Two minor typos:

(1) Proof of lemma 4: inequality should be reversed before “contradicting (3)”.

(2) Proof of lemma 5: the last inequality should be reversed… and it seems the coefficient before should be 2?

*[George: Fixed, thanks!]*

Thanks for your suggestion. I will have a reread through the argument when I have some time, and update accordingly.

]]>For , the -system property immediately gives . Note first that by definition of , we have for any .

As is the minimal d-system generated by and is another d-system, we obtain , which already implies , but so far only for .

Now consider more generally and additionally an . Because of what we just established, we know that , i.e. , which just means that . By symmetry, this implies , thus (because was arbitrary), . By the same argument as above (minimality of the d-system ), this again yields , this time for all .

This statement implies that for all , we know that , i.e. , which is the definition of being a -system. By lemma 3 (make link), is also a -algebra.

[Rest as in the text]

]]>No, it should be an equality, basically it is the parallelogram law extended to arbitrarily many terms, and is an identity in any Hilbert space.

]]>Indeed you are right! Thanks for this nice proof and your insight into these special cases!

In your second equation line, where you bound the sum by $K^2$, should not the first equality be replaced by an inequality? Since you take the sum over all permutations of +-1, you will get some extra terms. Of course, this will change nothing and your elegant argument still holds.

]]>No reference, this is something I came up with while writing the post. Actually, I did consider if the result holds for all reflexive Banach spaces. Maybe it does, but I am not sure, and did not want to spend too long on what was a side-comment and not the focus of the post. For a Hilbert space it is easy. If the norm of elements of the set is bounded by K, then

and it follows that . Uniformly convex spaces are a little trickier. Consider , where are chosen inductively to maximize (given the values of for k less than n). It can be seen that is non-decreasing and, as it is bounded by assumption, tends to a limit K (we have not used uniform convexity yet…). The uniform convex property can be used to show that for each fixed positive , there exists a such that

whenever , which is a contradiction unless for all large n, showing that it converges to zero.

It extends to Bochner spaces (defined wrt a Hilbert space H), you can generalise the Khintchine inequality to a_n lying in an arbitrary Hilbert space, so the result should still hold.

]]>Fixed. Thanks!

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