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

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

(See also Problem 12 of §2.)

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 $f\left(p^{+}\right) \neq f(p). \square$