# Showing convergence to $\delta$ distribution

Introduction

Let $f \in C(\mathbb{R}^n)$, supp $f \subset \lbrace x : |x| <1 \rbrace$, and $\int_\mathbb{R} f(x) dx = 1$.

Consider the sequence $f_k(x) = k^n f(kx)$, $k \in \mathbb{R}$.

Show that $f_k(x) \rightarrow \delta(x)$ as $k \rightarrow \infty$ in $\mathit{D'(\mathbb{R}^n)}$.

My main problem here is with the integral. $f_k(x)$ takes $x \in \mathbb{R}^n$ but in the integral in the first line, we only integrate over $\mathbb{R}$.

So when I start the proof I have $\int_\mathbb{R^n}k^nf(kx)\varphi(x) dx$ for $\varphi \in \mathit{D}(\mathbb{R}^n)$.

I was thinking of maybe having n integrals, so something like $\int_\mathbb{R} kf(kx) \int_\mathbb{R} kf(kx) \dots \int_\mathbb{R} kf(kx)\varphi(x) dx_1\dots dx_n$ but I don't think this is right since we have $dx$ and not $dx_i$ in the integral given at the beginning.

I'm sure I'm making a silly mistake somewhere, any hints would be appreciated!

• That's a typo, the integral should be over $\mathbb{R}^n$. Perhaps the author first intended to treat only the case $n = 1$ and later decided to do the general case, forgetting to update that occurrence. – Daniel Fischer Oct 27 '16 at 12:41
• @DanielFischer Great, had a sneaking suspicion that it might have been. As a follow-up, do you know how I could calculate $\int_\mathbb{R^n} k^n f(kx) dx$? – Ronique Hossain Oct 27 '16 at 18:03
• Change of variables: $y = kx$. – Daniel Fischer Oct 27 '16 at 18:06
• @DanielFischer I tried that, but I get $k^{n-1}$ and I can't see the way forward for the rest of my proof. Thanks for your help so far! – Ronique Hossain Oct 27 '16 at 18:09
• The Jacobian determinant of the map $x \mapsto kx$ is $k^n$, not $k$. – Daniel Fischer Oct 27 '16 at 18:11

Remark:

1. Here is the usual condition. cf. also "Mathematics for Physics and Physicists" (2007), Walter Appel, the notion of Dirac sequence, Thm 8.18 p.232 (proved p.601) and proposition 8.21 p.233.
2. I checked several books (ex. Real and Functional Analysis by Serge Lang), and a Dirac sequence is required to be positive but this is in fact not necessary for (weak) converge to the delta distribution. However, without positivity, one must impose finiteness of $$\lVert f\rVert_1 =\int_{\mathbb{R}^n} \lvert f(x) \rvert \, dx < +\infty$$

With the assumptions of the OP and $$\lVert f\rVert_1 =C < +\infty$$: we want to show that for any test function $$\varphi\in \mathcal{C}^{\infty}_c(\mathbb{R}^n)$$ $$\lim_{k\to\infty} \langle f_k, \varphi\rangle :=\lim_{k\to\infty} \int_{\mathbb{R}^n} f_k(x)\, \varphi(x)\, dx = \varphi(0)$$

Indeed we do a change of variable: $$\int_{\mathbb{R}^n} f_k(x)\, \varphi(x)\, dx := \int_{\mathbb{R}^n} k^n f(kx)\, \varphi(x)\, = \begin{bmatrix} y = kx \\ dy = k^n dx\end{bmatrix} = \int_{\mathbb{R}^n} f(y)\, \varphi(y/k)\, dy$$

By continuity of $$\varphi$$ at $$0$$: $$\forall\ \epsilon > 0,\ \exists\ \delta > 0,\enspace\forall\ u\in \mathcal{B}(0,\delta),\ \big\lvert \varphi(u) -\varphi(0)\big\rvert < \epsilon$$ For $$k$$ large enough, $$y/k$$ is in the ball $$\mathcal{B}(0,\delta)$$ if $$y\in \mathcal{B}(0,1)$$ and since $$f$$ has support in the unit ball, one needs only integrates on it. Let us now write what we wanted to prove: $$\left\lvert \int_{\mathbb{R}^n} f(y)\, \varphi(y/k)\, dy -\varphi(0) \right\rvert= \left\lvert\int_{\mathcal{B}(0,1)} f(y)\, \varphi(y/k)\, dy - \varphi(0) \int_{\mathcal{B}(0,1)} f(y)\, dy \right\rvert$$ $$\leq \left\lvert\int_{\mathcal{B}(0,1)} f(y)\, \Big( \varphi(y/k) -\varphi(0)\Big)\, dy \right\rvert \leq \int_{\mathcal{B}(0,1)} \left\lvert f(y)\, \Big( \varphi(y/k) -\varphi(0)\Big) \right\rvert\, dy$$ $$\leq \epsilon \int_{\mathcal{B}(0,1)} \lvert f(y) \rvert \, dy = \epsilon C \tag{*}\label{*}$$

One thus concludes that $$\lim_{k\to\infty} \langle f_k, \varphi\rangle = \varphi(0)= \langle \delta ,\varphi \rangle$$

Remark: If one had positivity of $$f$$ then the absolute value is unecessary in (*), and that integral was supposed to be equal to 1, (hence bounded).