
4.5: Monotone Function


A function $$f : A \rightarrow E^{*},$$ with $$A \subseteq E^{*},$$ is said to be nondecreasing on a set $$B \subseteq A$$ iff

$x \leq y\text{ implies } f(x) \leq f(y)\text{ for } x, y \in B.$

It is said to be nonincreasing on $$B$$ iff

$x \leq y\text{ implies } f(x) \geq f(y)\text{ for } x, y \in B.$

Notation: $$f \uparrow$$ and $$f \downarrow(\text { on } B),$$ respectively.

In both cases, $$f$$ is said to be monotone or monotonic on $$B.$$ If $$f$$ is also one to one on $$B$$ (i.e., when restricted to $$B$$), we say that it is strictly monotone (increasing if $$f \uparrow$$ and decreasing if $$f \downarrow$$).

Clearly, $$f$$ is nondecreasing iff the function $$-f=(-1) f$$ is nonincreasing. Thus in proofs, we need consider only the case $$f \uparrow$$. The case $$f \downarrow$$ reduces to it by applying the result to $$-f.$$

Theorem $$\PageIndex{1}$$

If a function $$f : A \rightarrow E^{*}\left(A \subseteq E^{*}\right)$$ is monotone on $$A,$$ it has a left and a right (possibly infinite) limit at each point $$p \in E^{*}$$.

In particular, if $$f \uparrow$$ on an interval $$(a, b) \neq \emptyset,$$ then

$f\left(p^{-}\right)=\sup _{a<x<p} f(x)\text{ for } p \in(a, b]$

and

$f\left(p^{+}\right)=\inf _{p<x<b} f(x)\text{ for } p \in[a, b).$

(In case $$f \downarrow,$$ interchange "sup" and "inf.")

Proof

To fix ideas, assume $$f \uparrow$$.

Let $$p \in E^{*}$$ and $$B=\{x \in A | x<p\} .$$ Put $$q=\sup f[B]$$ (this sup always exists in $$E^{*} ;$$ see Chapter 2, §13). We shall show that $$q$$ is a left limit of $$f$$ at $$p$$ (i.e., a left limit over $$B$$).

There are three possible cases:

(1) If $$q$$ is finite, any globe $$G_{q}$$ is an interval $$(c, d), c<q<d,$$ in $$E^{1}$$ . As $$c<q=\sup f[B], c$$ cannot be an upper bound of $$f[B]$$ (why?, so $$c$$ is exceeded by some $$f\left(x_{0}\right), x_{0} \in B.$$ Thus

$c<f\left(x_{0}\right), x_{0}<p.$

Hence as $$f \uparrow,$$ we certainly have

$c<f\left(x_{0}\right) \leq f(x)\text{ for all } x>x_{0}\text{ }(x \in B).$

Moreover, as $$f(x) \in f[B],$$ we have

$f(x) \leq \sup f[B]=q<d,$

so $$c<f(x)<d ;$$ i.e., $$f(x) \in(c, d)=G_{q}$$.

We have thus shown that

$\left(\forall G_{q}\right)\left(\exists x_{0}<p\right)\left(\forall x \in B | x_{0}<x\right) \quad f(x) \in G_{q},$

so $$q$$ is a left limit at $$p$$.

(2) If $$q=+\infty,$$ the same proof works with $$G_{q}=(c,+\infty].$$ Verify!

(3) If $$q=-\infty,$$ then

$(\forall x \in B) \quad f(x) \leq \sup f[B]=-\infty,$

i.e., $$f(x) \leq-\infty,$$ so $$f(x)=-\infty$$ (constant) on $$B$$ . Hence $$q$$ is also a left limit at $$p$$ (§1, Example (a)).

In particular, if $$f \uparrow$$ on $$A=(a, b)$$ with $$a, b \in E^{*}$$ and $$a<b,$$ then $$B=$$ $$(a, p)$$ for $$p \in(a, b] .$$ Here $$p$$ is a cluster point of the path $$B$$ (Chapter 3, §14, Example (h)), so a unique left limit $$f\left(p^{-}\right)$$ exists. By what was shown above,

$q=f\left(p^{-}\right)=\sup f[B]=\sup _{a<x<p} f(x),\text{ as claimed.}$

Thus all is proved for left limits.

The proof for right limits is quite similar; one only has to set

$B=\{x \in A | x>p\}, q=\inf f[B] . \quad \square$

Note 1. The second clause of Theorem 1 holds even if $$(a, b)$$ is only a subset of $$A,$$ for the limits in question are not affected by restricting $$f$$ to $$(a, b).$$ (Why?) The endpoints $$a$$ and $$b$$ may be finite or infinite.

Note 2. If $$D_{f}=A=N$$ (the naturals), then by definition, $$f : N \rightarrow E^{*}$$ is a sequence with general term $$x_{m}=f(m), m \in N$$ (see §1, Note 2).  Then setting $$p=+\infty$$ in the proof of Theorem 1, we obtain Theorem 3 of Chapter 3, §15. (Verify!)

Example $$\PageIndex{1}$$

The exponential function $$F : E^{1} \rightarrow E^{1}$$ to the base $$a>0$$ is given by

$F(x)=a^{x}.$

It is monotone (Chapter 2, §§11-12, formula (1)), so $$F\left(0^{-}\right)$$ and $$F\left(0^{+}\right)$$ exist. By the sequential criterion (Theorem 1 of §2), we may use a suitable sequence to find $$F\left(0^{+}\right),$$ and we choose $$x_{m}=\frac{1}{m} \rightarrow 0^{+}.$$ Then

$F\left(0^{+}\right)=\lim _{m \rightarrow \infty} F\left(\frac{1}{m}\right)=\lim _{m \rightarrow \infty} a^{1 / m}=1$

(see Chapter 3, §15, Problem 20).

