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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?

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    $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Jan 10 '19 at 10:06
  • $\begingroup$ @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. $\endgroup$ – arnab Jan 10 '19 at 17:28
  • $\begingroup$ This result follows immediately from definition of uniform continuity. $\endgroup$ – Kavi Rama Murthy Jan 10 '19 at 23:10
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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].

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