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.

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

1 Answer 1


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 $f$.

  • 1
    $\begingroup$ Thanks for such a thorough answer!! $\endgroup$ Dec 13, 2022 at 18:06

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