Minimum Rule for Continuous Functions
Theorem
Let $\struct {S, \tau}$ be a topological space.
Let $f, g: S \to \R$ be continuous real-valued functions.
Let $\min \set {f, g}: S \to \R$ denote the pointwise minimum of $f$ and $g$.
Then:
- $\min \set {f, g}$ is continuous.
Proof
Let $x \in S$.
Let $\epsilon > 0$.
Without loss of generality, assume that $\map f x \le \map g x$.
Case 1 : $\map f x = \map g x$
Let $ y = \map f x = \map g x$
By definition of the min operation, $\min \set {\map f x, \map g x} = y$.
From Continuity Test for Real-Valued Functions applied to $f$:
- $\exists U \in \tau : x \in U : \forall z \in U : \map f z \in \openint {y - \epsilon} {y + \epsilon}$
Similarly, Continuity Test for Real-Valued Functions applied to $g$:
- $\exists V \in \tau : x \in V : \forall z \in V : \map g z \in \openint {y - \epsilon} {y + \epsilon}$
Thus for all $z \in U \cap V$:
- $\map f z \in \openint {y - \epsilon} {y + \epsilon}$
- $\map g z \in \openint {y - \epsilon} {y + \epsilon}$
By definition of the min operation, for all $z \in U \cap V$:
- $\min \set {\map f z, \map g z} \in \openint {y - \epsilon} {y + \epsilon}$
By the Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets:
- $U \cap V \in \tau$
Since $\epsilon > 0$ was arbitrary then:
- $\forall \epsilon > 0 : \exists W \in \tau : x \in W : \forall z \in W : \min \set {\map f z, \map g z} \in \openint {y - \epsilon} {y + \epsilon}$
From Continuity Test for Real-Valued Functions, $\min \set {f, g}$ is continuous at $x$.
$\Box$
Case 2 : $\map f x < \map g x$
Let $\map f x < \map g x$.
Then:
- $\min \set {\map f x, \map g x} = \map f x$
Let $\delta = \min \set {\epsilon, \dfrac 1 2 \paren {\map g x - \map f x} }$.
Then $\delta > 0$.
From Continuity Test for Real-Valued Functions applied to $f$:
- $\exists U \in \tau : x \in U : \forall z \in U : \map f z \in \openint {\map f x - \delta} {\map f x + \delta}$
Similarly, Continuity Test for Real-Valued Functions applied to $g$:
- $\exists V \in \tau : x \in V : \forall z \in V : \map g z \in \openint {\map g x - \delta} {\map g x + \delta}$
Thus for all $z \in U \cap V$:
- $\map f z \in \openint {\map f x - \delta} {\map f x + \delta}$
- $\map g z \in \openint {\map g x - \delta} {\map g x + \delta}$
Now:
| \(\ds \map f z\) | \(<\) | \(\ds \map f x + \delta\) | as $\map f z \in \openint {\map f x - \delta} {\map f x + \delta}$ | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \map f x + \frac 1 2 \paren {\map g x - \map f x}\) | as $\delta \le \dfrac 1 2 \paren {\map g x - \map f x}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {2 \map f x + \map g x - \map f x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {\map g x + \map f x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {\map g x + \map g x - \map g x + \map f x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {2 \map g x - \paren{\map g x - \map f x} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map g x - \frac 1 2 \paren {\map g x - \map f x}\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \map g x - \delta\) | as $-\dfrac 1 2 \paren {\map g x - \map f x} \le -\delta$ | |||||||||||
| \(\ds \) | \(<\) | \(\ds \map g z\) | as $\map g z \in \openint {\map g x - \delta} {\map g x + \delta}$ |
Thus:
- $\min \set {\map f z, \map g z} = \map f z$
As
- $\map f z \in \openint {\map f x - \delta} {\map f x + \delta}$
and
- $\map f x = \min \set {\map f x, \map g x}$
then
- $\min \set {\map f z, \map g z} \in \openint {\min \set {\map f x, \map g x} - \delta} {\min \set {\map f x, \map g x} + \delta}$
By the Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets:
- $U \cap V \in \tau$
Since $\epsilon > 0$ was arbitrary then:
- $\forall \epsilon > 0 : \exists W \in \tau : x \in W : \forall z \in W : \min \set{\map f z, \map g z} \in \openint {\min \set {\map f x, \map g x} - \delta} {\min \set {\map f x, \map g x} + \delta}$
From Continuity Test for Real-Valued Functions, $\min \set {f, g}$ is continuous at $x$.
$\Box$
In either case, $\min \set {f, g}$ is continuous at $x$.
Since $x \in S$ was arbitrary, then $\min \set {f, g}$ is everywhere continuous.
$\blacksquare$