2· Preliminaries

The next lemma is a multivariable variant of [Reference Käenmäki, Orponen and Venieri 13 , lemma 3·1] and [Reference Pramanik, Yang and Zahl 19 , lemma 3·5]. Recall the modulus of continuity $\alpha$ of a cinematic family F in Remark 1·11 (2).

Proof. Since $\mathrm{diam \, } U \lt \alpha(\eta/2)$ , for any $x,x' \in U$ , $|x-x'| \lt \alpha(\eta/2)$ . By Definition 1·10, we know that $|\nabla_\xi \nabla_\xi (f-g)(x)-\nabla_\xi \nabla_\xi (f-g)(x')| \lt\eta/2$ . Moreover, since F is contained in a ball of diameter K, we know $\|f-g\|_{C^{2}([0,1]^{n-1})} \le K$ . In particular, we know Lip $(f-g)(x) \le K$ and Lip $|\nabla(f-g)|(x)\le K$ for all $x \in U$ . Recalling from Definition 1·10 that $\alpha(s) \le K^{-1}s$ , we know

$$ |k(x)-k(x')| \le \frac{\eta}2, \quad \forall x,x' \in U, $$

where k is one of the three functions $|f-g|, |\nabla (f-g)|, |\nabla_\xi \nabla_\xi (f-g)|$ .

Next, fix $U \subset I^{n-1}$ with $\mathrm{diam \, } U \lt \eta/2$ , and suppose that the alternative (L) fails for $k = f - g$ . So there is $x_0 \in U$ such that $|k(x_0)| \lt \eta \|f-g\|_{C^2( I^{n-1})}/2$ . And for any $x \in U$ , since $|x-x_0| \lt \alpha(\eta/2)$ , by defining $l= k/\|f-g\|_{C^2(I^{n-1})}= k/t$ , we deduce that Lip $_{I^{n-1}}l \le 1$ and

$$ \frac{|k(x)|}{t} \le \frac{|k(x_0)|}{t}+ \left| \frac{|k(x)|}{t} - \frac{|k(x_0)|}{t}\right| \lt \frac{\eta}{2} + |l(x)-l(x_0)| \le \frac{\eta}{2} +\mbox{ [Lip$_{I^{n-1}}l$}] |x-x_0| \le \eta.$$

Thus the alternative (S) holds for k. The proof of the cases $k = |\nabla (f-g)|$ and $k= |\nabla_\xi \nabla_\xi (f-g)|$ are the same.

We denote by $\lambda U$ the cube with the same center as U and has side length $\lambda$ times as that of U for $\lambda\gt0$ . In the following proof, for technical reasons, given a cinematic family $F \subset C^{2}([0,1]^{n-1},[0,1])$ with cinematic constant K, we first extend F to be a cinematic family in $C^{2}([-{\alpha(K^{-1}/6)}/2,1+{\alpha(K^{-1}/6)/}2]^{n-1})$ . Thus F will have a new cinematic constant $\bar K \ge K$ and a new modulus of continuity $\bar \alpha \le \alpha$ . Since $\bar\alpha,\alpha$ are increasing functions, $\bar\alpha(\bar K^{-1}/6) \le \alpha(\bar K^{-1}/6) \le \alpha(K^{-1}/6)$ . Thus F is also a cinematic family in $C^{2}([-{\bar\alpha(\bar K^{-1}/6)/}2,1+{\bar\alpha(\bar K^{-1}/6)/}2]^{n-1})$ with cinematic constant $\bar K$ and modulus of continuity $\bar \alpha$ . By abuse of notation, we still write $\bar K=K$ and $\bar \alpha =\alpha$ . As a result, letting $I_F^{n-1}\; :\!=[-{\alpha(K^{-1}/6)/}2, 1+{\alpha(K^{-1}/6)/}2]^{n-1}$ , we know $F \subset C^{2}(I_F^{n-1})$ is a cinematic family of cinematic constant K and modulus of continuity $\alpha$ .

The above lemma gives the following corollary.

1. (A): at least one of the following three conditions hold:

2. (i) $|h(x)| \ge \frac{1}{3K} \|h\|_{C^2(I_F^{n-1}) }$ for all $x \in 2U$ ;

3. (ii) $|\nabla h(x)| \ge \frac{1}{3K} \|h\|_{C^2( I_F^{n-1}) }$ for all $x \in 2U$ ;

4. (iii)

   (2·1) \begin{equation} |\nabla_\xi \nabla_\xi h (x)| \ge \frac{1}{3K} \|h\|_{C^2( I_F^{n-1}) } \qquad \mbox{for all } x \in 2U, \xi \in S^{n-2}. \end{equation}

5. (B): Moreover, if $|\nabla_\xi \nabla_\xi h|$ satisfies (2·1), then there exists at most one point $x \in 2U$ such that $\nabla h(x)=0$ and h is strictly convex/concave in 2U.

To see the assertion (B), assume on the contrary that there exists two points $x,y \in 2U$ such that $\nabla h(x)=\nabla h(y)=0$ . In particular, letting $e= {x-y}/{|x-y|}$ , we have $\nabla_e h(x)=\nabla_e h(y)=0$ . Let $\gamma\,:\,[0,|x-y|] \to 2U$ , $\gamma(t)=(1-{t}/{|x-y|})x + {t}/{|x-y|} y$ . Then $h\circ \gamma \in C^2([0,|x-y|],\mathbb{R})$ , $(h\circ \gamma)'(t) = \nabla h(\gamma(t)) \cdot \dot \gamma(t) = \nabla_e h(\gamma(t))$ and $(h\circ \gamma)''(t) = \nabla_e\nabla_e h(\gamma(t))$ . By Taylor’s formula, there exists $t_0 \in [0,|x-y|]$ such that

$$ 0= \nabla_e h(x)-\nabla_e h(y) = |x-y| \nabla_e\nabla_e h(\gamma(t_0)). $$

Hence $\nabla_e\nabla_e h(\gamma(t_0))=0$ , which contradicts (2·1).

To see h is strictly convex/concave in 2U, recalling $h \in C^2(2U)$ , we know $\nabla_\xi \nabla_\xi h (x)$ is continuous with respect to x and $\xi$ . Thus the assumption (2·1) holds implies that $\nabla_\xi \nabla_\xi h (x) \gt {1}/{3K} \|h|_{\lambda U}\|_{C^2( 2 U) } \gt0 $ or $\nabla_\xi \nabla_\xi h (x) \lt - {1}/{3K} \|h|_{\lambda U}\|_{C^2( 2 U) } \lt0 $ for all $x \in 2U$ and $\xi \in S^{n-2}$ . If the first case happens, for all $\xi \in S^{n-2}$ and t sufficiently small, we have

$$ h(x+t\xi) = h(x) + t \nabla_\xi h(x) + t^2 \nabla_\xi \nabla_\xi h(x) + o(t^2) \gt h(x) + t \nabla_\xi h(x). $$

Thus h is strictly convex at all $x \in 2U$ . Similarly, the second case implies that h is concave. The proof is complete.

