Let K be a Hilbert-Schmidt kernel which is real and symmetric. The associated operator is
T(f)(x) = \int_Rd {K(x,y) f(y) dy} , f \in L^2(R^d)
with kernel K. T is compact and symmetric.
Let {\phi_k (x)} be the eigenvectors (with eigenvalues \lambda_k) that diagonalize T.
Then, show that :
a/ \sum_k {|\lambda_k|^2} < \infty
b/ K(x,y) ~ \sum_{k\geq 1} {\lambda_k \phi_k (x) \phi_k (y)} is the expansion of K in the basis {\phi_k (x) \phi_k (y)}
c/ Suppost T is a compact and symmetric operator.
Then, T is of Hilbert-Schmidt type if and only if \sum_k {|\lambda_k|^2} < \infty where {\lambda_k (x)} are the eigenvalues of T counted according to their multiplicities.
For a/, we know |\lambda_k| tends to 0 as k goes to infinity ; is it enough to conclude that \sum_{k}{|\lambda_k|^2} < \infty ?
For b/, we know K(x,y) has an expansion in the basis {{\phi_k (x) \phi_l (y)}} of the form :
K(x,y) ~ \sum_{k,l} {a_{k,l} \phi_k (x) \phi_l (y)} where {|a_{k,l}|^2} < \infty
but I dont know how to show that a_{k,l} = \lambda_k and that the indices k and l are the same.
For c/ one direction follows directly from a/, but I can't find how to show the converse.
Any idea ?
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