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Here is Definition 7.19 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Let $\left\{ f_n \right\}$ be a sequence of functions defined on a set $E$.

We say that $\left\{ f_n \right\}$ is pointwise bounded on $E$ if the sequence $\left\{ f_n (x) \right\}$ is bounded for every $x \in E$, that is, if there exists a finite-valued function $\phi$ defined on $E$ such that $$ \left\lvert f_n(x) \right\rvert < \phi(x) \qquad ( x \in E, n = 1, 2, 3, \ldots). $$

We say that $\left\{ f_n \right\}$ is uniformly bounded on $E$ if there exists a number $M$ such that $$ \left\lvert f_n(x) \right\rvert < M \qquad (x \in E, n = 1, 2, 3, \ldots). $$

Here is Example 7.21 in Baby Rudin:

Let $$ f_n (x) = \frac{x^2}{x^2+ (1-nx)^2} \qquad (0 \leq x \leq 1, n = 1, 2, 3, \ldots). $$ Then $\left\lvert f_n(x) \right\rvert \leq 1$, so that $\left\{ f_n \right\}$ is uniformly bounded on $[0, 1]$. Also $$ \lim_{n \to \infty} f_n(x) = 0 \qquad (0 \leq x \leq 1), $$ but $$ f_n \left( \frac{1}{n} \right) = 1 \qquad (n = 1, 2, 3, \ldots), $$ so that no subsequence can converge uniformly on $[0, 1]$.

I think I understand the assertions in this Example.

Here is Definition 7.22:

A family $\mathscr{F}$ of complex functions $f$ defined on a set $E$ in a metric space $X$ is said to be equicontinuous on $E$ if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $$ \lvert f(x) - f(y) \rvert < \varepsilon $$ whenever $d(x, y) < \delta$, $x \in E$, $y \in E$, and $f \in \mathscr{F}$. Here $d$ denotes the metric of $X$.

Now Rudin states the following:

It is clear that every member of an equicontinuous family is uniformly continuous. [This is clear to me.]

The sequence of Example 7.21 is not equicontinuous. [Why? This is what is not clear to me, although I do understand that this is a sequence of uniformly continuous functions.]

Saaqib Mahmood
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Clearly, $f_n(0) = 0$, and as stated, $f_n\left(\frac{1}{n}\right) = 1$. Therefore, for $\epsilon < 1$ and any $\delta > 0$, you can let $n > \frac{1}{\delta}$, and then $\frac{1}{n} < \delta$, but $\lvert f_n\left(\frac{1}{n}\right)-f_n(0)\rvert = 1 > \epsilon$.

Michael L.
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  • I might also state that what you have defined as equicontinuity might be more precisely called "uniform equicontinuity" in order to distinguish it from "pointwise equicontinuity" (where $\delta$ can depend on $\epsilon$ and $x$). – Michael L. Oct 05 '17 at 05:59
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    Furthermore, the proposition is trivial by the contrapositive of Arzelà-Ascoli. Since no subsequence converges uniformly, the family of functions cannot be equicontinuous. – Michael L. Oct 05 '17 at 06:02