Antti Käenmäki

This blog post exhibits a recent work with Tuomo Ojala and Eino Rossi. The purpose of the paper is to consider the behavior of quasisymmetric mappings on self-affine carpets. More precisely, we study what kind of restrictions does the structure of self-affine carpets $E$ and $F$ give to quasisymmetric mappings $f \colon E \to F$ . Furthermore, we show that such self-affine carpets are minimal for the conformal dimension.

If $(X,d)$ and $(Y,\varrho)$ are metric spaces and $\eta \colon [0,\infty) \to [0,\infty)$ is a homeomorphism, then a homeomorphism $f \colon X \to Y$ is $\eta$ –quasisymmetric if

$\displaystyle \frac{\varrho(f(x),f(y))}{\varrho(f(x),f(z))} \le \eta\biggl( \frac{d(x,y)}{d(x,z)} \biggr)$

for all $x,y,z \in X$ with $x \ne z$ . Quasisymmetric mappings generalize bi-Lipschitz mappings, but compared to arbitrary homeomorphisms, their local behavior has some control:

Lemma. If $f \colon X \to Y$ is an $\eta$ -quasisymmetric mapping, then for each $x \in X$ and $r \;\textgreater\; 0$ there is $t \;\textgreater\; 0$ such that

$\displaystyle B(f(x),t) \subset f(B(x,r)) \subset B(f(x),2\eta(1)t).$

Proof. Choose $z,y \in Y$ as in the following picture:

quasisymmetric

Let $t = \varrho(f(x),z)$ . Then we clearly have

$\displaystyle B(f(x),t) \subset f(B(x,r)) \subset B(f(x),2\varrho(f(x),y)).$

This gives the claim since

$\displaystyle \varrho(f(x),y) = \frac{\varrho(f(x),y)}{\varrho(f(x),z)} t \le \eta\biggl( \frac{d(x,f^{-1}(y))}{d(x,f^{-1}(z))} \biggr) t \le \eta(1) t. \;\blacksquare$

A metric space is uniformly perfect if there exists a constant $D \ge 1$ such that $B(x,r) \setminus B(x,r/D) \ne \emptyset$ for all $x \in X$ and $r \;\textgreater\; 0$ whenever $X \setminus B(x,r) \ne \emptyset$ . The following lemma further describes properties of quasisymmetric mappings:

Lemma. If $X$ and $Y$ are uniformly perfect bounded metric spaces and $f \colon X \to Y$ is a quasisymmetric mapping, then there are $\Lambda \ge \lambda \;\textgreater\; 0$ such that

$\displaystyle cd(x,y)^\Lambda \le \varrho(f(x),f(y)) \le Cd(x,y)^\lambda$

for all $x,y \in X$ .

The Assouad dimension of a set $E$ , denoted by $\dim_{\mathrm{A}}(E)$ , is the infimum of all $s$ satisfying the following: there exists a constant $C \ge 1$ such that each set $E \cap B(x,R)$ can be covered by at most $C(R/r)^t$ balls of radius $r$ centered at $E$ for all $0 \;\textless\; r \;\textless\; R$ . The definition is due to Assouad (1983). Recall that the Hausdorff dimension of $E$ , $\dim_{\mathrm{H}}(E)$ , is defined by using arbitrary covers and the upper Minkowski dimension of $E$ , $\overline{\dim}_{\mathrm{M}}(E)$ , uses covers consisting of balls with a fixed radius. The upper Minkowski dimension is therefore related to the average small scale structure whereas the Assouad dimension depends on the extreme properties of the set and takes into account all scales. In fact, if $E \subset X$ is compact, then $\dim_{\mathrm{H}}(E) \le \overline{\dim}_{\mathrm{M}}(E) \le \dim_{\mathrm{A}}(E)$ . Furthermore, Assouad dimension is invariant under bi-Lipschitz mappings but, contrary to the Hausdorff dimension and the upper Minkowski dimension, it may increase under Lipschitz mappings.

