Is the sequence of functions $g_n=ng_1(nx)$ a Cauchy sequence?

Question

Given a function $g_1(x) \in \mathcal L^2(\Bbb R)$ that satisfies: $$\int_{-\infty}^{\infty}dx \space g_1(x)=1$$ one can define a sequence of functions $g_n=ng_1(nx)$. Does $g_n(x)$ define a Cauchy sequence?

This is what I have tried:

• Given that $\int_{-\infty}^{\infty}dx \space g_1(x)=1$ I "concluded" that $g_1(x)$ is $\delta$-function.
• Under that assumption, the scaling property of the Dirac-delta function (distribution) yields $g_1(nx)=\frac{1}{n}g_1(x) \implies g_n=ng_1(nx)\iff g_n=\frac{1}{n}ng_1(x)\iff g_n=g_1(x)=\delta(x)$

I looked up the definition of a Cauchy sequence on wikipedia. As far as I understand I have to show that there exists an $N\in \Bbb N$ such that for every $m \ge n \ge N$, $g_m-g_n< \epsilon$

I basically have three questions now:

1. Is my reasoning correct so far?
2. How can I continue this proof?
3. What is the difference between a convergent sequence and a Cauchy sequence?

Answer 1

No, the sequence $(g_n)$ need not be Cauchy in the $L^2$ topology.

Consider $g_1 = \begin{cases} 1, && x \in [0,1] \\ 0, && \text{otherwise}\end{cases}$. Then $g_n = \begin{cases} n, && x \in [0, \frac 1 n] \\ 0, && \text{otherwise}\end{cases}$.

One way of showing that $(g_n)$ is Cauchy is showing that $\| g_{n+p} - g_n \| \underset {p \to \infty} \longrightarrow 0 \ \forall n$. Let us show that this does not happen for our sequence.

Notice that $g_{n+p} - g_n = \begin{cases} p, && x \in [0, \frac 1 {n+p}] \\ -n, && x \in (\frac 1 {n+p}, \frac 1 n] \\ 0, && \text{otherwise} \end{cases}$. Then

$$\| g_{n+p} - g_n \| = \sqrt {\int \limits _{-\infty} ^\infty (g_{n+p} - g_n)^2} = \sqrt {\int \limits _{-\infty} ^\infty (g_{n+p} - g_n)^2} = \sqrt {\int \limits _0 ^{\frac 1 {n+p}} p^2 \ \textrm d x + \int \limits _{\frac 1 {n+p}} ^{\frac 1 n} (-n)^2 \ \textrm d x} = \\ \sqrt {\frac {p^2} {n+p} + \frac {n^2 p} {n(n+p)}} = \sqrt p$$

which clearly does not tend to $0$ for $p \to \infty$, therefore $(g_n)$ is not Cauchy in the topology of $L^2$.

---

On the other hand, if $g_1 \in L^1$ (not $L^2$, this is important!), then let us show that $g_n \to \delta$ in $\mathcal D '$ (the space of Schwartz distributions on $\Bbb R$). In particular, this will imply that $(g_n)$ is Cauchy in the topology of $\mathcal D '$.

Le $\varphi$ be a test function (i.e. a smooth function with compact support). Then

$$\lim \limits _n \int \limits _{-\infty} ^\infty g_n (x) \varphi (x) \ \textrm d x = \lim \limits _n \int \limits _{-\infty} ^\infty n g_1 (nx) \varphi (x) \ \textrm d x = \color{blue} {[\text{make } y = nx]} = \lim \limits _n \int \limits _{-\infty} ^\infty g_1 (y) \varphi \left( \frac y n \right) \ \textrm d x = \\ \int \limits _{-\infty} ^\infty \lim \limits _n \ g_1 (y) \varphi \left( \frac y n \right) \ \textrm d x = \int \limits _{-\infty} ^\infty g_1 (y) \varphi (0) \ \textrm d x = \varphi (0) \int \limits _{-\infty} ^\infty g_1 (y) \ \textrm d x = \varphi (0) \cdot 1 = \varphi (0) ,$$

which means exactly that $g_n \to \delta$ in $\mathcal D '$ (we have slipped the limit inside the integral with Lebesgue's dominated convergence theorem, and it is precisely here that you need $g_1 \in L^1$).