We recall the following facts from ODE. Consider the following dynamical system in a domain $\Omega \subset \mathbb{R}^{n-1}$ :

(2·2) \begin{equation} \dot x = f(x), \quad x \in \Omega, \end{equation}

where $f \in C^1(\Omega, \mathbb{R}^{n-1})$ . Define the flow of (2·2) by $\Phi(x,s)$ , i.e. for each $x \in \Omega$ , $\partial_s \Phi(x,s) = f(\Phi(x,s))$ and $\Phi(x,0)= x$ . Write $\nabla_x \Phi(x,s)= (\partial_{x_1}\Phi(x,s), \cdots, \partial_{x_{n-1}}\Phi(x,s))$ . By ODE theory (e.g. [Reference Walter 21 , theorem, p. 154]), we know $\Phi$ is $C^1$ and

$$ |\nabla_{x_i} \Phi(x,s)| \le \|f\|_{L^\infty(\Omega)} + \|\nabla f\|_{L^\infty(\Omega)} \int_{s_0}^s |\nabla_{x_i} \Phi(x,\tau)| \, d\tau \quad i=1,\ldots,n-1 $$

for any $s,s_0$ such that the flow $\Phi$ is defined. Using the Gronwall inequality one has

(2·3) \begin{equation} |\nabla_x \Phi(x,s)| = |(\partial_{x_1}\Phi(x,s), \cdots, \partial_{x_{n-1}}\Phi(x,s))| \lesssim \|f\|_{L^\infty(\Omega)} e^{\|\nabla f\|_{L^\infty(\Omega)}} \lesssim \|f\|_{C^1(\Omega)}.\end{equation}

We recall that $A^\circ$ is the interior of a set $A \subset \mathbb{R}^n$ .

Lemma 2·3. Let h be as in Corollary 2·2 and assume $\nabla h \ne 0$ in 2U. Consider the system

(2·4) \begin{equation} \dot \gamma(s) = \nabla h ( \gamma(s)) \quad in \ 2U. \end{equation}

Then each solution $\gamma$ of (2·4) can be extended to the boundary of 2U.

Proof. Given any $x \in (2U)^\circ$ , by the classical ODE theory, there exists a unique solution $\gamma_x\; :\; (-\epsilon,\epsilon) \to 2U$ with $\gamma_x(0)=x$ . We show $\gamma_x$ can be extended to a curve (still denote by $\gamma_x$ ) defined on some interval $[a_x,b_x]$ such that $\gamma_x(a_x), \gamma_x(b_x) \in \partial (2U)$ .

Indeed, recalling that $\|h\|_{C^2(2U)} \le K$ , we know $|h| \le K$ in 2U. On the other hand, by assumption, $\nabla h \ne 0$ in 2U implies that $\min_{x \in 2U} |\nabla h(x)| \ge a$ for some $a \gt0$ . Thus,

$$ 2K \ge h(\gamma_x(t)) - h(x) = \int_0^t \nabla h (\gamma_x(s)) \cdot \dot \gamma_x(s)\, ds = \int_0^t |\nabla h (\gamma_x(s))|^2 \, ds \ge a^2 t. $$

This implies that $b_x \le 2K/ a^2$ and $\gamma_x(b_x) \in \partial (2U)$ . Similarly, we obtain that $\gamma_x(a_x) \in \partial (2U)$ .

Lemma 2·4. Let h be as in Corollary 2·2 and the case (A)(ii) holds for h. Consider the system

(2·5) \begin{equation} \dot \gamma(s) = \frac{\nabla h ( \gamma(s))}{|\nabla h ( \gamma(s))|} \quad in \ U. \end{equation}

Set

$$ \partial U^+\;:\!= \{ x \in \partial U\; :\; \textit{there exists a maximal solution}\ \gamma_x \in C^2( [0,b_x], U)\ \textit{to (2.5) starting from x} \}.$$

Then the family $\{\gamma_x\}_{x \in \partial U^+}$ foliates U, that is,

$$ U= \bigcup_{x \in \partial U^+}\gamma_{x}, \text{ and } \gamma_{x_1} \cap \gamma_{x_2}=\emptyset \text{ if $x_1 \ne x_2$.} $$

Proof. Note that (A)(ii) in Corollary 2·2 holds implies that $\nabla h \ne 0$ in U. Thus ${\nabla h ( \gamma(s))}/{|\nabla h ( \gamma(s))|}$ is well defined in 2U. Moreover, up to a parametrisation, the maximal solution $\gamma_x$ is the same as the one of the following system

$$ \dot \gamma(s) = \nabla h ( \gamma(s)) \quad in \ U. $$

Thus for any $y \in U^\circ$ , there exists a unique solution $\gamma_y \,:\, (-\epsilon,\epsilon) \to U$ with $\gamma_y(0)=y$ . By Lemma 2·3, we know $\gamma_y$ extends to a maximal solution defined on $[a_y,b_y]$ (still denote by $\gamma_y$ ) such that $\gamma_y(a_y) \in \partial U$ . This implies $\gamma_y(a_y) \in \partial U^+$ . Let $x=\gamma_y(a_y)$ . We know $y \in \gamma_x$ . Thus, $U = \cup \{\gamma_x\}_{x \in \partial U^+}$ .

Remark 2·5. Under the assumption as in Lemma 2·4, note that the dyadic cube $U \subset \mathbb{R}^{n-1}$ has $2^{n-1}$ facets $\partial U_1, \ldots, \partial U_{2^{n-1}}$ . Let

$$ \partial U^+_i \;:\!= \partial U_1 \cap \partial U^+ \quad\mbox{ and }\quad U_i\;:\!= \bigcup_{x \in \partial U^+_i} \gamma_x \quad i=1,\ldots,2^{n-1}. $$

For each $\partial U_i$ , by possibly reordering and translating the coordinates, we can assume $\partial U_i \subset$ $\mathrm{span}\{x_1,\ldots,x_{n-2}\}$ . For each $x=(x_1,\ldots,x_{n-1}) \in \partial U_i$ , write $\bar x=(x_1,\ldots,x_{n-2})$ . Thus we can equip $U_i$ with a new coordinate by letting $y=(\bar x,s)$ for all $y \in U_i$ where $x \in \partial U_i^+$ and $s \in [0,b_x]$ is unique pair such that $y = \gamma_x(s)$ .

The following lemma is a slight variant of [Reference Pramanik, Yang and Zahl 19 , lemma 3·8] and the proof is essentially the same as [Reference Pramanik, Yang and Zahl 19 , lemma 3·8].

Lemma 2·6. Let $[0,a] \subset \mathbb{R}$ and $h \in C^2([0,a])$ with $\|h\|_{C^2([0,a])}=t$ . For $\delta\gt0$ , write

$$ E_{2\delta}=\{s \in [0,a]\, :\, |h(s)| \le 2\delta\}. $$

Then, for each $0 \lt \delta \lt c_1 t$ for some constant $0 \lt c_1\lt1$ and $4c_1 \le c_2\lt1$ :

1. (i) if

   $$ \lambda_1\;:\!= \inf_{s \in [0,a]}\{|h(s)| + |h'(s)|\} \ge c_2 t,$$
   then $|E_{2\delta}| \lesssim \delta/t$ ;

2. (ii) assume $\lim_{s \to 0+}h'(s)=0$ and

   $$ h''(s) \ge c_2 t, \quad s \in (0,a). $$
   Let
   $$ \lambda_2\;:\!= |h(0)|, $$
   then $E_\delta \subset [0,d]$ with $d \le c_2^{-1}\sqrt{(\lambda_2+\delta)/ t}$ . Moreover, $E_\delta$ is a closed interval of length $\lesssim \delta/\sqrt{(\lambda_2+\delta)t}$ .

