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In general, the indicator function of a set is not smooth; it is continuous if and only if its [[support (math)|support]] is a [[connected component (topology)|connected component]]. In the [[algebraic geometry]] of [[finite fields]], however, every [[affine variety]] admits a ([[Zariski topology|Zariski]]) continuous indicator function.<ref>{{Cite book|title=Course in Arithmetic|last=Serre|pages=5}}</ref> Given a [[finite set]] of functions <math>f_\alpha \in \mathbb{F}_q[x_1,\ldots,x_n]</math> let <math>V = \left\{ x \in \mathbb{F}_q^n : f_\alpha(x) = 0 \right\}</math> be their vanishing locus. Then, the function <math display="inline">P(x) = \prod\left(1 - f_\alpha(x)^{q-1}\right)</math> acts as an indicator function for <math>V</math>. If <math>x \in V</math> then <math>P(x) = 1</math>, otherwise, for some <math>f_\alpha</math>, we have <math>f_\alpha(x) \neq 0</math>, which implies that <math>f_\alpha(x)^{q-1} = 1</math>, hence <math>P(x) = 0</math>.
Although indicator functions are not smooth, they admit [[weak
and similarly the distributional derivative of <math display="block">G(x) := \mathbf{1}_{x < 0}</math> is <math display=block>\frac{d G(x)}{dx}=-\delta(x)</math>
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