# Is the composite of an uniformly continuous sequence of functions with a bounded continuous function again uniformly continuous?

Let $\{f_n\}$ be a sequence of functions $f_n: J\to \mathbb{R}$ that converges uniformly to $f:J\to \mathbb{R}$ where $J\subseteq \mathbb{R}$ is an interval.

It is clear that for a uniformly continuous function $g:\mathbb{R}\to\mathbb{R}$, the sequence $\{g\circ f_n\}$ converges uniformly to $g\circ f:J\to \mathbb{R}$. There is a counterexample, if $g$ is only continuous.

If $J$ is compact, there is no such counterexample because then every continuous function $g$ is uniformly continuous. If $J$ is not compact, bounded and continuous for $g$ does not imply uniformly continuous.

Let $g:\mathbb{R}\to\mathbb{R}$ be bounded and continuous and $\{f_n\}$ a sequence of functions $f_n: J\to \mathbb{R}$ that converges uniformly to $f:J\to \mathbb{R}$. Does the sequence $\{g\circ f_n\}$ converges uniformly to $g\circ f:J\to \mathbb{R}$? If not, what is a counterexample?

• I suppose the constant sequence $f_n(x) = x$ converges uniformly to the function $f(x) = x$, which is then a counterexample unless you are only considering non-constant sequences. – John Martin May 27 '13 at 13:33

Hint. $g(x) = \sin(x^2)$, $f(x)=x$, and find constants $a_n \to 0$ so that $f_n(x) = x+a_n$ does what you want. So that for each $n$ there is $x$ with $g(x)=\sin(2\pi n) = 0$ and $g(f_n(x))=\sin(2\pi(n+1/2)) = 1$.