Existence of quasiconformal maps with maximal stretching on any given countable set

The basic building block of $F_{\Lambda}$ will be the radial stretching map located at the origin:

$$S(\vec{x})=\vec{x}|\vec{x}|^{1/K-1}.$$

This is a $K$-quasiconformal map from $\mathbb{R}^{d}$ to $\mathbb{R}^{d}$ and stretches with exponent $1/K$ at the origin (and is bilipschitz on any compact set separated from the origin). We will need a detailed study of its differential matrix; fix a point $\vec{x}$ within the unit ball. The $(i,j)$ component of the matrix $DS$ is given by

	$\displaystyle[DS]_{i,j}$	$\displaystyle=\frac{\partial}{\partial x_{j}}(x_{1}^{2}+\cdots+x_{d}^{2})^{\frac{1-K}{2K}}x_{i}$
		$\displaystyle=\frac{1-K}{2K}(x_{1}^{2}+\cdots+x_{d}^{2})^{\frac{1-K}{2K}-1}\cdot 2x_{j}\cdot x_{i}+(x_{1}^{2}+\cdots x_{d}^{2})^{\frac{1-K}{2K}}\frac{\partial x_{i}}{\partial x_{j}}$
		$\displaystyle=\frac{1-K}{K}|\vec{x}|^{\frac{1-K}{K}-2}x_{i}x_{j}+|\vec{x}|^{\frac{1-K}{K}}\delta_{i,j}$
		$\displaystyle=|\vec{x}|^{1/K-1}\left(\delta_{i,j}-\left(1-\frac{1}{K}\right)\frac{x_{i}x_{j}}{|\vec{x}|^{2}}\right),$

where $\delta_{i,j}=1$ if $i=j$ and is zero otherwise. This gives a particularly special form to the matrix $DS$: it is essentially a perturbation of a (properly scaled) identity matrix via a rank one operator. To be specific, we have

(2.1)

$$DS(\vec{x})=|\vec{x}|^{1/K-1}\left(I-\left(1-\frac{1}{K}\right)\frac{\vec{x}\vec{x}^{T}}{|\vec{x}|^{2}}\right).$$

We need to establish a few properties of this matrix before continuing; note that $DS$ is a symmetric matrix. Next, and most important, is that this matrix is positive definite. To see this, observe that if $\vec{w}\in\mathbb{R}^{d}$, we have that

$\displaystyle\vec{w}^{T}(\vec{x}\vec{x}^{T})\vec{w}$

$\displaystyle=(\vec{x}^{T}\vec{w})^{T}(\vec{x}^{T}\vec{w})=|\vec{x}^{T}\vec{w}|^{2}$

is a nonnegative real number at most $|\vec{x}|^{2}\cdot|\vec{w}|^{2}$ via the Cauchy-Schwarz inequaltiy. This implies that

$$\left|\vec{w}^{T}\left(\left(1-\frac{1}{K}\right)\frac{\vec{x}\vec{x}^{T}}{|\vec{x}|^{2}}\right)\vec{w}\right|\leq\left(1-\frac{1}{K}\right)|\vec{w}|^{2}<|\vec{w}|^{2}$$

since $K>1$. The positive definiteness of $DS$ on the unit ball follows immediately from this upper bound.

We are now ready to construct the map $F_{\Lambda}$ from slightly modified versions of $S$. For a fixed point $\lambda$, define

$$S_{\lambda}(\vec{x})=(\vec{x}-\lambda)|\vec{x}-\lambda|^{1/K-1}$$

and set

(2.2)

$$F_{\Lambda}(\vec{x})=\sum_{n=1}^{\infty}\frac{1}{2^{n}}S_{\lambda_{n}}(\vec{x}).$$

Note that by translating (2.1),

$$DS_{\lambda}(\vec{x})=|\vec{x}-\lambda|^{1/K-1}\left(I-\left(1-\frac{1}{K}\right)\frac{(\vec{x}-\lambda)(\vec{x}-\lambda)^{T}}{|\vec{x}-\lambda|^{2}}\right).$$

