# Uniform continuous maps that converge uniformly to a function f. What can we say about f?

Here is the question:

Let $X$ and $Y$ be metric spaces.Suppose a sequence of uniformly continuous maps $f_n : X \rightarrow Y$ converges uniformly to a map $f: X \rightarrow Y$.Does that imply that f is continuous ? Uniformly continuous ($X$ is not necessarily compact)?

I proved (hopefully right) that $f:X \rightarrow Y$ is continuous , even without using uniform continuity of $f_n$ but just assuming $f_n$ are continuous. I can't prove or disprove that $f$ is uniformly continuous and whether adding the compactness condition would change the answer.

• In fact, such an $f$ is uniformly continuous. Perhaps you could post what you've got so far for a proof, and where you got stuck. – Nate Eldredge Apr 17 '13 at 13:24

## 1 Answer

Given an $\epsilon > 0$, first find an $N$ such that $d_Y(f(x),f_N(x))< \frac{\epsilon}{4}$ for all $x\in X$. This is possible because of uniform convergence.

Next, find a $\delta > 0$ so that for any $x, y \in X$ with $d_X(x, y) < \delta$ gives $d_Y(f_N(x), f_N(y)) < \frac{\epsilon}{2}$. This is possible by uniform continuity of $f_N$. We now have, for the same $x, y$ that $$d_Y(f(x), f(y)) \leq d_Y(f(x), f_N(x)) + d_Y(f_N(x), f_N(y)) + d_Y(f_N(y), f(y)) \\\\ < \frac{\epsilon}{4} + \frac{\epsilon}{2} + \frac{\epsilon}{4} = \epsilon$$ thus we have found a $\delta$ so that for any $x, y \in X$ with $d_X(x, y)<\delta$ we have $d_Y(f(x), f(y)) < \epsilon$. Our $\epsilon$ was arbitrary, and therefore $f$ must be uniformly continuous.