example of Dirac sequence

We can construct a Dirac sequence ${\{{\delta }_n\}}_{n\in {\mathbb{N}}_{+}}$ by choosing

$${\delta }_n(x)=\frac{n}{\pi (1+n^2{x}^2)}.$$

To show that conditions 1 and 3 in the definition of a Dirac sequence are satisfied is trivial and condition 2 is also fulfilled since

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\delta }_n(x)dx=\frac{1}{\pi }\cdot {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{n}{1+n^2{x}^2}dx=\left[\begin{array}{c}y=nx\hfill \\ dy=n\cdot dx\hfill \end{array}\right]=\frac{1}{\pi }\cdot {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{1+y^2}dy=\frac{1}{\pi }\cdot \arctan y{|}_{y=-\mathrm{\infty }}^{\mathrm{\infty }}=\frac{1}{\pi }\cdot \pi =1$$

for all $n\in {\mathbb{N}}_{+}$, hence ${\{{\delta }_n\}}_{n\in {\mathbb{N}}_{+}}$ is a Dirac sequence.

To prove that it actually converges in ${𝒟}^{\prime }(ℝ)$ (the space of all distributions on $𝒟(ℝ)$) to the Dirac delta distribution $\delta$, we must show that

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_{ℝ}{\delta }_n(x)\phi (x)𝑑x=\phi (0)$$

for any test function $\phi \in 𝒟(ℝ)$ (a topological vector space of smooth functions with compact support). Let us take an arbitrary test function $\phi \in 𝒟(ℝ)$ and assume that the closed and compact set $\mathrm{supp}(\phi )$ is contained in some open interval $(a,b)\subset ℝ$ ( and $b>0$). Using the triangle inequality and the fact that ${\int }_{ℝ}{\delta }_n(x)𝑑x=1$ for all $n\in {\mathbb{N}}_{+}$ we can write

$$\left|{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\delta }_n(x)\phi (x)𝑑x-\phi (0)\right|=\left|{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\delta }_n(x)(\phi (x)-\phi (0))𝑑x\right|\le$$

$$\le \underset{I_1}{\underbrace{\phi (0){\int }_{-\mathrm{\infty }}^a\left|{\delta }_n(x)\right|𝑑x}}+\underset{I_2}{\underbrace{{\int }_a^b\left|{\delta }_n(x)(\phi (x)-\phi (0))\right|𝑑x}}+\underset{I_3}{\underbrace{\phi (0){\int }_b^{\mathrm{\infty }}\left|{\delta }_n(x)\right|𝑑x}}$$

It is easy to see that ${lim}_{n\to \mathrm{\infty }}{\delta }_n(x)=0$, $\forall x\in (-\mathrm{\infty },a]\cup [b,\mathrm{\infty })$ and therefore${lim}_{n\to \mathrm{\infty }}{I}_1=0$and ${lim}_{n\to \mathrm{\infty }}{I}_3=0$.Finally we want to estimate $I_2$ when $n\to \mathrm{\infty }$.

$$I_2={\int }_a^b\left|{\delta }_n(x)\right|\underset{\le |x|\cdot sup|{\phi }^{\prime }(x)|}{\underbrace{\left|(\phi (x)-\phi (0))\right|}}𝑑x\le sup|{\phi }^{\prime }(x)|\cdot {\int }_a^b\left|{\delta }_n(x)x\right|𝑑x=$$

$$=sup|{\phi }^{\prime }(x)|\cdot \frac{1}{\pi }{\int }_a^b\left|\frac{nx}{1+{(nx)}^2}\right|𝑑x=sup|{\phi }^{\prime }(x)|\cdot \frac{1}{\pi }\left(-{\int }_a^0\frac{nx}{1+{(nx)}^2}𝑑x+{\int }_0^b\frac{nx}{1+{(nx)}^2}𝑑x\right)=$$

$$=sup|{\phi }^{\prime }(x)|\cdot \frac{1}{\pi }(-(\frac{1}{2n}\cdot \text{ln}|1+{(nx)}^2|{|}_{x=a}^0)+(\frac{1}{2n}\cdot \text{ln}|1+{(nx)}^2|{|}_{x=0}^b))=$$

$$=sup|{\phi }^{\prime }(x)|\cdot \frac{1}{\pi }\left(\frac{1}{2n}\cdot \text{ln}|1+{(na)}^2|+\frac{1}{2n}\cdot \text{ln}|1+{(nb)}^2|\right)$$

We now conclude that ${lim}_{n\to \mathrm{\infty }}{I}_2=0$. This means that ${lim}_{n\to \mathrm{\infty }}{I}_1+I_2+I_3=0$ which shows that ${\{{\delta }_n\}}_{n\in {\mathbb{N}}_{+}}$ converges to the Dirac delta distribution $\delta$.