Furthermore, because the series rapidly converges, $F_{\Lambda}$ lies in the Sobolev class $W^{1,d}(\mathbb{R}^{d})$. One way to argue this is to observe that $F_{\Lambda}$ is absolutely continuous along every vertical and horizontal line which does not intersect $\Lambda$ and study the classical difference quotients used to define its partial derivatives. In any case, we find that

$$DF_{\Lambda}(x)=\sum_{n=1}^{\infty}\frac{1}{2^{n}}DS_{\lambda_{n}}(\vec{x}).$$

Each $DS_{\lambda_{n}}$ is symmetric and positive definite, being just a translate of $DS$; therefore, $DF_{\Lambda_{n}}$ is also symmetric and positive definite and this shows that $F_{\Lambda}$ is injective. There are now two aspects left to verify: that $F_{\Lambda}$ satisfies the distortion inequality for $K$-quasiconformal maps and that it exhibits the correct stretching exponent on $\Lambda$.

In order to establish the distortion inequality (1.1), we rewrite the differential matrix with some extra notation. Starting with

$$DF_{\Lambda}(\vec{x})=\sum_{n=1}^{\infty}\frac{1}{2^{n}}|\vec{x}-\lambda_{n}|^{1/K-1}\left(I-\left(1-\frac{1}{K}\right)\frac{(\vec{x}-\lambda_{n})(\vec{x}-\lambda_{n})^{T}}{|\vec{x}-\lambda_{n}|^{2}}\right),$$

set $\vec{w}_{n}=(\vec{x}-\lambda_{n})/|\vec{x}-\lambda_{n}|$, $W(\vec{x})=\sum_{n=1}^{\infty}2^{-n}|\vec{x}-\lambda_{n}|^{1/K-1}$, and

$$\eta_{n}(\vec{x})=\frac{2^{-n}|\vec{x}-\lambda_{n}|^{1/K-1}}{W(x)},$$

defined for all $\vec{x}\notin\Lambda$. Note that for such $\vec{x}$, we have $\sum_{n=1}^{\infty}\eta_{n}(\vec{x})=1$. We then have

	$\displaystyle DF_{\Lambda}(\vec{x})$	$\displaystyle=\left(\sum_{n=1}^{\infty}\frac{1}{2^{n}}|\vec{x}-\lambda_{n}|^{1/K-1}\right)I-\left(1-\frac{1}{K}\right)\sum_{n=1}^{\infty}\frac{1}{2^{n}}|\vec{x}-\lambda_{n}|^{1/K-1}\vec{w}_{n}\vec{w}_{n}^{T}$
		$\displaystyle=W(\vec{x})I-\left(1-\frac{1}{K}\right)W(\vec{x})\sum_{n=1}^{\infty}\frac{2^{-n}|\vec{x}-\lambda_{n}|^{1/K-1}}{W(\vec{x})}\vec{w}_{n}\vec{w}_{n}^{T}$
		$\displaystyle=W(\vec{x})I-\left(1-\frac{1}{K}\right)W(\vec{x})\sum_{n=1}^{\infty}\eta_{n}(\vec{x})\vec{w}_{n}\vec{w}_{n}^{T}$
		$\displaystyle=W(\vec{x})\left(I-\left(1-\frac{1}{K}\right)\sum_{n=1}^{\infty}\eta_{n}(\vec{x})\vec{w}_{n}\vec{w}_{n}^{T}\right).$

For a final set of notation, set $\alpha=1-1/K\in(0,1)$ and

$$B(\vec{x})=\sum_{n=1}^{\infty}\eta_{n}(\vec{x})\vec{w}_{n}\vec{w}_{n}^{T},$$

noting that $B(\vec{x})$ is a matrix defined for each $\vec{x}\notin\Lambda$; in this new notation, $DF_{\Lambda}(\vec{x})=W(\vec{x})(I-\alpha B(\vec{x})).$

Our aim is to show that $\|DF_{\Lambda}\|^{d}\leq K\det(DF_{\Lambda});$ since both sides are homogeneous of the same degree and $W(\vec{x})$ is a scalar function, it is sufficient to establish that for each $\vec{x}\notin\Lambda$ we have

(2.3)

$$\left\|I-\alpha B(\vec{x})\right\|^{d}\leq K\det\left(I-\alpha B(\vec{x})\right).$$

Since $\Lambda$ has Lebesgue measure zero, this is sufficient to prove quasiconformality. To this end, we will show that the operator norm is at most $1$ and the determinant is at least $1/K$.

