# Uniformly continuous function on uniformly convergent sequence of functions

Let {$$f_{n} , n\in \mathbb{N}$$} be a uniformly convergent sequence of continuous real–valued functions defined on a metric space $$M$$ and let $$g$$ be a continuous function on $$\mathbb{R}$$. Define, for each $$n\in \mathbb{N}$$, $$h_{n}(x)=g(f_{n}(x))$$. We know if $$M= \mathbb{R}$$ then the sequence {$$h_{n}:n\in \mathbb{N}$$} may not be uniformly convergent. A very nice example is given here: Composition of a continuous function with functions that converge uniformly
Now if instead, $$h$$ is a bounded uniformly continuous function, can we say that {$$h_{n}:n\in \mathbb{N}$$} is uniformly convergent? Also, if I relax the condition on {$$f_{n}, n\in \mathbb{N}$$} and make it right continuous instead of continuous then will the answer remain same?

• If $g$ is uniformly continuous and $f_n \to f$ uniformly then $h_n$ converges uniformly. No continuity assumptions on $f_n$'s are required. If $g$ is just bounded and continuous then $h_n$ need not converge uniformly. – Kavi Rama Murthy Jan 10 '19 at 10:06
• @KaviRamaMurthy Thank you so much. Can you give me a reference? I have to cite a reference for my work. I have searched for it but not able to find a reference. – arnab Jan 10 '19 at 17:28
• This result follows immediately from definition of uniform continuity. – Kavi Rama Murthy Jan 10 '19 at 23:10

Suppose $$g:\mathbb R \to \mathbb R$$ is defined as follows: $$g(n)=g(n+\frac 2 n)=0$$, $$g(n+\frac 1 n)=1$$ and $$g$$ has a straightline graph on the intervals $$(n,n+\frac 1 n)$$ as well as $$(n+\frac 1 n,n+\frac 2 n)$$ for $$n=3,4,\cdots$$ and $$0$$ everywhere else. Then $$g$$ is a bounded continuous function. Let $$f_n(x)=x+\frac 1 n$$. Then $$f_n$$ converges uniformly on $$\mathbb R$$ to $$f(x)=x$$ but $$h_n$$ doesn't converge uniformly: $$\sup_x \{|g(f_n(x))-g(f(x))| \geq |g(f_n(n))-g(f(n))|=1$$ for all $$n$$. [My comment above has more information about the question].