Timeline for If $f(x_1,x_2)=f(x_2,x_1)$, $f(x_1,x_2)=\sum_k \lambda_k f_k(x_1)f_k(x_2)$?

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Dec 20, 2019 at 14:10	comment	added	Iosif Pinelis		@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.
Dec 20, 2019 at 7:56	comment	added	mathmetricgeometry		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).
Dec 19, 2019 at 12:34	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 135 characters in body
Dec 19, 2019 at 12:09	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 19 characters in body
Dec 19, 2019 at 12:03	history	answered	Iosif Pinelis	CC BY-SA 4.0