Recently, when I was working on a functional equation, I encountered something like an ordinary differential equation with boundary conditions!
Theorem. If the following holds for all $x \in \mathbb R$ $$\begin{align} f(2x) &=2 f(x) \\ f'(0) &=0 \end{align}$$ then $f(x)=0$ on $\mathbb R$.
Intuitively, it is evident for me that $f(x)=0$ but I cannot show this by a formal argument. In fact, I don't have any idea to work on it! :)
I will be thankful if you provide me a hint or help to show this with a nice formal proof.
Setting $x=0$ in $f(2x)=2f(x)$, $f(0)=0$. Now fix $x\neq 0$ and consider the values $f(x/2^n)$. By induction, $f(x/2^n)=f(x)/2^n$ for all $n\in\mathbb{N}$. But by the definition of the derivative, $$\frac{f(x/2^n)-f(0)}{x/2^n-0}=\frac{f(x)/2^n}{x/2^n}=\frac{f(x)}{x}$$ must converge to $f'(0)=0$ as $n\to\infty$. It follows that $f(x)=0$.