Fredholm convergence and far-field error for cell-volume product integration

Let $X=L^2(D;\mathbb{C})$. Since $\mathrm{\mid }D\mathrm{\mid }>0$, $X\mathrm{\ne }\{0\}$ (a normalized indicator of any positive-measure cell or subset is nonzero). By [F6] and AC, $X$ is a Banach space. [F2] gives compactness of $K_k$, and the assumed injectivity makes the Fredholm premise of [F3] applicable once the approximation hypotheses are checked.

By [F1] and [F4], $K_N=P_N{K}_k$ is bounded linear and $\mathrm{\parallel }{K}_N\mathrm{\parallel }\le \mathrm{\parallel }{K}_k\mathrm{\parallel }$. For every $v\in X$, [F1] applied to $K_{k}v$ gives $K_{N}v=P_N{K}_{k}v\to K_{k}v$ in norm.

Let $B_X$ be the closed unit ball and $C=\overline{K_k(B_X)}$. Compactness in [F2] makes $C$ compact. Strong convergence $P_{N}z\to z$ from [F1] is uniform for $z\in C$: given $ϵ>0$, cover $C$ by finitely many $ϵ\mathrm{/}3$ balls with centers $z_1,\dots ,z_s$, choose $N_0$ so that $\mathrm{\parallel }{P}_N{z}_j-z_j\mathrm{\parallel }<ϵ\mathrm{/}3$ for every center and $N\ge N_0$, then use $\mathrm{\parallel }{P}_N\mathrm{\parallel }\le 1$ to get ${\sup }_{z\in C}\mathrm{\parallel }{P}_{N}z-z\mathrm{\parallel }<ϵ$. To show the tail union is totally bounded, start with any radius $\delta >0$, apply this uniform estimate with $ϵ=\delta \mathrm{/}3$, and cover $C$ by finitely many $\delta \mathrm{/}3$ balls; each $P_{N}z$ with $N\ge N_0$ then lies within $2\delta \mathrm{/}3$ of one of their centers. Thus ${\bigcup }_{N\ge N_0}{P}_N(C)$ is totally bounded. Each of the finitely many earlier sets $P_N(C)$ is compact as a continuous image of $C$, hence totally bounded. Their union $U:={\bigcup }_{N\ge 1}{P}_N(C)$ is totally bounded. Its closure is totally bounded too: for any $\delta >0$, take a finite $\delta \mathrm{/}2$-net of $U$; every point of $\bar{U}$ is within $\delta \mathrm{/}2$ of some point of $U$, which is within $\delta \mathrm{/}2$ of a net center. The closure is complete: every Cauchy sequence in $\bar{U}$ converges in the Banach space $X$ from step 1.1, and closedness of $\bar{U}$ keeps that limit in $\bar{U}$. By [F6], $\bar{U}$ is compact. Since $K_N(B_X)=P_N{K}_k(B_X)\subset P_N(C)$, the family $(K_N)$ is collectively compact in the precise sense required by [F3].

Now all hypotheses of [F3] hold by steps 1.1, 1.2, and 2.1. Therefore there are $N_0$ and a uniform $C$ such that $I-K_N$ is invertible for $N\ge N_0$ and its solutions satisfy $\mathrm{\parallel }{u}_N-u{\mathrm{\parallel }}_2\le C(\mathrm{\parallel }(K_N-K_k)u{\mathrm{\parallel }}_2+\mathrm{\parallel }{f}_N-f{\mathrm{\parallel }}_2)=C(\mathrm{\parallel }(P_N-I)K_{k}u{\mathrm{\parallel }}_2+\mathrm{\parallel }(P_N-I)f{\mathrm{\parallel }}_2)$. The right side tends to zero by [F1].

For $N\ge N_0$, the full-space solution satisfies $u_N=f_N+P_N{K}_k{u}_N\in V_N$. In the normalized cell-indicator basis of [F4], its equation is exactly $(I-A^{(N)})c=b^{(N)}$. If that finite matrix had a nonzero null vector, its corresponding nonzero element of $V_N$ would be in the kernel of $I-K_N$, contradicting invertibility; thus the matrix is nonsingular and its reconstruction is the unique operator solution. Finally [F5] gives the stated far-field bound, and step 3.1 makes its right side tend to zero.