Proof. First, we show (i). Noticing that for $s \in E_{2\delta}$ , we have

$$ c_2t \le \lambda_1 \le \min_{s \in E_{2\delta}}\{|h(s)| + |h'(s)|\} \le 2\delta + \min_{s \in E_{2\delta}}\{|h'(s)|\}. $$

By $4c_1 \le c_2$ , we know $\min_{s \in E_{2\delta}}\{|h'(s)|\} \gtrsim t$ . Thus we get $|E_{2\delta}| \lesssim \delta/t$ as desired.

Next, we show (ii). Using $h''(s) \ge c_2 t$ , we deduce that for all $s \in [0,a]$ , h′(s) does not change sign and

\begin{align*} |h(s)| & \ge |\int_{0}^s h'(\tau) \, d \tau| -\lambda_2 = \int_{0}^s |h'(\tau)| \, d \tau - \lambda_2 \\[3pt] & = \int_{0}^s \int_{0}^\tau h''(\rho) \, d \rho \, d \tau - \lambda_2 \ge c_2 t s^2 - \lambda_2. \end{align*}

Noticing that $s \in E_{2\delta}$ implies $|h(s)| \le 2\delta$ , we know $\max E_{2\delta} \le \sqrt{(\lambda_2+\delta)/(c_2t)} \le c_2^{-1}\sqrt{(\lambda_2+\delta)/t}$ .

Finally, we estimate $|E_{2\delta}|$ . In the case $\lambda_2 \le \delta$ , it holds

$$ |E_{2\delta}| \le d \lesssim \sqrt{(\lambda_2+\delta)/t} \sim \frac{\lambda_2+\delta}{\sqrt{(\lambda_2+\delta)/t}} \sim \frac{\delta}{\sqrt{(\lambda_2+\delta)/t}}. $$

In the case $\lambda_2 \gt \delta$ , recalling $\lambda_2 =|h(0)|$ , for each $s \in E_{2\delta} \cap [0,a]$ , we deduce that

\begin{align*} 2\delta & \ge |h(s)| \ge \lambda_2-\int_{0}^s |h'(\tau)| \, d \tau \\[3pt] & \ge \lambda_2- \int_{0}^s \int_{0}^\tau |h''(\rho)| \, d \rho \, d \tau \ge \lambda_2 - t s^2, \end{align*}

which implies $s \gtrsim \sqrt{(\lambda_2 + \delta)/t} \sim \sqrt{\lambda_2 /t} $ for all $E_{2\delta} \cap [0,b]$ . Using $h''(s) \ge c_2 t$ for all $s \in (0,a)$ , we have

$$ |h'(s)|= |h'(s)- h'(0)| \gtrsim t|s| \ge \sqrt{\lambda_2 /t}, \quad s \in E_{2\delta} . $$

As a result, we get $|E_{2\delta}| \lesssim \delta/\sqrt{\lambda_2 /t} \sim \delta/\sqrt{(\lambda_2 + \delta)/t}$ as desired.

We end this section by recalling the following fact for convex sets (see for example [Reference Hug and Weil 10 , theorem 3·6 (b6)]).

Lemma 2·7. Let $A \subset B \subset \mathbb{R}^{n}$ be two compact convex sets with $\mathcal{L}^n(A)\gt0$ . Then

$$ \mathcal{H}^{n-1}(\partial A) \le \mathcal{H}^{n-1}( \partial B). $$

3· Intersections of functions in a cinematic family

We first show Lemma 1·17 in this section. Recall that for arbitrary $\eta\gt0$ and subdomain $D \subset I^{n-1}$ , the vertical $\eta$ -neighbourhood is

$$ f^{\eta}(D)\;:\!= \left\{(x,y)=(x_1,\cdots,x_{n-1},y) \in D\times\mathbb{R}\, :\, f(x)- \eta \le y \le f(x)+\eta\right \}.$$

Proof of Lemma 1·17. Write $t\;:\!=\|f-g\|_{C^2([0,1]^{n-1})}$ . Similar to the previous section, we may assume $F \subset C^{2}(I_F^{n-1})$ be a cinematic family with cinematic constant K and modulus of continuity $\alpha$ . By decomposing $[0,1]^{n-1}$ by dyadic cubes $\{U\}$ with ${\alpha(K^{-1}/6)/}4 \lt \mathrm{diam \, } U \lt {\alpha(K^{-1}/6)/}2$ , to prove (1·10), it suffices to show

$$ \mathcal{L}^{n}(f^\delta(U) \cap g^\delta(U)) \lesssim \frac{\delta^2}{t} $$

for each $U \subset[0,1]^{n-1}$ . Notice that for each $U \subset [0,1]^{n-1}$ , f, g are well-defined on 2U. In the following, we fix U and write $f^\delta \cap g^\delta = f^\delta(U) \cap g^\delta(U)$ for simplicity.

If $f^\delta \cap g^\delta = \emptyset$ , then there is nothing to prove. Also, since $t \in [\delta,1]$ , if $\delta \le t \le [{60K}/{\alpha(K^{-1}/6)}]^2\delta$ , then $\mathcal{L}^{n}(f^\delta \cap g^\delta) \le \mathcal{L}^{n}(f^\delta) \sim \delta \sim {\delta^2}/{t}$ . Thus in the following, we assume

(3·1) \begin{equation} t \gt 4\left[\frac{10^5K^4}{\alpha(K^{-1}/6)^2}\right]^2\delta.\end{equation}

Assume $f^\delta \cap g^\delta \ne \emptyset$ . Let $P_{f,g}$ be the orthogonal projection of $f^\delta \cap g^\delta$ to U. Since the vertical thickness of $f^\delta \cap g^\delta$ is no more than $2\delta$ , we need to show

(3·2) \begin{equation} \mathcal{L}^{n-1} (P_{f,g}) \lesssim \frac{\delta}{t}. \end{equation}

Write $h=f-g$ . By $f^\delta \cap g^\delta \ne \emptyset$ , we know there exists $x \in U$ such that $|h|\le 2\delta$ which implies $|h|$ does not enjoy Corollary 2·2 (A)(i). Hence either Corollary 2·2 (A)(ii) (the transversal case) or Corollary 2·2 (A)(iii) (the tangent case) holds. We consider these two cases separately.

Case 1. Corollary 2·2 (A)(ii) holds, that is

$$ |\nabla h(x)| \ge \frac{1}{3K} \|h\|_{C^2(I_F^{n-1})} \quad x \in 2U. $$

In particular, $\nabla h \ne 0$ in U.

