Notes - Analysis II HT23, Uniform continuity

Created: February 02, 2023

Flashcards

What’s the $\varepsilon$-$\delta$ definition for a function $f$ being continuous on $E$?
\[\forall p \in E\colon \forall \varepsilon > 0 \text{ } \exists \delta > 0 \text{ s.t. } \forall x \in E, (|x-p| < \delta \implies |f(x) - f(p)| < \varepsilon)\]

What’s the $\varepsilon$-$\delta$ definition for a function $f$ being uniformly continuous on $E$?
\[\forall \varepsilon > 0\colon \exists\delta>0 \text{ s.t. } \forall p \in E \text{ } \forall x \in E, (|x - p| < \delta \implies |f(x) - f(p)| < \varepsilon)\]

Can you state the $\varepsilon$-$\delta$ definition for a function $f$ being continuous on $E$, and then state it for being uniformly continuous on $E$?
\[\forall p \in E\colon \forall \varepsilon > 0 \text{ } \exists \delta > 0 \text{ s.t. } \forall x \in E, (|x-p| < \delta \implies |f(x) - f(p)| < \varepsilon)\] \[\forall \varepsilon > 0\colon \exists\delta>0 \text{ s.t. } \forall p \in E \text{ } \forall x \in E, (|x - p| < \delta \implies |f(x) - f(p)| < \varepsilon)\]

Can you state the useful theorem that relates the uniform continuity of a function to sequences?

$f : E \to \mathbb R$ is uniformly continuous if and only if for any $(x _ n), (y _ n)$ with $x _ n, y _ n \in E$ with $ \vert x _ n - y _ n \vert \to 0$, we have $ \vert f(x _ n) - f(y _ n) \vert \to 0$.

There is a theorem that states

$f : E \to \mathbb R$ is uniformly continuous if and only if for any $(x _ n), (y _ n)$ with $x _ n, y _ n \in E$ with $ \vert x _ n - y _ n \vert \to 0$, we have $ \vert f(x _ n) - f(y _ n) \vert \to 0$.

How does this provide a useful strategy for showing that a function is not uniformly continuous?

Show there exist two sequences $(x _ n)$, $(y _ n)$ where $ \vert x _ n - y _ n \vert \to 0$ but $ \vert f(x _ n) - f(y _ n) \vert \not\to 0$.

What theorem relates the continuity of functions on closed, bounded intervals to uniform continuity?

If $f : [a, b] \to \mathbb R$ is a continuous function on a closed, bounded interval, then $f$ is uniformly continuous.

When proving that continuity on a closed, bounded interval implies uniform continuity, you proceed by contradiction using the fact that this would imply

\[\exists \varepsilon > 0 \text{ s.t. } \forall \delta > 0 \text { } \exists x_n, y_n \in I \text{ s.t. } |x_n - y_n| < \delta \text{ but also } |f(x_n) - f(y_n)| \ge \epsilon\]

taking $\delta = \frac 1 n$ means you get sequences $(x _ n), (y _ n)$. How do you know $x _ n$ has a subsequence that converges?

The Bolzano-Weierstrass theorem.

When proving that continuity on a closed, bounded interval implies uniform continuity, you proceed by contradiction using the fact that this would imply

\[\exists \varepsilon > 0 \text{ s.t. } \forall \delta > 0 \text { } \exists x_n, y_n \in I \text{ s.t. } |x_n - y_n| < \delta \text{ but also } |f(x_n) - f(y_n)| \ge \epsilon\]

By doing some magic, you can construct two sequences $(x _ {s _ n}), (y _ {s _ n})$ that converge to some limit point $p$ of $[a, b]$. How can you use this to contradict $ \vert f(x _ n) - f(y _ n) \vert \ge \epsilon$?
\[\begin{aligned} |f(x_{s_n}) - f(y_{s_n})| &= |f(x_{s_n}) - f(p) + f(p) - f(y_{s_n})| \\\\ &\le |f(x_{s_n}) - f(p)| + |f(y_{s_n}) - f(p)| \\\\ &< \frac \varepsilon 2 + \frac \varepsilon 2 \\\\ &= \varepsilon \end{aligned}\]

When proving that any continuous periodic function is uniformly continuous on $\mathbb R$, on what interval do you know $f$ is already definitely uniformly continuous on?

$[0, p]$ where $p$ is the period, because continuity on closed, bounded intervals implies uniform continuity.

When proving that any continuous periodic function is uniformly continuous on $\mathbb R$, given an $\varepsilon > 0$, how do you construct a $\delta$ that means any $x, y$ with $ \vert x - y \vert < \delta$ satisfy $ \vert f(x) - f(y) \vert < \varepsilon$?
\[\delta = \min(\delta_1, p)\]

where $p$ is the period of the function, and $\delta _ 1$ is the delta that works for the guaranteed uniform continuity on the “base interval” $[0, p]$.

When proving that any continuous periodic function is uniformly continuous on $\mathbb R$, given an $\varepsilon > 0$, you construct a $\delta$ that means any $x, y$ with $ \vert x - y \vert < \delta$ satisfy $ \vert f(x) - f(y) \vert < \varepsilon$ like so:

\[\delta = \min(\delta_1, p)\]

where $p$ is the period of the function, and $\delta _ 1$ is the delta that works for the guaranteed uniform continuity on the “base interval” $[0, p]$. Now suppose $x, y$ do satisfy $ \vert x - y \vert < \delta$. What are the two cases that need to be considered?
  • $x, y \in [np, np+p]$, in the same interval
  • $x \in [np - p, np)$, $y \in [np, np+p]$, in different neighbouring intervals

Proofs

Prove:

$f : E \to \mathbb R$ is uniformly continuous if and only if for any $(x _ n), (y _ n)$ with $x _ n, y _ n \in E$ with $ \vert x _ n - y _ n \vert \to 0$, we have $ \vert f(x _ n) - f(y _ n) \vert \to 0$.

Todo.

Prove:

If $f : [a, b] \to \mathbb R$ is a continuous function on a closed, bounded interval, then $f$ is uniformly continuous.

Todo.

Prove that if $f$ is uniformly continuous on intervals $I$ and $J$, and $I \cap J \ne \{\}$, then $f$ is uniformly continuous on the interval $I \cup J$.

Todo.

Prove that any continuous periodic function is uniformly continuous on $\mathbb R$.

Todo.



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