Notes - Analysis II HT23, Monotonic functions
Non-examinable but still useful.
Flashcards
Can you state the theorem about one-sided limits of increasing functions?
Let $f : E \to \mathbb R$ be increasing. If $p \in E$ is a left limit point of $E$, then:
- The left-hand limit $f(p^-)$ of $f$ at $p$ exists and $f(p^-) = \sup \{f(x): x < p\} \le f(p)$
- The right hand limit $f(p^+)$ of $f$ at $p$ exists and $f(p^+) = \inf \{f(x): p < x\} \ge f(p)$
Can you state the theorem that describes the properties when an increasing function $f : E \to \mathbb R$ has a discontinuity at $p$?
If $f : E \to \mathbb R$ is increasing and discontinuous at $p \in E$, then either
- $p$ is a left limit point of $E$ and $f(p^-) < f(p)$
- $p$ is a right limit point of $E$ and $f(p^+) > f(p)$
What theorem relates monotonic functions defined on intervals and its continuity, given properties of its image?
Suppose $f : I \to \mathbb R$ is monotonic and $I$ is an interval. Then $f$ is continuous if and only if the image $f(I)$ is an interval.
There is a theorem that states:
Suppose $f : I \to \mathbb R$ is monotonic and $I$ is an interval. Then $f$ is continuous if and only if the image $f(I)$ is an interval.
How can you use this to prove that the inverse of a continuous function defined on an interval $I$ is also continuous (the hard bit of the continuous inverse function theorem)?
Suppose $f : I \to \mathbb R$ is monotonic and $I$ is an interval. Then $f$ is continuous if and only if the image $f(I)$ is an interval.
Consider $f^{-1} : f(I) \to I$. $f^{-1}$ is monotonic (by the first part of the continuous inverse function theorem), and $f(I)$ is an interval, so since $I$ is also an interval, $f^{-1}$ must also be continuous.