Solving the functional equation $f(x)=f\left(\frac{x}{2}\right)+\frac{x}{2}\cdot f'(x)$

Question

find all functiions $f:\mathbb{R}\to\mathbb{R}$ such that $f'$ exists and $$f(x)=f\left(\frac{x}{2}\right)+\frac{x}{2}\cdot f'(x),\forall x\in\mathbb{R}$$

Answer 1

Since $f'(x)$ exists, we can use the mean value theorem : for $x \neq 0$, there exists $c$ strictly between $x/2$ and $x$ such that $$ (x/2) f'(c) = f(x) - f(x/2) = (x/2) f'(x) \implies f'(c) = f'(x). $$ In particular, the fact that $$ f'(x) = \frac 1x [ 2 f(x) - 2 f'(x) ] $$ implies that $f$ is smooth for $x \neq 0$ (i.e. infinitely many times differentiable). Therefore, for $x > 0$, there exists $c$ with $x/2 < c < x$ such that $f'(c) = f'(x)$. Suppose that $$ \alpha \overset{def}= \inf \{ c > 0 \, | \, f'(c) = f'(x) \} > 0. $$ Then by the equation above, there is a $c$ between $\alpha$ and $\alpha/2$ such that $f'(c) = f'(\alpha)$. By continuity of $f'$, $f'(\alpha) = f'(x)$ (note that by definition of the infimum, there exists a sequence $c_n$ with $c_n \searrow \alpha$ and $f'(c_n) = f'(x)$), which contradicts the definition of $\alpha$ since $c < \alpha$ and $f'(c) = f'(x) = f'(\alpha)$. Therefore $\alpha = 0$.

By definition of $\alpha$, there exists a sequence $x_n > 0$ with $f'(x_n) = f'(x)$ and $x_n \searrow 0$, hence $f'(0) = \lim_{n \to \infty} f'(x_n) = f'(x)$. It follows that $f(x) = ax + b$ for $x > 0$, and by continuity for $x \ge 0$.

Conversely, for any real numbers $a,b$, $f(x) = ax+b$ is a solution to your equation for $x \ge 0$. Therefore this is precisely the set of all solutions in this case.

Repeat for $x \le 0$ to conclude that $f'(x) = cx + d$ when $x \le 0$. By continuity and differentiability at $0$, $a=c$ and $b=d$. This means the only solutions are the affine functions, i.e. the functions of the form $f(x) = ax+b$ for real numbers $a,b$.

Hope that helps,