= Solution
<Local isotropy of turbulence> means that joint small-scale <velocity increment> statistics are approximately unchanged by <rotations>, even when the large-scale flow is anisotropic. The <universal equilibrium range> contains separations much smaller than the <integral scale of turbulence>: their adjustment times are short compared with the outer <eddy turnover time>, and their normalized statistics are assumed independent of the detailed large-scale forcing. It includes the <inertial range> and the <dissipation range>.
The <Kolmogorov first similarity hypothesis> says that these locally isotropic statistics depend only on $r$, mean <viscous dissipation> $\epsilon$ and <kinematic viscosity> $\nu$. <Dimensional analysis> therefore expresses the second-order <longitudinal structure function> in terms of the <Kolmogorov microscales>:
$$
\boxed{S_2(r)=\langle(\Delta v)^2\rangle=v^2F(r/\eta),\qquad r\ll\ell}.
$$
The universality of $F$ is an assumption of the theory, not a consequence of <dimensional analysis> alone.
The <Kolmogorov second similarity hypothesis> removes dependence on <kinematic viscosity> when $\eta\ll r\ll\ell$. Only $\epsilon$ and $r$ remain, so the <Kolmogorov two-thirds law> is
$$
\boxed{S_2(r)=\beta(\epsilon r)^{2/3}}.
$$
The same reasoning gives the predicted order-$P$ <longitudinal structure function>:
$$
\boxed{\langle(\Delta v)^P\rangle=\beta_P(\epsilon r)^{P/3}}.
$$
These are signed <moments> for integer $P$; for noninteger orders, use $\langle|\Delta v|^P\rangle$. In particular, the signed third-order coefficient is negative, $\beta_3=-4/5$, in the <Kolmogorov four-fifths law>.
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