Exercise 81Calculus › Single-variable calculus › Continuity › Continuous and discontinuous functionsL1Added 2 Nov 2026

The absolute value of a continuous function

#continuity#absolute value#counterexamples

Statement

Show that if $f$ is continuous at $a$ , then $|f|$ is continuous at $a$ .
Show that the converse is false.

Hint

For part 1, use the reverse triangle inequality $\bigl||u|-|v|\bigr|\le|u-v|$ . For part 2, try a function that only takes the values $1$ and $-1$ .

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Solution

1. For all $x$ , $\bigl||f(x)|-|f(a)|\bigr|\le|f(x)-f(a)|$ . Given $\varepsilon>0$ , continuity of $f$ gives $\delta>0$ with $|f(x)-f(a)|<\varepsilon$ when $|x-a|<\delta$ ; the same $\delta$ works for $|f|$ .

2. Let $f(x)=1$ if $x$ is rational and $f(x)=-1$ if $x$ is irrational. Then $|f|=1$ is constant, hence continuous. But $f$ is discontinuous at every point: near any $a$ there are rationals (where $f=1$ ) and irrationals (where $f=-1$ ).

Remark. Taking the absolute value can only destroy sign jumps, never create new discontinuities. That is why $|f|$ is "more continuous" than $f$ .

The absolute value of a continuous function

Statement

Solution

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