Level-uniform 1D Poisson smoothing and approximation bounds

Extend a fine vector by $x_0=x_N=0$. Direct expansion gives $x^{\mathsf{T}}{A}_{h}x=h^{-1}{\sum }_{i=1}^N(x_i-x_{i-1}{)}^2\mathrm{.}$ This is positive for every nonzero $x$, since a zero sum forces all consecutive entries equal and the boundary entries are zero. The even rows of $P$ form the identity on coarse coordinates, so $P$ has full column rank. Thus [F2] ensures $A_H$ is invertible and $C$ is defined.

For any $x$, $(x_i-x_{i-1}{)}^2\le 2x_i^2+2x_{i-1}^2$ and the endpoints vanish, so $x^{\mathsf{T}}{A}_{h}x\le (4\mathrm{/}h)\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$. By [F4], $0<\lambda \le {\lambda }_{\max }(A_h)\le 4\mathrm{/}h$ for each eigenvalue. In an orthonormal eigenbasis, $S$ has eigenvalues $1-\lambda \mathrm{/}{\lambda }_{\max }\in [0,1)$. For $\xi =\lambda \mathrm{/}{\lambda }_{\max }\in (0,1]$, the finite geometric identity $\xi {\sum }_{k=0}^{2\nu }(1-\xi {)}^k=1-(1-\xi {)}^{2\nu +1}\le 1$ applies. Each summand is at least $(1-\xi {)}^{2\nu }$, so $\xi (1-\xi {)}^{2\nu }\le 1\mathrm{/}(2\nu +1)$. Expanding $e$ in that eigenbasis yields $\mathrm{\parallel }{A}_h{S}^{\nu }e{\mathrm{\parallel }}_2^2\le \frac{{\lambda }_{\max }(A_h)}{2\nu +1}\mathrm{\parallel }e{\mathrm{\parallel }}_{A_h}^2\le \frac{4}{h(2\nu +1)}\mathrm{\parallel }e{\mathrm{\parallel }}_{A_h}^2\mathrm{.}$

Let $J$ insert a coarse vector into the even fine indices and zero the odd indices. At an odd fine row, $A_{h}Pv=0$ because $2(Pv{)}_{2j+1}=(Pv{)}_{2j}+(Pv{)}_{2j+2}$. At an even row $2j$, $(A_{h}Pv{)}_{2j}=(2h{)}^{-1}(2v_j-v_{j-1}-v_{j+1}),$ with boundary values $v_0=v_M=0$. Direct multiplication gives $A_H$ diagonal entries $1\mathrm{/}h$, adjacent entries $-1\mathrm{/}(2h)$, and all other entries zero; hence $A_{h}P=J{A}_H$ and, by [F5], $P^{\mathsf{T}}{A}_h=A_H{J}^{\mathsf{T}}$. Substitution in [F1] yields $C=I-P{J}^{\mathsf{T}}$. Therefore $y=Cw$ has zero even entries. Also $A_{h}y=A_{h}w-J{A}_H{J}^{\mathsf{T}}w$, so the odd entries of $A_{h}y$ equal those of $A_{h}w$.

For odd $i$, the two neighbours of $y_i$ are even and zero, so $(A_{h}y{)}_i=2y_i\mathrm{/}h=(A_{h}w{)}_i$. Therefore $\mathrm{\parallel }Cw{\mathrm{\parallel }}_2^2={\sum }_{i\text{ odd}}{y}_i^2=\frac{h^2}{4}{\sum }_{i\text{ odd}}(A_{h}w{)}_i^2\le \frac{h^2}{4}\mathrm{\parallel }{A}_{h}w{\mathrm{\parallel }}_2^2\mathrm{.}$ Since each odd value is adjacent to two zero even values, the energy identity in step 1.1 gives $\mathrm{\parallel }y{\mathrm{\parallel }}_{A_h}^2=(2\mathrm{/}h)\mathrm{\parallel }y{\mathrm{\parallel }}_2^2$. Thus $\mathrm{\parallel }Cw{\mathrm{\parallel }}_{A_h}^2\le (h\mathrm{/}2)\mathrm{\parallel }{A}_{h}w{\mathrm{\parallel }}_2^2$.

Apply the approximation estimate of step 3.1 with $w=S^{\nu }e$ and then the smoothing estimate of step 1.2: $\mathrm{\parallel }C{S}^{\nu }e{\mathrm{\parallel }}_{A_h}^2\le \frac{h}{2}\mathrm{\parallel }{A}_h{S}^{\nu }e{\mathrm{\parallel }}_2^2\le \frac{2}{2\nu +1}\mathrm{\parallel }e{\mathrm{\parallel }}_{A_h}^2\mathrm{.}$ Because $\nu \ge 1$, $2\mathrm{/}(2\nu +1)\le 2\mathrm{/}3<1$, uniformly in $N$.