Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 22F (Linear Analysis)

A bounded linear operator $T:X\to Y$ is a compact operator when the image of the closed unit ball has compact closure, equivalently when every bounded sequence $(x_n)$ has a subsequence for which $(T{x}_n)$ converges.

Each $T_i$ is compact because its image lies in a finite-dimensional subspace. Let $B_X$ be the unit ball and fix $\epsilon >0$. Choose $i$ with $\Vert T-T_i\Vert <\epsilon /3$. Since $T_i(B_X)$ is relatively compact, it has a finite $\epsilon /3$-net $y_1,\dots ,y_N$. For each $x\in B_X$, choose $y_j$ with $\Vert T_{i}x-y_j\Vert <\epsilon /3$; thenThus $T(B_X)$ is totally bounded. Its closure is complete because $Y$ is a Banach space, so it is compact. This proves the norm-closedness of compact operators and hence

Let $K=\overline{T(B_X)}$, which is compact. Every $y^{\ast }$ in the unit ball of the continuous dual space $Y^{\ast }$ restricts to a function on $K$. These functions are uniformly bounded and equicontinuous, sinceBy the Arzela-Ascoli theorem, every sequence $(y_n^{\ast })$ in the unit ball has a subsequence that converges uniformly on $K$. Along this subsequence,The continuous dual space $X^{\ast }$ is complete, so $(T^{\ast }{y}_n^{\ast })$ converges in norm. Every sequence in the image of the unit ball therefore has a convergent subsequence, proving the forward direction of the Schauder theorem for compact operators:

Normalize the mutually orthogonal eigenvectors so that $(x_j)$ is an orthonormal sequence. If ${\lambda }_j$ did not tend to zero, some subsequence would satisfy $|{\lambda }_j|\ge \epsilon >0$. For distinct indices in this subsequence, orthogonality givesThus $(T{x}_j)$ would have no Cauchy subsequence, contradicting compactness of $T$. Therefore the orthogonal eigenvectors of a compact operator satisfy