# 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

 $\int_{-\infty}^{\infty}\delta_{n}(x)dx=\frac{1}{\pi}\cdot\int_{-\infty}^{% \infty}\frac{n}{1+n^{2}x^{2}}dx=\left[\begin{tabular}[]{ll}y=nx\\ dy=n\cdot dx\end{tabular}\right]=\frac{1}{\pi}\cdot\int_{-\infty}^{\infty}% \frac{1}{1+y^{2}}dy=\frac{1}{\pi}\cdot\arctan{y}\Big{\lvert}_{y=-\infty}^{% \infty}=\frac{1}{\pi}\cdot\pi=1$

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

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

 $\lim_{n\to\infty}\int_{\mathbb{R}}\delta_{n}(x)\varphi(x)dx=\varphi(0)$

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

 $\left|\int_{-\infty}^{\infty}\delta_{n}(x)\varphi(x)dx-\varphi(0)\right|=\left% |\int_{-\infty}^{\infty}\delta_{n}(x)(\varphi(x)-\varphi(0))dx\right|\leq$
 $\leq\underbrace{\varphi(0)\int_{-\infty}^{a}\left|\delta_{n}(x)\right|dx}_{I_{% 1}}+\underbrace{\int_{a}^{b}\left|\delta_{n}(x)(\varphi(x)-\varphi(0))\right|% dx}_{I_{2}}+\underbrace{\varphi(0)\int_{b}^{\infty}\left|\delta_{n}(x)\right|% dx}_{I_{3}}$

It is easy to see that $\lim_{n\to\infty}\delta_{n}(x)=0$, $\forall x\in(-\infty,a]\cup[b,\infty)$ and therefore $\lim_{n\to\infty}I_{1}=0$ and $\lim_{n\to\infty}I_{3}=0$. Finally we want to estimate $I_{2}$ when $n\to\infty$.

 $I_{2}=\int_{a}^{b}\left|\delta_{n}(x)\right|\underbrace{\left|(\varphi(x)-% \varphi(0))\right|}_{\leq|x|\cdot\sup{|\varphi^{\prime}(x)|}}dx\leq\sup{|% \varphi^{\prime}(x)|}\cdot\int_{a}^{b}\left|\delta_{n}(x)x\right|dx=$
 $=\sup{|\varphi^{\prime}(x)|}\cdot\frac{1}{\pi}\int_{a}^{b}\left|\frac{nx}{1+(% nx)^{2}}\right|dx=\sup{|\varphi^{\prime}(x)|}\cdot\frac{1}{\pi}\left(-\int_{a}% ^{0}\frac{nx}{1+(nx)^{2}}dx+\int_{0}^{b}\frac{nx}{1+(nx)^{2}}dx\right)=$
 $=\sup{|\varphi^{\prime}(x)|}\cdot\frac{1}{\pi}\left(-\left(\frac{1}{2n}\cdot% \text{ln}|1+(nx)^{2}|\Big{\lvert}_{x=a}^{0}\right)+\left(\frac{1}{2n}\cdot% \text{ln}|1+(nx)^{2}|\Big{\lvert}_{x=0}^{b}\right)\right)=$
 $=\sup{|\varphi^{\prime}(x)|}\cdot\frac{1}{\pi}\left(\frac{1}{2n}\cdot\text{ln}% |1+(na)^{2}|+\frac{1}{2n}\cdot\text{ln}|1+(nb)^{2}|\right)$

We now conclude that $\lim_{n\to\infty}I_{2}=0$. This means that $\lim_{n\to\infty}I_{1}+I_{2}+I_{3}=0$ which shows that $\{\delta_{n}\}_{n\in\mathbb{N}_{+}}$ converges to the Dirac delta distribution $\delta$.

Title example of Dirac sequence ExampleOfDiracSequence 2013-03-22 14:13:10 2013-03-22 14:13:10 Johan (1032) Johan (1032) 8 Johan (1032) Example msc 46F05 Distribution4 DeltaDistribution DiracDeltaFunction ConstructionOfDiracDeltaFunction