Notes - Analysis II HT23, Boundedness theorem
Flashcards
What does it mean for $f : E \to \mathbb R$ to be bounded on $E$?
Can you state the boundedness theorem in full?
If $f : [a, b] \to \mathbb R$ is continuous then $f$ is bounded:
\[\exists M \text{ s.t. } \forall x \in [a,b] : |f(x)| \le M\]and it attains its bounds:
\[\begin{aligned} &\exists \xi_1 \in [a,b] \text{ s.t. } f(\xi_1) = \inf \\{ f(x) : x\in[a,b] \\} \\\\ &\exists \xi_2 \in [a,b] \text{ s.t. } f(\xi_2) = \sup \\{ f(x) : x\in[a,b] \\} \end{aligned}\]What does it mean for a subset of $\mathbb R$ or $\mathbb C$ to be “compact”?
It is closed and bounded.
When proving the first part of the boundedness theorem, that if $f : [a, b] \to \mathbb R$ is continuous then $f$ is bounded, you construct a bounded sequence $(x _ n)$ (which therefore has a convergent subsequence, used in the rest of the proof). How do you do this?
Assume that $f$ is unbounded, so that for any $n \in \mathbb{N}$ there exists some $x _ n$ such that $ \vert f(x _ n) \vert > n$.
Suppose that some sequence $x _ {s _ n} \in [a, b]$ tends to a limit $p \in [a, b]$. Then if $f$ is a continuous function, why is $f(x _ {s _ n})$ bounded?
Because the series $f(x _ {s _ n})$ is convergent, and hence bounded.
Suppose $f$ is a continuous function on $[a, b]$ and let $M = \sup \{f(x) : x \in [a, b]\}$. Then how can you use the supremum approximation property to construct a fun sequence $x _ n$ that’s useful for proving that $f$ attains this supremum?
For all $n \ge 1$, then $\exists x _ n$ such that
\[M - \frac 1 n < f(x_n) \le M\]What contradiction is used in proving the “is bounded” part of the boundedness theorem?
If $f$ were unbounded, then there would exist a sequence $x _ n$ where $f(x _ n)$ was unbounded. But $x _ n$ must have a convergent subsequence $x _ {s _ n}$, and since $f$ is continuous $f(x _ {s _ n})$ must be bounded, a contradiction.
Proofs
Prove the boundedness theorem:
If $f : [a, b] \to \mathbb R$ is continuous then $f$ is bounded:
\[\exists M \text{ s.t. } \forall x \in [a,b] : |f(x)| \le M\]
and it attains its bounds:
\[\begin{aligned}
&\exists \xi_1 \in [a,b] \text{ s.t. } f(\xi_1) = \inf \\{ f(x) : x\in[a,b] \\} \\\\
&\exists \xi_2 \in [a,b] \text{ s.t. } f(\xi_2) = \sup \\{ f(x) : x\in[a,b] \\}
\end{aligned}\]
Todo.