We will prove (3·2) by showing that $P_{f,g}$ is the image of a set $\Omega$ under a Lipschitz map $\Psi$ with Lipschitz constant $\lesssim 1$ and $\mathcal{L}^{n-1}(\Omega) \lesssim \delta/t$ . To this end, we divide the proof into two steps. As mentioned in the introduction, in Step 1, we foliate $P_{f,g}$ by a family of curves $\gamma_x$ using Lemma 2·4. And we define a family of auxiliary functions $h_x$ derived from h and $\gamma_x$ which satisfies Lemma 1·2 (i). In Step 2, we apply Lemma 2·6 (i) to conclude (3·2).

Step 1. Consider the ODE

(3·3) \begin{equation} \dot \gamma(s) = \frac{\nabla h(\gamma)}{|\nabla h(\gamma)|} \quad \mbox{in } U. \end{equation}

Since $\nabla h \in C^1(U,\mathbb{R}^{n-1})$ and $\nabla h \ne 0$ in U, we know ${\nabla h}/{|\nabla h|} \in C^1(U,\mathbb{R}^{n-1})$ . By Lemma 2·4, we know, for any $x \in U$ , there exists a unique the maximal solution $\gamma_x \in C^2( [a,b], U)$ to (3·3) with $\gamma_x(0)=x$ , and $ \gamma_x(a),\gamma_x(b) \in \partial U$ . Denote

$$ \partial U^+\;:\!= \{ x \in \partial U\, : \,\textrm{there exists a solution}\ \gamma_x \in C^2( [0,b_x], U)\ \textrm{to (3.3) starting from x} \}.$$

Then, again by Lemma 2·4, we know $\{\gamma_x\}_{\xi \in \partial U^+}$ gives a foliation of U.

For each $x \in \partial U^+$ , define $h_x\;:\!= h\circ \gamma_x$ . Then $h_x \in C^2([0,b_x],U)$ . Our aim is to show $|\{ |h_x| \le 2\delta\}| \sim \delta/ t$ . To this end, we check that $h_x$ satisfies Lemma 2·6 (i). Define

$$ \lambda= \Delta(f,g) \;:\!= \inf_{x \in U} \{ |h(x)| + | \nabla h(x)|\}, $$

and

$$ \lambda_x\;:\!= \inf_{s \in [0,b_x]} \{ |h_x(s)| + | h_x'(s)|\} \text{ and } t_x\;:\!=\|h_x\|_{C^2([0,b_x])}. $$

We show

(3·4) \begin{equation} 3K^{-1} t \le \lambda \le \lambda_x \le t_x \le t \quad x \in \partial U^+. \end{equation}

In particular,

(3·5) \begin{equation} \lambda_x \ge \frac{1}{3K}t_x \quad x \in \partial U^+. \end{equation}

To see (3·4), we compute

$$ h_x'(s)= \nabla h(\gamma_{x}(s)) \cdot \dot \gamma_{x}(s) = \nabla h(\gamma_{x}(s))\cdot\frac{\nabla h(\gamma_{x}(s))}{|\nabla h(\gamma_{x}(s))|} = |\nabla h(\gamma_{x}(s))|.$$

This implies $\lambda_x \ge \lambda= \Delta(f,g)$ . Also, it is immediate that $t_x \le t$ . On the other hand, the fact that $|\nabla h|$ satisfies Corollary 2·2 (A)(ii) implies

$$ \lambda \ge 3K^{-1} \|h\|_{C^2(I_F^{n-1})} \ge 3K^{-1} \|h\|_{C^2([0,1]^{n-1})} = 3K^{-1} t.$$

Thus (3·4) and (3·5) hold.

Step 2. First, notice that (3·4) also gives $t_x \ge (3K)^{-1}t$ . Combining (3·1), we have $t_x \ge ({10^9K^7}/{\alpha(K^{-1}/6)^4})\delta$ for all $x \in \partial U^+$ . Applying Lemma 2·6 (1) to each $h_x$ with $c_1= ({\alpha(K^{-1}/6)^4}/{10^9K^7}) \le ({1}/{12K})$ and $c_2=3K^{-1} \ge 4c_1$ , we know

(3·6) \begin{equation} |I_x\;:\!=\{s \in [0,b_x]: |h_x(s)| \le 2\delta\}| \lesssim \delta/\sqrt{\lambda_x t_x} \sim \delta/ t. \end{equation}

In the following, by Remark 2·5, we can estimate $\mathcal{L}^{n-1}(P_{f,g})$ by estimating $\mathcal{L}^{n-1}(P_{f,g} \cap U_i)$ for each $i=1,\ldots 2^{n-1}$ . Thus we may without loss of generality assume that $P_{f,g} \subset U_i \subset$ $\mathrm{span}\{x_1,\ldots,x_{n-2}\}$ for some i and thus $\partial U^+ \subset \partial U_i$ . For any $x=(x_1,\ldots,x_{n-1}) \in \partial U^+ \subset 2U \subset \mathbb{R}^{n-1}$ , write $x=(\bar x,x_{n-1})$ as in Remark 2·5. Let $\Omega\;:\!= \cup_{x \in \partial U^+} \{(\bar x, I_x)\} \subset \mathbb{R}^{n-1}$ and define the map $\Psi\; :\; \Omega \to U$ by

$$ \Psi(\bar x,s) = \gamma_x(s). $$

Since $2U \subset I_F^{n-1}$ , the ordinary equation

(3·7) \begin{equation} \dot \gamma(s) = \frac{\nabla h(\gamma)}{|\nabla h(\gamma)|} \quad \mbox{in } 2U \end{equation}

is well-defined. Let $\Phi(x,s)$ be the flow of (3·7) with initial condition $\Phi(x,0)=x$ for all $x \in 2U$ . Then we know for all

(3·8) \begin{equation} \Psi(\bar x,s)=\Phi(x,s) \quad (\bar x,s) \in \Omega. \end{equation}

Note that ${\nabla h}/{|\nabla h|} \in C^1(2U,\mathbb{R}^{n-1})$ and

$$ \nabla\left(\frac{\nabla h}{|\nabla h|}\right)(x) = \frac{\nabla^2 h}{|\nabla h|}(x) - \frac{(\nabla^2 h \cdot \nabla h) \otimes \nabla h}{|\nabla h|^3}(x) \quad \forall x \in 2U . $$

Recall the standing assumption that $|\nabla h(x)| \ge ({1}/{3K}) \|h\|_{C^2(I_F^{n-1})} \ge ({1}/{3K}) \|h\|_{C^2(2U)}$ , we know

(3·9) \begin{equation} \|\nabla\left(\frac{\nabla h}{|\nabla h|}\right)\|_{L^\infty(2U)} \le 6K\frac{\|\nabla^2 h\|_{L^\infty(2U)} }{\|h\|_{C^2(2U)}} \lesssim 1 . \end{equation}

Recalling (2·3), one has

$$ |\nabla_x \Phi(x,s)| = |(\partial_{x_1}\Phi(x,s), \ldots, \partial_{x_{n-1}}\Phi(x,s))| \lesssim e^{\|\nabla(\frac{\nabla h}{|\nabla h|})\|_{L^\infty(2U)}} \lesssim 1 \quad (\bar x,s)\in \Omega. $$