In preparation for this, we need to analyze the spectrum of the matrix $I-\alpha B(\vec{x})$, starting with the symmetric matrix $B(\vec{x})$. If $y\in\mathbb{R}^{d}$ is a unit vector,

	$\displaystyle\vec{y}^{T}B(\vec{x})\vec{y}$	$\displaystyle=\sum_{n=1}^{\infty}\eta_{n}(\vec{x})\vec{y}^{T}\vec{w}_{n}\vec{w}_{n}^{T}\vec{y}$
		$\displaystyle=\sum_{n=1}^{\infty}\eta_{n}(\vec{x})|\vec{w}_{n}^{T}\vec{y}|^{2}.$

Estimating $|\vec{w}_{n}^{T}\vec{y}|\leq|\vec{w}_{n}|\cdot|\vec{y}|\leq 1$ and recalling that $\sum_{n}\eta_{n}(\vec{x})=1$, we have

$$\vec{y}^{T}B(\vec{x})\vec{y}\leq 1$$

for all unit vectors $\vec{y}$. On the other hand, each $\eta_{n}(\vec{x})$ is nonnegative and so $\vec{y}^{T}B(\vec{x})\vec{y}\geq 0$. Therefore, $B(\vec{x})$ is a positive semidefinite matrix with operator norm at most $1$. This implies that the spectrum $\sigma(B(\vec{x}))$ consists of $d$ eigenvalues $\{\sigma_{1},...,\sigma_{d}\}\subseteq[0,1]$.

There is another important fact about the spectrum of $B(\vec{x})$ which we will need: the sum of the eigenvalues is $1$. To see this, note that the trace of $B(\vec{x})$ is

$$\operatorname{Tr}B(\vec{x})=\sum_{n=1}^{\infty}\eta_{n}(\vec{x})\operatorname{Tr}(\vec{w}_{n}\vec{w}_{n}^{T})=\sum_{n=1}^{\infty}\eta_{n}(\vec{x})|\vec{w}_{n}|^{2}=1.$$

after using that each $w_{n}$ is a unit vector. Recalling that the trace of a matrix is the sum of its eigenvalues, the claim follows.

We also need to relate the spectrum of $B(\vec{x})$ to that of $I-\alpha B(\vec{x})$. When $\vec{v}_{i}$ is an eigenvector of $B(\vec{x})$ associated to the eigenvalue $\sigma_{i}$, it follows that $\vec{v}_{i}$ is also an eigenvector of $I-\alpha B(\vec{x})$ but now with eigenvalue $1-\alpha\sigma_{i}$. It follows that the spectrum of $I-\alpha B(\vec{x})$ is the set

$$\sigma(I-\alpha B(\vec{x}))=\{1-\alpha\sigma_{1},\dots,1-\alpha\sigma_{d}\}.$$

Because each $\sigma_{i}\in[0,1]$ and $\alpha\in(0,1)$, it follows that the eigenvalues of $I-\alpha B(\vec{x})$ all lie in the interval $(0,1]$.

We are now ready to bound the relevant quantities for the distortion inequality (2.3). Since $I-\alpha B(\vec{x})$ is a real symmetrix matrix, it is orthogonally diagonalizable; all its eigenvalues are at most $1$, and so

(2.4)

$$\|I-\alpha B(\vec{x})\|\leq\max\{\sigma_{1},...,\sigma_{d}\}\leq 1.$$

It therefore remains to bound the determinant from below; to this end, write the determinant as the product of the eigenvalues:

	$\displaystyle\det(I-\alpha B(\vec{x}))$	$\displaystyle=\prod_{i=1}^{d}(1-\alpha\sigma_{i})$
		$\displaystyle=\sum_{k=0}^{d}(-\alpha)^{k}\sum_{i_{1}<...<i_{k}}\sigma_{i_{i}}\cdots\sigma_{i_{k}}$
		$\displaystyle=1-\alpha(\sigma_{1}+\cdots+\sigma_{d})+\sum_{k=2}^{d}(-\alpha)^{k}\sum_{i_{1}<...<i_{k}}\sigma_{i_{1}}\cdots\sigma_{i_{k}}$
(2.5)			$\displaystyle=\frac{1}{K}+\sum_{k=2}^{d}(-\alpha)^{k}\sum_{i_{1}<...<i_{k}}\sigma_{i_{1}}\cdots\sigma_{i_{k}},$

where we have used the fact that the trace of $B(\vec{x})$ is $1$ together with the definition of $\alpha$ in the final line. We must show that the summation is nonnegative; the obstacle is its alternating nature. For notation, define the symmetric polynomials

$$P_{k}(\sigma_{1},...,\sigma_{d})=\sum_{i_{1}<...<i_{k}}\sigma_{i_{1}}\cdots\sigma_{i_{k}}.$$

