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I need to show that the space $C[a,b]$ is NOT reflexive. This is a question in which I have already showed a few things.

We have $S$ an infinite set and $\{s_n\}_{n=1}^{\infty}$ a sequence of distinct points in $S$. $X$ is a Banach space of continuous functions on $S$, supplied with $\|\cdot\|_{\infty}$. For all choices $s_1',\dots,s_k'$ of finitely many elements of $S$, and for all choices of scalars $\alpha_1,\dots,\alpha_k$ with $|\alpha_i|=1$ $\forall i$, there exists an element $f\in X$ such that $\|f\|_{\infty}\leq 1$ and $f(s_i')=\alpha_i$.

Now I have already shown that, for $x\in\ell^1$, $\phi_x(f)=\sum_{n=1}^{\infty}x_n f(s_n)$ is a bounded linear functional, and the map $x\mapsto \phi_x$ is an isometric embedding from $\ell^1$ into $X$. Also, $X$ is not reflexive.

Now I have to show that $C[a,b]$ is not reflexive if $a<b$. I thought because $X$ is not reflexive, $X'$ is not reflexive, also $\ell^1$ is not reflexive. We have an isometric embedding from $\ell^1\to X$. Maybe the space $C[a,b]$ is isomorphic to one of those spaces?

Or should I answer this question apart from the things that I have already shown?

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    $C[a,b]$ is separable, but its dual is not, and I leave you to see why. Therefore, its double dual is also not separable, by Hahn-Banach. That's all : there cannot be an isomorphism between a non-separable and a separable space. – Sarvesh Ravichandran Iyer Dec 12 '17 at 08:43
  • @астонвіллаолофмэллбэрг I can prove that the dual and double dual of $C[a,b]$ are not separable. So because there cannot be an isomorphism between $C[a,b]$ and its double dual, I can conclude that $C[a,b]$ is not reflexive? –  Dec 13 '17 at 09:18
  • @астонвіллаолофмэллбэрг Oh wait I already get it! Thank you for your help! –  Dec 13 '17 at 09:48
  • Good. It was a nice question, though. – Sarvesh Ravichandran Iyer Dec 14 '17 at 04:41

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