A metric space $X$ is doubling if there exists $N \in \mathbb{N}$ such that each ball $B(x,2r)$ can be covered by at most $N$ many balls of radius $r$ . If $X$ is a metric space, then $\dim_{\mathrm{A}}(X) \;\textless\; \infty$ if and only if $X$ is doubling.

Assouad’s embedding theorem states that for every $N \in \mathbb{N}$ and $\tfrac{1}{2} \;\textless\; \varepsilon \;\textless\; 1$ there are $D=D(N) \in \mathbb{N}$ and $L = L(N,\varepsilon) \;\textgreater\; 1$ such that if $(X,d)$ is an $N$ -doubling metric space, then $(X,d^\varepsilon)$ admits a $L$ -bi-Lipschitz embedding into $\mathbb{R}^D$ . This version of the theorem is due to Naor & Neiman (2012).

A set $E \subset \mathbb{R}^d$ is porous if there exists $\tfrac12 \;\textless\; \alpha \;\textless\; 1$ such that for each $x \in X$ and $r \;\textgreater\; 0$ there is $y$ such that $B(y,\alpha r) \subset B(x,r) \setminus E$ . If $E \subset \mathbb{R}^d$ , then $\dim_{\mathrm{A}}(E) \;\textless\; d$ if and only if $E$ is porous.

The conformal Assouad dimension of $E$ is

$\displaystyle \mathcal{C}\dim_{\mathrm{A}}(E) = \inf\{ \dim_{\mathrm{A}}(E') : E' \text{ is a quasisymmetric image of } E \}.$

The Assouad dimension above can of course be replaced by any set dimension, and this gives rise to other conformal dimensions. The definition is due to Pansu (1989). The conformal dimension measures the dimension of the “best shape” of the metric space. We say that a set $E$ is minimal for the conformal Assouad dimension if

$\displaystyle \mathcal{C}\dim_{\mathrm{A}}(E) \ge \dim_{\mathrm{A}}(E).$

Note that we trivially have $\mathcal{C}\dim_{\mathrm{A}}(E) \le \dim_{\mathrm{A}}(E)$ .

If $E \subset \mathbb{R}$ is porous, then $\mathcal{C}\dim_{\mathrm{H}}(E)=0$ . A nontrivial generalization of this is the result of Kovalev (2006) which states that a metric space $X$ with $\dim_{\mathrm{H}}(X) \;\textless\; 1$ has $\mathcal{C}\dim_{\mathrm{H}}(X)=0$ . On the other hand, Tyson (2000) has shown that for each $1 \le s \le d$ there is $E \subset \mathbb{R}^d$ such that $\mathcal{C}\dim_{\mathrm{H}}(E) = \dim_{\mathrm{H}}(E) = s$ , so $E$ is minimal for the conformal Hausdorff dimension.

Recall that bi-Lipschitz mappings preserve the dimension. By Marczewski’s result (1937), the topological dimension $\dim_{\mathrm{T}}(E)$ of a separable set $E$ is

$\displaystyle \dim_{\mathrm{T}}(E) = \inf\{ \dim_{\mathrm{H}}(E') : E' \text{ is a homeomorphic image } E \}.$

Therefore,

$\displaystyle \dim_{\mathrm{T}}(E) \le \mathcal{C}\dim_{\mathrm{H}}(E) \le \dim_{\mathrm{H}}(E).$

This observation gives an intuition that a quasisymmetric mapping defined on a minimal set behaves as an average like a bi-Lipschitz mapping, and defined on other kinds of sets, starts to resemble a general homeomorphism. Also, since $\dim_{\mathrm{T}}(\mathbb{R}^d) = d$ , it holds that $\mathcal{C}\dim_{\mathrm{H}}(\mathbb{R}^d) = \mathcal{C}\dim_{\mathrm{A}}(\mathbb{R}^d) = d$ .

Our main results are the following two theorems:

Theorem A (Ojala & Rossi & K., 2018). If $E,F \subset \mathbb{R}^2$ are horizontal self-affine carpets, then any quasisymmetric mapping $E \to F$ is quasi-Lipschitz.

