A Banach space $X$ is called a Grothendieck space if $\text{weak}^{*}$-null sequences in $X^{*}$ are weakly null. Some of the classical Grothendieck spaces are the $C(\Omega)$ spaces if $\Omega$ is extremally disconnected. However, $C[0,1]$ is not a Grothendieck space. This fact can easily be deduced from some well-known characterizations of Grothendieck spaces, such as, $C(\Omega)$ is a Grothendieck space if and only if it does not contain any complemented copy of $c_{0}$. But my concern is whether there is a direct proof of this fact, that is, a construction of a $\text{weak}^{*}$-null sequence $(\mu_{n})_{n}$ in $C[0,1]^{*}$ which is not weakly null.
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asked Dec 13, 2020 at 3:33
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1$\begingroup$ Replacing $C[0,1]$ with $C[-1,1]$ for convenience, let $h_n$ be the "obvious" function in $L^1[-1,1]$ whichh as norm $1$ and mean zero and is supported on the interval $[-1/n, 1/n]$. If we view each $h_n$ as a measure on $[-1,1]$ then I think this should have the required properties - let me know if this doesn't work dor some reason $\endgroup$– Yemon ChoiCommented Dec 13, 2020 at 3:59
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$\begingroup$ @YemonChoi Thanks, Yemon. Could you describe the function $h_{n}$ clearer? I do not know what mean zero is. $\endgroup$– Dongyang ChenCommented Dec 13, 2020 at 7:44
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$\begingroup$ @YemonChoi I guess that mean zero means that $\int h_{n}=0$. I can prove that $\int f h_{n}\rightarrow 0$ for each $f\in C[-1.1]$. But I can not prove that the sequence $(h_{n})_{n}$ is weakly null as a sequence of measures on $[-1,1]$. $\endgroup$– Dongyang ChenCommented Dec 13, 2020 at 8:39
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2$\begingroup$ @DieterKadelka: I'm not sure I follow. The sequence $(h_n)$ in $C([-1,1])^*$ does indeed weak${}^*$-converge to $0$, but it is not weakly convergent to any point. $\endgroup$– Jochen GlueckCommented Dec 13, 2020 at 18:22
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4$\begingroup$ @Dieter: If the sequence is weak$^*$ convergent to $0$ and weakly convergent to something, then something $=0$; so one just has to check weak convergence to $0$. $\endgroup$– Dirk WernerCommented Dec 13, 2020 at 19:34
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2 Answers
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Consider $\delta_{1/n}-\delta_0$; this defines a weak$^*$ null sequences which is not weakly null; e.g., $\langle \delta_{1/n}-\delta_0, \chi_{\{0\}} \rangle \not\to 0$. So if $\Omega$ contains a nontrivial convergent sequence, $C(\Omega)$ cannot be a Grothendieck space.
answered Dec 13, 2020 at 19:31
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3$\begingroup$ Nice - I like this argument better than my original suggestion, since it seems to pinpoint the key "largeness" property required for a given $\Omega$ to yield the Grothendieck space property for $C(\Omega)$ $\endgroup$ Commented Dec 13, 2020 at 19:51
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1$\begingroup$ More generally, if $K$ is a compact Hausdorff space for which $C(K)$ is a Grothendieck space, then every convergent sequence in $K$ is eventually constant. $\endgroup$ Commented Dec 14, 2020 at 14:38
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$\begingroup$ @Bill: I think I said so in my answer... $\endgroup$ Commented Dec 14, 2020 at 15:11
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Let us define $h_{n}(t)=\frac{n}{2}$, $t\in [0,\frac{1}{n}]$ and $h_{n}(t)=-\frac{n}{2}$, $t\in [-\frac{1}{n},0]$ and $h_{n}(t)=0$ otherwise. Then $\int f\cdot h_{n}\rightarrow 0$ for each $f\in C[-1,1]$. This means that $(h_{n})_{n}$ is $\text{weak}^{*}$-null.
Define $g(t)=-1$, $t\in [-1,0]$ and $g(t)=1$, $t\in [0,1]$. Then $\int g\cdot h_{n}=1$ for all $n$. This implies that $(h_{n})_{n}$ is not weakly null.
answered Dec 13, 2020 at 14:45