Home MATH 255 Wednesday, January 23, 2013

Wednesday, January 23, 2013 CC-BY-NC
Properties of monotone functions

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1Monotone functions have a countable number of discontinuities

What I'm about to say caused a lot of controversy in class. Given some monotone function on an interval, the number of discontinuities in this function is countable - that is, either finite or transfinite (there is a bijection mapping the discontinuities to the natural numbers - or, if you'd like, their cardinality is aleph-null). This fact is codified in the following theorem:

Theorem: Let $I \subseteq \mathbb{R}$ be an interval and $f : I \to \mathbb{R}$ be monotone on $I$. The set of discontinuities $D \subseteq I$ of $f$ is a countable set.

Proof: WLOG, let $f$ be monotone increasing on $I$ and let $I = [a,b]$. Define the set $D = \{x \in I : J_{f}(x) > 0\}$ as the set of jump discontinuities in $f$. Observe that for $a \leq x_1 < x_2 < \ldots < x_n \leq b$, $J_{f}(x_1) + J_{f}(x_2) + \ldots + J_{f}(x_n) \leq f(b) - f(a)$. Thus there are at most $k$ points in $I$ where the jump $J_{f}(x) \geq \frac{f(b) - f(a)}{k}$. So the set of discontinuities is countable.

This is technically all we need but it may be a bit unclear so we'll say this: notice that $\displaystyle D = \bigcup_{k=1}^{\infty} \left\{x \in I : J_{f}(x) \geq \frac{f(b) - f(a)}{k}\right\}$. Each set is finite, and since $k$ is an element of a countable set, $D$ is therefore countable. $\blacksquare$

Notice we can apply this to unbounded intervals such as $(-\infty, \infty)$ by noting that $\displaystyle(-\infty, \infty) = \bigcup_{N=1}^{\infty} [-N, N]$, and $(c, d)$ by noting that $\displaystyle(c, d) = \bigcup_{n=1}^{\infty}[c+\frac{1}{n}, d-\frac{1}{n}]$. $[a, \infty)$ and $(-\infty, d]$ follow similarly.

2The inverse of a strictly monotone function is strictly monotone

Theorem: Let $I \subseteq \mathbb{R}$ be an interval and $f : I \to \mathbb{R}$ be strictly monotone and continuous on $I$. Then $f^{-1}:f(I) \to I$ is strictly monotone and continuous.

Proof: First let's discuss the existence of $f^{-1}$. WLOG let $f$ be strictly increasing. Let $x_1, x_2 \in I$ be such that $x_1 \neq x_2$. Either $x_1 > x_2$ or $x_1 < x_2$ - thus either $f(x_1) > f(x_2)$ or $f(x_1) < f(x_2)$. In either case, $f(x_1) \neq f(x_2)$. Thus $f$ is one-to-one and onto $f(I)$, i.e. $f : I \to f(I)$ is bijective and therefore the inverse $f^{-1}$ exists.
Now we show that $f^{-1}$ is increasing: let $y_1, y_2 \in f(I)$ be such that $y_1 < y_2$. Then there exist $x_1, x_2 \in I$ such that $f(x_1) = y_1$ and $f(x_2) = y_2$. Since $y_1 < y_2$, $f(x_1) < f(x_2)$. Since $f$ is strictly increasing, this means $x_1 < x_2$ and therefore $f^{-1}(y_1)=x_1 < x_2=f^{-1}(y_2)$. Thus $f^{-1}$ must be strictly increasing as well.
Now we need to show continuity. Let $c \in f(I)$. Suppose $f^{-1}$ is not continuous at $c$. Then $J_{f^{-1}}(c) > 0$, i.e.

$$ \lim_{y \to c^-} f^{-1}(y) \neq \lim_{y \to c^+} f^{-1}(y) $$

As $f^{-1}$ is strictly increasing:

$$ \lim_{y \to c^-} f^{-1}(c) = m < M =\lim_{y \to c^+} f^{-1}(y) $$

Let $x \in (m, M)$ such that $x \neq f^{-1}(c)$. Then $x$ does not have a preimage $y_0$ such that $f^{-1}(y_0) = x$. If $y_0$ exists, $y_0 > c$. But if this is true, $y_0 > M$. Thus $y_0 \not\in (m, M)$. Thus $f^{-1}(f(I)) = I$

Basically what this is saying is since $f$ is continuous and defined on some interval, a jump discontinuity in $f^{-1}$ cannot happen as it would require $f(x)$ to be discontinuous for some $x \in I$.