5.2: Monotonic Functions
- Page ID
- 22665
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 \((a, b)\).
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
\[f(c+)=\sup \{f(x): c<x<b\}.\]
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.
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 \(a, 0<a<1, \lim _{x \rightarrow a} f(x)=f(a)\).