Theorem 13.7.4 (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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