Recalling that $\sigma_{1}+\cdots+\sigma_{d}=1$ and that each $\sigma_{i}\geq 0$, we have

	$\displaystyle P_{k}(\sigma_{1},...,\sigma_{d})$	$\displaystyle=P_{k}(\sigma_{1},...,\sigma_{d})\cdot(\sigma_{1}+\cdots+\sigma_{d})$
		$\displaystyle=\sum_{j=1}^{d}\sum_{i_{1}<...<i_{k}}\sigma_{i_{1}}\cdots\sigma_{i_{k}}\sigma_{j}$
		$\displaystyle=(k+1)\sum_{j_{1}<...<j_{k+1}}\sigma_{j_{1}}\cdots\sigma_{j_{k+1}}+\sum_{\text{duplicated indices}}\sigma_{\ell_{1}}\cdots\sigma_{\ell_{k+1}}$
		$\displaystyle\geq(k+1)P_{k+1}(\sigma_{1},...,\sigma_{d}).$

The third line follows from the fact that each $(k+1)$-tuple can be decomposed as a $k$-tuple and a single index in $(k+1)$ ways, and the fourth line uses the definition of $P_{k+1}$ together with the fact that every summand is nonnegative. In fact, we only need the weaker inequality $P_{k}(\sigma_{1},...,\sigma_{d})\geq P_{k+1}(\sigma_{1},...,\sigma_{d})$; since $\alpha\in(0,1)$ and the first term is nonnegative, it follows that

$$\sum_{k=2}^{d}(-\alpha)^{k}P_{k}(\sigma_{1},...,\sigma_{d})\geq 0.$$

Inserting this into (2.2) shows that $\det(I-\alpha B(\vec{x}))\geq\frac{1}{K}$, whose combination with with (2.4) gives

$$\|I-\alpha B(\vec{x})\|^{d}\leq 1\leq K\det(I-\alpha B(\vec{x}))$$

for all $\vec{x}\notin\Lambda$. This yields the distortion inequality (2.3) and completes the verification that $F_{\Lambda}$ is $K$-quasiconformal.

All that remains now is to study the stretching behavior on $\Lambda$; this will follow the lines of the argument in [4], but there are complications in working on $\mathbb{R}^{d}$ rather than $\mathbb{C}$. To begin, fix an index $N$; we will show that for all sufficiently small $r$ that

(2.6)

$$|F_{\Lambda}(\lambda_{N}+re_{1})-F_{\Lambda}(\lambda_{N})|\gtrsim r^{1/K}.$$

After translations, we may assume that $\lambda_{N}=\vec{0}$. Returning to the definition (2.2), we have

	$\displaystyle F_{\Lambda}(r\vec{e}_{1})-F_{\Lambda}(\vec{0})$	$\displaystyle=\sum_{n=1}^{\infty}\frac{1}{2^{n}}\left(S_{\lambda_{n}}(r\vec{e}_{1})-S_{\lambda_{n}}(\vec{0})\right)$
		$\displaystyle=\frac{1}{2^{N}}(r\vec{e}_{1})|r\vec{e}_{1}|^{1/K-1}-\sum_{n\neq N}\frac{1}{2^{n}}\left(S_{\lambda_{n}}(r\vec{e}_{1})-S_{\lambda_{n}}(\vec{0})\right)$
		$\displaystyle=\frac{r^{1/K}}{2^{N}}\vec{e}_{1}-\sum_{n\neq N}\frac{1}{2^{n}}\left(S_{\lambda_{n}}(r\vec{e}_{1})-S_{\lambda_{n}}(\vec{0})\right).$

We will show that if $r$ is sufficiently close to zero, then the latter term is negligible in comparison to the first term.

