Confused about this proof of Hilbert Schmidt theorem
math.stackexchange.com › questions › 2746426Apr 21, 2018 · 2. I need a really good explication with details of this proof of Hilbert - Schmidt theorem : Let ( H, , ) be a complex Hilbert space and let A: H → H be a bounded, compact, self-adjoint operator and ( λ n) n a sequence of non-zero real eigenvalues where each eigenvalue of A is repeated in the sequence according to its multiplicity, then there exists an orthonormal set ( v n) n of corresponding eigenfunctions, i.e.
Hilbert-Schmidt operators - KTH
people.kth.se › ~laptev › FA08Theorem 1. Let A : L2(R) → L2(R) be an integral operator Af(x) = Z R K(x,y)f(y)dy. A is Hilbert-Schmidt iff K ∈ L2(R×R) and kAk S 2 = kKk L2( R× ). Remark 3. Hilbert-Schmidt operators are compact. (For integral operators this fact has been proved before, see Lecture 7.) Fredholm-Riesz-Schauder theory let T : H → H be a compact operator.
Hilbert–Schmidt operator - Wikipedia
https://en.wikipedia.org/wiki/Hilbert–Schmidt_operatorIn mathematics, a Hilbert–Schmidt operator, named after David Hilbert and Erhard Schmidt, is a bounded operator that acts on a Hilbert space and has finite Hilbert–Schmidt norm where is an orthonormal basis. The index set need not be countable. However, the sum on the right must contain at most countably many non-zero terms, to have meaning. This definition is independent of the choice of the orthonormal basis. In finite-dimensional Euclidean space, the Hi…