= Solution
Use $S_C(r)=\langle(C'-C)^2\rangle$ for the <scalar structure function>, and assume <statistical homogeneity> as well as isotropy. Expanding the square gives
$$
S_C(r)=2\langle C^2\rangle-2\langle CC'\rangle.
$$
If the zero-mean scalar decorrelates at large separation, then
$$
\boxed{S_C(r)\longrightarrow2\langle C^2\rangle.}
$$
The decorrelation assumption is necessary: a spatially constant, zero-ensemble-mean random scalar is isotropic but has $S_C=0$ at every separation.
For the small-separation limit assume mean-square differentiability. Taylor expansion gives $C'-C=r_i\partial_iC+o(r)$ in mean square. Isotropy implies $\langle\partial_iC\partial_jC\rangle=\delta_{ij}\langle|\nabla C|^2\rangle/3$. Define the <scalar dissipation rate> in the half-variance convention, $\epsilon_c=\alpha\langle|\nabla C|^2\rangle$, so that <diffusion> removes $\langle C^2/2\rangle$ at rate $\epsilon_c$. It follows that
$$
\boxed{S_C(r)=\frac{\epsilon_c}{3\alpha}r^2+o(r^2).}
$$
If dissipation is instead defined as the loss rate of the full <variance>, the coefficient must be adjusted by a factor of two.
In the <inertial-convective scalar range>, $\max(\eta,\eta_c)\ll r\ll\ell$, suppose <scalar variance> cascades locally at an approximately constant flux $\epsilon_c$, while <velocity> increments have the <Kolmogorov 1941 theory> turnover time $t_r\sim\epsilon^{-1/3}r^{2/3}$. The <variance> associated with scale $r$ divided by that turnover time is of order its transfer rate, giving the <Obukhov-Corrsin theory> result
$$
\boxed{S_C(r)\sim\epsilon_ct_r\sim\epsilon_c\epsilon^{-1/3}r^{2/3}.}
$$
The assumptions include a passive scalar, high Reynolds number, local transfer, adequate scale separation, and statistical equilibrium or sufficiently slow variation of the supply. Isotropy alone does not provide a constant scalar flux.
For weak scalar <diffusion>, $\alpha\ll\nu$, the <velocity> is smooth below $\eta$, with characteristic strain rate $s\sim(\epsilon/\nu)^{1/2}$. <Diffusion> rate $\alpha/r^2$ balances this at the <Batchelor scalar microscale>
$$
\eta_c=\eta_B\sim(\alpha/s)^{1/2}=(\alpha^2\nu/\epsilon)^{1/4}.
$$
In the <viscous-convective scalar range>, the derivative of $S_C$ is assumed to depend only on $\epsilon_c$, $s$ and $r$. Their dimensions force
$$
\frac{dS_C}{dr}\sim\frac{\epsilon_c}{sr}.
$$
Integration from the <diffusion> cutoff gives
$$
S_C(r)=S_C(\eta_c)+A_B\frac{\epsilon_c}{s}\ln\left(\frac r{\eta_c}\right).
$$
Thus, well within a broad logarithmic range,
$$
\boxed{S_C(r)\sim\epsilon_c\sqrt{\frac\nu\epsilon}\ln(r/\eta_c).}
$$
The additive matching contribution is not exactly zero; the formula is a leading range estimate, not a boundary condition at $r=\eta_c$. Using $dS_C/dr$ isolates the contribution acquired on changing separation. The structure function itself contains <scalar variance> accumulated from smaller scales and must retain the inner cutoff through its integration constant. If one assumed that $S_C$ depended only on $\epsilon_c,s,r$ without that cutoff, <dimensional analysis> would incorrectly give an $r$-independent value proportional to $\epsilon_c/s$. This explains the use of the derivative with respect to $r$; the TeX aid changes that derivative to $t$ in its final sentence, but the original PDF consistently uses $r$.
Finally let $\alpha\gg\nu$. Balancing <diffusion> with the inertial eddy rate gives the <Corrsin scalar microscale> $\eta_c=\eta_C\sim(\alpha^3/\epsilon)^{1/4}\gg\eta$. For $\eta\ll r\ll\eta_c$, <diffusion> is faster than the eddy turnover rate. The scalar is smooth to leading order on those separations, even though the <velocity> still has inertial-range fluctuations. The <inertial-diffusive scalar range> therefore has
$$
\boxed{S_C(r)\simeq\frac{\epsilon_c}{3\alpha}r^2,\qquad\eta\ll r\ll\eta_c.}
$$
This continues the scalar's leading gradient-controlled quadratic behaviour; it is not a continuation of the inertial-convective two-thirds law below its <diffusion> cutoff.
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