Roughly speaking, there are two kinds of points $\lambda_{n}$ which we will consider. First, there are points with “small” indices $n$; these points can be treated as far away from $\lambda_{N}=\vec{0}$, in which case we use the Lipschitz nature of $S$. Secondly, there are points with “large” indices $n$ which may be arbitrarily close to $\lambda_{N}$; in this case, we use the exponential decay and a basic Hölder continuity estimate. We will take the cutoff point to an index $N^{*}$ to be specified later. We therefore need to establish a pair of inequalities: for each $\epsilon>0$, there exists an $r^{*}>0$ such that whenever $0<r<r^{*}$ we have

(2.7)

$$\sum_{\begin{subarray}{c}1\leq n<N^{*}\\ n\neq N\end{subarray}}\frac{1}{2^{n}}\left|S_{\lambda_{n}}(r\vec{e}_{1})-S_{\lambda_{n}}(\vec{0})\right|\leq\epsilon r^{1/K}$$

and

(2.8)

$$\sum_{n=N^{*}}^{\infty}\frac{1}{2^{n}}\left|S_{\lambda_{n}}(r\vec{e}_{1})-S_{\lambda_{n}}(\vec{0})\right|\leq\epsilon r^{1/K}.$$

In order to establish these inequalities, we need to study the precise behavior of the stretching maps $S_{\lambda}$. We begin with studying the stretch in a particular direction adapted to $\lambda$:

	$\displaystyle\left|S_{\lambda}\left(r\frac{\lambda}{|\lambda|}\right)-S_{\lambda}(\vec{0})\right|$	$\displaystyle=\left|\left(r\frac{\lambda}{|\lambda|}-\lambda\right)\left|r\frac{\lambda}{|\lambda|}-\lambda\right|^{1/K-1}-(-\lambda)|-\lambda|^{1/K-1}\right|$
		$\displaystyle=\left|\frac{\lambda}{|\lambda|}(r-|\lambda|)\cdot\left|\frac{\lambda}{|\lambda|}(r-|\lambda|)\right|^{1/K-1}+\frac{\lambda}{|\lambda|}|\lambda|^{1/K}\right|$
		$\displaystyle=\left|(r-|\lambda|)\cdot|r-|\lambda||^{1/K-1}+|\lambda|^{1/K}\right|$
		$\displaystyle=\left||\lambda|\left(\frac{r}{|\lambda|}-1\right)\cdot|\lambda|^{1/K-1}\left|\frac{r}{|\lambda|}-1\right|^{1/K-1}+|\lambda|^{1/K}\right|$
		$\displaystyle=|\lambda|^{1/K}\left|\left(\frac{r}{|\lambda|}-1\right)\left|\frac{r}{|\lambda|}-1\right|^{1/K-1}+1\right|$
(2.9)			$\displaystyle=|\lambda|^{1/K}\left|1+\frac{r/|\lambda|-1}{|r/|\lambda|-1|}\cdot|r/|\lambda|-1|^{1/K}\right|.$

Set $t=r/|\lambda|\in(0,\infty)$, in which case we have

$$f(t):=\left|S_{\lambda}\left(r\frac{\lambda}{|\lambda|}\right)-S_{\lambda}(0)\right|=|\lambda|^{1/K}\left|1+\frac{t-1}{|t-1|}|t-1|^{1/K}\right|.$$

We now separate into two cases based on the relative scale of $r$ and $|\lambda|$:

• •

  If $t\geq 1$, then

  $$f(t)=|\lambda|^{1/K}\left(|t-1|^{1/K}+1\right)\lesssim|\lambda|^{1/K}t^{1/K}.$$

• •

  If $t\in(0,1)$, then

  $$f(t)=|\lambda|^{1/K}(1-(1-t)^{1/K}).$$

  Taking the Taylor expansion around $0$, we have

  $$1-(1-t)^{1/K}=1-\left[1-\frac{1}{K}t+O(t^{2})\right]=\frac{1}{K}t+O(t^{2}).$$

  This leads to $f(t)\lesssim|\lambda|^{1/K}t$.

