# 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. Nov 1, 2011 at 14:59
• @Bill: Quick Google search for "Lp and Lq are not isomorphic" gives this book books.google.com/…, page 180.
– user6976
Nov 1, 2011 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. Nov 1, 2011 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.
– user6976
Nov 1, 2011 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 $L_q$? Nov 3, 2011 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. Nov 3, 2011 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 $Y$. Nov 3, 2011 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.) Nov 28, 2011 at 13:19