A mapping $f \colon X \to Y$ is quasi-Lipschitz if

$\displaystyle \frac{\log \varrho(f(x),f(y))}{\log d(x,y)} \to 1$

uniformly as $d(x,y) \to 0$ . If $X$ and $Y$ are separable, then quasi-Lipschitz mappings preserve the Hausdorff dimension. Therefore, by Theorem A, two horizontal self-affine carpets either have the same dimension or there is no quasi-symmetric mapping between them. This is in a huge contrast to the self-similar case: Wang, Wen, and Zhu (2010) have shown that if $E$ and $F$ are self-similar sets satisfying the strong separation condition, then there exists a quasisymmetric mapping $E \to F$ . So in fact, $\mathcal{C}\dim_{\mathrm{H}}(E)=0$ for all such self-similar sets $E$ . Furthermore, Bonk & Merenkov (2013) have proved that every quasisymmetric self-map of the standard Sierpiński carpet is an isometry.

Theorem B (Ojala & Rossi & K., 2018). Horizontal self-affine carpets are minimal for the conformal Assouad dimension.

A set $E \subset \mathbb{R}^2$ is a horizontal self-affine carpet if it satisfies the strong separation condition, the “construction rectangles” are horizontal, and each vertical line intersects at least two such rectangles:

horizontal

Mackay (2011) has shown that Gatzouras-Lalley carpets are minimal for the conformal Assouad dimension when they project to a line; otherwise the conformal Assouad dimension is zero. Theorem B strictly generalizes the result of Mackay. It should also be mentioned that since the codomains in the definition of the conformal dimension can be any metric spaces, Theorem A cannot be used to prove Theorem B.

The proof of Theorem A has the following three essential steps:

Theorem 1 (Ojala & Rossi & K., 2018). Let $X$ and $Y$ be uniformly perfect compact doubling metric spaces. If $f \colon X \to Y$ is a quasisymmetric mapping such that any quasisymmetric weak tangent mapping is bi-Lipschitz (with the same constants), then $f$ is quasi-Lipschitz.

Let $E \subset [0,1]^d$ be compact. We say that $W$ is a weak tangent of $E$ if there exists a sequence of similarity mappings $S_n \colon \mathbb{R}^d \to \mathbb{R}^d$ such that

$\displaystyle S_n(E) \to W$

in the Hausdorff metric as $n \to \infty$ . The above definition is given in $\mathbb{R}^d$ only for simplicity. The general case can be handled by the pointed Gromov-Hausdorff convergence and the Assouad embedding. Let $f \colon E \to E'$ be a quasisymmetric mapping. If $W'$ is a tangent set of $E'$ (and $S_n'$ the associated similarities), then, by the Arzelà-Ascoli theorem, $S_n' \circ f \circ S_n^{-1}$ converge (along a subsequence) to a quasisymmetric weak tangent mapping $\hat f \colon W \to W'$ .

The idea of the proof of Theorem 1 is as follows: If $f$ is not quasi-Lipschitz, then there exists a sequence of pairs of points in which $f$ obeys true Hölder behavior. For each pair there is a triplet of points (with comparable distances) in which $f$ still obeys true Hölder behavior. These triplets then allow us to define weak tangents so that the limit mappings are quasisymmetric and obey true Hölder behavior. This is a contradiction with the bi-Lipschitz assumption.

Let us sketch the outline of the proof. If $f$ is not quasi-Lipschitz, then there are $\kappa>0$ and sequences $(x_i)$ and $(y_i)$ such that

$\displaystyle \varrho(f(x_i),f(y_i)) \;\textless\; d(x_i,y_i)^{1+\kappa}$

for all $i$ and $d(x_i,y_i) \to 0$ as $i \to \infty$ . Relying on uniform perfectness, choose $z_i^j$ such that $z_i^0 = y_i$ and

$\displaystyle d(x_i,y_i) \approx \varepsilon d(x_i,z_i^1) \approx \varepsilon^2d(x_i,z_i^2) \approx \cdots$

