Combination Theorem for Continuous Mappings/Metric Space/Minimum Rule
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Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $\R$ denote the real numbers.
Let $f: M \to \R$ and $g: M \to \R$ be real-valued functions from $M$ to $\R$ which are continuous on $M$.
Let $\min \set {f, g}: M \to \R$ denote the pointwise maximum of $f$ and $g$.
Then:
- $\min \set {f, g}$ is continuous on $M$.
Proof
Let $a \in M$ be arbitrary.
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 Mappings on Metric Space:
- $\map f x - \map g x$ is continuous at $a$.
From Absolute Value Rule for Continuous Mappings on Metric Space:
- $\size {\map f x - \map g x}$ is continuous at $a$.
From Sum Rule for Continuous Mappings on Metric Space:
- $\map f x + \map g x$ is continuous at $a$
From Difference Rule for Continuous Mappings on Metric Space:
- $\map f x + \map g x - \size {\map f x - \map g x}$ is continuous at $a$
From Multiple Rule for Continuous Mappings on Metric Space:
- $\dfrac 1 2 \paren {\map f x + \map g x - \size {\map f x - \map g x} }$ is continuous at $a$.
As $a$ is arbitrary:
- $\min \set {f, g}$ is continuous on $M$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: Exercise $2.6: 14$