Mean-zero Hilbert–Schmidt operators on a compact boundary have square-integrable kernels

Let $c_0=\mathrm{\mid }M{\mathrm{\mid }}^{-1\mathrm{/}2}\mathbf{1}$ and choose a countable orthonormal basis $(c_j{)}_{j\ge 1}$ of $L_0^2(M)$, using [A1] and separability [F1]. Extend $T$ by $\tilde{T}{c}_0=0$ and $\tilde{T}{c}_j=T{c}_j$. The Hilbert–Schmidt condition means ${\sum }_{j\ge 1}\mathrm{\parallel }T{c}_j{\mathrm{\parallel }}_2^2<\mathrm{\infty }$. Choose measurable representatives and define ${\kappa }_N(x,y)={\sum }_{j=1}^N(T{c}_j)(x)\overline{c_j(y)}$. For $i\mathrm{\ne }j$, the summands are orthogonal in $L^2(M\times M)$ because their $y$-factors are orthogonal; Fubini [F3] gives $\mathrm{\parallel }{\kappa }_N-{\kappa }_{N^{\mathrm{\prime }}}{\mathrm{\parallel }}_2^2={\sum }_{j=N^{\mathrm{\prime }}+1}^N\mathrm{\parallel }T{c}_j{\mathrm{\parallel }}_2^2$ for $N>N^{\mathrm{\prime }}$. Thus $({\kappa }_N)$ is Cauchy and converges to ${\kappa }_T\in L^2(M\times M)$ with $\mathrm{\parallel }{\kappa }_T{\mathrm{\parallel }}_2^2={\sum }_{j\ge 1}\mathrm{\parallel }T{c}_j{\mathrm{\parallel }}_2^2=\mathrm{\parallel }T{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}^2$.

For a finite linear combination $f={\sum }_{j=0}^m{a}_j{c}_j$, orthonormality and the definition of ${\kappa }_N$ give ${\int }_M{\kappa }_N(x,y)f(y)d{S}_y={\sum }_{j=1}^{\min (N,m)}{a}_{j}T{c}_j(x)$. Passing to the limit in $N$ gives $\tilde{T}f$ in $L^2(M)$. By [F2], the integral operator $K_{{\kappa }_T}$ is bounded on $L^2(M)$ with norm at most $\mathrm{\parallel }{\kappa }_T{\mathrm{\parallel }}_2$; finite linear combinations of the basis are dense, so continuity of $K_{{\kappa }_T}$ and $\tilde{T}$ extends the equality to every $f\in L^2(M)$.

Since $\tilde{T}{c}_0=0$, step 2.1 with $f=c_0$ yields ${\int }_M{\kappa }_T(x,y)d{S}_y=0$ for almost every $x$. Since the range of $\tilde{T}$ lies in $L_0^2(M)$, for every $f\in L^2(M)$ Fubini gives $0={\int }_M\tilde{T}f(x)d{S}_x={\int }_{M}f(y)({\int }_M{\kappa }_T(x,y)d{S}_x)d{S}_y$. The inner section is in $L^2(M)$ by Cauchy–Schwarz and [F3]; its pairing with every $f$ vanishes, so it is zero almost everywhere. This proves the two zero-mean properties.

If ${\kappa }^{\mathrm{\prime }}$ is another $L^2$ kernel for $\tilde{T}$, let $q={\kappa }_T-{\kappa }^{\mathrm{\prime }}$. For every pair of basis vectors $c_i,c_j$, the coefficient ${\int }_{M\times M}q(x,y)c_j(y)\overline{c_i(x)}d{S}_{x}d{S}_y$ is zero because both kernels represent the same operator. For each fixed $j$, the section coefficient $g_j(x):={\int }_{M}q(x,y)c_j(y)d{S}_y$ belongs to $L^2(M)$ by Cauchy–Schwarz and Fubini, and its pairing with every $c_i$ is zero by the preceding product coefficient identity. Completeness of $(c_i)$ gives $g_j=0$ almost everywhere. Countable Choice supplies one conull set on which all these section coefficients vanish. For each $x$ in that set, $q(x,\cdot )$ is orthogonal to every member of the conjugate basis $(\overline{c_j}{)}_{j\ge 0}$, since $⟨q(x,\cdot ),\overline{c_j}⟩={\int }_{M}q(x,y)c_j(y)d{S}_y=0$. The conjugates of an orthonormal basis form an orthonormal basis, so $q(x,\cdot )=0$ there. Hence $q=0$ in $L^2(M\times M)$, proving uniqueness.

The kernel is unique, has the prescribed zero-mean convention, and has the exact norm in step 1.1. This establishes the claimed kernel representation without relying on the reverse implication of [F2].