Similarly, taking $$x_{m}=-\frac{1}{m} \rightarrow 0^{-},$$ we obtain $$F\left(0^{-}\right)=1.$$ Thus

$F\left(0^{+}\right)=F\left(0^{-}\right)=\lim _{x \rightarrow 0} F(x)=\lim _{x \rightarrow 0} a^{x}=1.$

Next, fix any $$p \in E^{1} .$$ Noting that

$F(x)=a^{x}=a^{p+x-p}=a^{p} a^{x-p},$

we set $$y=x-p.$$ (Why is this substitution admissible?) Then $$y \rightarrow 0$$ as $$x \rightarrow p,$$ so we get

$\lim _{x \rightarrow p} F(x)=\lim a^{p} \cdot \lim _{x \rightarrow p} a^{x-p}=a^{p} \lim _{y \rightarrow 0} a^{y}=a^{p} \cdot 1=a^{p}=F(p).$

As $$\lim _{x \rightarrow p} F(x)=F(p), F$$ is continuous at each $$p \in E^{1}.$$ Thus all exponentials are continuous.

Theorem $$\PageIndex{2}$$

If a function $$f : A \rightarrow E^{*}\left(A \subseteq E^{*}\right)$$ is nondecreasing on a finite or infinite interval $$B=(a, b) \subseteq A$$ and if $$p \in(a, b),$$ then

$f\left(a^{+}\right) \leq f\left(p^{-}\right) \leq f(p) \leq f\left(p^{+}\right) \leq f\left(b^{-}\right),$

and for no $$x \in(a, b)$$ do we have

$f\left(p^{-}\right)<f(x)<f(p)\text{ or } f(p)<f(x)<f\left(p^{+}\right) ;$

similarly in case $$f \downarrow$$ (with all inequalities reversed).

Proof

By Theorem 1, $$f \uparrow$$ on $$(a, p)$$ implies

$f\left(a^{+}\right)=\inf _{a<x<p} f(x)\text{ and } f\left(p^{-}\right)=\sup _{a<x<p} f(x);$

thus certainly $$f\left(a^{+}\right) \leq f\left(p^{-}\right).$$ As $$f \uparrow,$$ we also have $$f(p) \geq f(x)$$ for all $$x \in$$ $$(a, p);$$ hence

$f(p) \geq \sup _{a<x<p} f(x)=f\left(p^{-}\right).$

Thus

$f\left(a^{+}\right) \leq f\left(p^{-}\right) \leq f(p);$

similarly for the rest of (1).

Moreover, if $$a<x<p,$$ then $$f(x) \leq f\left(p^{-}\right)$$ since

$f\left(p^{-}\right)=\sup _{a<x<p} f(x).$

If, however, $$p \leq x<b,$$ then $$f(p) \leq f(x)$$ since $$f \uparrow$$. Thus we never have $$f\left(p^{-}\right)<f(x)<f(p).$$ similarly, one excludes $$f(p)<f(x)<f(x)<f\left(p^{+}\right) .$$ This completes the proof. $$\square$$

Note 3. If $$f\left(p^{-}\right), f\left(p^{+}\right),$$ and $$f(p)$$ exist (all finite), then

$\left|f(p)-f\left(p^{-}\right)\right|\text{ and } \left|f\left(p^{+}\right)-f(p)\right|$

are called, respectively, the left and right jumps of $$f$$ at $$p;$$ their sum is the (total) jump at $$p.$$ If $$f$$ is monotone, the jump equals $$\left|f\left(p^{+}\right)-f\left(p^{-}\right)\right|.$$

For a graphical example, consider Figure 14 in §1. Here $$f(p)=f\left(p^{-}\right)$$ (both finite $$),$$ so the left jump is $$0.$$ However, $$f\left(p^{+}\right)>f(p),$$ so the right jump is greater than $$0.$$ Since

$f(p)=f\left(p^{-}\right)=\lim _{x \rightarrow p^{-}} f(x),$

$$f$$ is left continuous (but not right continuous) at $$p$$.

Theorem $$\PageIndex{3}$$

If $$f : A \rightarrow E^{*}$$ is monotone on a finite or infinite interval $$(a, b)$$ contained in $$A,$$ then all its discontinuities in $$(a, b),$$ if any, are "jumps," that
is, points $$p$$ at which $$f\left(p^{-}\right)$$ and $$f\left(p^{+}\right)$$ exist, but $$f\left(p^{-}\right) \neq f(p)$$ or $$f\left(p^{+}\right) \neq f(p).$$

Proof

By Theorem 1, $$f\left(p^{-}\right)$$ and $$f\left(p^{+}\right)$$ exist at each $$p \in(a, b)$$.

If, in addition, $$f\left(p^{-}\right)=f\left(p^{+}\right)=f(p),$$ then

$\lim _{x \rightarrow p} f(x)=f(p)$

by Corollary 3 of §1, so f is continuous at $$p$$. Thus discontinuities occur only if $$f\left(p^{-}\right) \neq f(p)$$ or $$f\left(p^{+}\right) \neq f(p). \square$$