The signum function is the function $\signum\colon\R\to \R$ \begin{eqnarray*} \signum (x) &=& \left\{ \begin {array}{ll} -1 & \mbox{when}\,\, x<0, \\ 0 & \mbox{when}\,\, x=0,\\ 1 & \mbox{when}\,\, x>0. \\ \end{array}.

The following properties hold:

1. For all $x\in \R$ , $\signum(-x) = -\signum(x).$
2. For all $x\in \R$ , $|x|=\signum(x) x.$
3. For all $x\neq 0$ , $\frac{d}{dx}|x|=\signum(x)$ .

Here, we should point out that the signum function is often defined simply as $1$ for $x>0$ and $-1$ for $x<0$ . Thus, at $x=0$ , it is left undefined. See for example [1]. In applications such as the Laplace transform this definition is adequate, since the value of a function at a single point does not change the analysis. One could then, in fact, set $\signum(0)$ to any value. However, setting $\signum(0)=0$ is motivated by the above relations. On a related note, we can extend the definition to the extended real numbers $\overline{\mathbb{R}}=\mathbb{R}\cup\{\infty,-\infty\}$ by defining $\signum(\infty)=1$ and $\signum(-\infty)=-1$ .

A related function is the Heaviside step function defined as \begin{eqnarray*} H(x) &=& \left\{ \begin {array}{ll} 0 & \mbox{when}\,\, x< 0, \\ 1/2 & \mbox{when}\,\, x= 0,\\ 1 & \mbox{when}\,\, x> 0.\\ \end{array}. Again, this function is sometimes left undefined at $x=0$ . The motivation for setting $H(0)=1/2$ is that for all $x\in\R$ , we then have the relations \begin{eqnarray*} H (x) &=& \frac{1}{2}(\signum(x)+1), \\ H(-x) &=& 1-H(x). \end{eqnarray*}This first relation is clear. For the second, we have \begin{eqnarray*} 1-H(x) &=& 1-\frac{1}{2}(\signum(x)+1) \\ &=& \frac{1}{2}(1-\signum(x)) \\ &=& \frac{1}{2}(1+\signum(- x)) \\ &=& H(-x). \end{eqnarray*} Example Let $a<b$ be real numbers, and let $f:\R\to\R$ be the piecewise defined function \begin{eqnarray*} f (x) &=& \left\{ \begin {array}{ll} 4 & \mbox{when}\,\, x\in(a,b), \\ 0 & \mbox{otherwise.} \\ \end{array}. Using the Heaviside step function, we can write \begin{eqnarray} \label{almost} f(x) &=& 4\big(H(x-a) - H(x-b)\big) \end{eqnarray}almost everywhere. Indeed, if we calculate $f$ using equation we obtain $f(x)=4$ for $x\in(a,b)$ , $f(x)=0$ for $x\notin[a,b]$ , and $f(a)=f(b)=2$ . Therefore, equation holds at all points except $a$ and $b$ .

Signum function for complex arguments

Bibliography

1

E. Kreyszig, Advanced Engineering Mathematics, John Wiley & Sons, 1993, 7th ed.

2

G. Bachman, L. Narici, Functional analysis, Academic Press, 1966.