= Solution
Use a planar flux function $F=B_0z+a$ and write
$$
\mathbf B=(F_z,0,-F_x).
$$
This automatically satisfies the <solenoidal magnetic-field constraint>. There is a sign convention to settle: with the usual definition $\mathbf B=\nabla\times\mathbf A$, the <magnetic vector potential> is $\mathbf A=-F\widehat{\mathbf y}$. The positive sign printed for $A_y$ instead gives the opposite field, or corresponds to defining $\mathbf B=-\nabla\times\mathbf A$. We use the standard curl convention and the flux $F$ oriented along the prescribed positive $B_0$; the flux evolution and all transport identities below are unchanged by reversing the overall potential convention.
For planar <incompressible flow>, substitute this representation into the <resistive induction equation>. The two in-plane components imply
$$
F_t+\mathbf u\cdot\nabla F=\eta\nabla^2F+C(t).
$$
An additive time-dependent constant in $F$ leaves the field unchanged and removes $C(t)$. Since $\nabla^2(B_0z)=0$, this gives
$$
\boxed{a_t+\mathbf u\cdot\nabla a+B_0w=\eta\nabla^2a.}
$$
This is the scalar <advection-diffusion equation> underlying the <planar anti-dynamo theorem>, with the imposed-field source retained.
Put $M(z)=\bar a(z)$, $a=M+a'$ and $F_c(z)=\overline{wa'}$. Stationarity of the horizontal averages, $\bar w=0$, and incompressibility give
$$
\partial_zF_c=\eta M''.
$$
Here the horizontal derivative of the advective flux has zero average. Integration gives $F_c=\eta M'+C$. Averaging once more, with $\langle M'\rangle=0$, identifies $C=\langle wa'\rangle$. Thus
$$
\boxed{\overline{wa'}=\eta\bar a_z+\langle wa'\rangle.}
$$
Decay or periodic averaging removes the endpoint contribution from the mean derivative. The periodic assumption used later replaces decay at vertical infinity; a nonzero periodic function cannot also decay there.
Subtract the averaged equation from the scalar equation and multiply by $a'$. The average of the time derivative is zero by stationarity; the advective term becomes a boundary flux and vanishes by incompressibility and the stated boundary averaging. Terms depending only on $z$ multiply $\overline{a'}=0$. <Integration by parts> in the diffusion term therefore gives
$$
\boxed{\langle\overline{wa'}M'\rangle+B_0\langle wa'\rangle
=-\eta\langle|\nabla a'|^2\rangle.}
$$
Substituting $F_c=\eta M'+C$ and using $\langle M'\rangle=0$ yields the <Zeldovich planar flux balance>
$$
\boxed{B_0\langle wa'\rangle=-\eta\langle|\nabla a'|^2\rangle-\eta\langle(M')^2\rangle.}
$$
Both terms on the right are dissipative. In a nontrivial stationary forced problem this requires transport opposing the imposed flux gradient.
For finite-energy localized fields, enforcing decay at both vertical infinities literally gives $C=0$ from the mean-flux boundary values; the corresponding integral balance then forces both dissipation terms to vanish. The later nonzero-transport, periodic case therefore replaces that localized boundary setting rather than satisfying both simultaneously.
Now assume $B_0\ne0$ and a positive <turbulent magnetic diffusivity> $\eta_1$. Define
$$
V=\langle(f-1)^2\rangle=\langle f^2\rangle-1,\qquad
r_t=\frac{\eta_1}{\eta},\qquad
G=\langle|\nabla a'|^2\rangle.
$$
Since $F_c=-\eta_1B_0f$ and $\langle f\rangle=1$, its spatial mean is $C=-\eta_1B_0$. Hence
$$
M'=-r_tB_0(f-1).
$$
The <Zeldovich planar flux balance> becomes
$$
\eta_1B_0^2=\eta G+\frac{\eta_1^2}{\eta}B_0^2V,
\qquad
G=r_tB_0^2(1-r_tV).
$$
Because $C\ne0$, the fluctuation cannot vanish, and a nontrivial fluctuation has $G>0$ under the stated boundary or zero-mean periodic conditions. Therefore
$$
\boxed{\frac{\eta}{\eta_1}>\langle f^2\rangle-1.}
$$
If $\eta_1\gg\eta$, then $\|f-1\|_{L^2}^2<\eta/\eta_1\ll1$: \b[the transport profile must be nearly constant in mean square.] This does not, without additional regularity, imply a uniform pointwise bound on narrow spikes.
For the final estimate use the supplied <Poincaré inequality> in the periodic setting and the <Cauchy-Schwarz inequality>:
$$
\eta_1^2B_0^2=\langle wa'\rangle^2
\leq\langle w^2\rangle\langle a'^2\rangle
\leq\langle w^2\rangle\frac{d^2}{\pi^2}G.
$$
Insert the expression for $G$, cancel $B_0^2$, and multiply by $V$. Setting $y=r_tV\in[0,1)$ gives
$$
\eta_1^2V\leq\langle w^2\rangle\frac{d^2}{\pi^2}y(1-y)
\leq\langle w^2\rangle\frac{d^2}{4\pi^2}.
$$
Consequently the <variance bound for planar turbulent magnetic transport> is
$$
\boxed{\eta_1^2[\langle f^2\rangle-1]\leq\langle w^2\rangle\frac{d^2}{4\pi^2}.}
$$
The factor $1/4$ comes from maximizing $y(1-y)$, so the weaker supplied Poincare constant already suffices. Under a zero vertical mean at each horizontal location, the usual period-$d$ Fourier bound is even stronger, with $4\pi^2/d^2$ in place of $\pi^2/d^2$.
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