$\phi : \mathbb{R}\rightarrow \mathbb{R}$ is a function with the property that whenever a sequence $f_n : \mathbb{R}\rightarrow\mathbb{R}, n=1,2,…$ converges uniformly, so does $\phi\circ f_n$. Show $\phi$ is uniformly continuous.
Attempt) Suppose $\phi$ is not uniformly continuous. Then there is an $\varepsilon>0$ such that for every $\delta>0$, there exist $x,y\in\mathbb{R}$ with $\mid x-y\mid<\delta$ but $\mid\phi(x)-\phi(y)\mid\geq\varepsilon$. So for $n=1,2,…$, there exist $x_n,y_n$ such that $\mid x_n-y_n\mid<\frac{1}{n}$ but $\mid\phi(x_n)-\phi(y_n)\mid\geq\varepsilon$. I am trying to find a uniformly convergent sequence $f_n$ for which $\phi\circ f_n$ does not converge uniformly to draw a contradiction.
Best Answer
Let $f(x)=x$ and $f_n(x)=x+(x_n-y_n)$. Then $f_n \to f$ uniformly so $\sup_x |\phi(x+(x_n-y_n))-\phi (x)| \to 0$. In particular, (taking $x=y_n$) this gives $|\phi(x_n)-\phi(y_n)| \to 0$, a contradiction.
EDIT Identifying the limit of $\phi (f_n)$: by considering constant functions we see that $x_n \to x$ implies $\lim\phi(x_n)$ exists. Now look at $(x_1,x,x_2,x,...)$. This sequence also converges and hence $(\phi (x_1), \phi (x),\phi (x_2), \phi (x),...)$ converges. This implies that $\phi (x_n) \to \phi (x)$.