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
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense