Singular points of Hölder asymptotically optimally doubling measures

Abstract

We consider the question of how the doubling characteristic of a measure determines the regularity of its support. The question was considered by David et al. (Commun Pure Appl Math 54:385–449, 2001) for codimension 1 under a crucial assumption of flatness, and later by Preiss et al. (Calc Var PDE’s 35:339–363, 2009) in higher codimension. However, these studies leave open the question of what can be said about the geometry of the support of such measures in a neighborhood about a nonflat point. Here we answer the question for codimension-1 Hölder doubling measures in $\mathbb {R}^4$ by constructing parametrizations in a neighborhood of a nonflat point by the Kowalski–Preiss cone. These parametrizations extend to be $C^{1,\beta }$ diffeomorphisms on an open set. Some of our parametrization techniques build on ideas from: David et al. (Mem AMS 215, 2012; Commun Pure Appl Math 54:385–449, 2001), Preiss et al. (Calc Var PDE’s 35:339–363, 2009) and Taylor (Annals Math 103:489–539, 1976).

Appendix

Lemma 29

Suppose that $U,V\subseteq \mathbb {R}^n$ are open sets, $\Gamma \subseteq \mathbb {R}^n$, $0<m<n$, and $\psi \in C^2(U,V)$ is bijective and satisfies

$$\begin{aligned} 0<\lambda \leqslant \frac{|\psi (x)-\psi (y)|}{|x-y|}\leqslant \Lambda \quad \text{ for } \text{ all } x,y\in U \quad \text{ and } \quad ||D^2\psi ^a||_\infty = \sup _{x\in U}||D_x^2\psi ||<\infty . \end{aligned}$$

(298)

Let $z\in \Gamma \cap U$, $B(z,r)\subseteq U$, P be a plane of dimension m through z, and set $\widetilde{P} = D_z\psi (P-z) + \psi (z)$, $\widetilde{\Gamma }= \psi (\Gamma )$. Then

$$\begin{aligned} d^{\psi (z),\lambda r}\left( \widetilde{\Gamma },\widetilde{P}\right) \leqslant \frac{||D^2 \psi ||_\infty }{2\lambda }r + \frac{\Lambda }{\lambda }d^{z,r}\left( \Gamma ,P\right) . \end{aligned}$$

(299)

Proof

Without loss of generality, take $z = \psi (z) = 0$. Let $P\subseteq \mathbb {R}^n$ be an m-plane through 0 and set $d = d^{0,r}\left( \Gamma ,P\right) .$ Note that $\lambda \leqslant \frac{|D_0\psi (v) |}{|v|} \leqslant \Lambda$ for all $v\in \mathbb {R}^n {\setminus } \{0\}$ by (298). Further, note that we have

$$\begin{aligned} B(0,\lambda r)\subseteq \psi (B(0,r)). \end{aligned}$$

(300)

Let $y\in \widetilde{\Gamma }\cap B(0,\lambda r)$. Then by (300) and bijectivity, we have that there exists $x\in B(0,r)\cap \Gamma$ such that $y=\psi (x)$. By $d = d^{0,r}\left( \Gamma ,P\right)$, we get that there exists $p\in P$ such that $|p-x|\leqslant rd.$ Let $\widetilde{p} = D_0\psi (p)\in \widetilde{P}$. We compute

$$\begin{aligned} |\widetilde{p}-y|&= |D_0\psi (p) - \psi (x)|\leqslant |D_0\psi (p) - \psi (p)| + |\psi (p) - \psi (x)|. \end{aligned}$$

(301)

By (298) and Taylor expansion, we get that

$$\begin{aligned} |\widetilde{p}-y|\leqslant \frac{||D^2\psi ||_\infty }{2}|p-0|^2 + \Lambda |p-x|\leqslant \frac{||D^2\psi ||_\infty }{2}r^2 + \Lambda rd. \end{aligned}$$

(302)

Because for every $y\in B(0,\lambda r)\cap \widetilde{\Gamma }$, there exists $\widetilde{p}\in \widetilde{P}$ satisfying (302), we get that

$$\begin{aligned} \widetilde{d}^{0,\lambda r}\left( \widetilde{\Gamma },\widetilde{P}\right) \leqslant \frac{||D^2\psi ||_\infty }{2\lambda }r + \frac{\Lambda }{\lambda } d . \end{aligned}$$

(303)

By Lemma 2, we get that (303) tells us (299). $\square$

Lemma 30

For $a\in \mathcal {C}{\setminus } \{0\}$, and $A>0$ large enough, there exists a neighborhood $U\supseteq B(a, 2|a|/{A})$, $V\subseteq \mathbb {R}^3$ open, $I\ni 0$ an open interval, and a smooth coordinate map $\psi ^a: U\rightarrow V\times I$ such that $V\times \{0\} = \psi ^a(\mathcal {C}\cap U)$ and $\widetilde{\pi }= \psi ^a\circ \pi \circ (\psi ^a)^{-1}$ is orthogonal projection onto $\mathbb {R}^3\times \{0\}$ (where $\pi$ is the same map defined in Sect. 5; see (189)). Further, $\psi ^a$ satisfies the estimates

$$\begin{aligned} \frac{1}{2}\leqslant \frac{|\psi ^a(x)-\psi ^a(y)|}{|x-y|}\leqslant 2\quad \text{ for } \text{ all } \,\,x,y\in U \end{aligned}$$

(304)

and

$$\begin{aligned} ||D^2\psi ^a||_\infty = \sup _{x\in U} ||D^2_x\psi ^a|| \leqslant \frac{C}{|a|} \end{aligned}$$

(305)

for some C independent of a.

