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9

Let $A$ be a self-adjoint operator with domain $D(A)\subset\mathcal H$ ($\mathcal H$ is some Hilbert space). An operator $C$ with $D(A)\subset D(C)$ is called relatively compact with respect to $C$ if $C(A-zI)^{-1}$ is compact for some (hence all) $z\notin\sigma(A)$. Paraphrasing Corollary 2, page 113 Section XIII.4, in , we have If $C$ is relatively ...

4

Your operator $K$ is a Hilbert-Schmidt operator since its kernel belongs to $L^2$. As a result this is a compact operator whose spectrum contains a sequence of eigenvalues $\\{\lambda_k\not=0\\}$ with finite multiplicities such that $\lim_k\vert \lambda_k\vert=0$. To deal with the self-adjoint case, you can find an orthonormal set $\\{\mathbf e_k\\}$ such ...

4

It seems that I have found a counter example myself. For the Hilbert matrix $$H_\lambda:= \big( \frac{1}{1-\lambda+k+n} \big)_{k,n\geq 0}, \lambda < 1$$ Rosenblum in "On the Hilbert Matrix I, Proceedings of the AMS" proves that the pointspectrum considered as an operator on $\ell^p, p>2$ contains the set $$\{ \pi \sec(\pi u ) : | \Re ( u )| <... 2 Yes, the inequality is true. Let \lambda be large enough so (L+\lambda I) is positive definite. We have$$((L+\lambda I)y,y)=((L+\lambda I)^{1/2}y,(L+\lambda I)^{1/2}y)\ge C\|y\|^2_{H^{s/2}},$$since the domain of (L+\lambda I)^{1/2} is H^{s/2} by the general theory of interpolation spaces. It follows that$$(Ly,y)\ge C\|y\|^2_{H^{s/2}}-\lambda (y,y)....

1

In "On Majorization, Factorization, and Range Inclusion of Operators on Hilbert Space (1966)", R. G. Douglas proved the following result (Theorem 1 in the paper): Theorem. Let $C$ and $D$ be bounded linear operators on a real or complex Hilbert space $\mathcal{H}$; then the following are equivalent: (i) $C\mathcal{H} \subseteq D \mathcal{H}$. (ii) There ...

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