Let $p,q \in (1,\infty)$ with $p\neq q$. Are the Banach spaces $L^p(\mathbb{R})$, $L^q(\mathbb{R})$ isomorphic?

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    $\begingroup$ This is a perfectly reasonable question for a non expert to ask. Can anyone who voted to close prove that $L_4(0,1)$ is not isomorphic to $L_6(0,1)$? In practice, it is quite difficult to decide whether 2 Banach spaces are isomorphic. $\endgroup$ Nov 1, 2011 at 14:59
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    $\begingroup$ @Bill: Quick Google search for "Lp and Lq are not isomorphic" gives this book books.google.com/…, page 180. $\endgroup$
    – user6976
    Nov 1, 2011 at 15:18
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    $\begingroup$ Sure, Mark, and it is in other books as well (going back to Banach's classic). You will not find it in basic texts, though, and the result is certainly not obvious. I have been asked this exact question by famous people who work in a different part of functional analysis. $\endgroup$ Nov 1, 2011 at 15:52
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    $\begingroup$ @Bill: the key here is not that it is in a book, but that a 1 minute Google search is enough to answer that question. $\endgroup$
    – user6976
    Nov 1, 2011 at 16:17
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    $\begingroup$ I started a meta thread about this: tea.mathoverflow.net/discussion/1197/… $\endgroup$ Nov 3, 2011 at 13:04

1 Answer 1


The proof Fabian alludes to in the book reference Mark gave is a modern one using the notions of cotype and type. One way to prove that a Banach space $X$ is not isomorphic to a Banach space $Y$ is to exhibit a property that is preserved under isomorphisms that $X$ has but $Y$ does not. Type and cotype are examples of such properties. The (best) type and cotype of $L_p$ are calculated in many books. I suggest you look at Theorem 6.2.14 in the book of Albiac and Kalton. From the statement you see that if $p\not=q$, then $L_p$ and $L_q$ either have different (best) type or different (best) cotype.

Type and cotype depend only on the collection of finite dimensional subspaces of a space (we call such a property a local property). So you cannot use either to prove, e.g., that $L_p$ is not isomorphic to $\ell_p$ when $p\not= 2$. One way of proving this is to show that $\ell_2$ embeds isomorphically into $L_p$ but not into $\ell_p$ when $p\not=2$. These facts you can also find in Albiac-Kalton. You can also use infinite dimensional techniques to prove that $L_p$ and $L_q$ are not isomorphic when $p\not=q$. Banach knew this result through infinite dimensional considerations--the concepts of type and cotype came on the scene only 40 years ago.

You will also find in Albiac-Kalton a discussion of when $L_p$ or $\ell_p$ embeds isomorphically into $L_q$. That is more complicated and in fact Banach did not know everything. He called the question the problem of the linear dimension of $L_p$ spaces, IIRC.

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    $\begingroup$ @Bill: Out of curiosity, what was Banach's original argument for proving $L_p$ is not isomorphic to $L_q$? $\endgroup$
    – Yemon Choi
    Nov 3, 2011 at 8:49
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    $\begingroup$ Look at Chapter XII in Banach's book, Yemon, where he discusses the linear dimension of the $L_p$ and $\ell_p$ spaces. He shows that they are of incomparable linear dimension except possibly that $\ell_q$ or $L_q$ embeds into $L_p$ when $p<q<2$ or $2<q<p$ $^1$. Of course, we now know that $\ell_q$ and $L_q$ do not embed into $L_p$ when $2<q<p$ but do embed even isometrically when $p<q<2$. If you have Oeuvres vol. II version, there is a nice update to Banach's book written in 1979 by Pelczynski. 1. Well, except that $L_2$ isometrically embeds into all $L_p$ spaces. $\endgroup$ Nov 3, 2011 at 16:03
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    $\begingroup$ How do you make a line break in a comment? $\endgroup$ Nov 3, 2011 at 16:03
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    $\begingroup$ From linear dimension results in Banach's book, you can check the non isomorphism results even though the linear dimension problem was not completely solved until much later. BTW: When Banach says that the linear dimension of $X$ is less than the linear dimension of $Y$, he means that $X$ embeds isomorphically into $Y$. $\endgroup$ Nov 3, 2011 at 16:06
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    $\begingroup$ Completely tangential: software doesn't like line breaks in comments. You can cheat, however, by using an empty math environment. $$ $$ Like this. (Credit goes to Will Jagy for showing me it.) $\endgroup$ Nov 28, 2011 at 13:19

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