The concluding observation of Part I was that it’s hard to embed things into a Hilbert space: the geometry is the same in all directions, and the length of diagonals of parallelepipeds is tightly controlled. One may think that it should be easier to embed the Hilbert space itself into other things. And this is indeed so.
Let’s prove that the space contains an isomorphic copy of the Hilbert space , for every . The case can be immediately dismissed: this is a huge space which, by virtue of Kuratowski’s embedding, contains an isometric copy of every separable metric space. Assume .
Recall the Rademacher functions , They are simply square waves:
Define a linear operator by . Why is the sum in , you ask? By the Хинчин inequality:
The inequality tells us precisely that is an isomorphism onto its image.
So, even the indescribably ugly space contains a nice roundish subspace.
(Why is ugly? Every point of its unit sphere is the midpoint of a line segment that lies on the sphere. Imagine that.) One might ask if the same holds for every Banach space, but that’s way too much to ask. For instance, the sequence space for does not have any subspace isomorphic to . Informally, this is because the underlying measure (the counting measure on ) is not infinitely divisible; the space consists of atoms. For any given we can model the first Rademacher functions on sequences, but the process has to stop once we reach the atomic level. On the positive side, this shows that contains isomorphic copies of Euclidean spaces of arbitrarily high dimension, with uniform control on the distortion expressed by and . And this property is indeed shared by all Banach spaces: see Dvoretzky’s theorem.