When Is the Absolute Value of a Function Continuous?

The absolute value of a real-valued function is continuous wherever the original function is continuous. This follows because taking an absolute value is itself a continuous operation.

However, the converse is false in general: a continuous absolute-value function does not guarantee that the original function is continuous. The issue is that absolute value removes sign information.

The basic theorem

Let $f:D\to\mathbb{R}$ be a function, where $D$ is a subset of $\mathbb{R}$. If $f$ is continuous at $a\in D$, then the function

$$g(x)=|f(x)|$$

is continuous at $a$.

There are two standard ways to see this.

First, the function $u\mapsto |u|$ is continuous on $\mathbb{R}$. Therefore $g$ is the composition of two continuous functions:

$$g=|\cdot|\circ f.$$

Second, use the inequality

$$\bigl||f(x)|-|f(a)|\bigr|\leq |f(x)-f(a)|.$$

If $f(x)\to f(a)$ as $x\to a$, then the right-hand side tends to zero. Consequently,

$$|f(x)|\to |f(a)|,$$

which is exactly the continuity of $|f|$ at $a$.

Epsilon–delta proof

Suppose $f$ is continuous at $a$. To prove that $|f|$ is continuous at $a$, let $\varepsilon>0$.

Since $f$ is continuous at $a$, there exists $\delta>0$ such that

$$|x-a|<\delta \quad\Longrightarrow\quad |f(x)-f(a)|<\varepsilon.$$

The reverse triangle inequality gives

$$\bigl||f(x)|-|f(a)|\bigr|\leq |f(x)-f(a)|<\varepsilon.$$

Thus, whenever $|x-a|<\delta$,

$$\bigl||f(x)|-|f(a)|\bigr|<\varepsilon.$$

Therefore $|f|$ is continuous at $a$.

The same argument works when the domain is a metric space and the codomain is a normed vector space: if $F$ is continuous, then $x\mapsto \|F(x)\|$ is continuous.

Worked example

Take

$$f(x)=x^2-3x+1.$$

Because $f$ is a polynomial, it is continuous on $\mathbb{R}$. Therefore

$$g(x)=|x^2-3x+1|$$

is continuous on $\mathbb{R}$.

The formula for $g$ can be written piecewise by finding the zeros of $f$:

$$x^2-3x+1=0$$

has roots

$$x=\frac{3-\sqrt{5}}{2}\quad\text{and}\quad x=\frac{3+\sqrt{5}}{2}.$$

Thus $f$ changes sign at those two points, but $|f|$ remains continuous there. Explicitly,

$$|f(x)|= \begin{cases} x^2-3x+1, & x\leq \frac{3-\sqrt{5}}{2},\\ -(x^2-3x+1), & \frac{3-\sqrt{5}}{2}\leq x\leq \frac{3+\sqrt{5}}{2},\\ x^2-3x+1, & x\geq \frac{3+\sqrt{5}}{2}. \end{cases}$$

Although the derivative may fail to exist at a zero where the sign changes, continuity is preserved. For example, $|x|$ is continuous at $0$ but is not differentiable there.

The converse is false

It is tempting to claim

$$|f|\text{ continuous}\quad\Longrightarrow\quad f\text{ continuous},$$

but this is not true.

Define

$$f(x)= \begin{cases} 1, & x\geq 0,\\ -1, & x<0. \end{cases}$$

The function $f$ is discontinuous at $0$: its left-hand limit is $-1$, while its right-hand limit and value at $0$ are $1$.

But

$$|f(x)|=1$$

for every $x$. Hence $|f|$ is constant and therefore continuous everywhere.

The absolute value cannot detect this discontinuity because both $1$ and $-1$ have the same absolute value.

When can the converse hold?

The converse does hold under additional sign information. Suppose $|f|$ is continuous at $a$ and $f$ has a fixed sign in some neighborhood of $a$.

If $f(x)\geq 0$ near $a$, then $f(x)=|f(x)|$ near $a$, so $f$ is continuous at $a$ whenever $|f|$ is.

If $f(x)\leq 0$ near $a$, then $f(x)=-|f(x)|$ near $a$, so again $f$ is continuous at $a$.

The condition $f(a)\neq 0$ by itself is not sufficient. For example, define

$$f(x)= \begin{cases} 1, & x=0,\\ -1, & x\neq 0. \end{cases}$$

Then $|f(x)|=1$ for every $x$, so $|f|$ is continuous everywhere, but $f$ is discontinuous at $0$ even though $f(0)\neq 0$.

Thus, what matters is a fixed-sign condition, or another hypothesis that rules out sign switching, rather than merely the fact that $f(a)$ is nonzero. A sign change can occur without the function taking the value zero when no intermediate-value property is available.

Common mistakes

Confusing continuity and differentiability

If $f$ is continuous, then $|f|$ is continuous. But $|f|$ may not be differentiable where $f$ is zero and changes sign. The function $f(x)=x$ gives $|f(x)|=|x|$, which is continuous but not differentiable at zero.

Assuming absolute values preserve all information

The equality $|f(x)|=|f(a)|$ does not imply $f(x)=f(a)$. The values could have opposite signs. This is why continuity of $|f|$ alone cannot always establish continuity of $f$.

Ignoring the domain

Continuity is considered relative to the domain. For example, if a function is defined only on $[0,\infty)$, continuity at $0$ uses points from the domain and does not require a two-sided limit from all real numbers.

Forgetting the reverse triangle inequality

The estimate

$$\bigl||u|-|v|\bigr|\leq |u-v|$$

is often the quickest proof. It says that the absolute-value operation is Lipschitz continuous with constant $1$.

A useful generalization

The same reasoning applies to any continuous function $\phi:\mathbb{R}\to\mathbb{R}$. If $f$ is continuous, then $\phi\circ f$ is continuous. Taking $\phi(t)=|t|$ gives the result here.

More generally, if $f$ is continuous and $p$ is a positive integer, then $x\mapsto |f(x)|^p$ is continuous. For $p=1$, this is the absolute-value result; for $p=2$, it follows from $|f(x)|^2=f(x)^2$.