= Solution
Multiplying the vertical induction equation by $B$ and averaging eliminates advection and gives
$$
\frac12\frac{d}{dt}\langle B^2\rangle=\langle B(A_yU_x-A_xU_y)\rangle-\eta\langle|\nabla B|^2\rangle.
$$
Integrate the derivatives of $A$ by parts. Mixed derivatives of $U$ cancel, leaving
$$
\langle B(A_yU_x-A_xU_y)\rangle=\langle A(B_xU_y-B_yU_x)\rangle.
$$
The pointwise cross-product bound and the <Cauchy-Schwarz inequality> bound its absolute value by $Q\sqrt{\langle A^2\rangle\langle|\nabla B|^2\rangle}$, where $Q=\|\nabla U\|_\infty$. Hence
$$
\boxed{\frac12\frac{d}{dt}\langle B^2\rangle\leq Q\sqrt{\langle A^2\rangle\langle|\nabla B|^2\rangle}-\eta\langle|\nabla B|^2\rangle.}
$$
For $J=\langle|\nabla B|^2\rangle$ and $X=\langle A^2\rangle$, completing the square gives
$$
Q\sqrt{XJ}-\eta J=-\eta\left(\sqrt J-\frac{Q\sqrt X}{2\eta}\right)^2+\frac{Q^2X}{4\eta}.
$$
Apply part (i) and integrate, obtaining the <periodic two-coordinate anti-dynamo energy bound>
$$
\boxed{\langle B^2(t)\rangle-\langle B^2(0)\rangle
\leq\frac{Q^2A_0^2}{4\eta^2k^2}(1-e^{-2\eta k^2t})
\leq\frac{Q^2A_0^2}{4\eta^2k^2}.}
$$
\b[The printed bound has only one power of diffusivity and is false as written.] Both powers above are needed: one comes from maximizing against diffusion and the other from integrating the decay time of $A$.
For an explicit counterexample in a nondimensional $2\pi$-periodic box, take $u=(0,0,\sin y)$, $A(x,y,0)=\cos x$, $B(x,y,0)=0$, $k=1$. All means vanish and the velocity is solenoidal. The exact solution is
$$
A=e^{-\eta t}\cos x,\qquad
B=\frac1\eta(e^{-\eta t}-e^{-2\eta t})\sin x\cos y.
$$
Indeed $B_h\cdot\nabla U=e^{-\eta t}\sin x\cos y$ and this mode's Laplacian is $-2B$. Here $A_0^2=1/2$, $Q=1$, and at $t=(\log2)/\eta$,
$$
\langle B^2\rangle=\frac1{64\eta^2}.
$$
For $\eta=0.01$ this is $156.25$, exceeding the printed $1/(8\eta)=12.5$. The corrected bound is valid and finite.
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