In each case, we have that $f(t)\lesssim\min\{t,t^{1/K}\}.$ Returning to (2.3), we have established that

(2.10)

$$\left|S_{\lambda}\left(r\frac{\lambda}{|\lambda|}\right)-S_{\lambda}(\vec{0})\right|\lesssim|\lambda|^{1/K}\min\left\{\frac{r}{|\lambda|},\left(\frac{r}{|\lambda|}\right)^{1/K}\right\}.$$

In order to establish the inequalities (2.7) and (2.8), we need a slight extension of this, in that the unit vector $\lambda/|\lambda|$ needs to be replaced by an arbitrary unit vector $\vec{u}$. That is, we need to show that

(2.11)

$$|S_{\lambda}(r\vec{u})-S_{\lambda}(\vec{0})|\lesssim|\lambda|^{1/K}\min\left\{\frac{r}{|\lambda|},\left(\frac{r}{|\lambda|}\right)^{1/K}\right\}.$$

However, this follows immediately from the fact that a (global) quasiconformal map on $\mathbb{R}^{d}$ is also quasisymmetric.

We may now complete the analysis by selecting the radius scale $r^{*}$ and the cutoff $N^{*}$. Recall that by (2.11), there is a constant $C<\infty$ independent of $\lambda$ for which

(2.12)

$$\left|S_{\lambda}(r\vec{e}_{1})-S_{\lambda}(\vec{0})\right|\leq C\min\{r^{1/K},r\cdot|\lambda|^{1/K-1}\}.$$

Set the cutoff $N^{*}=N+A$, where $A$ is large enough that $C/2^{N+A-1}<\epsilon$; in this case, we use the first alternative of (2.12) to get

$\displaystyle\sum_{n=N^{*}}^{\infty}\frac{1}{2^{n}}|S_{\lambda_{n}}(r\vec{e}_{1})-S_{\lambda_{n}}(\vec{0})|$

$\displaystyle\leq\sum_{n=N+A}^{\infty}\frac{1}{2^{n}}\cdot Cr^{1/K}=\frac{2C}{2^{N+A}}r^{1/K}<\epsilon r^{1/K}.$

This establishes (2.7) as desired.

Finally, we need to handle small indices as well. For this, use the linear estimate available in (2.12):

	$\displaystyle\sum_{\begin{subarray}{c}1\leq n\leq N^{*}\\ n\neq N\end{subarray}}\frac{1}{2^{n}}|S_{\lambda_{n}}(r\vec{e}_{1})-S_{\lambda_{n}}(\vec{0})|$	$\displaystyle\leq\sum_{\begin{subarray}{c}1\leq n\leq N^{*}\\ n\neq N\end{subarray}}\frac{1}{2^{n}}Cr|\lambda_{n}|^{1/K-1}$
		$\displaystyle\leq Cr\max_{\begin{subarray}{c}1\leq n\leq N^{*}\\ n\neq N\end{subarray}}|\lambda_{n}|^{1/K-1}$
		$\displaystyle=Cr^{1/K}\max_{\begin{subarray}{c}1\leq n\leq N^{*}\\ n\neq N\end{subarray}}\left(\frac{r}{|\lambda_{n}|}\right)^{1-1/K}$
(2.13)			$\displaystyle=Cr^{1/K}\left(\frac{r}{\min\limits_{\begin{subarray}{c}1\leq n\leq N^{*}\\ n\neq N\end{subarray}}|\lambda_{n}|}\right)^{1-1/K}.$

Note that since $n\neq N$ and $N^{*}<\infty$, there is a uniform, positive lower bound on $|\lambda_{n}|$ for this context. Let $\rho>0$ be this minimum radius and set

$$r^{*}=\rho\left(\frac{\epsilon}{C}\right)^{1/(1-1/K)}.$$

Note that the selection of $\rho$ depends only on $\epsilon$, the set $\Lambda$, and the value $N^{*}$; but $N^{*}$ depends only on the constant in inequality (2.11), which ultimately only depends on $K$. That is, the selection of $r^{*}$ depends only on $\epsilon$, $\Lambda$, and $K$ and is independent of $N$. Inequality (2.8) now follows immediately from inserting the selection of $r^{*}$ into (2.3), completing the analysis of the stretching at $\Lambda_{N}$ and proving (2.6). Since $N\in\mathbb{N}$ was arbitrary, the proof is completed.