We assume that

$\displaystyle \varrho(f(x_i),f(z_i^{j-1})) \ge \varepsilon^{1+\kappa/2} \varrho(f(x_i),f(z_i^j))$

for all $j$ . This would mean that the distances $\varrho(f(x_i),f(z_i^j))$ are above the graph of $w \mapsto w^{1+\kappa/2}$ :

quasilip

By Hölder continuity and the estimate for $f(y_i)$ , this is a contradiction. Hence there exists $j$ such that

$\displaystyle \varrho(f(x_i),f(z_i^{j-1})) \;\textless\; \varepsilon^{1+\kappa/2} \varrho(f(x_i),f(z_i^j)).$

Let us denote $(x_i,z_i^{j-1},z_i^j)$ by $(a_i,b_i,c_i)$ . Define weak tangent sets and mapping $\hat f$ by using magnifications which take $d(a_i,c_i)$ and $\varrho(f(a_i),f(c_i))$ into unit lengths. Let $a,b,c$ be such that $a_i\to a$ , $b_i\to b$ , and $c_i\to c$ . Recalling that $\hat f$ is bi-Lipschitz, we get

$\displaystyle \varepsilon = \varepsilon d(a,c) \approx d(a,b) \approx \varrho(\hat f(a),\hat f(b)) \approx \frac{\varrho(f(a_i),f(b_i))}{\varrho(f(a_i),f(c_i))} \;\textless\; \varepsilon^{1+\kappa/2},$

which is a contradiction for small enough $\varepsilon \;\textgreater\; 0$ . $\blacksquare$

To prove Theorem A, the task is now to find geometric conditions under which the assumptions of Theorem 1 are satisfied. This is given by the result of Le Donne & Xie (2016):

Theorem 2 (Le Donne & Xie, 2016). If $X$ and $Y$ are fibered spaces and $f \colon X \to Y$ is a quasisymmetric mapping such that the fibers of $X$ go homeomorphically to fibers of $Y$ , then $f$ is bi-Lipschitz.

A metric space $X$ is fibered if $X = \bigcup_i F_i$ , where the fibers $F_i$ are unbounded geodesic spaces and they have positive distance. Furthermore, non-parallel fibers are assumed to diverge and parallel fibers are not isolated.

Theorem A follows now from Theorems 1 and 2 if weak tangents of horizontal self-affine carpets are unions of fibered spaces. This is the content of the last step of the proof:

Theorem 3 (Ojala & Rossi & K., 2018). If $E$ is horizontal self-affine carpet, then weak tangents of $E$ are of the form

$\displaystyle (-\infty,\omega] \times C_{\mathrm{left}} \cup [\omega,\infty) \times C_{\mathrm{right}},$

where $C_{\mathrm{left}}$ and $C_{\mathrm{right}}$ are uniformly perfect porous sets.

Theorem 3 generalizes the result of Bandt & K. (2013) which describes tangent sets at generic points under some assumptions. The result of Bandt & K. (2013) has been recently generalized for non-carpet self-affine sets by Koivusalo & Rossi & K. (2017).

The proof of Theorem B relies on the following two results. The first one guarantees that the weak tangents of a horizontal self-affine carpet are minimal.

Theorem 4 (Bishop & Tyson, 2001). If $C \subset \mathbb{R}^{d-1}$ is compact, then $[0,1] \times C \subset \mathbb{R}^d$ is minimal for the conformal Hausdorff dimension.

The second one states that we can always find a weak tangent set having maximal dimension.

Theorem 5 (Furstenberg, 2008). If $E \subset \mathbb{R}^d$ is compact, then there exists a weak tangent $W$ of $E$ such that

$\displaystyle \dim_{\mathrm{H}}(W) = \dim_{\mathrm{A}}(E).$

Proof of Theorem B. Let $E \subset \mathbb{R}^2$ be a horizontal self-affine carpet. By Theorem 5, there exists a weak tangent $W$ of $E$ such that

$\displaystyle \dim_{\mathrm{A}}(E) = \dim_{\mathrm{H}}(W).$

Recall that, by Theorem 3, $W$ is a union of fibered spaces. Therefore, by Theorem 4, $W$ is minimal for the conformal Hausdorff dimension, i.e.

