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$l^1$ has the Schur property (every weakly convergent sequence is norm convergent) and $L^1[0,1]$ does not, so the two spaces cannot be isomorphic.

Is this folklore, or is it credited to someone? (Also I wonder whether the original proof of non-isomorphism was this one.)

Edit: to clear up any confusion, I am asking about the nonexistence of an isomorphism, not the nonexistence of an isometric isomorphism (which is basic, as pointed out in the comments).

Edit 2: the comment section seems to have degenerated into attempts to find alternate proofs of this fact (and debate over common misconceptions, such as whether a Banach space which is isomorphic to a dual space must itself be a dual space). That's great, although I still think the Schur property argument is the easiest one. However, this was not my question. Who first proved this?

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    $\begingroup$ Another proof: $l^1$ has the Radon-Nikodym property, $L_1[0,1]$ does not. $\endgroup$ Commented May 3, 2016 at 17:42
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    $\begingroup$ @FedorPetrov: The question is about isomorphism, not isometric isomorphism. $\endgroup$
    – Nik Weaver
    Commented May 3, 2016 at 18:19
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    $\begingroup$ For instance, I found here a cute example of a Banach space which is isomorphic to $l^\infty$ but which is not a dual space. $\endgroup$
    – Nik Weaver
    Commented May 3, 2016 at 18:41
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    $\begingroup$ @NikWeaver, the cute example you refer to is indeed cute, but is not completely correct: the unit ball has many extreme points, namely all the elements $x \in \ell^\infty$ such that $|x_n|=\frac 1 2$ for all $n$. But this is still cute, because the closed convex hull of these extreme points is strictly smaller than the unit ball. $\endgroup$ Commented May 3, 2016 at 21:59
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    $\begingroup$ Let $K$ be a compact Hausdorff space and let $p$ be a point in $K$ that is not isolated. Then the continuous functions on $K$ that vanish at $p$ is a codimension one subspace of $C(K)$ whose unit ball has no extreme point. So any $C(K)$ that is isomorphic to its hyperplanes can be equivalently renormed so that its unit ball has no extreme point. $L_-\infty$ falls into this category. What is extraordinary is that, as Jerry pointed out, for any equivalent renorming of a separable conjugate space, the unit ball is the closed convex hull of its extreme points. $\endgroup$ Commented May 4, 2016 at 4:48

1 Answer 1

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This result was already published in the French (1932) edition of Banach's book "Theory of Linear Operations" (I do not know whether it was published in the Polish (1931) edition). On page 245 in the table you see two reasons why the spaces $\ell_1$ and $L_1(0,1)$ have different linear dimension:

  1. Schur property for $\ell_1$ and no Schur property for $L_1$.

  2. The fact that $\ell_1$ is isomorphically embeddable into each of its infinite dimensional subspaces and the same is not true for $L_1$.

Apparently for both statements about $L_1$ Banach means that they follow from the Khinchin's inequality (1923), but I did not find the corresponding place in the book.

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  • $\begingroup$ Great. Just what I was looking for! $\endgroup$
    – Nik Weaver
    Commented May 4, 2016 at 4:06
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    $\begingroup$ @NikWeaver To complement Mikhail's answer, let me point out that in the Polish edition of Banach's book on p. 183 the above argument is also used. See kielich.amu.edu.pl/Stefan_Banach/pdf/teoria-operacji-pol/08.pdf If you need a direct translation, I am happy to make you one. $\endgroup$ Commented May 4, 2016 at 9:36
  • $\begingroup$ @TomekKania: Thank you, I think I can see it. In number 4, right? $\endgroup$
    – Nik Weaver
    Commented May 4, 2016 at 11:52
  • $\begingroup$ @NikWeaver, yes :-) $\endgroup$ Commented May 4, 2016 at 12:34
  • $\begingroup$ I am curious - Schur's paper on the "Schur property" appeared in 1921. Did he not use it for this result? (Schur's paper is quite long, so I am a little worried about actually reading it...) $\endgroup$
    – Igor Rivin
    Commented May 5, 2016 at 10:55

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