= Solution
Multiply the separated <Sturm-Liouville problem> by real $R$ and apply <integration by parts>:
$$
\nu\int_0^1rR^2\,dr
=-\int_0^1R(r\kappa R')'\,dr
=-[r\kappa RR']_0^1+\int_0^1r\kappa R'^2\,dr.
$$
The outer endpoint term vanishes because $R(1)=0$. At zero, regularity gives vanishing radial flux. This can be obtained from the equation without assuming a bounded <derivative>: let $F=r\kappa R'$. Since $F'=-\nu rR$ and $R$ is bounded, $F$ has a finite limit at zero. If that limit were nonzero, positive bounded <thermal diffusivity> would make $R'$ have fixed sign and magnitude at least a constant times $1/r$ sufficiently near zero, contradicting bounded $R$. Hence $F(0)=0$, and in fact $F(r)=-\nu\int_0^r sR(s)\,ds=O(r^2)$. This also forces $RF\to0$ for bounded $R$.
The boundary terms therefore vanish, proving
$$
\boxed{\nu\int_0^1rR^2\,dr=\int_0^1r\kappa(r)R'(r)^2\,dr}.
$$
As usual for diffusion, $\kappa$ is nonnegative and positive for $0<r\leq1$; the later power-law case may vanish at the centre. The argument applies in that case as well.
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