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Can one view $\ell_{1}^{4}$, $\mathbb{R}^{4}$ equipped with the $\ell_{1}$-norm, as a space of continuous functions on any extremally disconnected space?

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No. The only possibility is a four point space. However, by looking at the combinatorial structure of the unit ball you'll see that $\ell^{1}(4)$ is not isometrically isomorphic to $\ell^{\infty}(4)$. Only $\ell^{1}(1)$ and $\ell^{1}(2)$ are isometrically isomorphic to the corresponding $\ell^{\infty}$-spaces, so only in dimensions $1$ and $2$ such a thing is possible. –  t.b. Oct 1 '11 at 14:45
Are this and your other question by any chance motivated by this question on MO? Then Bill Johnson already answered it in full... –  t.b. Oct 1 '11 at 14:53

1 Answer 1

  1. Since the function space is 4-dimensional, the underlying space $X$ would have to consist of 4 points.
  2. The norm of $C(X)$ is the $\ell_\infty$ norm on $\mathbb R^4$.
  3. For $d>2$, endowing $\mathbb R^d$ with $\ell_1$ and $\ell_\infty$ norms results in non-isometric normed spaces. See How to show that $\mathbb R^n$ with the $1$-norm is not isometric to $\mathbb R^n$ with the infinity norm for $n>2$?

(Based on t.b.'s comment)

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