Let $\mathscr S$ be the family of continuous real valued functions on $(0, \infty)$ defined by $$\mathscr S : = \left \{f : (0, \infty) \longrightarrow \Bbb R\ |\ f(x) = f(2x),\ \forall x \in (0, \infty) \right \}.$$ For each of the following statements, state whether it is true or false.
$(a)$ Any element $f \in \mathscr S$ is bounded.
$(b)$ Any element $f \in \mathscr S$ is uniformly continuous.
$(c)$ Any element $f \in \mathscr S$ is differentiable.
$(d)$ Any uniformly bounded sequence in $\mathscr S$ has a uniformly convergent subsequence.
I have proved that $(a)$ is true because for any $x \in (0,\infty)$ there exists $n \in \Bbb Z$ such that $2^n x \in [1,2].$ Since $[1,2]$ is compact it follows that for any $f \in \mathscr S$ there exists $M_f \gt 0$ such that for all $y \in [1,2]$ we have $|f(y)| \leq M_f.$ But then by the given hypothesis $|f (x)| = |f(2^n x)| \leq M_f,$ as required.
But I have no idea about $(b), (c)$ and $(d).$ I think $(b)$ and $(c)$ are false. But I am unable to construct an element $f \in \mathscr S$ which will work as a counter-example. Would anybody please help me in this regard?
Thanks for your time.