Prove or disprove $f'(x)=f(\dfrac{x}{2})$ and $\lim\limits_{x\to0}f(x)=0$ implies $f(x)\equiv 0$

Question

$f:\mathbb{R}^+\to\mathbb{R}$, $f'(x)=f(\dfrac{x}{2})$ and $\lim\limits_{x\to0}f(x)=0$, prove that:$f(x)\equiv 0$.

I have tried for it but I do not give an solution without any possible error.

First, I tried to integrate $f(x)$ for $$f(x)=\int_0^xf'(t)\mathrm{d}(t)=\int^x_0f(\dfrac{t}{2})\mathrm{d}t=2\int_0^{\frac{x}{2}}f(t)\mathrm{d}t$$

But I have no way to show the proposition. Then I want to analyse the monotony of $f$. I assume that $f(x)$ is increasing. To be honest, if $f(x)>0$ then we can prove $f$ is increasing clearly. Then $0=f'(x)-f(\dfrac{x}{2})>f'(x)-f(x)$, then consider $g(x)=e^{-x}f(x)$ and $g'(x)=e^{--x}(f'(x)-f(x))<0$ so $g(x)$ is decreasing then a contradiction clearly shows. But when I want to show that $f$ must be non-negative or non-positive, I found nothing to show it.

After that, I consider a counter-example, through my thoughts, I think the sum forms the linear combination of $e^{-c^nx}$, for a common c instead of $\dfrac{1}{2}$. But when I construct it, I found when $c<\dfrac{1}{2}$, the series with functions diverges.

All my efforts is useless. So if there is any solution to this problem?

Answer 1

There is no a priori reason to presume that $f$ is real analytic, so the previous two solutions need some justification. Here is a proof that does not rely on $f$ being analytic.

Note that $f$ is differentiable on $x>0$ hence continuous and hence $f$ is at least continuously differentiable.

Suppose $X \in (0,1)$ is such that $|f(x)| \le {1 \over 2}$ for $x \in (0,X]$. Then for $x\in (0,X]$ we have $|f(x)| \le \int_0^x |f({t \over 2})| dt \le {1 \over 2} x$ and repeating we get $|f(x)| \le \int_0^x {1 \over 2} {t \over 2} dt = {1 \over 2} {x^2 \over 2^2}$. We surmise and verify by induction that $|f(x)| \le {1 \over 2 n! 2^{1+2+\cdots+ (n-1)}} x^n$. In particular, since this is true for all $n$, we have $f(x) =0$ for $x \in (0,X]$.

Hence $f'(x) = 0$ for $x \in (0,2X]$ and so $f(x) = 0 $ for $x \in (0,2X]$. Repeating the previous argument (complete proof using induction) we get $f(x) = 0$ for all $x \in (0,2^nX]$ and hence for all $x>0$.

Answer 2

As shown here, here and here, the solution to the differential equation $f'(x)=f(ax)$ is $$ f(x)=f(0)\sum_{k=0}^{\infty}\frac{a^{\frac{k(k-1)}{2}}}{k!}x^k. \tag{1} $$ The series has infinite radius of convergence if $|a|\leq 1$, in particular if $a=\frac{1}{2}$. Therefore, if $\lim_{x\to 0}f(x)=0$, it follows from $(1)$ that $f(0)=0$, hence $f(x)\equiv 0$.

Answer 3

You can look at solutions that are power series,i.e.$f(x)=\sum_{n=0}^{\infty}a_nx^n$. You find that necessarily, $a_{n+1}=\frac{a_n}{2^n(n+1)}$ hence $$a_n=\frac{a_0}{2^{\frac{(n-1)n}{2}}n!}$$. Then the series $f=\sum a_nx^n$ $(*)$ has indeed a convergence radius $\infty$ so is defined on $\mathbb R$ and is solution of your equation. It's limit when $x$ goes to $0$ is $a_0$, so the only solution $f$ which is a power series with $\lim_{x\rightarrow 0}f(x)=0$ is indeed $f=0$. If any solution to the differential equation is given by $(*)$ then $f=0$ is indeed the only solution to your problem. I think it's true as the set of solutions to the equation $y'(x)=y(x/2)$ is a vector space (linear combinations of solutions are solutions) and similarly to degree one linear diffeential equations its dimension might be $1$ so if I'm right $(*)$ are the only solutions, spanned by the one obtained with $a_0=1$.