# Example of a compactly supported $\delta$-sequence.

Recall that a $$\delta$$-sequence can be defined as a sequence, $$(\phi_n)_{n\in\mathbb N}$$, of continuously differentiable, non-negative, real-valued functions for which $$\int_\mathbb R\phi_ndx=1$$ for all $$n\in\mathbb N$$. An example of a $$\delta$$-sequence often discussed on this website is that of, $$\phi_n(x)=\frac{n}{\sqrt\pi}e^{-n^2x^2},$$ where $$n\in\mathbb N$$ and $$x\in\mathbb R$$. Since $$\phi_n(x)\to0$$ as $$n\to\infty$$ for $$|x|\to\infty$$, we see that $$(\phi_n)_{n\in\mathbb N}$$ is a sequence of functions which vanish at infinity.

What is an example of a $$\delta$$-sequence which satisfies the stronger requirement of having compact support for each $$n\in\mathbb N$$?

Let $$\phi_n(x)=\frac 1 n f(\frac x n)$$ where $$f(x)=c(x+1)^{2}(1-x)^{2}$$ for $$-1\leq x \leq 1$$, with $$f(x)=0$$ for $$x \notin [-1,1]$$. Choose the positive constant $$c$$ such that $$\int f(x)\, dx=1$$.
• For this to be a $\delta$-sequence in the usual sense, shouldn't you take $\phi_n(x)=nf(nx)$? Your sequence satisfies the requirements of the opening post, though. – Mars Plastic Jan 31 '19 at 13:09
• For any continuous function $g\colon\mathbb R\to\mathbb R$, the term $\int_{\mathbb R} \phi_n(x)g(x) d(x)$ should converge to $g(0)$. In particular, if you think of $\phi_n$ as a regular distribution, it should converge strongly to the $\delta$-distribution in 0. This is the main reason why it's called a $\delta$-sequence. – Mars Plastic Jan 31 '19 at 16:29
• @MarsPlastic $\delta-$ sequence is not a standard terminology. Could you state all the properties you want from the sequence precisely? – Kavi Rama Murthy Jan 31 '19 at 23:14
• I don't think there is a general consensus about a precise definition (I may be wrong, though), but the term "$\delta$-sequence" itself suggests that it should at the very least converge to a $\delta$-distribution (pointwise when interpreted as a sequence of regular distributions). However, as I said: Concerning the requirements as stated in the opening post, your answer is undoubtedly correct. – Mars Plastic Feb 1 '19 at 12:20