Schauder compact adjoint theorem

Every bounded sequence in $\mathbb{K}$ has a convergent subsequence.

If $S$ is compact then $\mathrm{\Theta }$ is compact by [A1], so for every real $\delta >0$ there are finitely many points $x_0,\dots ,x_m\in {\bar{B}}_X$ with $\mathrm{\Theta }\subseteq {\bigcup }_{i\le m}B(S{x}_i,\delta )$: the balls $\mathrm{\Theta }\cap B(Sx,\delta )$, $x\in {\bar{B}}_X$, cover $\mathrm{\Theta }$ by [A5], compactness gives a finite subcover, and one index per member of that finite subcover may be chosen by [A4].

The dual $X^{\ast }$ is Banach by [A2], so the set $C:=\overline{S^{\ast }({\stackrel{‾}{B}}_{Y^{\ast }})}\subseteq X^{\ast }$ is a complete metric space by [A2].

If $K$ is a compact operator and $M$ is a closed subspace of its target containing $K(X)$, then the corestriction $K_0:X\to M$ is compact: for bounded $E\subseteq X$ the closure of $K_0(E)$ in $M$ equals $\overline{K(E)}\cap M$, a closed subset of the compact set $\overline{K(E)}$.

The space $J_Y(Y)$ is a closed subspace of $Y^{\ast \ast }$ and the inverse of $J_Y:Y\to J_Y(Y)$ is a bounded isometry, by [A6].

Under the hypothesis of [step 1.2], the union $D$ of the finite $(1\mathrm{/}(k+1))$-nets of $\mathrm{\Theta }$ obtained from [step 1.2] for $k\in \mathbb{N}$ is at most countable and dense in $\mathrm{\Theta }$, the nets being chosen together by ${\mathrm{A}\mathrm{C}}_{\omega }$ and their union counted by [A4]; hence there is a surjection $\mathbb{N}\to D$ listing $D$ as $(d_j)$.

If $(g_n)$ is a sequence in ${\bar{B}}_{Y^{\ast }}$ and $D=(d_j)$ is a countable subset of $Y$, then there are a strictly increasing index map $n:\mathbb{N}\to \mathbb{N}$ and scalars to which $g_{n_j}(d_k)$ converges for all $j,k$: for each fixed $d_k$ the scalar sequence $g_n(d_k)$ is bounded by $\mathrm{\parallel }{d}_k\mathrm{\parallel }$ and has a convergent subsequence by [step 1.1], and the standard diagonal selection of nested subsequences is licensed by $\mathrm{D}\mathrm{C}$.

Assume $S$ compact and let $(g_n)$ be a sequence in ${\bar{B}}_{Y^{\ast }}$. With $D$ as in [step 2.1], [step 2.2] gives a subsequence $(g_{n_j})$ with $g_{n_j}(d)$ convergent for every $d\in D$. Given a real $\epsilon >0$, choose $k$ with $1\mathrm{/}(k+1)<\epsilon \mathrm{/}4$ and let $F_k\subseteq {\bar{B}}_X$ be the finite net of [step 1.2] for $\delta =1\mathrm{/}(k+1)$; convergence on the finite set $F_k$ gives $J$ with $\mathrm{\mid }(g_{n_j}-g_{n_l})(S{x}_i)\mathrm{\mid }<\epsilon \mathrm{/}2$ for all $j,l\ge J$ and all $i\le m$, and for $x\in {\bar{B}}_X$ one has $\mathrm{\parallel }Sx-S{x}_i\mathrm{\parallel }<\epsilon \mathrm{/}4$ for some $i$, so $\mathrm{\mid }(S^{\ast }{g}_{n_j}-S^{\ast }{g}_{n_l})(x)\mathrm{\mid }\le \mathrm{\mid }(g_{n_j}-g_{n_l})(S{x}_i)\mathrm{\mid }+\mathrm{\parallel }{g}_{n_j}-g_{n_l}\mathrm{\parallel }\mathrm{\parallel }Sx-S{x}_i\mathrm{\parallel }<\epsilon \mathrm{/}2+2\epsilon \mathrm{/}4=\epsilon$; hence $(S^{\ast }{g}_{n_j})$ is Cauchy in $X^{\ast }$.

Under the hypothesis of [step 3.1] the Cauchy sequence $(S^{\ast }{g}_{n_j})$ converges in the complete space $X^{\ast }$ by [step 1.3], and its limit lies in $C$ because every $S^{\ast }{g}_{n_j}\in S^{\ast }({\bar{B}}_{Y^{\ast }})\subseteq C$ and $C$ is closed; so every sequence in $S^{\ast }({\bar{B}}_{Y^{\ast }})$ has a subsequence converging in $C$.

Under the hypothesis of [step 3.1], every sequence $(y_n)$ in $C$ has a subsequence converging in $C$: choosing $g_n\in {\bar{B}}_{Y^{\ast }}$ with $\mathrm{\parallel }{y}_n-S^{\ast }{g}_n\mathrm{\parallel }<1\mathrm{/}(n+1)$ for every $n$ is a countable selection licensed by [A3], and applying [step 4.1] to $(g_n)$ yields a subsequence with $S^{\ast }{g}_{n_j}\to y\in C$, whence $y_{n_j}\to y$.

Under the hypothesis of [step 3.1] the space $C$ is sequentially compact by [step 5.1], hence compact by [A3], and then $S^{\ast }$ is compact by [A1].

Suppose now that $T^{\ast }:Y^{\ast }\to X^{\ast }$ is compact. Both $X^{\ast }$ and $Y^{\ast }$ are Banach by [A2], so [step 6.1] applied to the bounded linear operator $T^{\ast }$ between Banach spaces gives that $(T^{\ast }{)}^{\ast }=T^{\ast \ast }$ is compact; with [step 1.5] and [A6], $T^{\ast \ast }{J}_X=J_{Y}T$, so $T^{\ast \ast }{J}_X$ is compact by [A7], and $T=J_Y^{-1}\circ (T^{\ast \ast }{J}_X)$ is the composite of the corestriction of $T^{\ast \ast }{J}_X$ to the closed subspace $J_Y(Y)$ — compact by [step 1.4] — with the bounded operator $J_Y^{-1}$, hence compact by [A7].

Conversely, if $T$ is compact then $T^{\ast }$ is compact by [step 6.1]; and if $T^{\ast }$ is compact then $T$ is compact by [step 7.1]; this is the asserted equivalence.