Consider a symmetric function $$ f(x_1,x_2):R^n \times R^n \to R $$ satisying $f(x_1,x_2)=f(x_2,x_1)$. Are there functions $f_k:R^n \to R$ such that $$ \int_{x\in R^n}f_k(x)f_l(x)dm=\delta_{kl}, $$ and $$ f(x_1,x_2)=\sum_{k=1}^{\infty} \lambda_k f_k(x_1)f_k(x_2). $$ Where $m$ is a probability measure.
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1$\begingroup$ Ummm. Are there any hypotheses you’d like us to know about? $\endgroup$– Anthony QuasCommented Dec 19, 2019 at 9:34
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$\begingroup$ Do you want these $f_k$ to be square integrable functions? $\endgroup$– Ben McKayCommented Dec 19, 2019 at 9:37
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$\begingroup$ @BenMcKay: The conditions given above show that it's square integrable $\endgroup$– mathmetricgeometryCommented Dec 19, 2019 at 10:02
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$\begingroup$ If $f$ is the kernel of, say, a Hilbert–Schmidt operator on $L^2(dm)$, then of course yes. In general, the answer depends on your notion of convergence of the series, but most likely it is "not necessarily". If I am not mistaken, $m(dx) = e^{-x^2} dx$ and $f(x,y) = e^{x^2+y^2}$ is a simple counter-example under reasonable notions of convergence. $\endgroup$– Mateusz KwaśnickiCommented Dec 19, 2019 at 11:38
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1$\begingroup$ In what sense is the series’ convergence supposed to be? $\endgroup$– Francois ZieglerCommented Dec 19, 2019 at 12:20
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1 Answer
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If the function $f$ is Borel-measurable and $m$ is a probability measure on $R^n$ such that $$\iint_{R^n\times R^n}f(x,y)^2 m(dx)m(dy)<\infty, \tag{1} $$ then your desired (spectral) sum-of-products decomposition of $f$ exists, where the convergence of the series is in $L^2(R^n\times R^n,m\otimes m)$.
It is easy to see that condition (1) is, not only sufficient, but also necessary for the existence of such a decomposition.
answered Dec 19, 2019 at 12:03
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$\begingroup$ Thank you. Assume (1), then $Tg(x)=\int f(x,y)g(y) dy$ is a bounded, self-adjoint operator from $L^2(R^n,m)$ to $L^2(R^n,m)$. Then there exist orthonormal basis $\phi_k$, s.t. $Tg(x)=\sum_k \lambda_k \phi_k$. But I don't know why $f(x,y)=\sum \lambda_k \phi_k(x)\phi_k(y)$? Another question is if we assume $f(x,y)$ is bounded, can we get that $\phi_k$ are bounded or $\phi_k \in L^p(R^n,m)$ for some $p>2$? (I think it's wrong, but I can't get a counter example). $\endgroup$ Commented Dec 20, 2019 at 7:56
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$\begingroup$ @mathmetricgeometry : Concerning your question about the spectral decomposition of $f$: Have you looked at the article linked in my answer and at further references there (in particular, [4])? Briefly, here is is a possible way to reason: Let $K_1(x,y):=\sum_k\lambda_k\phi_k(x)\phi_k(y)$ and let $T_1$ be the integral operator with kernel $K_1$. Then $T\phi_j=\lambda_j\phi_j=T_1\phi_j$ for all $j$, whence $Tg=T_1g$ for all $g\in L^2(R)$, whence $K=K_1$ $m\otimes m$-almost everywhere. As for additional questions you may have, it is better to ask them in separate posts. $\endgroup$ Commented Dec 20, 2019 at 14:10