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Suppose $K:[0,1]^2 \times [0,1]^2 \to \mathbb{R}$ is continuous and positive semi-definite, and define the corresponding Hilbert-Schmidt integral operator

$$ [Cu](x) = \int_{[0,1]^2} K(x,y) u(y) dy.$$

Suppose further that $K(x,y)=k_1(x_1,y_1) k_2(x_2,y_2)$, and that one has a complete set of eigenfunctions and eigenvalues $\{ (f_n^{i},\lambda_n^{i}) \}_{n=1}^\infty$for the corresponding one-dimensional operators

$$ [C_i v](x_i) = \int_0^1 k_i(x_i,y_i)v(y_i)dy_i$$

for $i=1,2$. Clearly $u_{ij}(x)=f_i^1(x_1)f_j^2(x_2)$ is an eigenfunction of $C$ with eigenvalue $\lambda_i^1 \lambda_j^2$ for any $i,j \in \mathbb{N}$, due to the structure of the kernel and the decoupling of the integral.

My question is, does this exhaust all possibilities? Namely, can every eigenfunction of $C$ be obtained in this way? Impose any conditions that you'd like... for me, $k_1$ and $k_2$ are continuous and piecewise smooth. A reference would be greatly appreciated, as well as any responses... also I am happy to clarify if anything is unclear. Thanks in advance!

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1 Answer 1

Welcome to Math.SE! Not every eigenfunction of $C$ will be a product; you may have to allow sums of products. For example, if $\lambda_i^1\lambda_j^2 = \lambda_j^1\lambda_i^2$ then $u(x_1,x_2)=f_i^1(x_1)f_j^2(x_2)+f_j^1(x_1)f_i^2(x_2)$ is an eigenfunction for the eigenvalue $\lambda_i^1\lambda_j^2$, but it has no reason to factor into a function of $x_1$ and a function of $x_2$.

In the positive direction: since $C$ is self-adjoint, we have $K(x,y)=K(y,x)$. This implies $k_i(x_i,y_i)=k_i(y_i,x_i)$; thus the operators $$[C_iu](x) = \int_0^1 k_i(x_i,y_i) u(y) dy_1$$ are also self-adjoint. Notice that I consider $C_i$ as acting on the space $L^2([0,1]^2)$, not on $L^2([0,1])$. The operators $C_1$ and $C_2$ commute: $C_1C_2=C_2C_1=C$. Commuting self-adjoint compact operators can be simultaneously diagonalized: there is an orthonormal basis of common eigenvectors. This ought to make the picture clear. Let $V^i(\lambda)$ be the $\lambda$-eigenspace of $C_i$. Then the $\lambda$-eigenspace of $C$ is the span of the union of all intersections $V^1(\lambda^1)\cap V^2(\lambda^2)$; the union is taken over all pairs $\lambda^1,\lambda^2$ such that $\lambda^1\lambda^2=\lambda$. There are only finitely many such pairs, since eigenvalues tend to $0$.

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