One-dimensional nonaligned micro sampling resonance bound

For either $b=a$ or $b(y)=a(\theta +ry)$, define $c_b=({\int }_0^1{b}^{-1}{)}^{-1}$ and $h_b=c_b\mathrm{/}b-1$. The ellipticity bounds make $h_b\in L^{\mathrm{\infty }}(0,1)$, and ${\int }_0^1{h}_b=0$. Put ${\chi }_b(t)={\int }_0^t{h}_b(s)ds$. For every smooth periodic $\phi$, Fubini [F5] gives ${\int }_0^1{\chi }_b(t){\phi }^{\mathrm{\prime }}(t)dt={\int }_0^1{h}_b(s)(\phi (1)-\phi (s))ds=-{\int }_0^1{h}_b(s)\phi (s)ds$, so $h_b$ is its periodic weak derivative. By [F4], the symmetric Fourier partial sums of ${\chi }_b$ are real trigonometric polynomials and converge in $H^1(0,1)$, since both ${\chi }_b$ and $h_b$ are in $L^2$ and their Fourier coefficients satisfy the weak-derivative relation; after subtracting its mean, the limit belongs to $V_{\mathrm{\#}}$ by the closure definition. Since $b(1+h_b)=c_b$ almost everywhere, for every $v\in V_{\mathrm{\#}}$ we have ${\int }_0^{1}b(1+{\chi }_b^{\mathrm{\prime }})v^{\mathrm{\prime }}=c_b{\int }_0^1{v}^{\mathrm{\prime }}=0$ by [F6]. Thus it solves the cell equation. If two mean-zero solutions existed, testing their difference equation by the difference would give $\int b\mathrm{\mid }{\chi }^{\mathrm{\prime }}-{\tilde{\chi }}^{\mathrm{\prime }}{\mathrm{\mid }}^2=0$; the periodic Poincare inequality [F6] then makes their difference zero. This proves both harmonic-mean formulas, including the nonperiodic-in-$y$ sampled coefficient under periodic coupling.

Put $g=1\mathrm{/}a$ and $I={\int }_0^{1}g$. Write $r=N+s$ with integer $N\ge 1$ and $0\le s<1$. Periodicity gives ${\int }_0^{r}g(\theta +t)dt=NI+R,R={\int }_{\theta }^{\theta +s}g(t)dt,$ so ${\int }_0^{1}g(\theta +ry)dy=I+(R-sI)\mathrm{/}r$. Since $0<g\le 1\mathrm{/}\lambda$, both $R$ and $sI$ lie in $[0,s\mathrm{/}\lambda ]$, hence $\mathrm{\mid }R-sI\mathrm{\mid }\le s\mathrm{/}\lambda \le 1\mathrm{/}\lambda$. In particular the weaker bound $\mathrm{\mid }{I}_r-I\mathrm{\mid }\le 2\mathrm{/}(\lambda r)$ holds. Also $I,I_r\ge 1\mathrm{/}\mathrm{\Lambda }$. Therefore $\mathrm{\mid }{A}_r-A^{\ast }\mathrm{\mid }=\frac{\mathrm{\mid }{I}_r-I\mathrm{\mid }}{I_{r}I}\le \frac{2}{\lambda r}{\mathrm{\Lambda }}^2,$ as claimed.