$\displaystyle \dim_{\mathrm{H}}(W) \le \mathcal{C}\dim_{\mathrm{H}}(W) \le \mathcal{C}\dim_{\mathrm{A}}(W).$

Recalling the definition of the Assouad dimension, it is easy to see that $\dim_{\mathrm{A}}(W) \le \dim_{\mathrm{A}}(E)$ and, similarly,

$\displaystyle \mathcal{C}\dim_{\mathrm{A}}(W) \le \mathcal{C}\dim_{\mathrm{A}}(E).$

The proof follows now by combining these inequalities. $\blacksquare$

The area of the unit circle can be estimated from below and from above by rectangles and octagons as illustrated in the following picture:

ball-area-estimate

This idea can of course be continued to obtain better and better estimates. But how to calculate the area precisely? In other words, how to derive the familiar formula for the area of a circle?

Besides learning how to write LaTeX in WordPress, the purpose of this first blog post is to demonstrate how the exact area can be calculated. Mathematics used here should be accessible to anyone familiar with basic trigonometric identities and concepts such as continuity, derivative, and integral.

Recall that the integral $\int_a^b f(t) \,\mathrm{d}t$ of a positive function $f$ is, by definition, the area beneath the graph of $f$ between $a$ and $b$ . Loosely speaking, the integral is defined to be the limit of the area of better and better approximating rectangles. The definition is therefore unsuitable for calculating the exact values of integrals in most cases.

fundamental-theorem-of-calculus

Assuming $a<x<x+h$ and $f$ to be a continuous function as in the above picture, we see that

$\displaystyle hf(x) \le \int_a^{x+h} f(t) \,\mathrm{d}t - \int_a^x f(t) \,\mathrm{d}t \le hf(x+h)$

and therefore, by the continuity of $f$ ,

$\displaystyle 0 \le \frac{1}{h}\biggl(\int_a^{x+h} f(t) \,\mathrm{d}t - \int_a^x f(t) \,\mathrm{d}t\biggr) - f(x) \le f(x+h) - f(x) \to 0$

as $h \to 0$ . So, writing $G(x) = \int_a^x f(t) \,\mathrm{d}t$ , we have

$\displaystyle f(x) = \lim_{h \to 0} \frac{G(x+h) - G(x)}{h} = G'(x).$

Let us formulate our observation as a theorem. Note that, for simplicity, we assumed $f$ to be positive and increasing on a neighborhood of $x$ . To get rid of these assumptions is left as an exercise.

Theorem 1 (Fundamental Theorem of Calculus). Let $f \colon [a,b] \to \mathbb{R}$ be a continuous function. If $G \colon [a,b] \to \mathbb{R}$ is a function such that

$\displaystyle G(x) = \int_a^x f(t) \,\mathrm{d}t$

for all $x \in [a,b]$ , then $G$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and

$\displaystyle G'(x) = f(x)$

for all $x \in (a,b)$ .

The following corollary introduces us a method to calculate exact values of integrals.

Corollary 2. Let $f \colon [a,b] \to \mathbb{R}$ be a continuous function. If there exists a continuous function $F \colon [a,b] \to \mathbb{R}$ that satisfies $F'(x) = f(x)$ for all $x \in (a,b)$ , then

$\displaystyle \int_a^b f(t) \,\mathrm{d}t = F(b) - F(a).$

Proof. Let $G(x) = \int_a^x f(t) \,\mathrm{d}t$ for all $x \in [a,b]$ . By the fundamental theorem of calculus, also $G'(x) = f(x)$ for all $x \in (a,b)$ . Therefore, by the linearity of the derivative operation,

$\displaystyle (F-G)'(x) = F'(x) - G'(x) = 0$

for all $x \in (a,b)$ . The mean value theorem and the continuity of $F$ and $G$ now imply that there exists a constant $c$ such that $(F-G)(x) = c$ for all $x \in [a,b]$ . Choosing $x=a$ , we see that

$\displaystyle F(a) - c = G(a) = \int_a^a f(t) \,\mathrm{d}t = 0$

and therefore, $c = F(a)$ . In other words, $(F-G)(x) = F(a)$ and $G(x) = F(x) - F(a)$ for all $x \in [a,b]$ , and so

