_1)=t_0h(t_1/2)$. By induction, $h(t_0)=t_0t_1/2t_2/4\cdots t_{k-1}/2^{k-1}h(t_{k}/2^k)$ where $t_k$ is between $0$ and $t_{k-1}$. This means that:

$$|h(t_0)|\leq\frac{|t_0|^k}{2^{k(k-1)/2}}|h(t_k/2^k)|$$

Taking the limit as $k\to\infty$ we see that $|h(t_0)|=0$.

So the only solutions are the analytic solutions.

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.