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

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.

• Sep 28 '19 at 18:53

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.
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 $$\sup_x |x^3 \phi(x)|$$.