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