# Proving $\phi : \mathbb{R}\rightarrow \mathbb{R}$ with a property is uniformly continuous.

$$\phi : \mathbb{R}\rightarrow \mathbb{R}$$ 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.

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 $$\phi (x_n) \to \phi (x)$$.
• Does $\phi\circ f_n$ converge to $\phi\circ f$?