Prove the absolute value function of a continuous function is continuous

Question

Suppose that $f$ is a continuous function defined on an interval $I$. Prove that $|f|$ is continuous on $I$.

Our definition of continuity: Let $I$ be an interval, let $f:I\rightarrow\Bbb{R}$, and let $c\in I$. The function $f$ is continuous at $c$ if for each $\epsilon>0$ there exists $\delta>0$ such that $|f(x)-f(c)| < \epsilon$ for all $x\in I$ that satisfy $|x-c|<\delta$. The function $f$ is continuous on $I$ if $f$ is continuous at each point of $I$.

Do you already know that the composition of continuous functions is continuous? — Daniel Fischer, Oct 27 '13 at 23:37
or that $||x|-|y|| \leqslant |x-y|$ for $x, y \in \mathbf{R}$? — knsam, Oct 27 '13 at 23:37
You can prove the continuity of $|x|$ use the reverse triangle inequality. — Hanul Jeon, Oct 28 '13 at 00:50

score 7 · Answer 1 · answered Jul 15 '14 at 13:22

Use the fact that $|x|$ is continuous and hence $|f|$ is continuous.(Composition of continuous functions)
$||x|-|y||\le|x-y|$ therefore for every $\varepsilon>0$ there exists a $\delta>0$ such that $|x-t|<\delta \implies ||f(x)|-|f(t)||\le |f(x)-f(t)|<\varepsilon$.

Prove the absolute value function of a continuous function is continuous

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