$C[0,1]$ fails the property (K)

Question

Recall that a Banach space $X$ has the property (K) if every $w^{*}$-convergent sequence in $X^{*}$ admits a convex block subsequence which converges with respect to the Mackey topology. The property (K) was invented by S. Kwapien to provide an alternative approach to some results of N. Kalton and A. Pelczynski on subspaces of $L_{1}[0,1]$. A. Pelczynski noted that $L_{1}(\mu)$ ($\mu$ is a finite measure) has the property (K). Further, Schur spaces, Grothendieck spaces and strongly weakly compactly generated spaces enjoy the property (K). It is known that $c_{0}$ and $C[0,1]$ fail the property (K). But it seems not easy to check that $C[0,1]$ fails the property (K).

Question. How to check that $C[0,1]$ fails the property (K)?

Answer 1

Let $(x_{n}^{*})_{n}$ be a weak*-null sequence in $X^{*}$. The following are equivalent:

(1)$\sup\limits_{x\in K}|\langle x^{*}_{n},x\rangle|\rightarrow 0$ for each weakly compact subset $K$ in $X$;

(2)$|\langle x^{*}_{n},x_{n}\rangle|\rightarrow 0$ for each weakly null sequence $(x_{n})_{n}$ in $X$.

Indeed, if (1) is false, there exist a subsequence $(x^{*}_{k_{n}})_{n}$ of $(x^{*}_{n})_{n}$, a sequence $(x_{n})_{n}$ in $K$ and $\epsilon_{0}>0$ so that $|\langle x^{*}_{k_{n}},x_{n}\rangle|>\epsilon_{0}$ for all $n$. Since $K$ is weakly compact, there is a subsequence $(x_{n_{m}})_{m}$ of $(x_{n})_{n}$ that converges weakly to $x\in K$. Let us define a weakly null sequence $(z_{n})_{n}$ in $X$ by $z_{k_{n_{m}}}=x_{n_{m}}-x$ and $z_{n}=0$ otherwise. By (2), $\langle x^{*}_{n},z_{n}\rangle\rightarrow 0$. Note that $\langle x^{*}_{n},x\rangle\rightarrow 0$. This implies that $\langle x^{*}_{k_{n_{m}}},x_{n_{m}}\rangle\rightarrow 0$, a contradiction.

It follows from the above fact that my property (K) is equivalent to the Kalton-Pelczynski version of property (K).