Conjugate transpose kernels give adjoints

Factor swap is measurable on the product sigma-algebra because the inverse image of $E\times F$ is $F\times E$. For a nonnegative product-measurable $q$, Tonelli applied in both orders gives $\iint q(y,x)d\mu (y)d\mu (x)=\iint q(x,y)d\mu (y)d\mu (x)$. In particular it preserves null sets. Thus it is measurable and measure-preserving on the completion as well: a completed measurable set differs from a product-measurable set by a subset of a product-null set, whose swapped set is still null. Consequently $k^{\ast }$ is a well-defined completed $L^2$ class with $\mathrm{\parallel }{k}^{\ast }{\mathrm{\parallel }}_2=\mathrm{\parallel }k{\mathrm{\parallel }}_2$. F1 gives its compact bounded operator $L$. AC supplies the countable-choice hypotheses here and in F1.

For $f,g\in L^2$, Tonelli gives $\mathrm{\parallel }f(y)\overline{g(x)}{\mathrm{\parallel }}_{L^2(X^2)}=\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2$. Product Cauchy–Schwarz bounds the absolute integral of $k(x,y)f(y)\overline{g(x)}$ by $\mathrm{\parallel }k{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2$. Hence Fubini applies, and conjugating the inner integral gives $⟨Kf,g⟩=\int f(y)\overline{\int \stackrel{‾}{k(x,y)}g(x)d\mu (x)}d\mu (y)=⟨f,Lg⟩$. Therefore $L=K^{\ast }$ by adjoint uniqueness. The a.e. section conventions are those of F1.

Applying the formula twice gives $(k^{\ast }{)}^{\ast }=k$ and hence $(K^{\ast }{)}^{\ast }=K$. For every $z\in L^2$, Cauchy–Schwarz gives ${\sup }_{\mathrm{\parallel }g\mathrm{\parallel }\le 1}\mathrm{\mid }⟨z,g⟩\mathrm{\mid }\le \mathrm{\parallel }z\mathrm{\parallel }$; equality follows by $g=z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel }$ if $z\mathrm{\ne }0$, and both sides are zero if $z=0$. Thus $\mathrm{\parallel }Kf\mathrm{\parallel }={\sup }_{\mathrm{\parallel }g\mathrm{\parallel }\le 1}\mathrm{\mid }⟨f,K^{\ast }g⟩\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }{K}^{\ast }\mathrm{\parallel }$. Taking the supremum over the unit ball gives $\mathrm{\parallel }K\mathrm{\parallel }\le \mathrm{\parallel }{K}^{\ast }\mathrm{\parallel }$. Apply this inequality to $K^{\ast }$ and use its double adjoint to obtain the reverse inequality. Compactness of both operators was supplied by F1 and step 1.1.