= Solution
Multiply the <Kármán-Howarth equation> for $C(r,t)$ by $4\pi r^2$ and integrate in $r$. <Integration by parts> gives the exact boundary-flux expression
$$
\boxed{\frac{dL}{dt}=4\pi\left[\frac1r\frac{d}{dr}(r^4u^3K)+2\nu r^2\frac{dC}{dr}\right]_{0}^{\infty}}.
$$
At the origin smooth isotropic correlations give $K=O(r)$ or smaller and $C'=O(r)$, so both boundary terms vanish. At infinity the usual regular power-law decay $K=O(r^{-4})$, with the corresponding derivative decay, makes the first term tend to zero. A sufficiently smooth integrable trace correlation makes $r^2C'\to0$ as well. Under these physical far-field conditions,
$$
\boxed{dL/dt=0}.
$$
The exact required conditions are the limits of these two boundary fluxes. A bare big-$O$ condition on $K$ without control of its derivatives is not by itself a rigorous sufficient condition: the smooth tail $K(r)=r^{-4}\sin(r^2)$ is $O(r^{-4})$, but $r^{-1}(r^4K)'=2\cos(r^2)$ has no zero limit. The intended power-law assumption excludes such increasingly rapid far-field oscillations; this qualification is needed in a literal proof.
Physically, $L$ is the large-volume <momentum> <variance> divided by volume. Internal advective and <pressure> forces exchange <momentum> among eddies but do not provide a net <impulse> to the whole unforced field. For a large spherical region, the <variance> balance is driven by its boundary <momentum> flux. The stated correlation decay makes that contribution negligible per unit volume as the region grows. Thus \b[the conservation law preserves <momentum>-fluctuation density, not <kinetic energy>]. <kinematic viscosity> can remove <kinetic energy> while leaving this integral invariant.
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