Editing Weak derivative (section)

== Examples ==
*The [[absolute value]] function <math>u : \mathbb{R} \rightarrow \mathbb{R}_+, u(t) = |t|</math>, which is not differentiable at <math>t = 0</math> has a weak derivative <math>v: \mathbb{R} \rightarrow \mathbb{R}</math> known as the [[sign function]], and given by <math display="block">
v(t) = \begin{cases} 1 & \text{if } t > 0; \\[6pt] 0 & \text{if } t = 0; \\[6pt] -1 & \text{if } t < 0. \end{cases}</math> This is not the only weak derivative for ''u'': any ''w'' that is equal to ''v'' [[almost everywhere]] is also a weak derivative for ''u''. For example, the definition of ''v''(0) above could be replaced with any desired real number. Usually, the existence of multiple solutions is not a problem, since functions are considered to be equivalent in the theory of [[Lp space|''L''<sup>''p''</sup> spaces]] and [[Sobolev space]]s if they are equal almost everywhere.
*The [[indicator function|characteristic function]] of the rational numbers <math> 1_{\mathbb{Q}} </math> is nowhere differentiable yet has a weak derivative.  Since the [[Lebesgue measure]] of the rational numbers is zero, <math display="block"> \int 1_{\mathbb{Q}}(t) \varphi(t) \, dt = 0.</math> Thus <math> v(t)=0 </math> is a weak derivative of <math> 1_{\mathbb{Q}} </math>.  Note that this does agree with our intuition since when considered as a member of an Lp space, <math> 1_{\mathbb{Q}} </math> is identified with the zero function.
*The [[Cantor function]] ''c'' does not have a weak derivative, despite being differentiable almost everywhere. This is because any weak derivative of ''c'' would have to be equal almost everywhere to the classical derivative of ''c'', which is zero almost everywhere. But the zero function is not a weak derivative of ''c'', as can be seen by comparing against an appropriate test function <math>\varphi</math>. More theoretically, ''c'' does not have a weak derivative because its [[distributional derivative]], namely the [[Cantor distribution]], is a [[singular measure]] and therefore cannot be represented by a function.