Distributional Derivative of Heaviside Step Function times Sine

Theorem

Let $H$ be the Heaviside step function.

Let $\delta$ be the Dirac delta distribution.

Then in the distributional sense:

$T'_{\map H x \sin x} = T_{\map H x \cos x} $

Proof

$x \stackrel f {\longrightarrow} \map H x \sin x$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.

By Differentiable Function as Distribution we have that $T'_f = T_{f'}$.

Moreover:

$x < 0 \implies \paren {{\map H x} \map \sin x}' = 0$.

$x > 0 \implies \paren {{\map H x} \map \sin x}' = \cos x$.

Altogether:

$\forall x \in \R \setminus \set 0 : \paren {{\map H x} \map \sin x}' = \map H x \cos x$

Furthermore:

$\ds \map f {0^+} - \map f {0^-} = \map H {0^+} \map \sin {0^+} - \map H {0^-} \map \sin {0^-} = 0$

The result follows from the Jump Rule.

$\blacksquare$

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