# Uniformly bounded and Convergence subsequence

Let $$\{f_n\}$$ be a sequence of real valued continuous function on $$(0,\infty)$$ such that $$f_n(x) = f_n(2x)$$ for all $$x$$ for all $$n$$. Suppose $$\{f_n\}$$ is uniformly bounded can we conclude that $$\{f_n\}$$ has a uniformly converging subsequence?

Any such function $$f_n$$ is defined by its value in the interval $$[1,2]$$. I have tried to construct an example contradicting this statement. But I could not process further. Kindly give some hints.

Thank you.

• Try construction $f_n$ having a peak at $1+\frac{1}{n}$ and decreasing to $0$ elsewhere very fast, such that at $\frac{1}{n\pm 1}$ it's already $0$. You'll then obtain a sequence $f_n$ whose point-wise limit is $0$, and whose convergence is not uniform. Apr 22, 2020 at 9:17
• @Learning Thank you. I can understand your idea. But how to make sure that such a function satisfies $f_n(x) = f_n(2x)$? Apr 22, 2020 at 9:36
• As you said, the function is determined by its value on the interval $[1,2]$, therefore as long as you construct such a sequence of continuous $f_n$ on $[1,2]$ with $f_n(1)=f_n(2)$, it defines a sequence $f_n$ on $(0,\infty)$ which satisfies the condition. A more geometric way to think of it is that $(0,\infty)/\times 2$ is a circle, no different from $\mathbb{R}/\mathbb{Z}$, via the function $\log$. Apr 22, 2020 at 19:52
• @Learning Thank you. What is the meaning of the notation $(0,\infty)/ \times 2$ ? Apr 24, 2020 at 3:52
• Also, We are trying to define $f_n$ only on $[1,2]$ then why do we need to worry about $\frac{1}{n \pm 1}$. I am not getting this point. Please explain. Apr 24, 2020 at 4:03