# Can $L^p(\mathbb{R})$ and $L^q(\mathbb{R})$ be isomorphic?

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

• 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. – Bill Johnson Nov 1 '11 at 14:59
• @Bill: Quick Google search for "Lp and Lq are not isomorphic" gives this book books.google.com/…, page 180. – Mark Sapir Nov 1 '11 at 15:18
• 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. – Bill Johnson Nov 1 '11 at 15:52
• @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. – Mark Sapir Nov 1 '11 at 16:17

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.
• @Bill: Out of curiosity, what was Banach's original argument for proving $L_p$ is not isomorphic to $L_q$? – Yemon Choi Nov 3 '11 at 8:49
• 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. – Bill Johnson Nov 3 '11 at 16:03
• 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 $Y$. – Bill Johnson Nov 3 '11 at 16:06
• 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.) – Willie Wong Nov 28 '11 at 13:19