How find this function $f(x)$
Let $f:\mathbb{R}\to\mathbb{R}$ be continuous such that $\dfrac{f(x)}{x}=f'\left(\dfrac{x}{2}\right)$. Find $f(x)$.
(2):if $f(x)$ on $x\in[a,b] $ be continuous,find all $f(x)$?
I think this is an ODE.
Let $\alpha_0 = 1, \alpha_1 = 2$ and let $\alpha_2, \alpha_3, \ldots$ be the complete set of roots in the complex plane of the equation:
$$\alpha = 2^{\alpha-1} \quad\text{ subject to constraint }\quad \Re{\alpha} > 1, \Im{\alpha} > 0$$
For $k \ge 2$, write $\alpha_k$ as $\beta_k + \gamma_k i$ where $\beta_k, \gamma_k \in \mathbb{R}$.
For any set of real constants $A_0, A_1, \ldots$ and $\delta_2, \delta_3, \ldots \in \mathbb{R}$. It is easy to check the function defined by:
$$\begin{align}f(x) &= A_0 x + A_1 x^2 + \sum_{k=2}^{\infty} \Re{\left( A_k e^{i\delta_k} x^{\alpha_k}\right)}\\ &= A_0 x + A_1 x^2 + \sum_{k=2}^{\infty} A_k x^{\beta_k} \cos(\gamma_k\log(x) + \delta_k) \end{align}$$
is a solution of the functional equation over $[0,\infty)$ provided this series and the series associated with its derivative converge.
The condition $\Re{\alpha_k} > 1$ for $k \ge 2$ ensure $f(x)$ falls off to zero fast enough so that $f'(0+)$ is defined and equal to $A_0$.
Over $(-\infty,0]$, we can define another function $\tilde{f}(x)$ in a similar manner:
$$\begin{align}\tilde{f}(x) &= \tilde{A}_0 x + \tilde{A}_1 x^2 + \sum_{k=2}^{\infty} \Re{\left( \tilde{A}_k e^{i\tilde{\delta}_k} |x|^{\alpha_k}\right)}\\ &= \tilde{A}_0 x + \tilde{A}_1 x^2 + \sum_{k=2}^{\infty} \tilde{A}_k |x|^{\beta_k} \cos(\gamma_k\log|x| + \tilde{\delta}_k) \end{align}$$
Once again, this is a solution for the functional equation over $(-\infty,0]$ provided this series and the series associated with its derivative converge. Furthermore, $f'(0-)$ exists and equal to $\tilde{A}_0$.
If $A_0 = \tilde{A}_0$, we can paste this two solution together to construct a $C^{1}$ solution of the functional equation over whole $\mathbb{R}$.
For an example, $\alpha_2 \sim 4.545364930374021 + 10.75397517526888 i$ implies $$f(x) \sim |x|^{4.545364930374021} \cos(10.75397517526888 \log|x|)$$ will be a non-trivial $C^1$ solution for the functional equation.
