Let $p: \mathbb{R} \to \mathbb{R}$ and $p \in \mathcal{C}^{\infty}(\mathbb{R})$.

$p(x)=p(-x) \quad \forall x \in \mathbb{R}$ and $\int_{\mathbb{R}}p(x)dx=1$. Let $$\delta_{x_0}^{(n)}(\phi) := \int_\mathbb{R} np[n(x-x_0)]\phi(x)dx \quad \forall \phi \in \mathcal{D}(\mathbb{R})$$

Show $$ \delta_{x_0}^{(n)} \xrightarrow{\mathcal{D}'(\mathbb{R})} \delta_{x_0} $$.

Where $\mathcal{D}(\mathbb{R}) $ is the space of compactly supported smooth functions.

I tried to show it in many different ways but I can't finish the proof.

Attempt of proof:

Let be $P : P'=p$ (it exist because $p$ is continuous).

It is trivial to show that $\max_{y \in \mathbb{R}}(P) < + \infty$.

$$\lim_{n \to +\infty}\delta_{x_0}^{(n)}(\phi) =\lim_{n \to +\infty} \int_\mathbb{R} np[n(x-x_0)]\phi(x)dx = \lim_{n \to +\infty} \int_\mathbb{R} p(y)\phi(\frac{y}{n}+x_0)dy = \lim_{n \to +\infty} [P(y)\phi(\frac{y}{n}+x_0)]_{-\infty}^{+\infty} - \lim_{n \to +\infty} \int_\mathbb{R} \frac{P(y)}{n}\phi'(\frac{y}{n}+x_0)dy = \phi(x_0) - \lim_{n \to +\infty} \int_\mathbb{R} \frac{P(y)}{n}\phi'(\frac{y}{n}+x_0)dy $$

The problem is to show that the last integral converges to $0$; I tried with Lebesgue's dominated convergence theorem and with Holder inequality but I can't get it. The trouble is that $\text{supp}(\phi'(\frac{y}{n}+x_0)) \to +\infty$ when $n \to +\infty$.

Thank you!

  • $\begingroup$ With $\mathcal D(\Bbb R)$ do you mean compactly supported smooth functions? Your current definition allows every function. $\endgroup$ – s.harp Dec 29 '16 at 17:06
  • $\begingroup$ yes, sorry. fixed! $\endgroup$ – Bremen000 Dec 29 '16 at 17:08
  • $\begingroup$ I mean $p: \mathbb{R} \to \mathbb{R}$ and $p \in \mathcal{C}^{\infty}(\mathbb{R})$. $\endgroup$ – Bremen000 Dec 29 '16 at 17:32

Let $P(x) = \int_{-\infty}^x p(y)dy$ which is assumed to be well-defined and bounded and $\lim_{x \to -\infty}P(x) = 0 , \lim_{x \to+\infty}P(x)= 1$.

If $\phi $ is $C^1$ and supported on $[a,b]$ then

$$\lim_{n \to \infty}\int_{-\infty}^\infty n p(nx) \phi(x)dx =\lim_{n \to \infty}\int_{a}^b np(nx) \phi(x)dx=\lim_{n \to \infty}-\int_a^b P(nx) \phi'(x)dx $$ $$= -\int_a^b \lim_{n \to \infty} P(nx) \phi'(x)dx = -\int_0^b \phi'(x)dx = \phi(0)-\phi(b)=\phi(0)$$ And hence $n p(nx)$ (and all its distributional derivatives) converges to $\delta(x)$ (and its distributional derivatives) in the sense of distributions.


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