Applied linear operators and spectral methods/Differentiating distributions

Differentiation of a distribution
If $$f(x)$$ is a differentiable function in $$\mathbb{R}$$ whose first derivative is locally integrable, then the derivative $$f'(x)$$ defines its own distribution

\left\langle f', \phi \right\rangle := \int_{-\infty}^{\infty} f'(x)~\phi(x)~\text{d}x $$ Integrating by parts and noting that at the boundaries $$x = \pm \infty$$, $$\phi(x) = 0$$ because of compact support, we have

\left\langle f', \phi \right\rangle := -\int_{-\infty}^{\infty} f(x)~\phi'(x)~\text{d}x = : \left\langle f, \phi' \right\rangle $$ This suggests defining the derivative of any distribution $$f$$ via
 * $$ \text{(1)} \qquad

{ \left\langle f', \phi \right\rangle = - \left\langle f, \phi' \right\rangle } $$ We need to check that this definition satisfies the properties of a distribution. We observe that if $$\phi$$ is a test function then so is $$-\phi'$$. Linearity is obvious. What about continuity? If $$\{\phi_n\}$$ is a zero sequence, so is the sequence $$\{-\phi_n\}$$ and thus $$\left\langle f, -\phi'_n \right\rangle$$ tends to zero as $$n \rightarrow \infty$$. Hence (1) defines a distribution.

 Remark: All distributions are infinitely differentiable is the class of test functions includes only infinitely differentiable functions.

 Comment: One can take test functions which are only $$C^n$$ differentiable. Then the distribution will only be $$n$$ times differentiable. Thus, enlarging the class of test functions reduces the class of distributions.

More generally, if $$f^{(n)}$$ is the $$n$$th derivative of the distribution $$f$$,

{ \left\langle f^{(n)}, \phi \right\rangle = - \left\langle f, \phi^{(n)} \right\rangle } $$

Distributions can be generated by functions which are not differentiable in the ordinary sense. However, we can differentiate them in the  distribution sense.

For example, the derivative of the delta distribution gives us the dipole distribution:

\left\langle \delta'_0, \phi \right\rangle = - \left\langle \delta_0, \phi' \right\rangle = - \phi'(0) $$

Also, the derivative of the Heaviside function is given by

\left\langle H', \phi \right\rangle = - \left\langle H, \phi' \right\rangle = - \int_0^{\infty} \phi'(x)~\text{d}x $$ Since $$\phi(\infty) = 0$$, from the fundamental theorem of calculus,

\left\langle H', \phi \right\rangle = - \phi(0) = \left\langle \delta_0, \phi \right\rangle $$ Therefore the derivative of the Heaviside function is the delta function.