$\displaystyle \int_a^b f(t) \,\mathrm{d}t = G(b) = F(b) - F(a). \;\blacksquare$

Example 3. Let $F(x) = \tfrac{1}{2}(x + \sin x \cos x)$ and observe that

$\displaystyle F'(x) = \tfrac{1}{2}(1-\sin^2x+\cos^2x) = \cos^2x.$

Therefore,

$\displaystyle \int_0^{\pi/2} \cos^2t \,\mathrm{d}t = F(\pi/2) - F(0) = \pi/4.$

In the previous example, by first defining $F$ , it was easy to find out its derivative $f$ and then calculate the integral of $f$ . We are, however, almost always interested in calculating the integral of a given function $f$ . The challenge thus is to find its antiderivative $F$ . For example, if $f(x) = \cos^2x$ , then how do we find $F$ for which $F' = f$ ?

To overcome this problem one often transfers the integral into a form which is easier to compute. The following substitution method is one way to implement this idea.

Theorem 4. Let $\varphi \colon [a,b] \to [c,d]$ be continuous on $[a,b]$ and differentiable on $(a,b)$ such that its derivative is continuous. If $f \colon [c,d] \to \mathbb{R}$ is continuous, then

$\displaystyle \int_{\varphi(a)}^{\varphi(b)} f(s)\,\mathrm{d}s = \int_a^b f(\varphi(t)) \varphi'(t) \,\mathrm{d}t.$

Proof. By the fundamental theorem of calculus, there exists continuous $F \colon [c,d] \to \mathbb{R}$ such that $F'(x) = f(x)$ for all $x \in (c,d)$ . Corollary 2 thus implies

$\displaystyle \int_{\varphi(a)}^{\varphi(b)} f(s)\,\mathrm{d}s = F(\varphi(b)) - F(\varphi(a)) = (F \circ \varphi)(b) - (F \circ \varphi)(a).$

Since, by the chain rule,

$\displaystyle (F \circ \varphi)'(t) = F'(\varphi(t)) \varphi'(t) = f(\varphi(t)) \varphi'(t),$

Corollary 2 also gives

$\displaystyle \int_a^b f(\varphi(t)) \varphi'(t) \,\mathrm{d}t = \int_a^b (F \circ \varphi)'(t) \,\mathrm{d}t = (F \circ \varphi)(b) - (F \circ \varphi)(a).$

This finishes the proof. $\blacksquare$

Example 5. Let $\varphi \colon [0,\pi/2] \to [0,1]$ , $\varphi(t) = \sin t$ , and

$\displaystyle f \colon [0,1] \to \mathbb{R}$ , $f(s) = \sqrt{1-s^2}.$

Since $f(\varphi(t)) = \sqrt{1-\sin^2t}$ and $\varphi'(t) = \cos t$ , Theorem 4 shows that

$\displaystyle \int_0^1 \sqrt{1-s^2} \,\mathrm{d}s = \int_0^{\pi/2} \sqrt{1-\sin^2t} \cos t \,\mathrm{d}t = \int_0^{\pi/2} \cos^2t \,\mathrm{d}t.$

We determined the above integral in Example 3, so we conclude that

$\displaystyle \int_0^1 \sqrt{1-s^2} \,\mathrm{d}s = \pi/4.$

quarter-circle

Observe that $(x,y)$ is contained in the unit circle if and only if $x^2+y^2 \leq 1$ . For $x,y \in [0,1]$ , this is equivalent to $y \le \sqrt{1-x^2}$ . Therefore, the integral $\int_0^1 \sqrt{1-s^2} \,\mathrm{d}s$ is the area of the intersection of the unit circle and the first quadrant. Since, by symmetry, the area of the unit circle is four times this number, an application of Example 5 shows that the area of the unit circle is $\pi$ .

To finish the discussion let us remark that this method can easily be adapted to show that the area of a circle of radius $r>0$ is $\pi r^2$ . This is also left as an exercise.

Rigidity of quasisymmetric mappings on self-affine carpets

Fundamental theorem of calculus and the area of the unit circle