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