NBHM 2020 PART C Question 28 Solution ($\mathcal F = \{f :(0,\infty) \to \Bbb R \mid f(x) = f(2x) \,\text{for all}\, x \in (0,\infty)\}$)

Let $\mathcal F = \{f :(0,\infty) \to \Bbb R \mid f(x) = f(2x) \,\text{for all}\, x \in (0,\infty)\}$. Which of the following  statements are true?
1. $f \in \mathcal F$ implies $f$ is bounded,
2. $f \in \mathcal F$ implies $f$ is uniformly continuous,
3. $f \in \mathcal F$ implies $f$ is differentiable,
4. Every uniformly bounded sequence in $\mathcal F$ has a uniform convergence subsequence.
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Solution:
Option 1: (True) Let $f \in \mathcal F$, then for all $x \in (0,\infty)$ we have $$f(x) = f(2x).$$ We note that $f(1) = f(2)$ and $$f([1,2]) = f([2,2^2]) = f([2^2,2^3]) = \cdots$$ In the other direction the given condition can be written as $$f(\frac{x}{2}) = f(x)$$ and
$$f([1,2]) = f([\frac{1}{2},1]) = f([\frac{1}{2^2},\frac{1}{2}]) = \cdots$$
This shows that $f$ is determined by its value in the interval $[1,2]$ with the additional condition that $f(1) = f(2)$. This also shows that $\text{Range}(f) = f([1,2])$, Since $f$ is continuous and $[1,2]$ is compact, $f$ is bounded and option 1 is true.
Option 3:(False) From the above observations, we can start with any function $g : [1,2] \to \Bbb R$ satisfying $g(1) = g(2)$ and can be extended to a function $\widetilde g \in \mathcal F$. Note that $\widetilde g$ restricted to $[1,2]$ is $g$ itself.
Define $g: [1,2] \to \Bbb R$ by $g(x) = |x - \frac{3}{2}|$ then $g(1) = g(2) = \frac{1}{2}$ and $g$ is not differentiable at $x = \frac{3}{2}$ ($|x|$ is not differentiable at $0$ proof works here also). Since $g$ is not differentiable at $\frac{3}{2}$ its extension $\widetilde g$ is also not differentiable at $\frac{3}{2}$ ($\widetilde g$ restricted to $[1,2]$ is $g$ itself). So option 3 is false
Option 2:(False) Consider the function $g:(0,\infty) \to \Bbb R$ defined by $$g(x) = sin(\frac{2\pi\text{log }x}{\text{log }2}).$$
$$g(2x) = sin(\frac{2\pi\text{log }2x}{\text{log }2}) = sin(\frac{2\pi(\text{log }2+\text{log x})}{\text{log }2}) = sin(\frac{2\pi\text{log }x}{\text{log }2} + 2\pi) \\ = g(x).$$
Therefore $g \in \mathcal F$.
Result: $f$ is uniformly continuous on $(0,\infty)$ if and only if $f$ can be extended to a continuous function on $[0,\infty)$. In other words, $\lim_{x \to 0}f(x)$ exists.
In our example, since $\lim_{x \to 0}g(x)$ doesn't exists (because of log x), $g$ cannot be extended to a continuous function on $[0,\infty)$. Therefore $g$ is not uniformly continuous.
option 4:(False) I made a mistake in the argument of this option. I am correcting it. I will update this option shortly. Sorry for the inconvenience.
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