signum function

The signum function is the function $\mathop{\mathrm{sgn}}\colon\mathbb{R}\to\mathbb{R}$

\displaystyle\mathop{\mathrm{sgn}}(x)

\displaystyle=

\displaystyle\left\{\begin{array}[]{ll}-1&\mbox{when}\,\,x<0,\\ 0&\mbox{when}\,\,x=0,\\ 1&\mbox{when}\,\,x>0.\\ \end{array}\right.

The following properties hold:

1.

For all $x\in\mathbb{R}$ , $\mathop{\mathrm{sgn}}(-x)=-\mathop{\mathrm{sgn}}(x).$
2.

For all $x\in\mathbb{R}$ , $|x|=\mathop{\mathrm{sgn}}(x)x.$
3.

For all $x\neq 0$ , $\frac{d}{dx}|x|=\mathop{\mathrm{sgn}}(x)$ .

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 $\mathop{\mathrm{sgn}}(0)$ to any value. However, setting $\mathop{\mathrm{sgn}}(0)=0$ 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 $\mathop{\mathrm{sgn}}(\infty)=1$ and $\mathop{\mathrm{sgn}}(-\infty)=-1$ .

A related function is the Heaviside step function defined as

\displaystyle H(x)

\displaystyle=

\displaystyle\left\{\begin{array}[]{ll}0&\mbox{when}\,\,x<0,\\ 1/2&\mbox{when}\,\,x=0,\\ 1&\mbox{when}\,\,x>0.\\ \end{array}\right.

Again, this function is sometimes left undefined at $x=0$ . The motivation for setting $H(0)=1/2$ is that for all $x\in\mathbb{R}$ , we then have the relations

	$\displaystyle H(x)$	$\displaystyle=$	$\displaystyle\frac{1}{2}(\mathop{\mathrm{sgn}}(x)+1),$
	$\displaystyle H(-x)$	$\displaystyle=$	$\displaystyle 1-H(x).$

This first relation is clear. For the second, we have

$\displaystyle 1-H(x)$	$\displaystyle=$	$\displaystyle 1-\frac{1}{2}(\mathop{\mathrm{sgn}}(x)+1)$
	$\displaystyle=$	$\displaystyle\frac{1}{2}(1-\mathop{\mathrm{sgn}}(x))$
	$\displaystyle=$	$\displaystyle\frac{1}{2}(1+\mathop{\mathrm{sgn}}(-x))$
	$\displaystyle=$	$\displaystyle H(-x).$

Example Let $a<b$ be real numbers, and let $f:\mathbb{R}\to\mathbb{R}$ be the piecewise defined function

\displaystyle f(x)

\displaystyle=

\displaystyle\left\{\begin{array}[]{ll}4&\mbox{when}\,\,x\in(a,b),\\ 0&\mbox{otherwise.}\\ \end{array}\right.

Using the Heaviside step function, we can write

\displaystyle f(x)

\displaystyle=

\displaystyle 4\big{(}H(x-a)-H(x-b)\big{)}

(1)

almost everywhere. Indeed, if we calculate $f$ using equation 1 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 1 holds at all points except $a$ and $b$ . $\Box$

1 Signum function for complex arguments

For a complex number $z$ , the signum function is defined as [2]

\displaystyle\mathop{\mathrm{sgn}}(z)

\displaystyle=

\displaystyle\left\{\begin{array}[]{ll}0&\mbox{when}\,\,z=0,\\ {z}/{|z|}&\mbox{when}\,\,z\neq 0.\\ \end{array}\right.

In other words, if $z$ is non-zero, then $\mathop{\mathrm{sgn}}z$ is the projection of $z$ onto the unit circle $\{z\in\mathbb{C}\mid|z|=1\}$ . Clearly, the complex signum function reduces to the real signum function for real arguments. For all $z\in\mathbb{C}$ , we have

z\mathop{\mathrm{sgn}}\overline{z}=|z|,

where $\overline{z}$ is the complex conjugate of $z$ .

References

1 E. Kreyszig, Advanced Engineering Mathematics, John Wiley & Sons, 1993, 7th ed.
2 G. Bachman, L. Narici, Functional analysis, Academic Press, 1966.

Title	signum function
Canonical name	SignumFunction
Date of creation	2013-03-22 13:36:41
Last modified on	2013-03-22 13:36:41
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	11
Author	yark (2760)
Entry type	Definition
Classification	msc 30-00
Classification	msc 26A06
Related topic	ModulusOfComplexNumber
Related topic	HeavisideStepFunction
Related topic	PlusSign
Related topic	SineIntegralInInfinity
Related topic	ListOfImproperIntegrals
Defines	Heavyside step function
Defines	step function