Convergence of eigenvalues and spaces of sequence of compact, szmmetric and positive-semi-definite operators in Hilbert spaces.

Question

I have a compact operator, which is uniformly approximated by (finite-dimensional) compact operators and I am concerned with the question, how the eigenvalues and eigenspaces of the former are approximated by the ones of the latter. Specifically, let $K_m,K:X\to X$ be compact symmetric and positive semi-definite with $X$ a separable Hilbert-space and $||K_m-K||\to 0$ as $m\to\infty$. Let $\lambda_1\ge\lambda_2\ge...$ and $\lambda_1^m\ge \lambda_2^m\ge ...\ge \lambda_m^m$ be the eigenvalues of $K$ and $K_m$ and $v_1,v_2,...$ and $v_1^m,...,v_m^m$ corresponding Eigenvectors, such that $(v_j)$ form a basis of $\mathcal{N}(K)^\perp$ (the orthogonal komplement of the kernel of $K$) and $(v_j^m)$ forms a (finite) basis of $\mathcal{N}(K_m)^\perp$. Define iteratively $I_1:=[k~:\lambda_k=\lambda_1]$ and $I_j:= [ k~:~\lambda_k=\lambda_{\max(I_{j-1})+1}]$ for $j\ge 2$. Define the subspaces $E_i:=span(v_k~:~k\in I_i)$ and $E_i^m:=span(v_k^m~:~k\in I_i)$ (thus $E_i$ is the space spanned by the eigenvectors of the i-th largest eigenvector of K) and $E_i^m$ is only defined for $m$ sufficiently large. Denote by $P_{E_i}$ and $P_{E_i^m}$ the orthogonal projections onto the subspaces $E_i$ and $E_i^m$. I cannot find reference for the following statements, which I belive should be true:

1: For all $i\in\mathbb{N}$ there holds $\lim_{m\to\infty}\lambda_i^m = \lambda_i$.

2: For all $i\in\mathbb{N}$ there holds $\lim_{m\to\infty}||P_{E_i}-P_{E_i}^m||=0$.

May someone provide a proof or a counter example for the statements?

Does someone have a proof at least for the finite-dimensional case?

Answer 1

Your intuition is correct and the claim follows from the following statement: Let $A\in L(X)$ be a bounded linear operator on a Banach space $X$ and $\lambda\in {\Bbb C}$ an isolated eigenvalue of finite algebraic multiplicity $d$. Let $P_0$ be the spectral projection associated.

Then there is $r>0$ and $\delta>0$ so that for any $\Delta\in L(X)$ with $\|\Delta\|<\delta$ the operator $A+\Delta$ has precisely $d$ eigenvalues (repeated according to multiplicity) $\lambda_1(\Delta),...,\lambda_d(\Delta)$ in $B=B(\lambda,r)$ which depends continuously upon $\Delta$. The associated projection operator $P_\Delta$ has rank $d$ and depends continuously (in fact, analytically) upon $\Delta$. In finite dimensions this is shown e.g. in II.1.4, T. Kato, Perturbation theory of linear operators.

The proofs rely on the study of the resolvent $R(z,A)=(z-A)^{-1}$. Under the hypothesis there is $r>0$ such that $R_z=R(z,A)$ is analytic in ${B}^*=B(\lambda,r)^*=B(\lambda,r)\setminus\{\lambda\}$ and with $M=\sup_{z\in \partial B(\lambda,r)} \|R(z,A)\|<+\infty$. Then a Neumann series expansion $$ R(z,A+\Delta) = R_z + R_z\Delta R_z + ...$$ converges uniformly for $\|\Delta\|< \delta=1/M$ and $$P_\Delta = \oint_{\partial B} R(z,A+\Delta) \frac{dz}{2\pi i}$$ gives the desired projection onto the eigenvalues in $B$. This operator depends smoothly upon $\Delta$ (when $\|\Delta\|< \delta$). Much more can be said, but useful only if you are acquainted with the above type of integral. The book of T. Kato gives you one source of information, Dunford and Schwartz (Linear operators) another.

Answer 2

Thank you for the answer! I am still not sure, if I am able to use the statement to show the claim for compact operators. I'll try first the finite-dimensional case and assume for simplicity that the multiplicity of each eigenvalue is one ($\lambda_1>\lambda_2>\lambda_3>...$). I prove $\lambda_i^m\to \lambda$ as $m\to\infty$ via induction over $i\in\mathbb{N}$. For $i=1$ this holds, since $\lambda_1^m=||K_m||\to ||K|| = \lambda_1$ as $m\to\infty$. Now assume that $\lambda_j^m\to\lambda_j$ as $m\to\infty$ for all $j\le i-1$. Let $\varepsilon<\lambda_i/2$ be arbitrary. For $m$ large enough, there holds $|\lambda_j - \lambda_j^m|<\varepsilon$ for all $j\le i-1$. Moreover, by your statement, for $m$ sufficiently large there exists an eigenvalue $\mu^m$ of $K_m$ such that $|\mu^m-\lambda_i^m|<\varepsilon$ and $\mu^m$ is the only eigenvalue of $K_m$ which has this property. It remains to show, that for $m$ large enough there holds $\mu^m=\lambda_i^m$. We argue by contradiction and assume, that for infinitely many $m\in\mathbb{N}$ there are $j_m\neq i$ with $\mu^m=\lambda_{j_m}^m$. Clearly, we then must have $j_m<i$ because of the induction hypothesis and the choice of $\varepsilon$. Then it follows, that for such $m$ we have $\lambda_i^m<\lambda_{i-1}-\varepsilon$ and $\lambda_i^m>\lambda_i+\varepsilon$. W.l.o.g. (restrict to a subsequence of the $\lambda_i^m$ if necessary) there is a $\lambda$ with $\lambda_i+\varepsilon <\lambda<\lambda_{i-1}-\varepsilon$ and $\lambda_i^m\to\lambda$. Let $v_i^m$ be such that $K_m v_i^m = \lambda_i^m v_i^m$. W.l.o.g., there is a $v$ with $v_i^m\to v$ as $m\to\infty$ (because of finite dimension). Finally,

$$||Kv - K_m v_i^m|| \le || (K-K_m)v|| + ||K_m(v-v_i^m)|| \to 0$$

as $m\to\infty$, thus $Kv=\lambda v$. This is a contradiction, since $\lambda_i+\varepsilon<\lambda<\lambda_{i-1}+\varepsilon$ implies that $\lambda$ is not an eigenvector of $K$. Thus it holds that $\mu^m=\lambda_i^m$ for $m$ sufficiently large.

Is the above reasoning correct? If yes, how to handle the infinite-dimensional case? The proof feels awkward to me...