Notes - Analysis II HT23, Uniform continuity
Flashcards
What’s the $\varepsilon$-$\delta$ definition for a function $f$ being continuous on $E$?
What’s the $\varepsilon$-$\delta$ definition for a function $f$ being uniformly continuous on $E$?
Can you state the $\varepsilon$-$\delta$ definition for a function $f$ being continuous on $E$, and then state it for being uniformly continuous on $E$?
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?
$f : E \to \mathbb R$ 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$.
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$?
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$?
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$.
$f : E \to \mathbb R$ 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.
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.