Collectively compact operator approximation

Remarks

Every operator in a collectively compact set is compact, and the limit $K$ in the definition is compact as well. The main result assumes that $K_n$ is a collectively compact approximation of $K$ and that $I - K$ is injective. Then for large $n$ the operators $I - K_n$ are invertible with uniformly bounded inverses. The solutions of $\varphi_n - K_n\varphi_n = f$ then satisfy $\norm{\varphi_n - \varphi} \le C \norm{K_n\varphi - K\varphi}$. The proof uses that $\norm{(K_n - K) K} \to 0$ and $\norm{(K_n - K) K_n} \to 0$, although $\norm{K_n - K}$ need not tend to $0$.

Example: let $X = C[0, 1]$, let $k$ be continuous on $[0, 1]^2$, and let $(K_n\psi)(x) = \frac{1}{n}\sum_{j=1}^{n} k(x, \frac{j}{n}) \psi(\frac{j}{n})$. For $\psi \in B$ one has $\abs{K_n\psi(x)} \le \max \abs{k}$ and $\abs{K_n\psi(x) - K_n\psi(x')} \le \max_t \abs{k(x, t) - k(x', t)}$, so the family is uniformly bounded and equicontinuous, hence relatively compact by the Arzelà–Ascoli theorem. Riemann sums converge, so $K_n\psi \to K\psi$ uniformly with $(K\psi)(x) = \int_0^1 k(x, t) \psi(t) \, dt$. In general $\norm{K_n - K} \not\to 0$ here: for $k = 1$, functions $\psi$ in $B$ that equal $-1$ at the nodes and have integral close to $1$ show $\norm{K_n - K} = 2$ for every $n$.

Non-example: on $X = \ell^2$, let $K_n$ be the projection onto the first $n$ coordinates. Then $K_n \psi \to \psi$ for every $\psi$, but the set $\set{K_n e_n} = \set{e_n}$ has no convergent subsequence, so clause 2 fails; indeed the limit $I$ is not compact.