Weak Compactness

Prove Corollary 13.2: if $\mathscr {X}$ is a Banach space and $A \subseteq \mathscr {X}$, then $A$ is weakly compact if and only if $A \cap \mathscr {M}$ is weakly compact for every separable subspace $\mathscr {M}$ of $\mathscr {X}$.

If $\mathscr {X}$ is a Banach space and $K$ is a compact subset of $\mathscr {X}$, prove that $\overline{\co }(K)$ is compact. (This will be proved in Theorem VI.4.8 below.)

In the proof of (13.4) (where $K$ is a weakly compact subset of a Banach space $\mathscr {X}$, $M(K) = C(K)^*$, and $T : M(K) \to \mathscr {X}$ is defined by $T(\mu ) = x_\mu$, the element of $\mathscr {X}$ such that $\left\langle x_\mu , x^* \right\rangle = \int_K \left\langle x, x^* \right\rangle \, d\mu (x)$ for all $x^*$ in $\mathscr {X}^*$), if $\mathscr {P} =$ the probability measures on $K$, show that $T(\mathscr {P}) = \overline{\co }(K)$.

Prove the Eberlein-Smulian Theorem in the setting of Hilbert space: if $\mathscr {H}$ is a Hilbert space and $A \subseteq \mathscr {H}$, then the following statements are equivalent: (a) each sequence of elements of $A$ has a subsequence that is weakly convergent; (b) each sequence of elements of $A$ has a weak cluster point; (c) the weak closure of $A$ is weakly compact.

Refer to the notation of Exercise 11.10. Prove that there is a bounded linear functional $L : AP(G) \to \mathbb {F}$ such that $L(1) = 1$, $L(f) \geq 0$ if $f \geq 0$, and $L(f_x) = L(f)$ for all $f$ in $AP(G)$ and $x$ in $G$.