When $J(x)$ varies, the radial wave equation recursively generates local derivative terms before the normalizable response. Substituting
$$
\phi=J+z^2\phi_{(2)}+z^4\phi_{(4)}+z^5A+\cdots
$$
into
$$
z^2\phi_{zz}-4z\phi_z+z^2\Box_5\phi=0
$$
gives
$$
\phi_{(2)}=\frac16\Box_5J,
\qquad
\phi_{(4)}=\frac1{24}\Box_5^2J.
$$
Because the boundary dimension is odd, no logarithmic term is required. <Holographic renormalization> subtracts these source-dependent pieces, leaving
$$
\boxed{
\mathcal O(x)=
\lim_{z\to0}z^{-5}
\left[
\phi(z,x)-J(x)
-\frac{z^2}{6}\Box_5J(x)
-\frac{z^4}{24}\Box_5^2J(x)
\right]}
$$
in the normalization $\mathcal O=A$. Equivalently,
$$
\boxed{
\mathcal O(x)=\frac15\lim_{z\to0}z^{-4}
\left[
\partial_z\phi
-\frac z3\Box_5J
-\frac{z^3}{6}\Box_5^2J
\right]}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 354 / 1 / b / iii / Solution

{wiki}

Holographic renormalization subtracts local boundary counterterms from the regulated on-shell bulk action. It gives finite correlation functions and identifies the renormalized normalizable coefficient with an operator expectation value.


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 Incoming links: Holographic renormalization