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.

$$f([1,2]) = f([\frac{1}{2},1]) = f([\frac{1}{2^2},\frac{1}{2}]) = \cdots$$

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,

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

$$g(2x) = sin(\frac{2\pi\text{log }2x}{\text{log }2}) = sin(\frac{2\pi(\text{log }2+\text{log x})}{\text{log }2}) = sin(\frac{2\pi\text{log }x}{\text{log }2} + 2\pi) \\ = g(x).$$

Therefore $g \in \mathcal F$.

In our example, since $\lim_{x \to 0}g(x)$ doesn't exists (

Click 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.

**I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.****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.**here**for more problems.

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