I'm failing to reproduce an (indirect) result in a paper, namely
$${δF[g]\overδg(x,y,z)}={r^4\over\ell^5} $$ where $F[g]=\iiint \frac{2dxdydz}{\ell g(x,y,z)} $ and $g(x,y,z)=-{\ell^2 \over r^2} $. Note that $\ell$ and $r$ do not depend on $x$, $y$ or $z$.
I'm not terribly familiar with functional derivatives, but it seems to me that this is a simple example, since no derivatives of $g$ appear in $F$, and hence, when $F$ is of the form $F[ρ]=\int f(ρ(\mathbf x)) d\mathbf x $, the following should hold[e.g. from wiki]: $${δF[f]\overδρ(\mathbf x)}=\frac {\partial f}{\partial ρ}$$ In this particular case, this gives $$\begin{align} {δF[g]\overδg(x,y,z)} &= \frac\partial{\partial g}\left(\frac {2}{\ell g(x,y,z)}\right)\\ &=- \frac {2} {\ell g^2(x,y,z)}\\ &=-2\frac {r^4} {\ell^5} \end{align}$$
This isn't what the authors find. The weird thing about this derivation is that the function derived with respect to, $g$, doesn't depend on its arguments but on $r$, and so does the functional. It is as if I'm looking for a generalized relation for when $F$ of the form $F[ρ]=\int f(ρ(\mathbf x, r)) d\mathbf x $ rather than of the form $F[ρ]=\int f(ρ(\mathbf x)) d\mathbf x $, $${δF[f]\overδρ(\mathbf x, r)}=?$$
What am I doing wrong, and more generally, does this dependence on $r$ even make a difference?
This is the paper I'm referring to, page 11, eq. (46) and line below. The calculation is the $AdS_4$ case, with $F \to S_{ct}=-\frac 2\ell\int dtdx_idx_i \sqrt{-γ}, g \to γ^{tt}=-\frac {\ell^2}{r^2}$ and $\sqrt {-γ}=-(γ^{tt})^{-1}$.
They claim $8πGT_{tt}=0=-2\frac {r^2}{\ell^3}+\frac 2{\sqrt{-γ}} \frac {δS_{ct}}{δγ^{tt}}$ from which I get the 'indirect' statement (at top): $\frac {δS_{ct}}{δγ^{tt}} =\frac {r^4}{ \ell^5}$
I) We have an answer up to an overall sign. From the $AdS_4$ Poincare bulk metric
$$ ds^2~=~ \frac{\ell^2}{r^2}dr^2 +\frac{r^2}{\ell^2}\left(-dt^2+dx^2_1+dx^2_2 \right) \tag{44}$$
in Ref. 1, we deduce that the 3D $\gamma_{\mu\nu}$ boundary metric is diagonal and
$$-\gamma_{tt}~=~ \gamma_{x_1x_1}~=~ \gamma_{x_2x_2}~=~\frac{r^2}{\ell^2}. \tag{A} $$
[This interpretation is further strengthen by comparing with eq. (17) for the $AdS_3$ case.] In other words,
$$\gamma_{\mu\nu}~=~\frac{r^2}{\ell^2}\eta_{\mu\nu},\tag{B} $$
where
$$\eta_{\mu\nu}~:=~{\rm diag}(-1,1,1).\tag{C}$$
II) Eventually, we want to vary the boundary metric $\gamma_{\mu\nu}$. To have a matrix close to a positive definite matrix, define
$$ M^{\mu}{}_{\nu}~:=~ \gamma^{\mu\lambda}\eta_{\lambda\nu}.\tag{D}$$
III) The counterterm action reads
$$ S_{ct}~=~\left(- \frac{2}{\ell}\right)\int \! dt ~dx_1 ~dx_2 ~\sqrt{-\gamma},\tag{46d} $$
where
$$ \gamma ~:=~\det\gamma_{\mu\nu}, \qquad \gamma^{-1} ~=~\det\gamma^{\mu\nu}.\tag{E} $$
IV) The functional derivative
$$\frac {\delta S_{ct}}{\delta \gamma^{tt}}\tag{F} $$
is proportional to
$$ \left(- \frac{\ell}{2}\right)\frac{2}{\sqrt{-\gamma}} \frac {\delta S_{ct}}{\delta \gamma^{tt}} ~=~\frac{2}{\sqrt{-\gamma}} \frac {\partial \sqrt{-\gamma}}{\partial \gamma^{tt}} ~=~\frac{1}{\gamma} \frac {\partial \gamma}{\partial \gamma^{tt}} ~=~ -\frac{1}{\gamma^{-1}} \frac {\partial \gamma^{-1}}{\partial \gamma^{tt}}$$ $$~=~-\frac {\partial \ln \det M}{\partial \gamma^{tt}} ~=~-\frac {\partial {\rm tr}\ln M}{\partial \gamma^{tt}} ~=~-{\rm tr} \left( M^{-1}\frac {\partial M}{\partial \gamma^{tt}}\right)$$ $$~=~-\gamma_{\mu\nu}\frac {\partial \gamma^{\nu\mu}}{\partial \gamma^{tt}} ~=~-\gamma_{tt}~\stackrel{(A)}{=}~ \frac{r^2}{\ell^2}, \tag{G}$$
and hence
$$\frac 2{\sqrt{-\gamma}} \frac {\delta S_{ct}}{\delta \gamma^{tt}} ~\stackrel{(G)}{=}~-\frac {2r^2}{\ell^3},\tag{H} $$
which is precisely the opposite of what is needed to make
$$ 8\pi G T_{tt}~=~-\frac {2r^2}{\ell^3}+\frac 2{\sqrt{-\gamma}} \frac {\delta S_{ct}}{\delta \gamma^{tt}} \tag{46a} $$
zero. We would be interested in learning the culprit of the missing sign. It might be a matter of conventions.
References: