How to show this sequence is a delta sequence?

Question

Consider the sequence $(\phi_n)_{n\in \mathbb{N}}$ of test-functions $\phi_n\in \mathcal{D}(\mathbb{R})$ defined by

$$\phi_n(x) = \dfrac{n}{\sqrt{\pi}}e^{-n^2x^2}.$$

I want to show that this is a delta sequence, in the sense that:

$$\lim_{n\to\infty}(\phi_n,\psi)=(\delta,\psi)=\psi(0), \quad \forall \psi\in \mathcal{D}(\mathbb{R}).$$

In that case we know that we have

$$(\phi_n,\psi)=\dfrac{n}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-n^2x^2}\psi(x)dx.$$

The problem here is to compute this integral. The function $x\mapsto e^{-n^2x^2}\psi(x)$ has no singularities, hence contour integration wouldn't help. Integration by parts or substitution also doesn't seem to be of great help.

The major problem is that there is a function $\psi$ inside the integral about which we don't know really anything apart from the fact that it is $C^\infty$ and with compact support.

How can we compute this integral and in the end show that the sequence $(\phi_n)$ is a delta sequence?

Just to make clear, I'm here doing all of this in the context of the Riemann integral.

Answer 1

Do the substitution $y=nx$. You will get $$\dfrac{n}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-n^2x^2}\psi(x)dx=\dfrac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-y^2}\psi\left(\frac{y}{{n}}\right)dy.$$ Now apply Lebesgue convergence theorem to obtain $$\lim_{n\to +\infty} \dfrac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-y^2}\psi\left(\frac{y}{{n}}\right)dy =\psi(0)\dfrac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-y^2}dy=\psi(0).$$

Answer 2

I thought it might be instructive to present a solution that does not rely on the Dominated Convergence Theorem. To that end, we proceed.

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First, noting that $\int_{-\infty}^\infty \phi_n(x)\,dx=1$ for all $n$, we can write the integral of interest as

$$\begin{align} \int_{-\infty}^\infty \phi_n(x)\psi(x)\,dx&=\psi(0)\int_{-\infty}^\infty \phi_n(x)\,dx+\int_{-\infty}^\infty \phi_n(x)(\psi(x)-\psi(0))\,dx\\\\ &=\psi(0)+\int_{-\infty}^\infty \phi_n(x)(\psi(x)-\psi(0))\,dx\tag1 \end{align}$$

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Second, since $\psi$ is in the space of smooth functions with compact support, then certainly $\psi(x)$ is continuous at $x=0$. Therefore, given any $\epsilon>0$, there exists a number $\delta>0$ such that $|\psi(x)-\psi(0)|<\epsilon$ whenever $|x|<\delta$.

We keep this $\epsilon>0$ and $\delta>0$ pair and write the integral on the right-hand side of $(1)$ as

$$\begin{align} \int_{-\infty}^\infty \phi_n(x)(\psi(x)-\psi(0))\,dx&=\int_{|x|>\delta}\phi_n(x)(\psi(x)-\psi(0))\,dx+\int_{|x|<\delta}\phi_n(x)(\psi(x)-\psi(0))\,dx\tag 2 \end{align}$$

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Third, inasmuch as $\psi(x)$ is a test function, it is smooth with compact support. And since $\phi_n(x)\to 0$ uniformly for $|x|>\delta>0$, the limit of the first integral on the right-hand side of $(2)$ is

$$\begin{align} \lim_{n\to \infty}\int_{|x|>\delta}\phi_n(x)\psi(x)\,dx&=\int_{|x|>\delta}\left(\lim_{n\to \infty} \phi_n(x)\right)\,\psi(x)\,dx\\\\ &=0\tag 3 \end{align}$$

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Fourth, we have the estimate for the second integral in the right-hand side of $(2)$

$$\begin{align} \left|\int_{|x|<\delta}\phi_n(x)(\psi(x)-\psi(0))\,dx\right|&\le \epsilon\int_{|x|<\delta}\phi_n(x)\,dx<\epsilon \tag 4 \end{align}$$

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Putting it all together, we see that for any $\epsilon>0$

$$\lim_{n\to \infty}\left|\int_{-\infty}^\infty \phi_n(x)(\psi(x)-\psi(0))\,dx\right|<\epsilon$$

Since $\epsilon$ is arbitrary, we find that

$$\lim_{n\to \infty}\int_{-\infty}^\infty \phi_n(x)\psi(x)\,dx=\psi(0)$$

And we are done!

Answer 3

Assuming that the expansion is possible, substitute the following into the integral: $$\psi(x)=\psi(0)+\psi'(0)x+\psi''(0)\frac{x^2}{2!} \dots$$ $$\frac{n}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-n^2x^2}\psi(x)dx=\frac{n}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-n^2x^2}\{\psi(0)+\psi'(0)x+\psi''(0)\frac{x^2}{2!} \dots\}dx$$ $$=\psi(0)\frac{n}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-n^2x^2}dx+\psi'(0)\frac{n}{\sqrt{\pi}}\int_{-\infty}^{\infty}xe^{-n^2x^2}dx+\frac{\psi''(0)}{2}\frac{n}{\sqrt{\pi}}\int_{-\infty}^{\infty}x^2e^{-n^2x^2}\dots$$ Note that the original integral has been converted into a series of standard (solvable) integrals. Particularly: $$\int_{-\infty}^{\infty}e^{-n^2x^2}dx=\frac{\sqrt\pi}{n}$$ $$\int_{-\infty}^{\infty}x^{2k+1}e^{-n^2x^2}dx=0; k=0,1,2 \dots$$ $$\int_{-\infty}^{\infty}x^{2k}e^{-n^2x^2}dx=\frac{\sqrt\pi}{2^k}(2k-1)(2k-3)\dots(3)(1)n^{-2k-1}; k=1,2,\dots$$ Using these, we get: $$(\phi_n,\psi)=\psi(0)+\frac{n}{\sqrt\pi}\sum_{k=1}^{\infty}\frac{\psi^{(2k)}(0)}{(2k)!}\frac{\sqrt\pi}{2^k}(2k-1)(2k-3)\dots(3)(1)n^{-2k-1}$$ Or more insightfully: $$(\phi_n,\psi)=\psi(0)+\sum_{k=1}^{\infty}O(n^{-2k})$$ And hence: $$\lim_{n\to\infty}(\phi_n,\psi)=\psi(0)$$