14.8 Compact Operators
Definition 14.8.1 (Compact Operator).label Let $E, F$ be locally convex spaces over $K \in \RC$ and $T \in L(E; F)$, then $T$ is compact if there exists $U \in \cn_{E}(0)$ such that $T(U)$ is relatively compact in $F$.
The set $\mathcal{K}(E; F)$ is the space of compact operators from $E$ to $F$.
Proposition 14.8.2.label Let $E$ be a normed space over $K \in \RC$ and $F$ be a complete Hausdorff topological vector space over $K$, then $\mathcal{K}(E; F)$ is a closed subspace of $L_{b}(E; F)$.
Proof. Let $T \in \ol{\mathcal{K}(E; F)}$ and $U \in \cn_{F}(0)$ be circled, then there exists $S \in \mathcal{K}(E; F)$ such that $Sx - Tx \in U$ for all $x \in B_{E}(0, 1)$. Since $S$ is compact, there exists $Y \subset F$ finite with $S(B_{E}(0, 1)) \subset Y + U$. In which case, $T(B_{E}(0, 1)) \subset Y + U + U = Y + 2U$. Therefore $T(B_{E}(0, 1))$ is totally bounded, and as $F$ is complete, relatively compact in $F$.$\square$
Theorem 14.8.3 (Schauder).label Let $E, F$ be normed vector spaces over $K \in \RC$ with $F$ being complete and $T \in L(E; F)$, then $T$ is compact if and only if $T^{*} \in L(F^{*}; E^{*})$ is compact.
Proof. ($\Rightarrow$): Let $\cf \subset F^{*}$ be bounded, then $\cf$ is equicontinuous and hence relatively compact in the weak* topology by the Banach-Alaoglu Theorem. Since $F$ is complete, $\ol{T(B_E(0, 1))}$ is compact. By the Arzelà-Ascoli Theorem, $\cf$ is relatively compact with respect to the topology of uniform convergence on $\ol{T(B_E(0, 1))}$. Thus $T^{*}(\cf) = \bracsn{\phi \circ T|\phi \in \cf}$ is relatively compact with respect to the topology of uniform convergence on $B_{E}(0, 1)$. In other words, $T^{*}(\cf)$ is relatively compact in $E^{*}$, and $T^{*}$ is a compact operator.
($\Leftarrow$): By Proposition 12.13.8, $E^{*}$ is complete. The preceding case then implies that $T^{**}\in L(E^{**}; F^{**})$ is compact. As such, its restriction to $E$, being identified with $T$, is also compact.$\square$
Theorem 14.8.4 (Gantmacher).label Let $E, F$ be Banach spaces over $K \in \RC$, and $T \in L(E; F)$, then the following are equivalent:
- (1)
$T(B_{E}(0, 1))$ is relatively $\sigma(F, F^{*})$-compact.
- (2)
$T^{**}(E^{**}) \subset F \subset F^{**}$.
Proof. Let $B_{E}$ be the closed unit ball of $E$, and $B_{E^{**}}$ be the closed unit ball of $E^{**}$. By Goldstine’s Theorem, $B_{E}$ is $\sigma(E^{**}, E^{*})$-dense in $B_{E^{**}}$. Since $T^{**}$ is a $\sigma(E^{**}, E^{*})$-$\sigma(F^{**}, F^{*})$-continuous extension of $T$,
by Proposition 5.5.3. On the other hand, $B_{E^{**}}$ is $\sigma(E^{**}, E^{*})$-compact by the Banach-Alaoglu Theorem. Hence Proposition 5.16.3 implies that $T^{**}(B_{E^{**}})$ is $\sigma(F^{**}, F^{*})$-closed, so
The above equality shows that the following five statements are equivalent:
- (i)
$T(B_{E})$ is relatively $\sigma(F, F^{*})$-compact.
- (ii)
$\ol{T(B_E)}^{\sigma(F^{**}, F^*)}= \ol{T(B_E)}^{\sigma(F, F^*)}$.
- (iii)
$\ol{T(B_E)}^{\sigma(F^{**}, F^*)}\subset F$.
- (iv)
$T^{**}(B_{E^{**}}) \subset F$.
- (v)
$T^{**}(E^{**}) \subset F$.
where (i) is equivalent to (1), and (v) is equivalent to (2).$\square$
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