If $|f(x)|$ is continuous at $a$, is $f(x)$ continuous at $a$?

I tried doing it using composite functions. If $g(x)= |x|$, then $g\circ f(x)= |f(x)|$. Since $g(x)$ and $g\circ f(x)$ are continuous, $f(x)$ is continuous.

I don't know if this is correct. Please help.

  • 16
    $\begingroup$ composition of continuous functions doesn't work that way. If it did, we could let $g = 0$ be a constant function, and then $g\circ f$ is also a constant function, so $g$ and $g \circ f$ are continuous, and thus every function is continuous! $\endgroup$
    – jxnh
    Oct 3, 2014 at 15:29
  • $\begingroup$ It does work the other way around though: if $f(x)$ is continuous, so is $|f(x)|$. But that's quite trivial to show, as you only need to consider the points where $f(x) = 0$. $\endgroup$
    – CompuChip
    Oct 4, 2014 at 7:39

4 Answers 4


Let $f(x)=-1$ if $x$ is rational, and let $f(x)=1$ if $x$ is irrational.

Or else more modestly let $f(x)=-1$ if $x\lt 17$, and $f(x)=1$ for $x\ge 17$.

  • $\begingroup$ I wonder why you chose 17 particularly. $\endgroup$ May 14, 2020 at 20:24

This is not always true. If $f(x)$ is a piecewise function such that

$f(x)=1$ for $x<a$ or $x=a$ and

$f(x)=-1$ for $x>a$

Then $|f(x)|$ is continuous at $a$ but $f(x)$ is not.


Maybe you are misunderstanding the logic behind it:

Let $g(x)$ be continuous, OK.

Now, saying

$g \circ f(x)$ is continous if $f(x)$ is continuous

doesn't imply that

$g \circ f(x)$ is not continuous if $f(x)$ is not continuous

Also, it's never true to say that $g \circ f(x)$ is continuous because $f(x)$ is continuous.

The single (and obvious) conclusion that we can take is that if $f(x)$ is continuous, then $g \circ f(x)$ certainly is continuous too. But if $f(x)$ is discontinuous, then $g \circ f(x)$ may be, or may be not.

So you can deduce nothing about $f(x)$ from the behaviour of $g \circ f(x)$.


In programming speak as I don't know how to express this in math speak:

f(x) = odd(x) ? x : -x;

Ed.: that would be

$$ f(x) = \begin{cases} x \quad \text{if $x$ is odd} \\ -x \quad \text{otherwise} \end{cases} $$

  • $\begingroup$ This function is continuous. (I assume its domain is the integers?) $\endgroup$ Oct 5, 2014 at 0:39
  • 2
    $\begingroup$ Why would you call it continuous? $\endgroup$ Oct 5, 2014 at 3:32
  • $\begingroup$ +1. After thinking about it, I can see no reason to presume that its domain is the integers. I believe that the original edit to express it in math speak was in error, since in the programming language version there is no mention of evenness. $\endgroup$ Oct 5, 2014 at 17:05
  • $\begingroup$ I assumed the domain was the integers because the way I see it, a number $x$ has to be an integer for the statement '$x$ is odd' to make sense. I guess that's a matter of convention, though. The wording 'if $x$ is an odd integer', plus an indication of what the domain is, would make the answer clearer. $\endgroup$ Oct 23, 2014 at 15:23

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