5.2: Monotonic Functions

Definition

Suppose $D \subset \mathbb{R}, f: D \rightarrow \mathbb{R},$ and $(a, b) \subset D .$ We say $f$ is increasing on $(a, b)$ if $f(x)<f(y)$ whenever $a<x<y<b ;$ we say $f$ is decreasing on $(a, b)$ if $f(x)>f(y)$ whenever $a<x<y<b ;$ we say $f$ is nondecreasing on $(a, b)$ if $f(x) \leq f(y)$ whenever $a<x<y<b ;$ and we say $f$ is nonincreasing on $(a, b)$ if $f(x) \geq f(y)$ whenever $a<x<y<b .$ We will say $f$ is monotonic on $(a, b)$ if $f$ is either nondecreasing or nonincreasing on $(a, b)$ and we will say $f$ is strictly monotonic on $(a, b)$ if $f$ is either increasing or decreasing on $(a, b)$.

Proposition $\PageIndex{1}$

If $f$ is monotonic on $(a, b),$ then $f(c+)$ and $f(c-)$ exist for every $c \in(a, b)$.

Proof

Suppose $f$ is nondecreasing on $(a, b) .$ Let $c \in(a, b)$ and let

$$\lambda=\sup \{f(x): a<x<c\}.$$

Note that $\lambda \leq f(c)<+\infty .$ Given any $\epsilon>0,$ there must exist $\delta>0$ such that

$$\lambda-\epsilon<f(c-\delta) \leq \lambda .$$

Since $f$ is nondecreasing, it follows that

$$|f(x)-\lambda|<\epsilon$$

whenever $x \in(c-\delta, c) .$ Thus $f(c-)=\lambda .$ A similar argument shows that $f(c+)=\kappa$ where

$$\kappa=\inf \{f(x): c<x<b\}.$$

If $f$ is nonincreasing, similar arguments yield

$$f(c-)=\inf \{f(x): a<x<c\}$$

and

$$f(c+)=\sup \{f(x): c<x<b\}.$$

Proposition $\PageIndex{2}$

If $f$ is nondecreasing on $(a, b)$ and $a<x<y<b,$ then

$$f(x+) \leq f(y-).$$

Proof

By the previous proposition,

$$f(x+)=\inf \{f(t): x<t<b\}$$

and

$$f(y-)=\sup \{f(t): a<t<y\}.$$

Since $f$ is nondecreasing,

$$\inf \{f(t): x<t<b\}=\inf \{f(t): x<t<y\}$$

and

$$\sup \{f(t): a<t<y\}=\sup \{f(t): x<t<y\}.$$

Thus

$$f(x+)=\inf \{f(t): x<t<y\} \leq \sup \{f(t): x<t<y\}=f(y-).$$

Q.E.D.

Exercise $\PageIndex{1}$

Let $\varphi: \mathbb{Q} \cap[0,1] \rightarrow \mathbb{Z}^{+}$ be a one-to-one correspondence. Define $f:[0,1] \rightarrow \mathbb{R}$ by

$$f(x)=\sum_{q \in \mathbb{Q} \cap[0,1]_{q \leq x}} \frac{1}{2^{\varphi(q)}}.$$

a. Show that $f$ is increasing on $(0,1)$.

b. Show that for any $x \in \mathbb{Q} \cap(0,1), f(x-)<f(x)$ and $f(x+)=f(x)$.

c. Show that for any irrational $a, 0<a<1, \lim _{x \rightarrow a} f(x)=f(a)$.