# $C[0,1]$ is not a Grothendieck space

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.

• 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 Dec 13, 2020 at 3:59
• @YemonChoi Thanks, Yemon. Could you describe the function $h_{n}$ clearer? I do not know what mean zero is. Dec 13, 2020 at 7:44
• @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 $[-1,1]$. Dec 13, 2020 at 8:39
• @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. Dec 13, 2020 at 18:22
• @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 $0$. Dec 13, 2020 at 19:34

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.
• 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 $C(\Omega)$ Dec 13, 2020 at 19:51
• 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. Dec 14, 2020 at 14:38
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.