## The displacement set of nonlinear maps in vector spaces

Given a vector space ${V}$ and a map ${f\colon V\to V}$ (linear or not), consider the displacement set of ${f}$, denoted ${D(f) = \{f(x)-x\colon x\in V\}}$. For linear maps this is simply the range of the operator ${f-I}$ and therefore is a subspace.

The essentially nonlinear operations of taking the inverse or composition of maps become almost linear when the displacement set is considered. Specifically, if ${f}$ has an inverse, then ${D(f^{-1}) = -D(f)}$, which is immediate from the definition. Also, ${D(f\circ g)\subset D(f)+D(g)}$.

When ${V}$ is a topological vector space, the maps for which ${D(f)}$ has compact closure are of particular interest: these are compact perturbations of the identity, for which degree theory can be developed. The consideration of ${D(f)}$ makes it very clear that if ${f}$ is an invertible compact perturbation of the identity, then ${f^{-1}}$ is in this class as well.

It is also of interest to consider the maps for which ${D(f)}$ is either bounded, or is bounded away from ${0}$. Neither case can occur for linear operators, so this is essentially nonlinear analysis. In the nonlinear case, the boundedness assumption for linear operators is usually replaced by the Lipschitz condition. Let us say that ${f}$ is ${(L, \ell)}$-bi-Lipschitz if ${\ell\|x-y\|\le \|f(x)-f(y)\|\le L\|x-y\|}$ for all ${x, y}$ in the domain of ${f}$.

Brouwer’s fixed point theorem fails in infinite-dimensional Hilbert spaces, but it not yet clear how hard it can fail. The strongest possible counterexample would be a bi-Lipschitz automorphism of the unit ball with displacement bounded away from 0. The existence of such a map is unknown. If it does not exist, that would imply that the unit ball and the unit sphere in the Hilbert space are not bi-Lipschitz equivalent, because the unit sphere does have such an automorphism: ${x\mapsto -x}$.

Concerning the maps with bounded displacement, here is a theorem from Patrick Biermann’s thesis (Theorem 3.3.2): if ${f}$ is an ${(L, \ell)}$-bi-Lipschitz map in a Hilbert space, ${L/\ell < \pi/\sqrt{8}}$, and ${f}$ has bounded displacement, then ${f}$ is onto. The importance of bounded displacement is illustrated by the forward shift map ${S(x_1, x_2, \dots) = (0, x_1, x_2, \dots)}$ for which ${L=\ell=1}$ but surjectivity nonetheless fails.

It would be nice to get rid of the assumption ${L/\ell < \pi/\sqrt{8}}$ in the preceding paragraph. I guess any bi-Lipschitz map with bounded displacement should be surjective, at least in Hilbert spaces, but possibly in general Banach spaces as well.

## Orthonormal bases formed by translation of sequences

The standard orthonormal basis (ONB) in the Hilbert space ${\ell^2(\mathbb N_0)}$ consists of the vectors
(1, 0, 0, 0, …)
(0, 1, 0, 0, …)
(0, 0, 1, 0, …)

Let S be the forward shift operator: ${S(x_0, x_1, \dots) = (0, x_0, x_1, \dots)}$. The aforementioned ONB is precisely the orbit of the first basis vector ${e_0}$ under the iteration of S. Are there any other vectors x whose orbit under S is an ONB?

If one tries to construct such x by hand, taking some finite linear combination of ${e_k}$, this is not going to work. Indeed, if the coefficient sequence has finitely many nonzero terms, then one of them, say, ${c_m}$, is the last one. Then ${S^m x}$ is not orthogonal to ${x}$ because the inner product is ${\overline{c_0} c_m}$ and that is not zero.

However, such vectors x exist, and arise naturally in complex analysis. Indeed, to a sequence ${(c_n)\in\ell^2(\mathbb N_0)}$ we can associate a function ${f(z) = \sum_{n=1}^\infty c_n z^n}$. The series converges in the open unit disk to a holomorphic function which, being in the Hardy space ${H^2}$, has boundary values represented by an square-integrable function on the unit circle ${\mathbb T}$. Forward shift corresponds to multiplication by ${z}$. Thus, the orthogonality requires that for every ${k\in \mathbb N}$ the function ${z^kf}$ be orthogonal to ${f}$ in ${L^2(\mathbb T)}$. This means that ${|f|^2}$ is orthogonal to all such ${z^k}$; and since it’s real, it is orthogonal to ${z^k}$ for all ${k\in \mathbb Z\setminus \{0\}}$ by virtue of conjugation. Conclusion: |f| has to be constant on the boundary; specifically we need |f|=1 a.e. to have a normalized basis. All the steps can be reversed: |f|=1 is also sufficient for orthogonality.

