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The Banach space $L^\infty[0,1]$ is isomorphic to all its hyperplanes (closed linear subspaces of codimension one). One way of seeing this is the following:

  1. all hyperplanes of a Banach space are mutually isomorphic
  2. $\ell_\infty$ is isomorphic to its hyperplane $\{x\in\ell_\infty\colon x(1)=0\}$.
  3. $L^\infty[0,1]$ is isomorphic to $\ell_\infty$.

In the above list the step (2) can be made explicit. On the other hand, the third item (3) cannot; see this question and its answer by Robert Israel.

Is there a direct way of proving the above statement which allows for an explicit construction?

In particular, I would be interested in an answer to the question of whether it is possible to construct an explicit isomorphism between $L^\infty[0,1]$ and its hyperplane $\{f\in L^\infty[0,1]\colon \int_{0}^{1}f(t)\,\mathrm{d}t=0\}$. This question is related to this question on mathoverlow.

Christian
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  • How do you define a hyperplane in this case? – Asaf Karagila Apr 03 '20 at 11:32
  • @AsafKaragila linear subspace of codimension one – Christian Apr 03 '20 at 11:34
  • Also, when you say "isomorphism", do you mean as vector spaces, or as Banach spaces (i.e. a linear isometry, or a linear homeomorphism, or what?) Because the only way I immediately see why (1) is true, is by appealing to Hahn–Banach, which also uses choice. – Asaf Karagila Apr 03 '20 at 11:45
  • @AsafKaragila I mean linear homeomorphism and you are right - the standard proof of this step needs Hahn-Banach, I will fix this part of the question – Christian Apr 03 '20 at 11:47
  • I suppose then that you want to define a hyperplane as a closed linear subspace of codimension one? – Nate Eldredge Apr 03 '20 at 12:36
  • @NateEldredge: yes, of course. I am sorry for creating such a chaos in the comments. – Christian Apr 03 '20 at 12:39

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