This is the inaugural blog post containing notes from my lectures given at UVa in the Fall of 2016. This first lecture was on Friday September 9th. Please comment or email if you have any questions or corrections.
Let be the space of all finitely supported sequences of real numbers. After defining a norm on the resulting Banach space will be the completion of with respect to this norm. The standard unit vectors are denoted and will be a Schauder basis for this Banach space. We begin with the definition of the Schreier familes of finite order.
Let (the singleton subsets of ) and for each let
Here means the maximum of is less than the minimum of (by convention ). For each define the following norm on
Let be the completion of with the norm . It is known that every (always infinite dimensional) subspace of contains an isomorphic copy of (this is called -saturated). This proof is somewhat complicated but we can note that for the following block sequence is isomorphic to : Let , , , etc. It is (relatively) easy to show that for each we have .
As a warm up exercise we will prove that does not contain . We need the following theorem due to R.C. James. Our general Banach space is the completion of with respect to a norm and having as a Schauder basis .
Theorem 1 The Banach space contains an isomorphic copy of if and only if for each there is a normalized block sequence in so that for all we have
Moreover the analogous statement holds for .
We have included the proof of the above theorem at the end of this post. We first prove the following.
Theorem 2 does not contain .
This follows from the statement above that is saturated, however, it is not hard to prove so we will is supply the proof.
Proof: Let be the normalized block sequence equivalent to with . Let and . Note that by our assumption . If then . Thus . In this case
This contradicts the lower bound of .
We will now define Tsirelson’s norm and prove Tsirelson’s space is distortable for each . Recall the definition of distortable.
Definition 3 Let be a Banach space with a norm . Then is -distortable (for ) if there is an equivalent norm on so that in every infinite dimensional subspace of there are vectors with and .
The theorem of R.C. James from above can be rephrased as stating the and are not distortable for any (or simply not distortable). Also note that if is -distortable for some then it every subspace is -distortable. Therefore in order to show that a space does not contain is suffices to show that it is distortable (although it is a bit overkill!).
We will now construct Tsirelson’s space. We construct a norm on by constructing a set of vectors in are that will constitute the so-called norming set. A set is called a norming set if it contains (the standard unit vectors) and satisfies implies . If is a norming set then is a norm on . Here is the dot product and will also be denoted . Norming sets offer an generic way for constructing a space whose basis is the unit vector basis .
We now construct the norming set of Tsirelson’s space . Let and
Then . We list is few important facts regarding . The Tsirelson norm is .
- (i) If then either or there is a block seqquence so that and . Such block sequences are called admissible.
- (ii) If is an admissible block sequence in then .
It is useful for us to record the following somewhat technical notation. If such that is an admissible block we say (note that this is not uniquely defined). Suppose further that is a normalized block sequence and define the following two sets:
For let . As a convention, we will try to keep as indices for the vectors and as indices for functionals. Then
Observe that ( is clear and because there is at least one unique for each in ). In addition,
In the other case apply the triangle inequality to see
Theorem 4 Tsirelson’s space does not contain for any or .
Proof: If a normalized block was isomorphic to for then, in particular, However we can find for each so that and . Then and
This is a contradiction of sufficiently large .
In the case of we suppose that is a normalized block sequence with and isomorphic to with . Then applying the above equation for an arbitrary with length
If then the estimate is . If then and so we have the estimate which is a contradiction.
The final theorem for this post is to show that is distortable.
Theorem 5 Tsirelson’s space is distortable for each .
Consider the following equivalent norm on
It is clear that . By letting be the norming functional of and we can see that . Therefore each norm is equivalent.
Fix and consider a block basic sequence in . Krivine’s Theorem states that for any Banach space with a basis there is a so that is block finitely representable in the space. For Tsirelson’s space it follows from our arguments from before that no other than can be a Krivine . Thus for each and there is a block that is equivalent to the unit vector basis of . Let . Then . We need to estimate from above. Of course we have
Choosing and , we have the .
We build what is called a rapidly increading sequence (RIS) of averages . That is, is an average with constant and
Let . Then it is easily seen that . We need an upper estimate on . Consider an arbitrary functional of length . Suppose is the smallest index so that intersects . Then
Then for each . Then using the inequality (3)
Simplifying and summing up yields
Now for sufficiently large we have . Let . Then . To finish the proof note that and
for sufficiently small .
In the above we use two facts: (1) For fixed we can find an large so that and average is not normed by a functional of length . It misses it by the most it can which is . (2) For this fixed we can build a vector that is normed by and functional of length . This vector is an average of averages each of which are not normed by a functional of length .
Proof of James’ Non-Distortion of Theorem: Let . Suppose that there are positive constants and is in a block sequence in a Banach space that satisfies
for all . Let . Then is increasing to some . Find a block sequence that witnesses this stabilization. Let with satisfying and for all . Let . Then,
For the James’ the distortion theorem we we consider the values
and proceed somewhat analogously.