Distributional derivative of multiple step function i am learning about distributions and distributional derivatives, and would like to check my understanding.
suppose we have a step function that jumps by a unit at every integer:
$$H(x)=
\begin{cases}
 \lfloor x\rfloor, & x>0; \\
 0, & x<0; 
\end{cases} $$
my goal (just for understanding) is to take the distributional derivative and return something that looks like a dirac comb- a delta at each integer.
Question 1. In order to create a distribution we can simply convolve, $H(x)$ against a smooth, compactly supported test function, $\phi(x)$,
$$ H(x)*\phi(x)$$, where $*$ is the convolution operator.
Is this a valid operation here? most of the material i've found on convolution deals with two functions with compact support. i read elsewhere that as long as $\phi(x)$ has compact support and $H(x)$ is locally integrable, then the convolution is valid - i haven't seen a proof of this.
Question 2. To obtain a distributional derivative, can we simply convolve $H(x)$ with $\phi'(x)$?
$$ H(x)*\phi'(x) $$
thank you
 A: Possibly you are making the issue more complicated than it needs to be.
E.g., if the goal is to take the distributional derivative of your step function, it is easy enough to think in terms of the "locality" of the differentiation operator. So the local constant-ness of your function away from integers entails that its derivative (in any sense you like) is locally $0$ away from integers.
There is a general exercise one can do to see/show that the derivative of a piecewise-differentiable function with jump discontinuities, but/and with left and right derivatives at the jumps, is the "obvious thing" away from jumps, and is a multiple of Dirac delta at the jumps, with the multiple telling how large the jump is. To my taste, this is more forthright than expressing the derivative as a convolution...
A: If $\phi\in C_C^\infty$, then we have
$$\begin{align}
\langle H',\phi \rangle &=-\langle H, \phi'\rangle \tag1\\\\
&=-\int_{-\infty}^\infty H(x)\phi'(x)\,dx\tag2 \\\\
&=-\int_{-\infty}^\infty \lfloor x \rfloor \phi'(x)\,dx\tag3 \\\\
&=-\sum_{n=-\infty}^\infty \int_n^{n+1}\lfloor x \rfloor \phi'(x)\,dx\tag4\\\\
&=\sum_{n=-\infty}^\infty n(\phi(n)-\phi(n+1))\tag5\\\\
&=\underbrace{\sum_{n=-\infty}^\infty (n\phi(n)-(n+1)\phi(n+1))}_{\text{Telescopes to }\,0}+\sum_{n=-\infty}^\infty \phi(n+1)\tag6\\\\
&=\sum_{n=-\infty}^\infty \phi(n)\tag7
\end{align}$$

Therefore, in distribution we assert that the floor function is equal to a Dirac Comb given by
$$\bbox[5px,border:2px solid #C0A000]{\lfloor x\rfloor =\sum_{n=-\infty }^\infty \delta(x-n)}$$

NOTES:
The equality in $(1)$ is the definition of the distributional derivative.
Since $H\phi'\in L^1$, the distribution in $(2)$ is an integral over $\mathbb{R}$.
In going from $(2)$ to $(3)$, we used $H(x)=\lfloor x\rfloor$.
In arriving at $(4)$ we divided the integral into the sum of integrals on intervals for which $\lfloor x\rfloor$ is constant.
In going from $(4)$ to $(5)$, we integrated $\phi'(x)$ on the interval $[n,n+1]$.
We simply write $n(\phi(n)-\phi(n+1))=(n\phi(n)-(n+1)\phi(n+1)) +\phi(n+1)$ to obtain $(6)$.
In going from $(6)$ to $(7)$, we summed the telescoping series and used the fact that $\phi\in C_C^\infty$ so that $\lim_{n\to \pm \infty}\phi(n)=0$.  We also shifted the index from $n$ to $n-1$.
