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
Best Answer
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 $n-1$.