Show a function's distributional derivative as the summation of delta function

Question

9) Is $$ \sum_{n=1}^\infty \delta_n \tag{7.10.1} $$ a well-defined distribution? Note, to be a well-defined distribution, its action on any test function should be a finite number. Provide an example of a function $f(x)$ whose derivative in the sense of distributions is $(7.10.1)$

Hello, I want to find a distribution whose distributional derivative as the summation of the delta function ($\delta_1$ to $\delta_k$). I find the distributional derivative of the summation of the shift of the Heaviside Function $H(x-a)$ is equal to the summation of the delta function. However, I have trouble of finding the convergence of the summation of the shift of the Heaviside function in the sense of the distribution. If I can find this convergence, and then , by the theorem, the derivative of the convergence is also the convergence of the summation of the delta function in the sense of distribution.

Answer 1

It is a well-defined distribution when the test functions have compact support. Formally, it is an element of $C_c(\mathbb R)^*$, the dual of the space of continuous functions with compact support. Given such a function $f$, the result of applying your distribution is $$ \sum_{n=1}^{\infty}f(n), $$ which is finite since the set $\mathbb N\cap \textrm{supp}(f)$ is both compact and discrete, hence finite - so the sum above is finite.

Answer 2

What is the derivative of $\lfloor x \rfloor 1_{x > 0}$ ? It is $L^1_{loc}$ thus it is a distribution, it has polynomial growth thus it is a tempered distribution.

On a bounded interval $[-N,N]$ $$\lfloor x \rfloor 1_{x > 0}=\sum_{n \ge 0} 1_{x-n > 0}=\sum_{n=0}^N 1_{x-n > 0}$$ is a finite sum thus the convergence in the sense of distributions is obvious.

The convergence of $\sum_{n \ge 0} 1_{x-n > 0}$ in the sense of tempered distributions follows from the semi-norms definition of the Schwartz space comprising $\sup_x |x^3 \phi(x)|$.