Continuous kernel integral operator is compact on c of an interval

For every $f\in C([a,b],\mathbb{K})$ the function $\mathcal{K}f$ is well defined and continuous, and $\mathrm{\parallel }\mathcal{K}f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2M(b-a)\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$: for real scalars the sharper bound without the factor $2$ is the uniform integral estimate of [A2] applied to $y\mapsto k(x,y)f(y)$; for complex scalars the real and imaginary parts of the integrand are continuous, each has absolute value at most $M\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, and [A2] bounds each real integral by $M(b-a)\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, so the complex triangle inequality gives the displayed factor $2$. Continuity in $x$ follows from the uniform continuity of $k$ on the square and the same real-component estimates. Linearity follows componentwise from real Riemann-integral linearity (Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and ${\int }_a^b(\lambda f+\mu g)=\lambda {\int }_a^{b}f+\mu {\int }_a^{b}g$), so this bound also makes $\mathcal{K}$ a bounded linear operator.

For all $f$ and all $x,x^{\mathrm{\prime }}\in [a,b]$ one has $\mathrm{\mid }\mathcal{K}f(x)-\mathcal{K}f(x^{\mathrm{\prime }})\mathrm{\mid }\le 2(b-a)\omega (x,x^{\mathrm{\prime }})\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, where $\omega (x,x^{\mathrm{\prime }})={\sup }_{y\in [a,b]}\mathrm{\mid }k(x,y)-k(x^{\mathrm{\prime }},y)\mathrm{\mid }$ and $\omega (x,x^{\mathrm{\prime }})\to 0$ as $\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }\to 0$ uniformly in $y$, by the uniform continuity of $k$ on the square; the factor $2$ covers the complex real-and-imaginary-part estimate and is harmless in the real case. In particular the family $F:=\{\mathcal{K}f:\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\}$ is equicontinuous (with complex modulus in the complex case, so both real component families satisfy [A3]) and pointwise bounded with $\mathrm{\mid }\mathcal{K}f(x)\mathrm{\mid }\le 2M(b-a)$.

In the real case $\mathbb{K}=\mathbb{R}$ the closure of $F$ in the supremum metric is compact by [A3] and [step 2.1], hence $\mathcal{K}$ is compact by [A5].

In the complex case every sequence $(f_j)$ with $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ has a subsequence for which $\mathcal{K}{f}_j$ converges uniformly: the real functions $x\mapsto \mathrm{Re}\mathcal{K}{f}_j(x)$ form an equicontinuous pointwise bounded sequence in $C([a,b],\mathbb{R})$ by [step 2.1] and [A2], so by [A3] they have a uniformly convergent subsequence; within that subsequence the imaginary parts, which are again equicontinuous and pointwise bounded, have a further uniformly convergent subsequence; along that further subsequence $\mathcal{K}{f}_j$ converges uniformly because the complex modulus is at most the sum of the moduli of the real and imaginary parts by [A2].

In the complex case the closure of $F$ is sequentially compact: given a sequence $(g_j)$ in the closure, [A4] chooses $f_j$ with $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ and $\mathrm{\parallel }\mathcal{K}{f}_j-g_j{\mathrm{\parallel }}_{\mathrm{\infty }}<1\mathrm{/}(j+1)$ for every $j$, and [step 3.2] applied to $(f_j)$ gives a subsequence along which $\mathcal{K}{f}_j$ converges, hence $g_j$ converges to the same limit; by [A3] the closure of $F$ is compact, and $\mathcal{K}$ is compact by [A5].

In both cases $\mathbb{K}=\mathbb{R}$ and $\mathbb{K}=\mathbb{C}$ the operator $\mathcal{K}$ is compact, which is the assertion.