= Solution
Let $C(r)=\langle\mathbf u(\mathbf x)\cdot\mathbf u(\mathbf x+\mathbf r)\rangle$ in zero-mean <isotropic turbulence>. Define the scalar <Saffman integral> $L=\int_{\mathbb R^3}C(r)\,d^3r=4\pi\int_0^\infty r^2C(r)\,dr$. The spectral transform is
$$
E(k)=\frac1\pi\int_0^\infty C(r)kr\sin(kr)\,dr.
$$
For an integrable $r^2C(r)$, dominated convergence applied to $\sin(kr)/(kr)$ gives
$$
\boxed{E(k)=\frac{L}{4\pi^2}k^2+o(k^2)}\quad(k\to0).
$$
A further $O(k^4)$ expansion requires an additional correlation <moment> and is not automatic from finite $L$. The normalization agrees with $\int E\,dk=\langle|\mathbf u|^2\rangle/2$.
For a control volume $V$ put $\mathbf P_V=\int_V\mathbf u\,dV$. <Statistical homogeneity> gives the exact finite-volume identity
$$
\langle|\mathbf P_V|^2\rangle
=\int_V\int_V C(\mathbf y-\mathbf x)\,d^3x\,d^3y
=\int_{\mathbb R^3}C(\mathbf r)|V\cap(V-\mathbf r)|\,d^3r.
$$
For geometrically similar volumes increasing without bound, the overlap fraction $|V\cap(V-\mathbf r)|/|V|$ tends to one at every fixed separation. If $C$ is absolutely integrable, dominated convergence gives
$$
\boxed{L=\lim_{|V|\to\infty}\frac{\langle|\mathbf P_V|^2\rangle}{|V|}}.
$$
The printed finite-large-volume expression is the corresponding asymptotic approximation, not an exact identity for an arbitrary finite volume. For instance, positive nonzero $C$ with a finite correlation length gives an overlap fraction smaller than one and hence a strictly smaller finite-volume value. The limit also shows $L\ge0$; a nonzero <Saffman integral> is positive.
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