Suppose $f$ is continuous from $(0,\infty)\to\mathbb{R}$, and $f(x)=f(2x)$. Then, can we conclude that $f$ is differentiable and uniformly continuous?
I think yes, because, it is clear that $f(x)=f(2^kx)$ for any $x$ and $k\in\mathbb{N}$. Also, this implies that for any $\epsilon>0$, $f(x+\epsilon)=f(x)$, for $x=\frac{\epsilon}{2^k-1}$. This implies that $f$ is constant. Hence, the differentiability and uniform continuity is trivial. any counterexamples to this? Am I right here? Thanks beforehand.
An example of function satisfying the given properties is $f(x)=\sin (\frac {2 \pi} {\ln 2} \ln (|x|)$. So there are non-constant continuous functions with $f(x)=f(2x)$. The function in this example is not uniformly continuous because it does not extend to a continuous function on $[0,\infty)$. Hence uniform continuity need not be true. To show that $f$ need not be differentiable we can take a non-differentiable continuous function on $[1,2]$ with $f(1)=f(2)$ and the extend it to the intervals $[2^{n}, 2^{n+1}]$ and $[2^{-n-1}, 2^{-n}]$ using the condition $f(x)=f(2x)$. [ This last part is a repetition of the answer by Empy2 and I am just quoting it here for completeness.