So, all we need is a holomorphic function f on the unit disk such that almost all boundary values are unimodular and f(0) is nonzero; the latter requirement comes from having to span the entire space. In addition to the constant 1, which yields the canonical ONB, one can use

• A Möbius transformation ${f(z)=(z-a)/(1-\bar a z)}$ where ${a\ne 0}$.
• A product of those (a Blaschke product), which can be infinite if the numbers ${a_k}$ converge to the boundary at a sufficient rate to ensure the convergence of the series.
• The function ${f(z) = \exp((z+1)/(z-1))}$ which is not a Blaschke product (indeed, it has no zeros) yet satisfies ${|f(z)|=1}$ for all ${z\in \mathbb T\setminus \{-1\}}$.
• Most generally, an inner function which is a Blaschke product multiplied by an integral of rotated versions of the aforementioned exponential function.

Arguably the simplest of these is the Möbius transformation with ${a=1/2}$; expanding it into the Taylor series we get
${\displaystyle -\frac12 +\sum_{n=1}^\infty \frac{3}{2^{n+1}} z^n }$
Thus, the second simplest ONB-by-translations after the canonical one consists of
(-1/2, 3/4, 3/8, 3/16, 3/32, 3/64, …)
(0, -1/2, 3/4, 3/8, 3/16, 3/32, …)
(0, 0, -1/2, 3/4, 3/8, 3/16, …)
and so on. Direct verification of the ONB properties is an exercise in summing geometric series.

What about the exponential one? The Taylor series of ${\exp((z+1)/(z-1))}$ begins with
${\displaystyle \frac{1}{e}\left(1 - 2z +\frac23 z^3 + \frac23 z^4 +\frac25 z^5 + \frac{4}{45} z^6 - \frac{10}{63} z^7 - \frac{32}{105}z^8 - \cdots \right) }$
I don’t know if these coefficients in parentheses have any significance. Well perhaps they do because the sum of their squares is ${e^2}$. But I don’t know anything else about them. For example, are there infinitely many terms of either sign?

Geometrically, a Möbius transform corresponds to traversing the boundary circle once, a Blaschke product of degree n means doing it n times, while the exponential function, as well as infinite Blaschke products, manage to map a circle onto itself so that it winds around infinitely many times.

Finally, is there anything like that for the backward shift ${S^*(x_0, x_1, x_2, \dots) = (x_1, x_2, \dots)}$? The vector ${(S^*)^k x}$ is orthogonal to ${x}$ if and only if ${x}$ is orthogonal to ${S^kx}$, so the condition for orthogonality is the same as above. But the orbit of any ${\ell^2(\mathbb N_0)}$ vector under ${S^*}$ tends to zero, thus cannot be an orthonormal basis.

## Weak convergence in metric spaces

Weak convergence in a Hilbert space is defined as pointwise convergence of functionals associated to the elements of the space. Specifically, ${x_n\rightarrow x}$ weakly if the associated functionals ${f_n(y) = \langle x_n, y \rangle}$ converge to ${f(y) = \langle x_n, y \rangle}$ pointwise.

How could this idea be extended to metric spaces without linear structure? To begin with, ${f_n(y)}$ could be replaced with ${\|x_n-y\|^2-\|x_n\|^2}$, since this agrees with original ${f_n}$ up to some constant terms. Also, ${\|x_n\|^2}$ here could be ${\|x_n-z\|^2}$ for any fixed ${z}$; to avoid introducing another variable here, let’s use ${x_1}$ for the purpose of fixed reference point ${z}$. Now we have a metric space version of the weak convergence: the functions
${f_n(y) = d(x_n,y)^2 - d(x_n,x_1)^2}$
must converge pointwise to
${f(y) = d(x,y)^2 - d(x,x_1)^2}$

The triangle inequality shows that strong convergence ${d(x_n,x)\rightarrow 0}$ implies weak convergence, as expected. And the converse is not necessarily true, as the example of a Hilbert space shows.

