# If $(f_n)$ converges pointwise to $f$, then $f$ is uniformly continuous

I am preparing for my Analysis $$2$$ final and am looking over some problems on past homeworks and exams that I could not solve. This is one of them:

Prove that if $$(f_n)_{n\in\mathbb{N}}$$ is equicontinuous and converges pointwise to some function $$f: D \to \mathbb{R}$$, then $$f$$ is uniformly continuous on D.

My solution earned $$4/7$$ points and now looking back on it, my professor was quite generous giving me even that many points.

He said we were supposed to use an "$$\varepsilon/3$$" argument, which I have not seen before, however, it seems there are many similar questions on the SE regarding uniform convergence, but I have not seen any with pointwise convergence.

• A three-epsilon argument succeeds under the assumption that the convergence is not merely pointwise, but is uniformly pointwise. Dec 12, 2022 at 23:11

## Fixing the problem statement

Here's a Wikipedia link with definitions of various types of equicontinuity. Normally, if you just say a sequence of functions is "equicontinuous" then it means they're pointwise equicontinuous. That assumption is not sufficient here and I'll provide a counterexample below.

However, the problem will work out fine if we instead assume the $$\{f_n\}$$ are uniformly equicontinuous, so I'm guessing that was the intended assumption. As to why it got miscommunicated - maybe your professor just defined "equicontinuous" to mean "uniformly equicontinuous", or maybe you made a typo in your question here, or maybe the professor made a typo in the problem statement.

## Counterexample if we only assume pointwise equicontinuity

The problem as currently stated doesn't work. As a counterexample, let $$f$$ be some function that's continuous but not uniformly continuous; say $$f(x) = \frac 1 x$$ with $$D = (0,1]$$. Let $$f_n = f$$ for all $$n$$. Then the $$f_n$$ are continuous, and the sequence $$\{f_n\}$$ is equicontinuous (because all the functions are the same anyway), and clearly $$f_n \to f$$ pointwise. So the problem's assumptions are fulfilled but the claim doesn't hold.

## Proof if we assume uniform equicontinuity

Choose some $$\varepsilon > 0$$. To show $$f$$ is uniformly continuous, we need to produce $$\delta > 0$$ such that if $$a, b \in D$$ with $$|b-a|<\delta$$ then $$|f(b)-f(a)|<\varepsilon$$.

To accomplish that using the $$\frac{\varepsilon}{3}$$ trick, the idea is to write \begin{align} |f(b) - f(a)| &= |f(b) - f_n(b) + f_n(b) - f_n(a) + f_n(a)-f(a)| \\ &\le |f(b) - f_n(b)| + |f_n(b) - f_n(a)| + |f_n(a)-f(a)| \end{align} where this inequality holds for any $$n \in \mathbb{N}$$. Now we aim to show that each of the three RHS terms can be made small.

For the 1st term, we know $$f_n \to f$$ pointwise, so there exists some $$N_1 \in \mathbb{N}$$ such that $$|f(b) - f_n(b)| < \frac{\varepsilon}{3}$$ for all $$n \ge N_1$$. Similarly, for the 3rd term we can find some $$N_2 \in \mathbb{N}$$ such that $$|f_n(a) - f(a)| < \frac{\varepsilon}{3}$$ for all $$n \ge N_2$$. That means if we pick $$n \ge \max\{N_1, N_2\}$$ then both of these terms will be $$< \frac \varepsilon 3$$.

To handle the 2nd term, we use the assumption that $$\{f_n\}$$ is a uniformly equicontinuous sequence. That means we can find $$\delta > 0$$ such that if $$|x_2 - x_1| < \delta$$ then $$|f_n(x_2) - f_n(x_1)| < \frac{\varepsilon}{3}$$ for all $$n$$. So as long as we require $$|b-a| < \delta$$, the 2nd term in our breakdown will be $$< \frac{\varepsilon}{3}$$.

Finally, we combine our estimates. For $$n \ge \max\{N_1,N_2\}$$, and for any $$a,b \in D$$ satisfying $$|b-a| < \delta$$, we have:

\begin{align} |f(b) - f(a)| &\le |f(b) - f_n(b)| + |f_n(b) - f_n(a)| + |f_n(a)-f(a)| \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} \\ &= \varepsilon \end{align}

which is what we needed to show for uniform continuity of $$f$$.

• Thanks for such a thorough answer!! Dec 13, 2022 at 18:06