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. 1.

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

2. 2.

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

3. 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 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