Composite midpoint Nyström operators are collectively compact and consistent

Given: AC, the continuous kernel $k$, its integral operator $K$, and the composite midpoint operators $K_N$ on $X=C([0,1],\mathbb{R})$.

The Axiom of Choice and AC supplies the countable and dependent choices used in Banach integration: AC supplies both Countable Choice and Dependent Choice, as required by the local compactness and Arzelà–Ascoli suppliers.

Nyström quadrature operators and their reconstructed equations: $K_{N}u(x)=h{\sum }_{j=0}^{N-1}k(x,(j+\frac{1}{2})h)u((j+\frac{1}{2})h)$.

A continuous kernel defines a compact integral operator on C of an interval: under Countable Choice and Dependent Choice, the continuous-kernel integral operator $K:X\to X$ is bounded and compact.

Composite midpoint rule on a uniform grid: the midpoint rule on $N$ uniform panels uses $h=1\mathrm{/}N$, one midpoint per panel, and weight $h$.

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: the kernel $k$ and, for each fixed $u\in X$, the function $(x,y)\mapsto k(x,y)u(y)$ are uniformly continuous on the compact square.

Arzelà--Ascoli for real $C(K)$ under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded: under Countable Choice and Dependent Choice, a pointwise bounded equicontinuous family in $C([0,1],\mathbb{R})$ has compact closure.

For $a<c<b$: $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$, and then ${\int }_a^{b}f={\int }_a^{c}f+{\int }_c^{b}f$; with the oriented form for arbitrary $a,b,c$, If $f\le g$ on $[a,b]$ and both are integrable then ${\int }_a^{b}f\le {\int }_a^{b}g$; and $m(b-a)\le {\int }_a^{b}f\le M(b-a)$, and Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error: midpoint panel integrals add, and a uniformly bounded continuous panel error has integral bounded by panel length times its supremum.

$C(K,\mathbb{R})$ is complete in the supremum metric for every nonempty compact metric space $K$: the continuous kernel is bounded on the compact square, so $M:=\mathrm{\parallel }k{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$.

The interval $[0,1]$ and square $[0,1{]}^2$ with their Euclidean metrics are nonempty compact metric spaces (Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

On the nonempty compact interval $[0,1]$ (Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), let $X=C([0,1],\mathbb{R})$ with its supremum norm. Pointwise addition and real scalar multiplication preserve continuity. The supremum is finite by $C(K,\mathbb{R})$ is complete in the supremum metric for every nonempty compact metric space $K$; the pointwise absolute-value inequalities give $\mathrm{\parallel }u+v{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $\mathrm{\parallel }cu{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\mid }c\mathrm{\mid }\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}$, and $\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}=0$ holds exactly when $u=0$. Thus this is a real norm (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). Its induced metric is the complete supremum metric of $C(K,\mathbb{R})$ is complete in the supremum metric for every nonempty compact metric space $K$, so $X$ is Banach (Banach space).

A linear map between normed spaces is bounded when one constant bounds its output norm by that constant times the input norm (A bounded linear operator between normed spaces).

Additivity and scalar homogeneity of finite sums make the midpoint finite-sum formula linear in $u$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).