Then by (3·8), for $(\bar x,s)\in \Omega$ , write $\nabla_{\bar x} \Psi(\bar x,s)=(\partial_{x_1}\Psi(\bar x,s), \ldots, \partial_{x_{n-2}}\Psi(\bar x,s))$ . We have

$$ |\nabla_{\bar x} \Psi(\bar x,s)| = |(\partial_{x_1}\Psi(\bar x,s), \ldots, \partial_{x_{n-2}}\Psi(\bar x,s))|= |(\partial_{x_1}\Phi(x,s), \ldots, \partial_{x_{n-2}}\Phi(x,s))| \lesssim 1 . $$

On the other hand, we have $|\partial_s \Psi(\bar x,s)| = |\dot \gamma_x(s)| =1 $ for all $(x,s) \in \Omega$ . Thus, writing $\nabla\Psi=(\nabla_{\bar x}\Psi,\partial_s\Psi )$ , we have

$$ \|\mbox{Det } \nabla\Psi\|_{L^\infty(\Omega)} = \|\mbox{Det } (\nabla_{\bar x}\Psi,\partial_s\Psi )\|_{L^\infty(\Omega)} \lesssim 1.$$

Combining (3·6), we conclude,

\begin{align*} \mathcal{L}^{n-1}(\{y \in U: & |h(y)| \le 2\delta\}) = \int_U \chi_{\{|h(y)| \le 2\delta\}}(y) \, dy \\[3pt] & \le \int_{x \in \partial U^+} \int_{I_x}\chi_{\{ |h(\gamma_x(s))| \le 2\delta \}} (s)\mbox{ $|$Det } \nabla \Psi(\bar x,s)|\, ds \, d\mathcal{H}^{n-2}(x) \\[3pt] & \lesssim \int_{x \in \partial U^+} |I_x| \, d\mathcal{H}^{n-2}(x) \\[3pt] &\lesssim \mathcal{H}^{n-2}(\partial U^+) \delta/ t \lesssim \delta/ t. \end{align*}

Since $ P_{f,g}= \{ |h| \le 2\delta\}, $ we arrive at (3·2).

Case 2. Corollary 2·2 (A)(iii) holds, that is,

(3·10) \begin{equation} |\nabla_\xi \nabla_\xi h (x)| \ge \frac{1}{3K} \bar t \;:\!= \frac{1}{3K} \|h\|_{C^2(I_F^{n-1}) } \qquad \mbox{for all } x \in 2U, \xi \in S^{n-2}. \end{equation}

Recall

$$ \Delta=\Delta(f,g)= \inf_{x \in U}\{ |f(x)-g(x)| + |\nabla f(x)- \nabla g(x)| \} = \inf_{x \in U}\{ |h(x)| + |\nabla h(x)| \}.$$

Subcase 2-1. Assume $\Delta(f,g) \ge {\alpha(K^{-1}/6)^2}/{10^5K^4}\bar t= {\alpha(K^{-1}/6)^2}/{10^5K^4} \|f-g\|_{C^2(I^{n-1}_F)}$ .

Applying the same argument in Case 1, Step 1, instead of (3·4) and (3·5), we have

\begin{equation*} \frac{\alpha\left(K^{-1}/6\right)^2}{10^5K^4}\bar t \le \lambda \le \lambda_x \le t_x \le \bar t \quad x \in \partial U^+. \end{equation*}

In particular,

\begin{equation*} \lambda_x \ge \frac{\alpha\left(K^{-1}/6\right)^2}{10^5K^4} t_x \quad\mbox{ and }\quad t_x \ge \frac{\alpha\left(K^{-1}/6\right)^2}{10^5K^4}\bar t \quad x \in \partial U^+. \end{equation*}

Combining (3·1), we have $t_x \ge 4 ({10^5K^4}/{\alpha(K^{-1}/6)^2})\delta$ for all $x \in \partial U^+$ . We apply Lemma 2·6 (i) to each $h_x$ with $c_1=(1/4) ({\alpha(K^{-1}/6)^2}/{10^5K^4})$ and $c_2= {\alpha(K^{-1}/6)^2}/{10^5K^4} \ge 4c_1$ . Then use the same argument as in Case 1, Step 2, we have $ \mathcal{L}^{n-1}(P_{f,g}) \lesssim \delta/ \bar t$ and noting that $t \le \bar t$ , we arrive at (3·2).

Subcase 2 (ii). Assume ${\Delta}(f,g) \lt ({\alpha(K^{-1}/6)^2}/{10^5K^4})\|h\|_{C^2(I^{n-1}_F)} = ({\alpha(K^{-1}/6)^2}/{10^5K^4}) \bar t$ .

By Corollary 2·1 (B), we may assume h is strictly convex in 2U (otherwise we consider $h=g-f$ ). In this subcase, we will deeply use the convexity of h and divide the proof into three steps. In Step 1, we first find the unique point $x_M \in 2U$ such that $\nabla(x_M)=0$ by constructing an auxiliary family of functions $\{h_\xi\}$ . In Step 2, we define another family $\{\bar h_\xi\}$ and show that $\{\bar h_\xi\}$ satisfies Lemma 2·6 (ii). Then, in Step 3, we apply Lemma 2·6 (ii) to deduce (3·2).

Step 1. Let $x_{\Delta} \in U$ be the point such that ${\Delta}=|h(x_{\Delta})| + |\nabla h(x_{\Delta})|$ . Then $|h(x_{\Delta})| \le {\Delta}$ and $|\nabla h(x_{\Delta})| \le {\Delta}$ . We will show that there exists a point $x_M$ such that $\nabla h(x_M)=0$ and

(3·11) \begin{equation} x_M \in \frac65 U \mbox { and } |h(x_M)| \le 10K^2\Delta. \end{equation}

First, if $\nabla h(x_{\Delta})=0$ , since Corollary 2·2 (A)(iii) holds for h, by Corollary 2·2 (B), we know $x_M=x_{\Delta}$ . Thus apparently, (3·11) holds.

Next, assume $\nabla h(x_{\Delta}) \ne 0$ . For each $\xi \in S^{n-2}$ , let

$$ E_\xi=\{s \in \mathbb{R}\; :\; x_{\Delta} + s\xi \in 2U \}$$

and

(3·12) \begin{equation} h_\xi(s)\;:\!= h(x_{\Delta} + s\xi) \quad s \in E_\xi. \end{equation}

Thus by the strict convexity of h, we know $h_\xi: E_\xi \to \mathbb{R}$ is a $C^2$ strictly convex function for each $\xi \in S^{n-2}$ . Note that

(3·13) \begin{equation} h_\xi'(s)= \nabla_\xi h (x_{\Delta} + s\xi) \mbox{ and } h_\xi''(s)= \nabla_\xi \nabla_\xi h (x_{\Delta} + s\xi). \end{equation}

In particular, $|h_\xi'(0)| \le |\nabla h (x_{\Delta})| \le \Delta$ .

