3.2 The Gram operator
For a square-integrable map \(b : \Omega \to H\) into a Hilbert space the Gram matrix becomes the Gram operator \(J = \int _\Omega b(x) b(x)^* \, d\mu (x)\). It is positive, since its quadratic form is nonnegative, and on a separable space its trace is \(\int _\Omega \| b\| ^2 \, d\mu \), by Parseval and monotone convergence (Lemma 3.3(a)). The only genuine assumption on \(J\) is therefore that it is injective, which is what positive definiteness of the Gram matrix says for a finite family, and Theorem 1.1 assumes nothing else.
Let \((\Omega , \mu )\) be a measure space, let \(H\) be a complex Hilbert space, let \(b : \Omega \to H\) be square-integrable, and let \(J = \int _\Omega b(x) b(x)^* \, d\mu (x)\).
(a) The operators \(b(x) b(x)^*\) are Bochner integrable, and \(J\) is positive with \(\langle u, J u \rangle = \int _\Omega |\langle u, b(x) \rangle |^2 \, d\mu (x)\) for every \(u \in H\).
(b) Let \(\Lambda \in \mathbb {R}\). Then \(J \preceq \Lambda \cdot 1\) if and only if \(\int _\Omega |\langle u, b(x) \rangle |^2 \, d\mu (x) \leq \Lambda \| u\| ^2\) for every \(u \in H\).
(c) \(J\) is injective if and only if every \(u \in H\) with \(\langle u, b(x) \rangle = 0\) for almost every \(x\) is zero.
(a) The operator \(b(x) b(x)^*\) has norm \(\| b(x)\| ^2\), which is integrable. Evaluation at \(u\) and the inner product with \(u\) are continuous linear maps, so both pass under the integral, and \(\langle u, b(x) b(x)^* u \rangle = |\langle u, b(x) \rangle |^2\). This quadratic form is real and nonnegative, so \(J\) is positive.
(b) By (a) both sides say that the quadratic form of the self-adjoint operator \(\Lambda \cdot 1 - J\) is nonnegative.
(c) The quadratic form vanishes at \(u\) exactly when \(\langle u, b(\cdot ) \rangle \) vanishes almost everywhere. If \(J u = 0\) it vanishes. Conversely, \(J\) is positive, and \(\langle u, J u \rangle = \| J^{1/2} u\| ^2 = 0\) gives \(J^{1/2} u = 0\) and hence \(J u = 0\).
Let \((\Omega , \mu )\) be a measure space, let \(H\) be a separable complex Hilbert space, let \(b : \Omega \to H\) be square-integrable, and let \(J = \int _\Omega b(x) b(x)^* \, d\mu (x)\).
(a) \(J\) is of finite trace, and \(\operatorname {Tr}J = \int _\Omega \| b(x)\| ^2 \, d\mu (x)\).
(b) For every positive bounded operator \(Q\) on \(H\),
(a) Choose a Hilbert basis \((e_k)\) of \(H\). Its index set is countable, because the open balls of radius \(1/2\) around the basis vectors are pairwise disjoint and a separable space holds only countably many pairwise disjoint nonempty open sets. By Tonelli and Lemma 2.4(a), \(\sum _k \langle e_k, J e_k \rangle = \sum _k \int _\Omega |\langle e_k, b \rangle |^2 \, d\mu = \int _\Omega \| b\| ^2 \, d\mu {\lt} \infty \), which is the finite trace and its value.
(b) Choose a countable Hilbert basis \((e_k)\) as in (a). Pointwise \(\operatorname {Re}\langle b, Q b \rangle = \| Q^{1/2} b\| ^2\), which Lemma 2.4(a) turns into \(\sum _k |\langle Q^{1/2} e_k, b \rangle |^2\). Sum and integral may be exchanged because all terms are nonnegative and their integrals sum to a finite value, and the quadratic form of \(J\) turns the result into \(\sum _k \operatorname {Re}\langle Q^{1/2} e_k, J \, Q^{1/2} e_k \rangle \). That equals \(\operatorname {Tr}(J Q)\) because both are the squared Hilbert–Schmidt norm of \(Q^{1/2} J^{1/2}\), once read through the operator and once through its adjoint, by Lemma 2.4(b) and Lemma 2.6(b). Countability of the index set of the basis enters here, in the interchange.