Setup: Let $(\mathcal{X},d_{\mathcal{X}})$ and $(\mathcal{Y},d_{\mathcal{Y}})$ be two separable metric spaces. Let $M^1(\mathcal{X})$ be the space of Borel probability measures on $\mathcal{X}$ with finite first moment, i.e. a Borel probability measure $\mu$ on $\mathcal{X}$ is in $M^1(\mathcal{X})$ if $\int d_{\mathcal{X}}(x,o) d\mu(x)<\infty$ for any $o\in\mathcal{X}$. The space $M^1(\mathcal{Y})$ is defined in similar fashion.

Fix $\mu\in M^1(\mathcal{X})$ and $\nu\in M^1(\mathcal{Y})$ and define $$ d_\mu(x_1,x_2)= d_{\mathcal{X}}(x_1,x_2) -\int d_{\mathcal{X}}(x_1,x)\, d\mu(x) - \int d_{\mathcal{X}}(x_2,x)\, d\mu(x) + \int d_{\mathcal{X}}(x,x')\, d\mu^2(x,x'), $$ and a similar definition of $d_\nu:\mathcal{Y}\times \mathcal{Y}\to\mathbb{R}$.

Now let $S:L^2(\mathcal{X}\times \mathcal{Y},\mathcal{B}(\mathcal{X})\otimes \mathcal{B}(\mathcal{Y}),\mu\times \nu) \to L^2(\mathcal{X}\times \mathcal{Y},\mathcal{B}(\mathcal{X})\otimes \mathcal{B}(\mathcal{Y}),\mu\times \nu),$ be a Hilbert-Schmidt operator given by $$ S(f)(x,y) = \int d_\mu(x,x')d_\nu(y,y') f(x',y') d\mu\times \nu(x',y'). $$ and let $\{\lambda_i\}_{i\geq 1}$ denote the non-zero eigenvalues of $S$ repeated according to multiplicity.

Question: How do I prove the following identity: $$\sum_{i=1}^\infty\lambda_i=\int d_\mu(x,x)d_\nu(y,y) \, d\mu\times \nu(x,y).$$ I tried but failed to show that $S$ is of trace class, since they under certain conditions (which I also can't verify in this setup) satisfy that $$ Trace(S)=\int d_\mu(x,x)d_\nu(y,y) \, d\mu\times \nu(x,y), $$ which yields the result since $Trace(S)=\sum_{i=1}^\infty\lambda_i$ (if it were of trace class).

The identity $\sum_{i=1}^\infty\lambda_i=\int d_\mu(x,x)d_\nu(y,y) \, d\mu\times \nu(x,y)$ is stated in Distance covariance in metric spaces by Russell Lyons (2013) Theorem 2.7, without proof, so my approach with traces is only an idea.

If another way of proving the identity appears or if there is a counterexample, that would more than satisfy my needs. In the case of a counterexample i would very much appreciate stronger initial conditions rendering the identity true.

Please bear in mind that i am a novice in the theory of operators and trace class operators (only went down this road to explain the equality above), so references would be much appriciated.

Update: A counterexample to the operator being of trace-class is presented by Russell Lyons in the errata to the mentioned paper. Furthermore a proof that the formula holds and that the operator is of traceclass, whenever the marginal spaces posses additional nice proporties (isometric embeddability into hilbert spaces), is also presented in this errata.

  • $\begingroup$ If not possible in the above setup, can the equality be shown for when $\mathcal{X}$ and $\mathcal{Y}$ are separable Hilbert spaces? $\endgroup$ – Martin Nov 29 '16 at 19:14
  • $\begingroup$ math.stackexchange.com/questions/185587/… $\endgroup$ – Renart Dec 3 '16 at 11:35
  • $\begingroup$ @Renart Yes that is indeed one of the questions that i have read, suggesting the trace approach. Though this is a integral kernel over $\mathbb{R}$, and i'm quite sure that problems arise when integrating over arbitrary metric spaces. $\endgroup$ – Martin Dec 3 '16 at 13:58
  • $\begingroup$ Well i don't have tried it myself but... From what i read the answer of jonhatan doesn't really use the fact that it's on $\mathbf R$. The only thing that need justification is the permutation of a sum and an integral at the very end. Fubini works pretty independently of the space you're concidering. $\endgroup$ – Renart Dec 3 '16 at 17:21
  • $\begingroup$ @Renart I think there are more subtle problems. For example $y\mapsto K(x,y) = \sum_{n}e_n(y) \int e_n(z) K(x,z)dz,$ only in $L^2(\mathbb{R})$ for each fixed $x$, which by definition means that for any $x$ $\|K(x,\cdot)- \sum_{n=1}^k e_n(\cdot) \int e_n(z) K(x,z)dz\|_2\to_k 0.$ But in order to say that $\int \sum_{n=1}^\infty e_n(x)\int K(x,y)e_n(y)dy dx= \int K(x,x) dx$ we need stronger convergence. Anyways, all this assumes that the operator is of trace class, which i haven't even verified. $\endgroup$ – Martin Dec 3 '16 at 21:02

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