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!

  • $\begingroup$ 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
  • 1
    $\begingroup$ The Jacobian determinant of the map $x \mapsto kx$ is $k^n$, not $k$. $\endgroup$ – Daniel Fischer Oct 27 '16 at 18:11


  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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.