Sign Function
Summarize
Introduction to the sign function
The sign function assigns to each real number its sign, disregarding its magnitude. The function has the following piecewise definition:
$$ \mathrm{sgn}(x) = \begin{cases} -1 & \text{if } x < 0 \\[6pt] 0 & \text{if } x = 0 \\[6pt] 1 & \text{if } x > 0 \end{cases} \quad \forall \ x \in \mathbb{R} $$
Applying the definition we have:
$$ \mathrm{sgn}(-7) = -1 \qquad \mathrm{sgn}(0) = 0 \qquad \mathrm{sgn}(4) = 1 $$
The graph of $y = \mathrm{sgn}(x)$ consists of two horizontal rays and one isolated point. The ray on $y = -1$ extends over all $x < 0,$ the ray on $y = 1$ extends over all $x > 0,$ and the isolated point at the origin $(0, 0)$ lies on $y = 0.$ The two rays approach but do not intersect the y-axis.
The sign function is an odd function because it satisfies the identity:
$$ \mathrm{sgn}(-x) = -\mathrm{sgn}(x) \quad \forall \ x \in \mathbb{R} $$
The sign function is one of the simplest examples of a discontinuous function.
Properties
- Domain: $\mathbb{R}.$
- Range: $\{-1,\ 0,\ 1\}.$
- The function is odd, since $\mathrm{sgn}(-x) = -\mathrm{sgn}(x).$
- The function has exactly one root at $x = 0,$ since $\mathrm{sgn}(x) = 0$ only when $x = 0.$
- The function is constant on $(-\infty, 0)$ and on $(0, +\infty),$ so it never decreases on either piece, which makes it non-decreasing over $\mathbb{R}.$
- The function has a jump discontinuity at $x = 0$ and is continuous everywhere else.
- The function is not differentiable at $x = 0.$ It is differentiable, with zero derivative, at every other point.
- The function is multiplicative, since $\mathrm{sgn}(xy) = \mathrm{sgn}(x)\mathrm{sgn}(y)$ for all $x, y \in \mathbb{R}.$ Restricted to $\mathbb{R} \setminus \{0\},$ it is a homomorphism from the multiplicative group $(\mathbb{R}^*, \cdot)$ onto $\{-1, 1\}.$
The two one-sided limits at the origin are:
$$ \begin{align} \lim_{x \to 0^-} \mathrm{sgn}(x) &= -1 \\[6pt] \lim_{x \to 0^+} \mathrm{sgn}(x) &= 1 \end{align} $$
Since the two one-sided limits differ, the two-sided limit does not exist:
$$\lim_{x \to 0} \mathrm{sgn}(x)$$
Limits at infinity
As $x$ moves away from the origin in either direction, the sign function stays at a constant value:
$$ \begin{align} \lim_{x \to -\infty} \mathrm{sgn}(x) &= -1 \\[6pt] \lim_{x \to +\infty} \mathrm{sgn}(x) &= 1 \end{align} $$
These follow from the fact that $\mathrm{sgn}(x) = -1$ for all $x < 0$ and $\mathrm{sgn}(x) = 1$ for all $x > 0,$ so the function value does not change as $x$ moves further from zero.
Derivative and integral
On each of the two open half-lines where the sign function is constant, its derivative is zero:
$$ \frac{d}{dx} \mathrm{sgn}(x) = 0 \quad \text{for } x \neq 0 $$
At $x = 0$ the derivative does not exist because the function is discontinuous there. In the sense of distributions, however, the derivative of the sign function is:
$$ \frac{d}{dx} \mathrm{sgn}(x) = 2\delta(x) $$
$\delta(x)$ is the Dirac delta, a generalised function that is zero everywhere except at the origin and whose total integral over the real line equals one.
The Dirac delta can be pictured as a spike of unit area concentrated at a single point.
This distributional identity records a jump of amplitude $2$ at the origin, since:
$$ \begin{align} \lim_{x \to 0^-} \mathrm{sgn}(x) &= -1 \\[6pt] \lim_{x \to 0^+} \mathrm{sgn}(x) &= 1 \end{align} $$
The jump from $-1$ to $1$ accounts for the factor $2$ preceding the Dirac delta.
The indefinite integral of the sign function, computed away from the origin, gives back the absolute value:
$$ \int \mathrm{sgn}(x) \ dx = |x| + c $$
This agrees with the fact that the derivative of $|x|$ equals $\mathrm{sgn}(x)$ wherever the former is differentiable.
Relationship with the absolute value function
A direct algebraic relationship connects the sign function and the absolute value. For any $x \neq 0,$ the following identity holds:
$$ \mathrm{sgn}(x) = \frac{x}{|x|} $$
This agrees with the definition: when $x > 0,$ the ratio $x/|x| = x/x = 1.$ When $x < 0,$ the ratio $x/|x| = x/(-x) = -1.$ The formula is undefined at $x = 0,$ so the value $\mathrm{sgn}(0) = 0$ is assigned separately by convention.
Conversely, the absolute value can be written in terms of the sign function:
$$ |x| = x \cdot \mathrm{sgn}(x) $$
This identity holds for all $x \in \mathbb{R},$ including $x = 0,$ where both sides are zero.
Two further identities arise from this relationship. The first writes any real number as the product of its magnitude and its sign:
$$ x = |x| \cdot \mathrm{sgn}(x) $$
The second uses the equality $|x| = \sqrt{x^2},$ valid for all $x \in \mathbb{R},$ and gives an alternative representation of the sign function for $x \neq 0$:
$$ \mathrm{sgn}(x) = \frac{x}{\sqrt{x^2}} $$
Relationship with the Heaviside step function
The Heaviside step function $H(x)$ is defined as:
$$ H(x) = \begin{cases} 0 & \text{if } x < 0 \\[6pt] \dfrac{1}{2} & \text{if } x = 0 \\[8pt] 1 & \text{if } x > 0 \end{cases} $$
The sign function and the Heaviside step function are related by a linear transformation:
$$ \mathrm{sgn}(x) = 2H(x) - 1 $$
Equivalently, we have:
$$ H(x) = \frac{1 + \mathrm{sgn}(x)}{2} $$
The Heaviside function maps $(-\infty, 0)$ to $0$ and $(0, +\infty)$ to $1,$ and may therefore be read as a shifted and rescaled form of the sign function.