I was looking at this question is MSE.
Suppose $f_n:[0,1] \to \mathbb{R}$ is a sequence of uniformly continous functions and $f_n\to f$ uniformly.
These are the questions that I put myself.
- What can I say about the uniform continuity of the uniform limit $f$ ?
- Suppose that the uniform limit function $f$ is also uniformly continuous. Will the result hold good if I change the domain to $\mathbb{R}$ or any other subset $A$ of $\mathbb{R},$ which is not necessarily compact ?
My attempt:
- Given that $f_n:[0,1] \to \mathbb{R}$ is a sequence of uniformly continous functions. Then for every $\epsilon>0,$ there exists a $\delta_n(\epsilon)>0,$ such that $|f_n(x)-f_n(y)|<\frac{\epsilon}{3},$ whenever $|x-y|<\delta_n, $ with $x,y \in [0,1].$
Since, $f$ is the uniform limit of the sequence $f_n,$ for the same $\epsilon>0,$ that we took previously, there exists a natural number $N(\epsilon),$ such that $|f_n(x)-f(x)|<\frac{\epsilon}{3},$ whenever $n>N(\epsilon).$
Now, choose $\delta_1=max\{\delta_n,\delta\}.$
Then, we have that $|f(x)-f(y)|= |f(x)-f_n(x)+f_n(x)-f_n(y)+f_n(y)-f(y)| \le |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|=\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon,$ for any $x,y \in [0,1]$ such that $|x-y|<\delta_1.$ So, the uniform limit $f$ is also uniformly continuous.
Is this proof correct? Or have I made any mistakes. If yes, please point it out. This shall benefit me immensly.
As for the part 2), I feel like the same logic works well for any arbitrary subsets of $\mathbb{R}$ also. I am not able to proceed further. Please help.
