Is the Laplace distribution a nascent delta function?

Question

Define the normalized Laplace distribution as $$ L(x;a) = \frac{1}{2a} \exp\Big(-\frac{|x|}{a}\Big) \quad (a>0). $$ With the normalization out of the way, what we then need to prove is that for kind $f$ $$ \lim_{a\rightarrow 0} I_a[f] \equiv \lim_{a\rightarrow 0} \int_{-\infty}^\infty L(x;a) f(x) dx = f(0). $$ I think I found a way with Fourier theory. Denoting the Fourier transform operator as $\text{FT}$, we have $$ \begin{eqnarray} Λ(ω;a) &=& \text{FT}[L] &=& \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty L(x;a) \exp(i ω x) dx = \frac{1}{\sqrt{2\pi}} \frac{1}{1+a^2 ω^2} \\ F(ω) &=& \text{FT}[f] &=& \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x) \exp(i ω x) dx \end{eqnarray} $$ Now $I_a[f]$ can be written as a convolution of $f(x)$ and $L(x;a)$ evaluated at zero: $$ \begin{eqnarray} I_a[f] &=& \Big[\int_{-\infty}^\infty f(x) L(y - x;a) dx\Big]_{y=0} \\ &\equiv& (L * f)(0) \end{eqnarray} $$ With $\text{FT}^{-1}$ denoting the inverse Fourier transform operator, we then have $$ \begin{eqnarray} I_a[f] &=& \text{FT}^{-1}[F(ω) \times Λ(ω;a)](0) \\ &=& \frac{1}{\sqrt{2\pi}} \Big[\int_{-\infty}^\infty F(ω) \times \frac{1}{\sqrt{2\pi}} \frac{1}{1+a^2 ω^2} \exp(-iωy) dω \Big]_{y=0} \\ &=& \frac{1}{2\pi} \int_{-\infty}^\infty F(ω) \times \frac{1}{1+a^2 ω^2} dω. \end{eqnarray} $$ Now the limit $a\rightarrow 0$ is easy to take: $$ \lim_{a\rightarrow0} I_a[f] = \frac{1}{2\pi} \int_{-\infty}^\infty F(ω) dω = \frac{1}{\sqrt{2\pi}} f(0). $$ This seems to imply that the proper nascent delta function is actually $$ \sqrt{2\pi} \times L(x;a). $$ Question: is this correct? But what about the normalization condition then? Thanks for any help!

Answer 1

I think that you have missed a factor when you write $$ I_a[f] = \text{FT}^{-1}[F(ω) \times Λ(ω;a)](0)$$ See rule 109 and the "Fourier transform unitary, angular frequency" column in this table. It should be $$I_a[f] = (L*f)(0) = \sqrt{2\pi} \mathrm{FT}^{-1}[\Lambda \cdot F](0) = \cdots$$

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Using the Fourier transform is unnecessary. It's easier to use a direct calculation: $$ \langle L(\bullet;a), f \rangle = \int_{-\infty}^{\infty} \frac{1}{2a} e^{-|x|/a} \, f(x) \, dx = \{ x = ay \} = \int_{-\infty}^{\infty} \frac{1}{2} e^{-|y|} \, f(x/a) \, dx \\ \to \int_{-\infty}^{\infty} \frac{1}{2} e^{-|y|} \, f(0) \, dx = \left( \frac{1}{2} \int_{-\infty}^{\infty} e^{-|y|} \, dx \right) \, f(0) = f(0) = \langle \delta, f \rangle. $$ The limit is valid by the dominated convergence theorem.

Answer 2

If $p(x)$ is any pdf, $L(x;\,a):=\frac{1}{a}p\left(\frac{x}{a}\right)$ with $a>0$ is a nascent delta, viz. $$\lim_{a\to0^+}\int_{\Bbb R}\frac{1}{a}p\left(\frac{x}{a}\right)g(x)dx=\lim_{a\to0^+}\int_{\Bbb R}p(y)g(ay)dy=g(0).$$In your case, $p(x):=\frac12\exp -x$. The mistake is in how you handle the Fourier normalization. Your claim is equivalent to saying$$I_{a}\left[f\right] =\left.\color{red}{\frac{1}{\sqrt{2\pi}}}\int_{\mathbb{R}}F\left(\omega\right)\Lambda\left(\omega,\,a\right)\exp\left(-i\omega y\right)d\omega\right|_{y=0} \\=\left.\color{red}{\frac{1}{\sqrt{2\pi}}}\int_{\mathbb{R}^{3}}\color{blue}{\frac{1}{\sqrt{2\pi}}}f\left(x\right)\color{green}{\frac{1}{\sqrt{2\pi}}}L\left(x^{\prime},\,a\right)\exp\left(i\omega\left(x+x^{\prime}-y\right)\right)dxdx^{\prime}d\omega\right|_{y=0},$$where the red factor is from the inverse FT, the blue one is from the FT defining $F$, and the green one is from the FT defining $\Lambda$. Integrating over $\omega$ first, the result is $$\color{red}{\frac{1}{\sqrt{2\pi}}}\int_{\mathbb{R}^{2}}f\left(x\right)L\left(-x,\,a\right)dx,$$where the red factor survives after we use $$\int_{\Bbb R}\exp i\omega zd\omega=2\pi\delta(z).$$But the red factor was spurious to begin with. If you had instead defined $L(x,\,a)$ as $\delta(x)$, we'd have the same problem. Let's revisit the proof that a product of transforms is a transform of a convolution: the left-hand side is$$\frac{1}{\sqrt{2\pi}}\int_{\Bbb R}g(x)\exp(i\omega x)dx\frac{1}{\sqrt{2\pi}}\int_{\Bbb R}h(y)\exp(i\omega y)dy,$$while the right-hand side is$$\frac{1}{\sqrt{2\pi}}\int_{\Bbb R^2}g(x)h(y-x)\exp(i\omega y)dxdy=\sqrt{2\pi}\frac{1}{\sqrt{2\pi}}\int_{\Bbb R}g(x)\exp(i\omega x)dx\int_{\Bbb R}h(y)\exp(i\omega y)dy.$$So actually, the transformed convolution has an extra factor of $\sqrt{2\pi}$, which is just what we need.