How to solve $f'(x)=f'(\frac{x}{2})$

Question

How do we solve this given $f'(0)=-1$. It does not look separable. I can integrate both sides but end up with a functional equation with is not helpful.

Answer 1

I will assume that $f:\mathbb{R}\to\mathbb{R}$ is supposed to be $C^1$, so $f'$ exists and is continuous everywhere. Now note that for any $x\in \mathbb{R}$, $f'(x)=f'(x/2)=f'(x/4)=f'(x/8)=\cdots$. But $x/2^n$ converges to $0$ as $n\to\infty$, so continuity of $f'$ now implies $$-1=f'(0)=\lim_{n\to \infty}f'(x/2^n)=f'(x).$$ So $f'(x)=-1$ for all $x$, and thus $f(x)=-x+C$ for some constant $C$.

In fact, the assumption that the derivative is continuous can be eliminated, as explained in detail in Robert Israel's answer.

Answer 2

Obviously $f'(x)=-1$ solves this, taking the anti derivative we get the solution $f(x)=-x+k$.

Answer 3

This is a functional equation for $g(x)=f'(x)$ and the solution is $g(x)=\varphi(\log x)$ where $\varphi(x)=\varphi(x-\log 2)$ is an arbitrary periodic function with period $\log 2$.

In fact for the equation $g(ax)=bg(x)$ we have $g(aa^x)=bg(a^x)$ and $g(a^{x+1})=bg(a^x)$; then $g(a^x)=\varphi(x)b^x$, where $\varphi(x)$ is an arbitrary periodic function with unit period.

Finally $g(x)=\varphi(\log_ax)x^{\log_ab}$. For $a=1/2$ and $b=1$, $g(x)=\varphi(\log x)$ where $\varphi(x)=\varphi(x-\log 2)$ is an arbitrary periodic function with period $\log 2$.

For $\varphi(x)=C$ constant, we have the particular solution $g(x)=f'(x)=C$ that is $f(x)=Cx+k$ and from $f'(0)=-1$ we have $f(x)=-x+k$.

Answer 4

Suppose $f$ is differentiable everywhere on $\mathbb R$ and $f'(x) = f'(x/2)$ for all $x$. Now $$\dfrac{d}{dx} \left(f(x) - 2 f(x/2)\right) = f'(x) - f'(x/2) = 0$$ so $f(x) - 2 f(x/2)$ must be constant, say $f(x) - 2 f(x/2) = c$. $f$ is continous at $0$ since it is differentiable there, and taking the limit as $x \to 0$ we find $c = -f(0)$. Since the problem is invariant under addition of a constant, we may assume for convenience $f(0)=0$, so $f(x) = 2 f(x/2)$.

Defining $$ g(t) = 2^{-t} f(2^t)$$ we find $$ g(t) = 2^{-t} f(2^t) = 2^{1-t} f(2^{t-1}) = g(t-1) $$ i.e. $g$ is periodic with period $1$.

Now we need $f$ to be differentiable at $0$. We get $$ f'(0) = \lim_{x \to 0} \dfrac{f(x) - f(0)}{x} = \lim_{x \to 0+} g(\log_2 x) = \lim_{t \to -\infty} g(t)$$ But in order for this limit to exist, given that $g$ is periodic, $g$ must be constant; if $f'(0)=-1$ that constant is $-1$, and $f(x) = 2^{\log_2 x} g(\log_2 x) = -x$. Similarly, looking at $2^{-t} f(-2^t)$ we find that $f(x) = -x$ for $x < 0$. Removing the assumption $f(0)=0$, the general solution is $f(x) = -x + k$.