= Solution
For the proposed trial profile, $R'=-qr^{q-1}$. Its weighted <energy> is
$$
\int_0^1r^{n+1}R'^2\,dr=q^2\int_0^1r^{n+2q-1}\,dr=\frac{q^2}{n+2q}.
$$
The squared <weighted norm> is
$$
\int_0^1r(1-r^q)^2\,dr
=\frac12-\frac2{q+2}+\frac1{2q+2}
=\frac{q^2}{2(q+1)(q+2)}.
$$
Both are finite for $q>0$ and $n\geq0$, and $R(1)=0$. Therefore the <Rayleigh-Ritz variational principle> gives
$$
\boxed{\nu_{\mathrm{trial}}(q,n)=\frac{2(q+1)(q+2)}{n+2q},\qquad
\inf\operatorname{spec}\leq\nu_{\mathrm{trial}}}.
$$
When a <fundamental diffusion mode> exists, its decay rate is the infimum and is bounded above by this expression.
The existence qualification matters for the full range of $n$ printed in the question. For $n>2$, trial functions concentrated within radius $\varepsilon$ have quotient proportional to $\varepsilon^{n-2}$, proving the <loss of a spectral gap under strongly degenerate radial diffusion>. The infimum is zero and is not attained, because zero <energy> together with $R(1)=0$ forces the zero function. Thus there is no positive lowest-mode decay rate in that case. Even at $n=2$, the mode equation is $r^2R''+3rR'+\nu R=0$: its characteristic powers are $r^{-1\pm\sqrt{1-\nu}}$, and for $\nu>0$ every nonzero solution is unbounded at the centre. Hence the asserted bounded fundamental mode cannot be assumed for all $n\geq0$. The trial quotient remains valid, and the next part's $n=0$ case has the usual fundamental mode.
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