# Metric approach to limit operators IV

In this post I’ll finally get to the “condensation of singularities” argument that was invented by Lindner and Seidel in the (free abelian) group context and generalized by Spakula and Willett to metric spaces.  (Calling this “condensation of singularities” is my idea, but it does seem to me to get at what is going on.  I can’t help feeling that there should be a way of replacing some of the explicit constructions with an abstract argument involving the Baire category theorem.  But I have not yet been able to come up with one.)

Recall from post III the concepts of the lower norm and of a $$(\delta,s)$$-witness to an inequality for the lower norm.   To start things off, I will work on a metric space which is actually a discrete group $$\Gamma$$ (so this section is really an exposition of the relevant part of Lindner-Seidel).  For an operator $$A$$ we’ll say that a lower norm estimate $$\nu(A)\le\alpha$$ is $$(\delta,s)$$-centrally witnessed if there is a nonzero vector $$\xi$$ supported in the ball $$B(e;s)$$ such that

$\|A\xi\| \le (\alpha+\delta) \|\xi\|.$

Here $$e$$ denotes the identity element of the group; the difference between the notions of “central witness” and “witness” is that the central witness vector $$\xi$$ must be supported in a specific ball of radius $$s$$, namely, the one centered at the identity.  It is important to notice that this notion of “central witness” relates well to strong convergence:

Lemma: Suppose that $$C_n$$ is a norm-bounded sequence of operators converging strongly to $$C$$.  If there exist $$(\delta,s)$$ such that  for each $$n$$ the inequality $$\nu(C_n)\le \alpha$$ is $$(\delta,s)$$-centrally witnessed, then $$\nu(C)\le\alpha+\delta$$.

Proof: Let $$(\xi_n)$$ be a sequence of unit vectors that are $$(\delta,s)$$-central witnesses to $$\nu(C_n)\le \alpha$$.  Since they are all elements of the finite-dimensional vector space $$\ell^2(B(e;s))$$, there is a subsequence – which we may as well assume is the whole sequence – converging in norm to some unit vector $$\xi$$.  The bounded strong convergence of the operators now easily yields $$C_n\xi_n\to C\xi$$ (in norm). Since  $$\|C_n\xi_n\| \le (\alpha+\delta)$$, we obtain $$\|C\xi\| \le (\alpha+\delta)$$ giving the result.

Remark: If a lower norm inequality is witnessed for an operator $$B$$, then there is some translate $$C=B^\gamma$$ for which the corresponding inequality is centrally witnessed (with the same constants): the group element $$\gamma$$ is chosen to move the center of the support ball for the witness to the identity.

Now let’s get to the Big Question.  We want to prove that if the whole operator spectrum of a BDO $$A$$, that is all its limit operators $$A_\omega$$, consists of invertible operators, then these operators are automatically uniformly invertible.  To prove this what is needed is the “compactness type” statement that the lower bound for $$\nu(A_\omega)$$ is actually attained for some $$\omega\in\partial X$$.

Here’s how this is done.  Let $$a = \inf \{\nu(A_\omega): \omega\in\partial X\}.$$ Then, of course, for each $$n>0$$ there is an $$A_n$$ in the operator spectrum with $$\nu(A_n) \le a+ 2^{-n}$$.  The estimates of post III apply uniformly across the operator spectrum, so (under the assumption of property A) there is a certain sequence $$s_n$$, tending to infinity, such that for all $$n$$, and all $$m\le n$$, the (true) inequality

$\nu(A_n) \le a + 2^{-m}$

is $$(2^{-m},s_m)$$-witnessed.

By the Remark above, for each $$m,n$$ we can translate $$A_n$$ to an operator for which the above inequality is centrally witnessed with the same parameters.  I claim now that the translations can be chosen to depend only on $$n$$, not on $$m$$, at the expense of making the sequence of support radii increase faster.  In other words, there exists a sequence of group elements $$\gamma_n$$ and a sequence $$r_n$$ tending to infinity, such that, if $$C_n = A_n^{\gamma_n}$$, then for all $$n$$, and all $$m\le n$$, the (true) inequality

$\nu(C_n) \le a + 2^{-m}$

is $$(2^{-m},r_m)$$- centrally witnessed.   This is accomplished by an inductive construction, fitting a finite number of smaller witnesses together inside a not-too-large ball. Notice that each $$C_n$$ belongs to the operator spectrum of $$A$$, because that spectrum is closed under translation.

Now we are done. By compactness the $$C_n$$ have a strongly convergent subsequence (we may as well assume the whole sequence is strongly convergent).  Let $$C$$ be the strong limit (which is once again in the operator spectrum). By the Lemma above, $$\nu(C)\le a + 2\cdot 2^{-m}$$ for each $$m$$ and thus $$\nu(C)\le a$$.  That is, the lower norm attains its lower bound on the operator spectrum, which is what was to be shown.

In the next and hopefully final post, I will talk about what Spakula and Willett do to implement this argument without the group.