Since $|\nabla_\xi\nabla_\xi h(x)| \ge ({1}/{3K}) \bar t$ for all $x \in 2U$ and $\xi \in S^{n-2}$ , for each $\xi$ , there exists $s_\xi \in E_\xi$ with $|s_\xi| \le 3K{\Delta}/ \bar t \lt {\alpha(K^{-1}/6)}/{20} \lt {\mathrm{diam \,} U}/5$ such that $ h_\xi'(s_\xi)=0$ where we recall $\mathrm{diam \, } U \gt {\alpha(K^{-1}/6)}/{4}$ . Moreover, we have

$$ x_{\Delta}+ s_\xi\xi \in \frac{6}5 U \subset 2U, \quad \xi \in S^{n-2}.$$

In particular, $s_\xi \notin \partial E_\xi$ and $x_{\Delta}+ s_\xi\xi \notin \partial(2U)$ . In addition, by the strict convexity of $h_\xi$ , we know $s_\xi$ is the unique point in $E_\xi$ which reaches the minimum of $h_\xi$ . We show that there exists $\xi_0 \in S^{n-2}$ such that

(3·14) \begin{equation} h(x_{\Delta}+ s_{\xi_0}\xi_0) = \min_{\xi \in S^{n-2}} \{h(x_{\Delta}+ s_{\xi}\xi) \}. \end{equation}

To see (3·14), it suffices to show that $s_\xi$ continuously depends on $\xi$ . Indeed, consider the $C^1$ function, $\phi(\xi,s) =\nabla h(x_{\Delta}+ s\xi) \cdot \xi$ . Then $\phi(\xi,s_\xi)=0$ for all $\xi \in S^{n-2}$ and $\frac{\partial\phi}{\partial s}(\xi,s)= \nabla_\xi\nabla_\xi(x_{\Delta}+ s_\xi\xi) \ne 0$ at all $(\xi,s_\xi)$ . Applying the implicit function theorem for $\phi$ , we know $s_\xi$ is a continuous function of $\xi$ and (3·14) holds.

Define

$$ U_\epsilon\;:\!= \bigcup_{|\xi-\xi_0|\le \epsilon } \{x_{\Delta}+ s\xi\; :\; |s-s_\xi| \le \epsilon \}.$$

Choose an $\epsilon$ small enough such that $U_\epsilon \subset 2U^\circ$ , we know for each point $x \in U_\epsilon$ , $x= x_{\Delta}+ s_x\xi_x$ for some $\xi_x$ and $s_x$ . Thus

$$ h(x) = h(x_{\Delta}+ s_x\xi_x) = h_{\xi_x}(s_x) \ge h_{\xi_x}(s_{\xi_x}) = h(x_{\Delta}+ s_{\xi_x}\xi_x) \ge h(x_{\Delta}+ s_{\xi_0}\xi_0). $$

As a consequence, we know $x_{\Delta}+ s_{\xi_0}\xi_0$ reaches the local minimum of h in $U_\epsilon$ . Recalling that $h \in C^2(I_F^{n-1})$ , we know $\nabla h(x_{\Delta}+ s_{\xi_0}\xi_0) =0$ .

Let $x_M= x_{\Delta}+ s_{\xi_0}\xi_0$ . We have $\nabla h(x_M)=0$ and $x_M \in (6/5) U$ as desired. By Corollary 2·1 (B), we know $x_M$ is the unique point such that $\nabla h =0$ . We also have

$$ |h(x_M)- h(x_{\Delta})| \le \max_{s \in [0,1]}\{|\nabla_e\nabla_e h((1-s)x_M + s x_{\Delta} )|\}|x_M -x_{\Delta}|^2 \le \bar t(3K{\Delta}/\bar t)^2 \le 9K^2\Delta, $$

where $e = {x_M -x_{\Delta}}/{|x_M -x_{\Delta}|}$ . Recalling that $|h(x_{\Delta})|\le \Delta$ , we have $|h(x_M)| \le 10K^2\Delta$ . Thus (3·11) holds.

Step 2. We define the family $\{\bar h_\xi\}_{\xi \in S^{n-2}}$ and check $\{\bar h_\xi\}_{\xi \in S^{n-2}}$ satisfies Lemma 2·6 (ii).

Let

$$ I_\xi\;:\!= \{s \ge 0\; :\; x_M + s\xi \in 2U\}$$

and

$$ \bar h_\xi (s)\;:\!= h(x_M +s \xi), \quad s \in I_\xi, \ \xi \in S^{n-2}. $$

Since $x_M \in (6/5) U$ , we know $\min I_\xi =0$ for all $\xi \in S^{n-2}$ . Thus $\bar h_\xi(0)$ is well-defined and $\bar h_\xi(0)=h(x_M)$ . Similar to (3·13), we compute

(3·15) \begin{equation} \bar h_\xi'(s)= \nabla_\xi h (x_M + s\xi) \mbox{ and } h_\xi''(s)= \nabla_\xi \nabla_\xi h (x_M + s\xi) \quad s \in I_\xi^\circ. \end{equation}

Since $\nabla h(x_M)=0$ , by (3·15) and the standing assumption (3·10) of Case 2, we have

$$ \lim_{s \to 0+} \bar h_\xi'(s)=\lim_{s \to 0+}\nabla h(x_M+ s \xi) \cdot \xi = \nabla h(x_M) \cdot \xi =0 $$

and

$$ \bar h_\xi''(s)= \nabla_\xi\nabla_\xi \bar h(x_M+s\xi) \ge \frac{1}{3K} \bar t \quad s \in I_\xi^\circ. $$

Moreover, let $t_\xi\;:\!= \|\bar h_\xi\|_{C^2(I_\xi)}$ for all $\xi \in S^{n-2}$ . Then, it is easy to see $t_\xi=\|\bar h_\xi\|_{C^2(I_\xi)} \le \|h\|_{C^2(I_F^{n-1})} =\bar t$ . Thus

(3·16) \begin{equation} \bar h_\xi''(s) \ge \frac{1}{3K} \|\bar h_\xi\|_{C^2(I_\xi)} =\frac{1}{3K} t_\xi \quad s \in I_\xi^\circ.\end{equation}

As a consequence, we know $h_\xi$ satisfies the condition of Lemma 2·6 (ii) with constant $c_2= ({1}/{3K})$ for all $x \in S^{n-2}$ . We end this step by letting

$$ \bar \lambda\;:\!=h(x_M) \equiv \bar h_\xi(0). $$

Recalling (3·11), we have

(3·17) \begin{equation} \bar \lambda \le 10K^2\Delta.\end{equation}

Step 3. Notice that (3·16) gives

$$ \bar t \ge t_\xi = \|\bar h_\xi\|_{C^2(I_\xi)} \ge \|\bar h_\xi''\|_{L^\infty(I_\xi)} \ge \frac{1}{3K} \bar t, \quad \xi \in S^{n-2}.$$

Thus $t_\xi \sim \bar t$ . Combining (3·1), we have $t_\xi \ge 4 ({10^9K^7}/{\alpha(K^{-1}/6)^4})\delta$ for all $\xi \in S^{n-2}$ . Applying Lemma 2·6 (ii) to each $\bar h_\xi$ with $c_1= (1/4) ({\alpha(K^{-1}/6)^4}/{10^9K^7})$ and $c_2=(3K)^{-1}$ , we know

(3·18) \begin{equation} |\hat I_\xi\;:\!=\{s \in I_\xi: |\bar h_\xi(s)| \le 2\delta\}| \lesssim \delta/\sqrt{\bar \lambda t_\xi} \sim \delta/\sqrt{\bar \lambda \bar t} \end{equation}

and

(3·19) \begin{equation} \max \hat I_\xi \le c_2^{-1}\sqrt{\frac{\bar \lambda +\delta}{ \bar t}} = 3K\sqrt{\frac{\bar \lambda +\delta}{ \bar t}} \quad \xi \in S^{n-2}. \end{equation}

