Bulk-to-Boundary propagator

Question

How can I show that the bulk-to-boundary propagator $$ K(z,x;x')~=~\frac{z^{\Delta}}{[z^2+(x-x')^2]^{\Delta}} \tag{1} $$ goes as a Dirac delta function near the boundary $$ K(z,x;x')~\sim ~z^{d-\Delta}\delta^d(x-x') ~~?\tag{2} $$

Answer 1

In order to show that some function is the Delta distribution, you have to show that

1. It is zero except for where the argument of the function vanishes.
2. It integrates to 1 when integrated over the full coordinate range.

We can see these two properties explicitly.

1. If $x-x'\not=0$, then $$ \lim_{z\to0}K(z,x,x')=\lim_{z\to0}\frac{z^\Delta}{(x-x')^{2\Delta}}=0. $$ So near the boundary, $K(z,x,x')=0$ if $x-x'\not=0$.
2. We can do the integral $$ \begin{align} \int d^dx\, K(z,x,x')&=\int d^dx\, K(z,x,0)\\ &= \int d^dx\,\frac{z^\Delta}{(z^2+x^{2})^\Delta}\\ &= z^\Delta*S^{d-1}\int_0^\infty dr\,\frac{r^{d-1}}{(z^2+r^{2})^\Delta}\\ &= z^\Delta*\frac{2\pi^{\frac{d}{2}}}{\Gamma\left(\frac{d}{2}\right)}*z^{d-2\Delta}\frac{\Gamma\left(\frac{d}{2}\right)\Gamma\left(\Delta-\frac{d}{2}\right)}{2\Gamma(\Delta)}\\ &=\frac{\pi^{\frac{d}{2}}\Gamma\left(\Delta-\frac{d}{2}\right)}{\Gamma(\Delta)}z^{d-\Delta} \end{align} $$ In the first step we have shifted $x$. In the third step, we switched to spherical coordinates. As the integrad only depends on the radius, we simply get a factor of the surface of the unit sphere. The resulting integral can be explicitly done.

The result is $$ \lim_{z\to0}z^{\Delta-d}K(z,x,x')=\frac{\pi^{\frac{d}{2}}\Gamma\left(\Delta-\frac{d}{2}\right)}{\Gamma(\Delta)}\delta^d(x-x') $$ which is what is meant by the statement $$ K(z,x;x')~\sim ~z^{d-\Delta}\delta(x-x'). $$

Answer 2

Within distribution theory, a mathematically rigorous formulation of OP's eq. (2) is$^1$ $$ \lim_{z\to 0^+} z^{\Delta-d}K(z,x)~=~\delta^d(x), \tag{A}$$ where $$ K(z,x)~:=~ {\cal N}^{-1} \left( \frac{z}{z^2+x^2}\right)^{\Delta},\qquad x~\in~\mathbb{R}^d, \qquad \Delta >\frac{d}{2},\tag{B} $$ is the normalized bulk-to-boundary propagator, with normalization $$\begin{align} {\cal N}~:=~&\int_{\mathbb{R}^d} \frac{d^dx}{(1+x^2)^{\Delta}}~=~ {\rm Vol}(\mathbb{S}^{d-1})\int_{\mathbb{R}_+} \frac{r^{d-1}dr}{(1+r^2)^{\Delta}} \cr ~=~&{\rm Vol}(\mathbb{S}^{d-1})\frac{1}{2}B(\frac{d}{2},\Delta\!-\!\frac{d}{2})~=~\frac{\pi^{\frac{d}{2}}\Gamma(\Delta\!-\!\frac{d}{2})}{\Gamma(\Delta)},\end{align}\tag{C} $$ see e.g. Ref. 1. The proof of formula (A) is a straightforward application of e.g. Lebesgue's dominated convergence theorem after introducing test functions.

Conversely, this yields a useful representation of the $d$-dimensional Dirac delta distribution $$\delta^d(x)~=~\lim_{\epsilon\to 0^+} {\cal N}^{-1} \frac{\epsilon^{\Delta-d/2}}{(x^2+\epsilon)^{\Delta}}, \qquad \Delta > \frac{d}{2}.\tag{D}$$

References:

1. D.Z. Freedman, S.D. Mathur, A. Matusis & L. Rastelli, Correlation functions in the $CFT_d/AdS_{d+1}$ correpondence, Nucl. Phys. B546 (1999) 96, arXiv:hep-th/9804058; eqs. (11)-(12).

--

$^1$ For comparison, note that $$ K(z,x)~\longrightarrow~\left\{\begin{array}{ccl} 0\text{ almost everywhere} & \text{if} & \Delta~<~d \cr \delta^d(x) &\text{if} & \Delta~=~d\cr \text{too singular}& \text{if} & \Delta~>~d\end{array} \right\} \text{ for } z~\to~ 0^+. \tag{E}$$ The last case $\Delta>d$ is too singular to make sense as a distribution.