Absolute Value of Continuous Real Function is Continuous
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Theorem
Let $f: \R \to \R$ be a real function.
Let $f$ be continuous at a point $a \in \R$.
Then:
- $\size f$ is continuous at $a$
where:
- $\map {\size f} x$ is defined as $\size {\map f x}$.
Proof
Let $\epsilon \in \R_{>0}$ be a positive real number.
Because $f$ is continuous at $a$:
- $\exists \delta \in \R_{>0}: \size {x - a} < \delta \implies \size {\map f x - \map f a} < \epsilon$
From Reverse Triangle Inequality:
- $\size {\size {\map f x} - \size {\map f a} } \le \size {\map f x - \map f a}$
and so:
- $\exists \delta \in \R_{>0}: \size {x - a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$
$\epsilon$ is arbitrary, so:
- $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \size {x - a} < \delta \implies \size {\size {\map f x} - \size {\map f a} } < \epsilon$
That is, by definition:
- $\size f$ is continuous at $a$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 17$