Unclear on distributional derivative Suppose we have a distribution $\phi \in D'(\mathbb{R}^n)$. The distributional derivative of $\phi$ is defined as
$$\phi'(g) = -\phi(g'), \quad \forall g \in C^\infty_C(\mathbb{R}^n).$$
I am clear on this definition and how it is motivated. What has been confusing me is perhaps to due with terminology.
Suppose we look at the example of the Heaviside function:
$$H(x) \begin{cases} 1, \quad x>0\\ 0, \quad x\leq 0.\end{cases}$$
We can define a distribution $\phi_H$ as
$$\phi_H(g) =\langle H, g\rangle = \int H g dx.$$
In textbooks you see that the derivative of the Heaviside function is calculated as
$$\langle H', g \rangle = \int H' g dx = -\int Hg'dx = -\int_0^\infty g' dx = g'(0) = \langle\delta, g'\rangle$$
and so we conclude that $H' = \delta$, where $\delta$ is the Dirac-delta function.
My question is do we say that the function $H$ has derivative $\delta$, or that the distribution induced by $H$ (what I had called $\phi_H$) has derivative $\delta$? I assume it is the latter, but I see a lot of sources say statements such as "the derivative of the Heaviside function is the Dirac-delta function". Is this merely abuse of notation, or do they mean to say "the distribution induced by the Heaviside function is equal to the Dirac-delta function/distribution".
Further, is there some other connection where one implies the other? From my reading it seems that if a function $f$ induces a distribution $\phi_f$, and $f$ is differentiable, then the distributional derivative $\phi'_f$ is equal to the distribution induced by $f'$, $\phi_{f'}$. Can we say anything more?
 A: Let $\phi_H$ be the distribution induced by $H$. Then the statement is that $(\phi_H)’=\delta$. Also, the proof you gave is wrong. It should be stated as follows: for any $g\in \mathcal{D}(\Bbb{R})$,
\begin{align}
\langle (\phi_H)’, g\rangle &:=-\langle \phi_H,g’\rangle\\
&:=-\int_{\Bbb{R}}H(x)g’(x)\,dx\\
&=-\int_{0}^{\infty}g’(x)\,dx\\
&=g(0)\\
&=\langle \delta,g\rangle.
\end{align}
Hence, $(\phi_H)’=\delta$. And yes, it is the usual abuse of language: those who know what they’re talking about are reluctant to beat a dead horse, so will abbreviate things. So, whenever you see statements like “a function $f$ has distributional derivative ___”, you should immediately think “the associated distribution $\phi_f$ has distributional derivative ___”. Once we get used to things, we also omit mention of the canonical injection $L^1_{\text{loc}}(U)\hookrightarrow \mathcal{D}(U)$ (which you denoted as $\phi$).
In your ‘proof’ you say $\langle(\phi_H)’,g\rangle=\int_{\Bbb{R}}H’g\,dx$, but this is not even the definition of $(\phi_H)’$. On the other hand, $H$ is differentiable on $\Bbb{R}\setminus\{0\}$ with $H’=0$ here, so as an element of $L^1_{loc}(\Bbb{R})$, we can say $H’=0$. Hence, $\phi_{H’}=0$ is the zero distribution. This shows that $(\phi_H)’=\delta\neq 0=\phi_{H’}$. This is characteristic of jump discontinuities in the derivative of a function: the associated distributional derivatives differ by a multiple of $\delta$.
A: I think the answer to your question is ultimately that $L^1_{\text{loc}}(X)$ is a subset of the distributions on $X$ (assume $X$ is an open subset of a Euclidean space). This comes precisely from the "motivation": if $f$ is locally integrable and $g$ is continuous with compact support, its a hopefully simple proof to show that $fg$ is integrable.
Since test functions $g$ are defined to be smooth with compact support, this means that $D(f)(g)=\int_X fg dV$ defines $D(f)$ as a distribution for $f\in L^1_{\text{loc}}$ (you can check Schwarz estimates how you like).
Show that this is an injection (after identifying functions $f$ that are a.e. equal, of course) using bump functions for $g$ and I think that's what you're asking about? Good luck!
