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 $\epsilon > 0$.

, 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 :
 * $U \cap V \in \tau$

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 $x$.

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:

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 :
 * $U \cap V \in \tau$

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 $x$.

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.

Also see

 * Maximum Rule for Continuous Functions