# Uniform Convergence and $\frac{\epsilon}{3}$

### 2019-10-05

An important result in analysis is that the uniform limit of continous functions is a continuous function. Analysis proofs, as they are presented in textbooks, sometimes feel like they Come From Nowhere. This one is famous for its use of $\frac{\epsilon}{3}$. While it's not difficult to follow the individual steps of the proof, figuring out What's Really Going On Here can be more difficult. It turns out that there is a nice reason why $\frac{\epsilon}{3}$ is involved:

To be more precise, let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces and let $\{f_n\}$ be a uniformly convergent sequence of continuous functions from $X$ to $Y$. That each of the $f_n$ are continuous means that for each $x \in X$, and for each $\epsilon > 0$, there exists some $\delta > 0$ such that: $$d_X(x, y) < \delta \implies d_Y(f_n(x), f_n(y)) < \epsilon$$ Which is just to say that if $x$ and $y$ are close together, then $f_n(x)$ and $f_n(y)$ are close together. I'll introduce the notation $\approx$ for this, so if $f$ is continuous, then $x \approx y$ implies $f(x) \approx f(y)$. That $\{f_n\}$ converges uniformly means that there's some function $f : X \rightarrow Y$ such that for all $\epsilon > 0$, there exists an $N$ such that: $$n > N \implies d_Y(f_n(x), f(x)) < \epsilon$$ Which is to say that for big enough $n$, $f_n(x) \approx f(x)$ for any $x$.

We now want to show that $f$, the function that the $f_n$'s get closer and closer to, it itself continuous. We therefore want to show that for any $x \in X$, for all $\epsilon > 0$ there exists some $\delta > 0$ such that: $$d_X(x, y) < \delta \implies d_Y(f(x), f(y)) < \epsilon$$ In essence, we want that if $x \approx y$, then $f(x) \approx f(y)$. To see why $f(x) \approx f(y)$ we travel through two other points: $f_n(x)$ and $f_n(y)$ . The Basic Idea here is that if we can show that $f(x) \approx f_n(x)$ and $f_n(x) \approx f_n(y)$, and $f_n(y) \approx f(y)$, then $f(x) \approx f(y)$. Consider this picture: We can break up the distance between $f(x)$ and $f(y)$ into $3$ other distances: $d_Y(f(x), f_n(x))$, $d_Y(f_n(x), f_n(y))$, and $d_Y(f_n(y), f(y))$. From the above diagram, it is clear that the distance between $f(x)$ and $f(y)$ is less than or equal to the sum of the other 3, and this is What's Really Going On Here. $\frac{\epsilon}{3}$ shows up because of the three blue lines in the diagram. Since $\{f_n\}$ converges uniformly to $f$, we know that $f(x) \approx f_n(x)$ and that $f_n(y) \approx f(y)$. Because each $f_n$ is continuous, we know that $f_n(x) \approx f_n(y)$ if $x \approx y$. From continuity and uniform convergence, we can argue that $f(x) \approx f(y)$ by connecting them through two other points: So, to formalize the proof, first use the triangle inequality to say that: By uniform convergence, given an $\epsilon$ we can pick an $n > N$ such that $d_Y(f(x), f_n(x)) < \frac{\epsilon}{3}$ and $d_Y(f_n(y), f(y)) < \frac{\epsilon}{3}$. By continuity of $f_n$, given an $\epsilon$, there exists some $\delta$ such that $d_Y(f_n(x), f_n(y)) < \epsilon$ if $d_X(x, y) < \delta$. All together, we therefore have that if $d_X(x, y) < \delta$, then: $$d_Y(f(x), f(y)) < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$$ This completes the famous $\frac{\epsilon}{3}$ proof.