We follow
[16], Proposition 4.3(3). It is sufficient to find a Riemann-integrable function
\(F\) and then approximate it by a suitable smooth function. We make a further
ansatz of the form
\begin{equation*}
F(\rho_1,\dots,\rho_k) = \chi_{\Delta_k} \prod_{i=1}^k g(k\rho_i)
\end{equation*}
where \(g: [0, \infty) \to \RR\) is a smooth function supported on some interval \([0,T]\) with \(g(0)=1\) (for normalization purposes). To simplify notation, write \(I_k, J_k\) as shorthand for \(I_k(F), J_k^{(1)}(F)\text{.}\)
We first give an upper bound on \(I_k\text{.}\) Set \(\gamma = \int_{u=0}^\infty g(u)^2\,du > 0\text{;}\) then
\begin{equation*}
I_k = \idotsint_{\Delta_k} F(\rho_1,\dots,\rho_k)^2 d\rho_1\cdots d\rho_k
\leq \left( \int_0^\infty g(k\rho)^2 \,d\rho \right) = k^{-k} \gamma^k.
\end{equation*}
We next work on a lower bound for \(J_k\text{.}\) We first exploit the nonnegativity of squares to restrict the domain of integration for convenience:
\begin{equation*}
J_k \geq \idotsint_{\substack{\rho_2,\dots,\rho_k \geq 0}{\sum_{i=2}^k \rho_i \leq 1-T/k}}
\left( \int_0^{T/k} \left( \prod_{i=1}^k g(k\rho_i)\right) \,d\rho_1 \right)^2 d\rho_2\cdots d\rho_k.
\end{equation*}
We then do the opposite in order to separate variables: we rewrite this as \(J_k \geq J_k' - E_k\) where
\begin{align*}
J_k' \amp:= \idotsint_{\rho_2,\dots,\rho_k \geq 0}
\left( \int_0^{T/k} \left( \prod_{i=1}^k g(k\rho_i)\right) \,d\rho_1 \right)^2 d\rho_2\cdots d\rho_k\\
\amp = \left( \int_0^\infty g(k\rho_1)d\rho_1 \right)^2
\left(\int_0^\infty g(kt)^2\,dt \right)^{k-1} = k^{-k-1} \gamma^{k-1} \left( \int_0^\infty g(u)\,du\right)^2,\\
E_k \amp:= \idotsint_{\substack{\rho_2,\dots,\rho_k \geq 0}{\sum_{i=2}^k \rho_i > 1-T/k}}
\left( \int_0^{T/k} \left( \prod_{i=1}^k g(k\rho_i)\right) \,d\rho_1 \right)^2 d\rho_2\cdots d\rho_k\\
\amp= k^{-k-1} \left( \int_0^\infty g(u)\,du\right)^2
\idotsint_{\substack{u_2,\dots,u_k \geq 0}{\sum_{i=2}^k u_i > k-T}} \left( \prod_{i=2}^k g(u_i)^2 \right) du_2\cdots du_k.
\end{align*}
We first show that the error term \(E_k\) is small by comparison with a second moment, assuming that
\begin{equation*}
\mu := \frac{\int_0^\infty ug(u)^2\,du}{\int_0^\infty g(u)^2\,du} \lt 1 - \frac{T}{k}.
\end{equation*}
Define also
\begin{equation*}
\eta := \frac{k-T}{k-1} - \mu > 0.
\end{equation*}
We may then give an upper bound on \(E_k\) by multiplying the integrand by \(C := \mu^{-2} \left(\sum_{i=2}^k \frac{u_i}{k-1} - \mu \right)^2\) (this factor being at least 1) and dropping the restriction \(\sum_{i=1}^k u_i > k-T\) from the domain of integration. After expanding the square in \(C\text{,}\) we may further replace \(u_j^2 g(u_j)^2\) with \(T u_j g(u_j)^2\) thanks to the support restriction on \(g\text{.}\) We may then compute that
\begin{align*}
E_k \amp\leq \eta^{-2} k^{-k-1} \left( \int_0^\infty g(u)\,du \right)^2 \left( \frac{\mu T \gamma^{k-1}}{k-1} - \frac{\mu^2 \gamma^{k-1}}{k-1} \right)\\
\amp\leq \frac{\eta^{-2} \mu T k^{-k-1} \gamma^{k-1}}{k-1} \left( \int_0^\infty g(u)\,du \right)^2.
\end{align*}
Putting everything together and observing that \((k-1)\eta^2 \geq k(1-T/k-\mu)^2\) and \(\mu \leq 1\text{,}\) we get
\begin{equation*}
\frac{kJ_k}{I_k} \geq \frac{\left( \int_0^\infty g(u)\,du \right)^2}{\int_0^\infty g(u)^2\,du}
\left( 1 - \frac{T}{k(1-T/k - \mu)^2} \right).
\end{equation*}
By writing down an Euler–Lagrange equation, we may identify a good candidate for \(g(u)\text{:}\)
\begin{equation*}
g(u) = \frac{1}{1+Au}
\end{equation*}
for some suitable \(A>0\text{.}\) For this choice,
\begin{align*}
\int_0^T g(u)\,du \amp= \frac{\log(1+Au)}{A},\\
\int_0^T g(u)^2\,du \amp= \frac{1}{A} \left( 1 - \frac{1}{1+AT} \right),\\
\int_0^T ug(u)^2\,du \amp= \frac{1}{A^2} \left( \log(1+AT) - 1 + \frac{1}{1+AT} \right).
\end{align*}
We now fix \(A := \log k - 2 \log \log k\) and choose \(T\) so that \(1+AT = e^A\text{.}\) We then compute that
\begin{equation*}
\frac{kJ_k}{I_k} \geq (\log k - 2 \log \log k) \left( 1 - \frac{\log k}{(\log k)^2 + O(1)} \right)
\end{equation*}
which proves the claim.
