Let $\mathcal{F}$ be the set of functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(2x)=f(x)$ for all $x\in\mathbb{R}.$
[list=a]
[*]Determine all functions $f\in\mathcal{F}$ which admit antiderivatives on $\mathbb{R}.$
[*]Give an example of a non-constant function $f\in\mathcal{F}$ which is integrable on any interval $[a,b]\subset\mathbb{R}$ and satisfies \[\int_a^bf(x) \ dx=0\]for all real numbers $a$ and $b.$
[/list][i]Mihai Piticari and Sorin Rădulescu[/i]
1. **Determine all functions \( f \in \mathcal{F} \) which admit antiderivatives on \(\mathbb{R}\).**

   Let \( F \) be the antiderivative of \( f \) with \( F(0) = 0 \). Then, by the definition of the antiderivative, we have:
   \[
   F'(x) = f(x)
   \]
   Given that \( f(2x) = f(x) \) for all \( x \in \mathbb{R} \), we need to find \( F \) such that:
   \[
   F(2x) - 2F(x) = C
   \]
   where \( C \) is a constant. Evaluating at \( x = 0 \), we get:
   \[
   F(0) - 2F(0) = C \implies 0 = C
   \]
   Thus, \( F(2x) = 2F(x) \) for all \( x \in \mathbb{R} \).

   Define \( g(x) = \frac{F(x)}{x} \) for \( x \neq 0 \). Then:
   \[
   g(2x) = \frac{F(2x)}{2x} = \frac{2F(x)}{2x} = \frac{F(x)}{x} = g(x)
   \]
   This implies \( g(x) \) is constant for all \( x > 0 \). Let \( g(x) = c \). Then:
   \[
   F(x) = cx \quad \text{for} \quad x > 0
   \]
   Similarly, for \( x < 0 \), we can extend \( g(x) \) to be constant, so:
   \[
   F(x) = cx \quad \text{for} \quad x < 0
   \]
   Therefore, \( F(x) = cx \) for all \( x \in \mathbb{R} \). Differentiating \( F \), we get:
   \[
   f(x) = F'(x) = c
   \]
   Hence, \( f(x) \) must be a constant function. Therefore, the only functions \( f \in \mathcal{F} \) that admit antiderivatives on \(\mathbb{R}\) are constant functions.

2. **Give an example of a non-constant function \( f \in \mathcal{F} \) which is integrable on any interval \([a,b] \subset \mathbb{R}\) and satisfies \(\int_a^b f(x) \, dx = 0\) for all real numbers \( a \) and \( b \).**

   Consider the function \( f \) defined as:
   \[
   f(x) = \begin{cases} 
   1 & \text{if } x = 2^k \text{ for some } k \in \mathbb{Z} \\
   0 & \text{otherwise}
   \end{cases}
   \]
   This function is non-constant and satisfies \( f(2x) = f(x) \) for all \( x \in \mathbb{R} \). 

   To show that \( f \) is integrable on any interval \([a, b]\) and that \(\int_a^b f(x) \, dx = 0\), note that \( f(x) \) is zero almost everywhere except at the points \( 2^k \). Since the set of points where \( f(x) = 1 \) is countable, the integral of \( f \) over any interval \([a, b]\) is zero:
   \[
   \int_a^b f(x) \, dx = 0
   \]

The final answer is \( \boxed{ f(x) = c } \) for part (a) and the function defined above for part (b).