Min Operation on Continuous Real Functions is Continuous

Theorem
Let $f: \R \to \R$ and $g: \R \to \R$ be real functions.

Let $f$ and $g$ be continuous at a point $a \in \R$.

Let $h: \R \to \R$ be the real function defined as:


 * $\map h x := \map \min {\map f x, \map g x}$

Then $h$ is continuous at $a$.

Proof
From Min Operation Representation on Real Numbers
 * $\min \set{x, y} = \dfrac 1 2 \paren {x + y - \size {x - y} }$

Hence:
 * $\min \set {\map f x, \map g x} = \dfrac 1 2 \paren {\map f x + \map g x - \size {\map f x - \map g x} }$

From Difference Rule for Continuous Real Functions:
 * $\map f x - \map g x$ is continuous at $a$.

From Absolute Value of Continuous Real Function is Continuous:
 * $\size {\map f x - \map g x}$ is continuous at $a$.

From Sum Rule for Continuous Real Functions:
 * $\map f x + \map g x$ is continuous at $a$

and hence from Difference Rule for Continuous Real Functions again:
 * $\map f x + \map g x - \size {\map f x - \map g x}$ is continuous at $a$

From Multiple Rule for Continuous Real Functions:
 * $\dfrac 1 2 \paren {\map f x + \map g x - \size {\map f x - \map g x} }$ is continuous at $a$.

Also see

 * Max Operation on Continuous Real Functions is Continuous