Distributional Derivative of Absolute Value Function

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Theorem

Let $H: \R \to \closedint 0 1$ be the Heaviside step function.

Let $\size x$ be the absolute value of $x$.

Let $T_{\size x}$ be the distribution associated with $\size x$.


Then the distributional derivative of $T_{\size x}$ is $T_{2 H - 1}$


Proof

\(\ds \dfrac {\d \size x} {\d x}\) \(=\) \(\ds \begin{cases} 1 & : x > 0 \\ -1 & : x < 0 \end{cases}\)
\(\ds \) \(=\) \(\ds -1 + \begin{cases} 2 & : x > 0 \\ 0 & : x < 0 \end{cases}\)
\(\ds \) \(=\) \(\ds -1 + 2 \begin{cases} 1 & : x > 0 \\ 0 & : x < 0 \end{cases}\)
\(\ds \) \(=\) \(\ds 2 \map H x - 1\) Definition of Heaviside Step Function

Furthermore:

$\ds \lim_{x \mathop \to 0^+} \size x = \lim_{x \mathop \to 0^-} \size x = 0$

By the Jump Rule:

$T_{\size x}' = T_{2H - 1}$

$\blacksquare$


Also see


Sources