Aside: the above is not the only way to define weak convergence in metric spaces. Another approach is to think of ${\langle x_n , y\rangle}$ in terms of projection onto a line through ${y}$. A metric space version of this concept is the nearest-point projection onto a geodesic curve. This is a useful approach, but it is only viable for metric spaces with additional properties (geodesic, nonpositive curvature).

Also, both of these approaches take the Hilbert space case as the point of departure, and do not necessarily capture weak convergence in other normed spaces.

Let’s try this out in ${\ell^1}$ with the standard basis sequence ${e_n}$. Here ${f_n(y) = \|e_n-y\|^2 - 4 \rightarrow (\|y\| + 1)^2 - 4}$. Is there an element ${x\in \ell^1}$ such that

${\|x-y\|^2 - \|x-e_1\|^2= (\|y\|+1)^2 - 4}$ for all ${y}$?

Considering both sides as functions of one variable ${y_n}$, for a fixed ${n}$, shows that ${x_n=0}$ for ${n}$, because the left hand side is non-differentiable at ${y_n=x_n}$ while the right hand side is non-differentiable at ${y_n=0}$. Then the desired identity simplifies to ${\|y\|^2 - 1 = (\|y\|+1)^2 - 4}$ which is false. Oh well, that sequence wasn’t weakly convergent to begin with: by Schur’s theorem, every weakly convergent sequence in ${\ell^1}$ also converges strongly.

This example also shows that not every bounded sequence in a metric space has a weakly convergent subsequence, unlike the way it works in Hilbert spaces.

## Graphical embedding

This post continues the theme of operating with functions using their graphs. Given an integrable function ${f}$ on the interval ${[0,1]}$, consider the region ${R_f}$ bounded by the graph ${y=f(x)}$, the axis ${y=0}$, and the vertical lines ${x=0}$, ${x=1}$.

The area of ${R_f}$ is exactly ${\int_0^1 |f(x)|\,dx}$, the ${L^1}$ norm of ${f}$. On the other hand, the area of a set is the integral of its characteristic function,

$\displaystyle \chi_f = \begin{cases}1, \quad x\in R_f, \\ 0,\quad x\notin R_f \end{cases}$

So, the correspondence ${f\mapsto \chi_f }$ is a map from the space of integrable functions on ${[0,1]}$, denoted ${L^1([0,1])}$, to the space of integrable functions on the plane, denoted ${L^1(\mathbb R^2)}$. The above shows that this correspondence is norm-preserving. It also preserves the metric, because integration of ${|\chi_f-\chi_g|}$ gives the area of the symmetric difference ${R_f\triangle R_g}$, which in turn is equal to ${\int_0^1 |f-g| }$. In symbols:

$\displaystyle \|\chi_f-\chi_g\|_{L^1} = \int |\chi_f-\chi_g| = \int |f-g| = \|f-g\|_{L^1}$

The map ${f\mapsto \chi_f}$ is nonlinear: for example ${2f}$ is not mapped to ${2 \chi_f}$ (the function that is equal to 2 on the same region) but rather to a function that is equal to 1 on a larger region.

So far, this nonlinear embedding did not really offer anything new: from one ${L^1}$ space we got into another. It is more interesting (and more difficult) to embed things into a Hilbert space such as ${L^2(\mathbb R^2)}$. But for the functions that take only the values ${0,1,-1}$, the ${L^2}$ norm is exactly the square root of the ${L^1}$ norm. Therefore,

$\displaystyle \|\chi_f-\chi_g\|_{L^2} = \sqrt{\int |\chi_f-\chi_g|^2} = \sqrt{\int |\chi_f-\chi_g|} = \sqrt{\|f-g\|_{L^1}}$

In other words, raising the ${L^1}$ metric to power ${1/2}$ creates a metric space that is isometric to a subset of a Hilbert space. The exponent ${1/2}$ is sharp: there is no such embedding for the metric ${d(f,g)=\|f-g\|_{L^1}^{\alpha} }$ with ${\alpha>1/2}$. The reason is that ${L^1}$, having the Manhattan metric, contains geodesic squares: 4-cycles where the distances between adjacent vertices are 1 and the diagonal distances are equal to 2. Having such long diagonals is inconsistent with the parallelogram law in Hilbert spaces. Taking the square root reduces the diagonals to ${\sqrt{2}}$, which is the length they would have in a Hilbert space.

This embedding, and much more, can be found in the ICM 2010 talk by Assaf Naor.