Noting that

$$ 2U = \bigcup_{\xi \in S^{n-2}}\{x_M + \xi I_\xi\},$$

we know

$$ \{ y \in 2 U\; :\; |h(y)| \le 2\delta \} = \bigcup_{\xi \in S^{n-2}}\{x_M + \xi s \; :\; |\bar h_\xi(s)| \le 2\delta\} = \bigcup_{\xi \in S^{n-2}}\{x_M + \xi \hat I_\xi \}. $$

As a consequence, combining (3·19) we have

(3·20) \begin{equation} P_{f,g}= \{ y \in U\; :\; |h(y)| \le 2\delta \} \subset \bigcup_{\xi \in S^{n-2}}\{x_M + \xi \hat I_\xi \} \subset B\left(x_M, 3K\sqrt{\frac{\bar \lambda +\delta}{ \bar t}} \right). \end{equation}

Since the parameters $(\xi,s)$ forms a natural polar coordinate centered at $x_M$ , by considering $\bar\lambda \subset [0,\delta]$ and $\bar\lambda \gt\delta$ separately, in the case $\bar \lambda \gt \delta$ , noting that

$$ \max \hat I_\xi \le 3K \sqrt{(\bar\lambda +\delta) / \bar t} \le 6K\sqrt{\bar\lambda / \bar t} \mbox{ and } |\hat I_\xi| \sim \delta/\sqrt{(\bar\lambda +\delta)\bar t} \sim \delta /\sqrt{\bar\lambda \bar t},$$

we deduce

\begin{align*} \mathcal{L}^{n-1}(P_{f,g}) &= \int_U \chi_{\{y \in U: |h(y)| \le 2\delta\}}(y) \, dy \\[3pt] & \le \int_{\xi \in S^{n-2}} \int_{0}^{6K\sqrt{\bar\lambda / \bar t}}\chi_{\{ |\bar h_\xi(s)| \le 2\delta \}} (s)s^{n-2}\, ds \, d\mathcal{H}^{n-2}(\xi) \\[3pt] & \lesssim \int_{\xi \in S^{n-2}} \int_{\hat I_\xi} \left(\sqrt{\bar\lambda / \bar t}\right)^{n-2}\, ds \, d\mathcal{H}^{n-2}(\xi) \\[3pt] & \lesssim \mathcal{H}^{n-2}(S^{n-2}) |\hat I_\xi| \left(\sqrt{\bar\lambda / \bar t}\right)^{n-2} \\[3pt] &\lesssim \frac{\delta}{\sqrt{\bar\lambda \bar t}} \left(\sqrt{\bar\lambda / \bar t}\right)^{n-2} = \frac{\delta\sqrt{\bar\lambda}^{n-3}}{\sqrt{\bar t}^{n-1}} \lesssim \frac{\delta}{\bar t}, \end{align*}

where in the last inequality we recall $\bar\lambda = h(x_M) \le \|h\|_{C^2(2U)} \le \bar t$ and $n\ge 3$ .

And in the case $\bar \lambda \in [0,\delta]$ , noting that

$$ \max \hat I_\xi \le 3K\sqrt{(\bar\lambda +\delta) / \bar t} \le 6K\sqrt{\delta / \bar t}, $$

we deduce that

(3·21) \begin{equation} P_{f,g} \subset B(x_M, 6K\sqrt{\delta / \bar t}), \end{equation}

which implies

\begin{align*} \mathcal{L}^{n-1} (P_{f,g}) & \le \mathcal{L}^{n-1} \left[B(x_M, 6K\sqrt{\delta / \bar t} )\right] \lesssim (\delta/ \bar t )^{(n-1)/2} \le \delta/ \bar t . \end{align*}

Finally, recalling that $\bar t =\|h\|_{C^2(I_F^{n-1})} \ge \|h\|_{C^2([0,1]^{n-1})} \ge t$ , we arrive at (3·2). The proof is complete.

The following lemma will be used in Section 4.

Lemma 3·1. Under the same assumption as in Lemma 1·17, let $\epsilon\gt0$ , $0 \lt s \le 1$ and $E \subset [0,1]^{n-1}$ be a $(\delta,n-2+s,\delta^{-\epsilon})$ -set with $|E| \le \delta^{-(n-2+s)}$ . Denote by $P_{f,g}$ the orthogonal projection of $f^\delta \cap g^\delta$ to $[0,1]^{n-1}$ . Then

(3·22) \begin{equation} |E \cap P_{f,g}|_\delta \lesssim \delta^{-\epsilon} \frac{\delta^{-(n-2)}}{\|f-g\|_{C^2([0,1]^{n-1})}^s}. \end{equation}

Proof. If $P_{f,g} = \emptyset$ and $\delta \le t\;:\!=\|f-g\|_{C^2([0,1]^{n-1})} \le 4[{10^5K^4}/{\alpha(K^{-1}/6)^2}]^2\delta$ , it is easy to see (3·22) holds. In the following we assume $t\gt4[{10^5K^4}/{\alpha(K^{-1}/6)^2}]^2\delta$ .

Let $\{U\}$ be the dyadic cubes in the above proof. Similar to the proof of the above lemma, to show (3·22), it suffices to prove

(3·23) \begin{equation} |E \cap P_{f,g} \cap U|_\delta \lesssim \delta^{-\epsilon} \frac{\delta^{-(n-2)}}{\|f-g\|_{C^2([0,1]^{n-1})}^s} \end{equation}

for each U. To this end, we fix $U \subset [0,1]^{n-1}$ and write $P_{f,g}=P_{f,g} \cap U$ for simplicity. In the following, let all the notations be same as those in the proof of Lemma 1·17.

If we are in Case 1 (and also Subcase 2(i)), then (by considering $\partial U^+$ in each facet of $\partial U$ ), we know $ P_{f,g} = \Psi(\Omega) $ where $\Omega= \cup_{x \in \partial U^+}\{(\bar x, I_x)\} \subset \mathbb{R}^{n-1}$ . Recall from (3·6) that $|I_x| \sim {\delta}/{\sqrt{\lambda t}} \sim {\delta}/{t}$ . Thus $\Omega$ can be covered by $C ({t}/{\delta})^{n-2}$ many balls $B^{n-1}(x_i,{\delta}/{t})$ of radius ${\delta}/{t}$ . Since $\|\mbox{Lip }\Psi\|_{L^\infty(\Omega)} \lesssim 1$ , we know $P_{f,g}$ can be covered by $\bar C ({t}/{\delta})^{n-2}$ many balls $B^{n-1}(y_i,{\delta}/{t})$ of radius ${\delta}/{t}$ .

