convergence of sequence in a Distribution

Question

Let $\varphi \in D$ be a test function on $\Bbb{R}$. Is the sequence $f_n(x)=\frac{1}{n}\varphi(\frac{x}{n})$ convergent in the test function space $D$? What is the limit? please a hint to start.

Answer 1

Well, you need to check two things:
- Is there a compact set $K$ satisfying $\bigcup \limits_{n \in \mathbb{N}} supp( f_n) \subseteq K$
- Do all derivatives converge uniformly towards a test function and its respective derivatives?

The second statement should be obvious. A candidate limit function is the zero function. By the boundedness of $\varphi$ and the suppression factor $\frac{1}{n^{k+1}}$ for the $k$th partial derivative of $f_n$ uniform convergence towards zero is guaranteed.
The first statement however becomes rather troublesome in this case.
If we let $K_n = supp(f_n)$, it should be apparent that the $K_n$ will successively get bigger and bigger. (Make a drawing of the first few $f_n$)
As a matter of fact you need to realize that (unless $\varphi = 0$) the $K_n$ get so big, they will grow without bound! Hence there can be no compact set $K$ containing all $K_n$ and thus convergence in $D$ fails.

Also contrast this with the notion of pointwse convergence.