# Is $L^2(\mathbb R)$ homeomorphic to $L^1(\mathbb R)$?

Is $L^2(\mathbb R)$ homeomorphic to $L^1(\mathbb R)$?

More generally, are there instances of surprising homeomorphisms between non-isomorphic Banach spaces?

• Just to add some extra buzzwords: the homeomorphism mentioned by Nate Eldredge in his answer is sometimes known as the Mazur map - apparently this goes back to a paper of Mazur in Studia Math. from 1929 – Yemon Choi May 8 '17 at 15:23

## 1 Answer

According to Are $L^\infty(\Bbb R)$ and $L^2(\Bbb R)$ homeomorphic?, the map $f \mapsto \operatorname{sgn}(f) |f|^2$ is a homeomorphism from $L^2$ to $L^1$.

It's clearly a bijection. Suppose $f_n \to f$ in $L^2$; set $g_n = \operatorname{sgn}(f_n) |f_n|^2$ and $g$ likewise. Passing to a subsequence we can assume $f_n \to f$ a.e., so that $g_n \to g$ a.e. also. And we have $\|g_n\|_1 = \|f_n\|_2^2 \to \|f\|_2^2 = \|g\|_1$. By this lemma due to Riesz (a clever use of Fatou's lemma) we have $g_n \to g$ in $L^1$. So the map is continuous. The same argument shows the inverse is also continuous. This works over any measure space.

Thanks to Yemon Choi for pointing out that this observation is apparently due to S. Mazur, Studia Math., 1929 (EuDML).

Much more powerfully, all separable infinite-dimensional Banach spaces are homeomorphic.

MR0209804 Kadec, M. I. A proof of the topological equivalence of all separable infinite-dimensional Banach spaces. (Russian) Funkcional. Anal. i Priložen. 1 1967 61–70. (Reviewer: G. L. Krabbe)

English translation at: M.I. Kadets, Functional Analysis and Its Applications 1 (1967) p 53. (https://doi.org/10.1007/BF01075865)

The MathSciNet review points out that, combined with other contemporaneous work, all separable infinite-dimensional Fréchet spaces are homeomorphic.

I found this reference in a Math.SE post by Tomek Kania, where Tomek says that two infinite-dimensional Banach spaces are homeomorphic iff they have dense subsets of the same minimal cardinality.