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Definition: Uniformly Continuous

Let $f $ be a real-valued function defined on a set $S \subseteq \R $.

$f $ is uniformly continuous on $S $ if

$$\forall \epsilon > 0, \exists \delta > 0\text{ s.t. }x, y \in S, \abs{ x - y } < \delta \implies \abs{ f(x) - f(y)} < \epsilon$$


We say $f$ is uniformly continuous if $f$ is uniformly continuous on $\dom{ f } $.

Note

It makes no sense to say a function is uniformly continuous at each point.

Theorem 19.2

If $f $ is continuous on a closed interval $[a, b] $, then $f $ is uniformly continuous on $[a, b] $.

Proof [expand]

Assume $f $ is not uniformly continuous on $[a, b] $. Then there exists $\epsilon > 0 $ such that for each $\delta > 0 $ the implication

$$\abs{ x - y } < \delta \implies \abs{ f(x) - f(y)} < \epsilon$$

fails.

That is,

$$\forall \delta > 0, \exists x, y \in [a, b], \abs{ x - y } < \delta \implies \abs{ f(x) - f(y)} \geq \epsilon $$

Thus, $\forall n \in \N, \exists x_n, y_n \in [a, b] $ s.t. $\abs{ x_n - y_n } < \frac{ 1 }{ n } \implies \abs{ f(x_n) - f(y_n)} \geq \epsilon $.

By the Bolzano-Weierstrass Theorem a subsequence $(x_{n_k}) $ of $(x_n) $ converges.

Moreover, if $x_0 = \limu{ k}{\infty} x_{n_k}$ then $x_0$ belongs to $[a, b]$. Since $f$ is continuous at $x_0$.

We also have $x_0 = \limu{ k }{ \infty } y_{n_k}$.

We have

$$f(x_0) = \limu{ k }{ \infty } f(x_{n_k}) = \limu{ k }{ \infty } f(y_{n_k}) $$

so,

$$\limu{ k }{ \infty } [f(x_{n_k}) - f(y_{n_k})] = 0$$

Since $\abs{ f(x_{n_k}) - f(y_{n_k})} \geq \epsilon $ for all $k $, we have a contradiction. We conclude $f$ is uniformly continuous on $[a, b]$.

Note

A convergent sequence in $[a, b]$ converges to element in $[a,b] $.

Corollary 19.2

If $f $ is continuous on a closed and bounded set $S $, then $f $ is uniformly continuous on $S $.

Theorem 19.4

If $f$ is uniformly continuous on a set $S$ and $(s_n)$ is a Cauchy sequence in $S$, then $(f(s_n))$ is a Cauchy sequence.

Proof [expand]

Let $(s_n) $ be a Cauchy sequence in $S$ and let $\epsilon > 0 $. Since $f $ is uniformly continuous on $S $, there exists $\delta > 0 $ so that

$x, y \in S$ and $\abs{ x- y } < \delta \implies \abs{ f(x) - f(y)} < \epsilon$

Since $(s_n)$ is a Cauchy sequence, there exists $N $ so that

$m, n > N \implies \abs{ f(s_n) - f(s_m)} < \epsilon$

This proves $(f(s_n))$ is also a Cauchy sequence.

Definition: Extended Function

A functions $\fnext{ f }$ is an extension of a function $f $ if $\dom{ f } \subseteq \dom{ \fnext{ f }}$and $f(x) = \fnext{ f} (x) $for all $x \in \dom{ f } $.

Theorem 19.5

Let $f $ be a real-valued function on $(a, b)$.

$f $ is uniformly continuous on $(a, b) \iff \exists \fnext{ f } $, $\fnext{ f } $ is continuous on $[a,b] $.

Theorem 19.6

Let $f $ be a continuous function on an interval $I$ ($I$ may be bounded or unbounded). Let $I^{\circ}$ be the interval obtained by removing from $I$ any endpoints that happen to be in $I$.

$f $ is differentiable on $I^{\circ} $ and $f’$ is bounded on $I^{\circ} \implies f $ is uniformly continuous on $I$.