# Metric approach to limit operators III

This is a continuation of my posts on the Spakula-Willett paper Metric approach to limit operators (see part I and part II).  In this post I will talk about “lower norm witnesses” on spaces with property A.  (This is quite close to what is done in my earlier post here, though using direct geometric tools rather than the functional analysis tricks I suggested, which only work in the Hilbert space case.)  Then in the next post I will talk about the “condensation of singularities” argument that completes the proof.

So at this point the authors have defined the “operator spectrum” of a band dominated operator on a space $$X$$, and they have the theorem that a BDO is Fredholm if and only if the operators in its spectrum are uniformly invertible.  What remains – the “Big Question” – is to remove the uniform invertibility hypothesis.

Recall the definition of the lower norm of an operator $$A$$:

$\nu(A) = \inf \{ \|A\xi\| : \|\xi\| = 1 \}.$

(Throughout this post I shall assume that we are working on the Hilbert space $$\ell^2(X)$$: this already includes the essential ideas.)  An inequality such as $$\nu(A)\le \alpha$$ is of course proved by exhibiting, for each $$\delta>0$$, a nonzero vector $$\xi$$ such that

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

such a $$\xi$$ is called a $$\delta$$-witness to the inequality.  If, in addition, $$\xi$$ is supported within some ball of radius $$s$$ we say it is a $$(\delta,s)$$-witness to the inequality $$\nu(A)\le \alpha$$.   If the inequality is true, then for each $$\delta>0$$ there is $$s>0$$ such that there is a $$(\delta,s)$$-witness to it.  This is trivial.  What is much deeper is the uniform control provided by property A:

Proposition (7.5 in Spakula-Willett) Suppose that $$X$$ has the metric sparsification property (equivalent to property A).  Then for all positive $$\delta, r, M$$ there is a positive $$s$$ such that every operator $$A$$ with propagation at most $$r$$ and norm at most $$M$$ has a $$(\delta,s)$$-witness to its lower norm (that is, to the true inequality $$\nu(A)\le\nu(A)$$).

Proof  By definition of the lower norm, there is $$\xi\neq 0$$ such that $$\|A\xi\| \le \left( \nu(A) + \delta/2\right) \|\xi\|$$.  Apply the metric sparsification property to the Borel measure whose mass at each $$x\in X$$ is $$|\xi(x)|^2$$.  This gives the following: for each $$c<1$$ (we’ll choose $$c$$ in a moment) there is $$s>0$$, depending only on $$c$$ and not on $$\xi$$, for which we can find a subset $$\Omega$$ of $$X$$ such that:

• $$\Omega$$ supports most of the mass of $$\xi$$: specifically, $$\sum_{x\in\Omega} |\xi(x)|^2 \ge c \|\xi\|^2$$;
• $$\Omega$$ is a disjoint union of “components” – subsets $$\Omega_j$$, $$j=1,2,\ldots$$,  that have diameter $$\le s$$ and are mutually separated by distance at least $$2r+1$$.

Let $$\xi_j$$, $$j=1,2,\ldots$$, be the restriction of $$\xi$$ to $$\Omega_j$$ and let $$\xi_0$$ be what’s left, that is the restriction of $$\xi$$ to the complement of $$\Omega$$.  Note that $$\xi_0$$ is small, its norm being at most $$(1-c)^{1/2}$$ times the norm of $$\xi$$.  I claim that one of the $$\xi_j$$ is the desired witness.

To see this, pretend for a moment that $$\xi_0$$ is actually zero.   Because of the propagation condition, the $$A\xi_j$$ have disjoint supports.  Therefore we have

$\|\xi\|^2 = \sum_j \|\xi_j\|^2, \qquad \|A\xi\|^2 = \sum_j \|A\xi_j\|^2,$

and since $$\|A\xi\| \le \left( \nu(A) + \delta/2\right) \|\xi\|$$ it follows that one of the $$\xi_j$$ must satisfy the same inequality and must therefore be a $$(\delta/2,s)$$-witness to the lower norm.  Of course our assumption that $$\xi_0=0$$ is not completely accurate but correcting for this just introduces a small error term on the right hand side which can be absorbed by the remaining $$\delta/2$$ if we make $$c$$ close enough to 1.   This completes the proof.

In the final post I will explain how this lower norm estimate allows one to solve the “Big Question”.