But before that, let's list some basic properties which are more evocative of elementary real analysis than anything else. For a distribution $\langle f, \phi\rangle$:
- Linearity, i.e. $f(a\phi_1+\phi_2)$ for any real constant $a$ and test functions $\phi_1,\ \phi_2$;
- There exists a sequence of test functions $\{\phi_n\}$ such that $\phi_n \to f$
The real magic starts when we attempt to translate the distribution. Recall that any function can be translated $y$ units by taking $f(x-y)$ instead of $f(x)$; the same thing can be done for generalized functions by considering $\lim_{n\to\infty}\langle \phi_n(x-y),\phi(x)\rangle$. (Let's define the translation function tau as $\tau_y\phi(x)=\phi(x-y)$.) Using some simple $u$-substitution magic, \begin{align}\langle\tau_yT_f,\phi\rangle &=\lim_{n\to\infty}\langle \phi_n(x-y),\phi(x)\rangle\\&=\lim_{n\to\infty}\int_{\mathbb{R}}\phi_n(x-y)\phi(x)\;dx;\qquad u = x-y\\&=\lim_{n\to\infty}\int_{\mathbb{R}}\phi_n(u)\phi(u+y)\;du\\&=\langle T_f,\tau_{-y}\phi\rangle.\end{align} We have essentially found that any distribution can be translated by applying the opposite translation to every test function in $\mathcal{D}$. To reiterate:$$\langle\tau_yT_f,\phi\rangle =\langle T_f,\tau_{-y}\phi\rangle.$$ Hooray!
Differentiating a distribution works in much the same way as translation in that the operation gets pawned off onto the test function but with an extra minus sign. However, it does involve an extra technique: integration by parts. I assume that nobody who is reading this is unfamiliar with the practice, but, for the sake of cute mnemonics, a friend of my fiancé's refers to $$\int u\;dv = uv - \int v \;du$$as "sudv uv svidoo."
Let's take a moment to appreciate how adorable that is.
The actual fancy differentiation trick can be proved in essentially one integration-by-parts step:\begin{align}\left\langle \frac{d}{dx}T_f,\phi\right\rangle &= \lim_{n\to\infty}\int_\mathbb{R}\left(\frac{d}{dx}\phi_n(x)\right)\phi(x)\;dx\\&= \lim_{n\to\infty}-\int_\mathbb{R}\phi_n(x)\left(\frac{d}{dx}\phi(x)\right)\;dx \\ &=\left\langle T_f, -\frac{d}{dx}\phi(x)\right\rangle\end{align}(brownie points if you've already figured out what happened to the $uv$ term). This identity is essential for a crazy number of distributional calculus proofs.
For example, we can directly use this identity to prove the Dirac delta function is the distributional derivative of the Heaviside function in two seconds. Let $T_H= \int_0^\infty \phi(x)\,dx$ represent the Heaviside distribution. Now, from the above identity, we conclude $$\left\langle \frac{d}{dx}T_H(x),\phi\right\rangle=\left\langle T_H(x),\frac{d}{dx}\phi\right\rangle=-\int_0^\infty \phi(x)\,dx=\phi(0)-\phi(\infty)=\phi(0),$$ that is, because $\phi$ is zero at infinity. Yet $\phi(0)=\langle \delta, \phi\rangle$ by definition! We're done here.
As super awesome as that is, there should be some material on how all this pertains to weak solutions of DEs up on Thursday. Woooo! This is basically my definition of a party!
* The MCT happened here. Shhh.
