A continuous kernel defines a compact integral operator on C of an interval

Given: Countable Choice, Dependent Choice, $X=C([0,1],\mathbb{R})$, and a continuous kernel $k$ on the compact square $[0,1{]}^2$.

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$) and The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain: these are the two explicit choice hypotheses in the real Arzelà–Ascoli theorem used below.

A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: every continuous real-valued function on a compact interval is Riemann integrable.

Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and ${\int }_a^b(\lambda f+\mu g)=\lambda {\int }_a^{b}f+\mu {\int }_a^{b}g$: finite sums and scalar multiples of Riemann-integrable functions are integrable and their integrals are the corresponding finite sums and scalar multiples.

Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error: if two integrable functions differ everywhere by at most $\eta$, then their integrals on an interval of length $L$ differ by at most $L\eta$.

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: a continuous function on a compact metric space is uniformly continuous.

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, an equicontinuous pointwise bounded family in $C(K,\mathbb{R})$ on a compact metric space has compact closure in the supremum metric.

$C(K,\mathbb{R})$ is complete in the supremum metric for every nonempty compact metric space $K$: every continuous real-valued function on a compact metric space is bounded, so $M$ is finite.

Compact linear operator: a bounded linear operator is compact when the image of its closed unit ball has compact closure.

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 its values satisfy a uniform norm bound (A bounded linear operator between normed spaces).