I am trying to solve the following problem:
Let $H$ be an infinite-dimensional Hilbert space. $A_\varepsilon, A$ are compact operators in $L(H,H)$ and $A_\varepsilon \to A$ in $L(H,H)$ as $\varepsilon\to 0^+$. We know that $\dim N(I-A_\varepsilon)$ and $\dim N(I-A)$ are finite, by Fredholm - Riesz theory. Prove that there exists a positive integer $n$ and a positive real number $\delta$ such that $$\dim N(I-A_\varepsilon)\le n\quad \forall\ 0<\varepsilon<\delta.$$
I try to prove by a proof by contradiction, which leads to the existence of a sequence of compact operators $A_n$ such that $A_n\to A$, $\dim N(I-A_n)\to\infty$, but I can not get a contradiction, since I have no idea about the relation between $\dim N(I-A_n)$ and $\dim N(I-A)$.
Could anyone help me, please. Thanks in advance.
Since $A$ is compact, there exist orthonormal systems $\{\varphi_i\}, \{\psi_i\}$ in $H$ and a sequence $\lambda_i\to 0$ such that: $$Ax=\sum_{i=1}^\infty \lambda_i\langle x,\varphi_i\rangle \psi_i. $$ Let $N$ be a positive integer such that $|\lambda_i|<1/4$ for every $i> N$. Set $$Bx=\sum_{i=1}^N \lambda_i\langle x,\varphi_i\rangle \psi_i.$$ Then we have: $$\|Ax-Bx\|^2=\sum_{i=N}^\infty \lambda_i^2\langle x,\varphi_i\rangle^2\le 1/16\|x\|^2$$ Set $Rx=Ax-Bx$, we have $\|B\|\le 1/4$. Since $A_\varepsilon \to A$, there exists $\delta>0$ such that $\|A_\varepsilon -A\|\le 1/4$ for every $\varepsilon \in (0,\delta)$. We have \begin{align*} \|(I-A_\varepsilon)x\|&=\|(I-B)x + (A-A_\varepsilon)x-Rx\|\\ &\ge \|(I-B)x\|-\|Rx\|-\|(A-A_\varepsilon)x\|\\ &\ge\|(I-B)x\|-\frac{1}{2}\|x\|. \end{align*} Let $V={\rm span}(\varphi_1,...,\varphi_N)$. We have that $Bx=0$ for every $x\in V^\perp$, hence that $\ker(I-A_\varepsilon)\cap V^\perp=\{0\}$ and that $\dim N(I-A_\varepsilon)\le N$ (QED).