Proof

First, we fix an $a\in \mathcal {C}$, $|a| = 1$. We define $\psi ^a$ by defining its inverse. Choose orthonormal coordinates $(z_1, z_2, z_3)$ on $T_a\mathcal {C}$ centered at a. Let p be orthogonal projection from $\mathcal {C}$ onto $T_a\mathcal {C}$, and take $U'\supseteq B(a, 8/A)$ to be an open set where $p^{-1}$ is defined. Identifying $T_a\mathcal {C}$ with $\mathbb {R}^3$ under the z coordinates, let $V = U'\cap T_a\mathcal {C}$. Let $I = (-8/A,8/A)$. Define for $(z,t)\in V\times I$

$$\begin{aligned} (\psi ^a)^{-1}(z,t) = p^{-1}(z) + t\eta _{p^{-1}(z)}. \end{aligned}$$

(306)

Assume that $A\geqslant 16$, so that $8/A\leqslant 1/2$. Note that $\psi ^a$ is bijective onto $U = (\psi ^a)^{-1}(V\times I)$, because $\eta$ is a smooth vector field (away from the $x_4$ axis), and the point (z, t) is the flow after time t of the point $p^{-1}z$ along the integral curves of $\eta$. Further, the same comments imply that it is smooth. Then we note that because $p^{-1}(z)\in \mathcal {C}$, $V\times \{0\} = \psi ^a(\mathcal {C}\cap U)$. Further, $\widetilde{\pi }(z,t) = \psi ^a(\pi ( p^{-1}(z) + t\eta _{p^{-1}(z)})) = \psi ^a(p^{-1}(z)) = z$, and so $\widetilde{\pi }$ is orthogonal projection onto $V\times \{0\}$.

We now show that (304) holds for $\psi ^a$ as long as $U'$ is chosen small enough and 1 / A is chosen small enough. Continuing the identifcation of $T_a\mathcal {C}$ with $\mathbb {R}^3$, we set $e_i$ to be the coordinate vector of $z_i$, and note that $D_a\psi ^a$ is the map

$$\begin{aligned} D_a\psi ^a(e_i) = e_i,\quad D_a\psi ^a \eta _a = e_4\quad \text{ for } \, i =1,2,3. \end{aligned}$$

(307)

Because the $z_i$ are orthonormal and $\measuredangle (\eta _a, T_a\mathcal {C}) = \pi /4$, we get that

$$\begin{aligned} \langle e_i, e_j\rangle = \delta _{ij}, \quad |\langle e_i, \eta _a\rangle |\leqslant 1/{\sqrt{2}}. \end{aligned}$$

(308)

From (307) and (308), as well as recalling that $|\nu _a| = 1$, we get that

$$\begin{aligned} \frac{1}{\sqrt{2}} \leqslant \frac{|D_a\psi ^a v|}{|v|}\leqslant \sqrt{2}\quad \text{ for } \, v\in \mathbb {R}^4 {\setminus } \{ 0\}. \end{aligned}$$

(309)

Because $\psi ^a$ is smooth, it follows from (309) that for $U'$ small enough,

$$\begin{aligned} \frac{1}{2}\leqslant \frac{|\psi ^a(x) - \psi ^a(y)|}{|x-y|} \leqslant 2 \quad \text{ for } x,y\in U'. \end{aligned}$$

(310)

Thus if A is large enough, $B(a, 8/A)\subseteq U'$. By definition of U and (310), we have that $B(a, 2/A)\subseteq U$. Because $\psi$ is $C^2$, by potentially restricting to a compactly contained open set, we may assume

$$\begin{aligned} C = ||D^2\psi ^a||_\infty < \infty . \end{aligned}$$

(311)

Let $b\in \mathcal {C}$, $|b| = 1$. Then there is a rotation $O\in O(4)$ taking b to a and fixing $\mathcal {C}$. Define $\psi ^b = O^{-1}\circ \psi ^a\circ O$. For $b\in \mathcal {C}{\setminus } \{0\}$, we define $\psi ^b = |b|\psi ^{\frac{b}{|b|}}(\cdot /{|b|})$. Note that $\psi ^{\frac{b}{|b|}}$ satisfies (310) and (311), we have that $\psi ^b$ satisfies (304) and (305). $\square$