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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!

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  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Oct 27 '16 at 12:41
  • $\begingroup$ @DanielFischer Great, had a sneaking suspicion that it might have been. As a follow-up, do you know how I could calculate $\int_\mathbb{R^n} k^n f(kx) dx$? $\endgroup$ – Ronique Hossain Oct 27 '16 at 18:03
  • $\begingroup$ Change of variables: $y = kx$. $\endgroup$ – Daniel Fischer Oct 27 '16 at 18:06
  • $\begingroup$ @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! $\endgroup$ – Ronique Hossain Oct 27 '16 at 18:09
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    $\begingroup$ The Jacobian determinant of the map $x \mapsto kx$ is $k^n$, not $k$. $\endgroup$ – Daniel Fischer Oct 27 '16 at 18:11
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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).

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