Since ${\delta}/{t} \ge \delta$ and E is a $(\delta,n-2+s,\delta^{-\epsilon})$ -set, we have

\begin{align*} |E \cap P_{f,g}|_\delta & \le \sum_{i=1}^{\bar C (t\delta^{-1})^{n-2}} |E \cap B^{n-1}\left(y_i, \frac{\delta}{t}\right)|_\delta \\[5pt] & \le \bar C\left(t\delta^{-1}\right)^{n-2}\delta^{-\epsilon}\left(\frac{\delta}{t}\right)^{n-2+s} |E| \\[5pt] & \lesssim \delta^{-\epsilon} \frac{\delta^{-(n-2)}}{t^s}=\delta^{-\epsilon} \frac{\delta^{-(n-2)}}{\|f-g\|_{C^2([0,1]^{n-1})}^s}\end{align*}

which gives (3·22).

Assume we are in Subcase 2 (ii). Recall $\bar t\;:\!= \|f-g\|_{C^2(I_F^{n-1})} \ge t=\|f-g\|_{C^2([0,1]^{n-1})}$ .

We first consider the case $\bar \lambda \in [0,\delta] $ . In this case, noting that $\sqrt{(\bar\lambda +\delta) / \bar t} \le 2\sqrt{\delta / \bar t}$ and recalling (3·21), we know we know $ P_{f,g} \subset B(x_M, 6K\sqrt{\delta / \bar t})$ . As a result, noting that $n-2+s \ge 1+s \ge 2s$ , we have

\begin{align*} |E \cap P_{f,g}|_\delta & \le |E \cap B(x_M, 6K\sqrt{\delta / \bar t})|_\delta \\[5pt] & \le \delta^{-\epsilon}\left(6K\frac{\delta}{\bar t }\right)^{n-2+s} |E| \\[5pt] & \lesssim \delta^{-\epsilon} \frac{\delta^s}{\bar t^s}\delta^{-(n-2+s)} \le \delta^{-\epsilon} \frac{\delta^{-(n-2)}}{\|f-g\|_{C^2([0,1]^{n-1})}^s}\end{align*}

which gives (3·22).

For the case $\bar \lambda \gt \delta $ , noting that $\sqrt{(\bar\lambda +\delta) / \bar t} \le 2\sqrt{\bar\lambda / \bar t}$ and recalling (3·20), we know $ P_{f,g} \subset B(x_M, 6K\sqrt{\bar\lambda / \bar t})$ . Moreover, recalling the standing assumption in Subcase 2(ii) that $\Delta ={\Delta}(f,g) \lt \frac{\alpha(K^{-1}/6)^2}{10^5K^4}\|h\|_{C^2(I^{n-1}_F)} = \frac{\alpha(K^{-1}/6)^2}{10^5K^4} \bar t$ and inequality (3·17) that $\bar \lambda \le 10K^2 \Delta$ , we know

$$ 6K\sqrt{\bar\lambda / \bar t} \le 6K \frac{\alpha(K^{-1}/6)}{100K} \le \frac{\alpha(K^{-1}/6)}{16} \le \frac1{4}\mathrm{diam \, } U $$

where we recall ${\alpha(K^{-1}/6)}/4 \lt \mathrm{diam \, } U \lt {\alpha(K^{-1}/6)}/2$ . By further recalling $x_M \in (6/5) U$ , we deduce that

$$ P_{f,g} \subset B\left(x_M, 6K\sqrt{\bar\lambda / \bar t}\right) \subset \frac32 U \subset 2U. $$

On the other hand, by Corollary 2·2 (B), we can assume h is strictly convex in 2U. Thus

$$ \Lambda_{2\delta}(h)\;:\!=\{x \in 2U\; :\; h(x) \le 2\delta\}$$

is a convex set. Since $x_M$ is the unique point in 2U such that $\nabla h(x) =0$ , we know $x_M$ is the unique point reaches the minimum of h. And we may assume $h(x_M) \lt 2\delta$ (otherwise $h(x) \gt h(x_M) \ge 2\delta$ for all $x \in 2U$ and $P_{f,g}=\{x_M\}$ ). In particular, $x_M \notin \partial \Lambda_{2\delta}(h)= \{x \in 2U\; :\; h(x) =2\delta\}$ . Hence for all $x \in \partial \Lambda_{2\delta}(h)$ , we have $\nabla h(x) \ne 0$ . Applying the preimage theorem for h, we know $\partial \Lambda_{2\delta}(h)$ is an $(n-2)$ -dimensional manifold. Moreover, $\{x \in 2U\; :\; h(x) \lt2\delta\}$ is a nonempty open set and $\mathcal{L}^{n-1}(\Lambda_{2\delta}(h))\gt0$ .

Noting that $\Lambda_{2\delta}(h) \subset B(x_M, 6K\sqrt{\bar\lambda / \bar t})$ (since $\partial \Lambda_{2\delta}(h) \subset P_{f,g} \subset B(x_M, 6K\sqrt{\bar\lambda / \bar t})$ and $\Lambda_{2\delta}(h)$ is convex), applying Lemma 2·7, we have

(3·24) \begin{equation} \mathcal{H}^{n-2}(\partial \Lambda_{2\delta}(h)) \le \mathcal{H}^{n-2}\left(\partial B\left(x_M, 6K\sqrt{\bar\lambda / \bar t}\right)\right) \lesssim \sqrt{\bar\lambda / \bar t}^{n-2}. \end{equation}

For any $x \in P_{f,g}= \{x \in 2U\; :\; |h(x)| \le 2\delta\} \subset \Lambda_{2\delta}(h)$ , we know $x=x_M+\xi_x \tau$ for some $\xi_x \in S^{n-2}$ and $\tau \in \hat I_{\xi_x}$ . We also know that there exists $\tau' \in \hat I_{\xi_x}$ such that $x_M+\xi_x \tau' \in \partial \Lambda_{2\delta}(h)= \{x \in 2U\; :\; h(x) = 2\delta\}$ . Using $|\tau -\tau'| \le |\hat I_{\xi_x}| \lesssim\delta /\sqrt{ \bar\lambda\bar t}$ , we know

\begin{equation*} \mathrm{dist \, }(x, \partial \Lambda_{2\delta}(h) ) \le |\tau -\tau'| \lesssim\delta /\sqrt{ \bar\lambda\bar t}, \quad x \in P_{f,g}. \end{equation*}

Consequently, we know $P_{f,g} \subset [\partial \Lambda_{2\delta}(h)]^{C_1\delta /\sqrt{ \bar\lambda\bar t}}$ for some constant $C_1\gt0$ where $[\partial \Lambda_{2\delta}(h)]^a$ is the a-neighbourhood of $\partial \Lambda_{2\delta}(h)$ . Combining (3·24) and by an elementary computation, we know $P_{f,g}$ can be covered by $C_2 ({\bar \lambda}/{\delta})^{n-2}$ many balls $B^{n-1}(x_i,\delta/\sqrt{\bar \lambda \bar t})$ of radius $\delta/\sqrt{\bar \lambda \bar t}$ for some constant $C_2\gt0$ .

Since $\delta/\sqrt{\bar \lambda \bar t} \ge \delta$ and E is a $(\delta,n-2+s,\delta^{-\epsilon})$ -set, recalling that $\bar\lambda \le \bar t$ and $t = \|f-g\|_{C^2([0,1]^{n-1})}\le \bar t$ , we have

\begin{align*} |E \cap P_{f,g}|_\delta & \le \sum_{i=1}^{C_2(\bar \lambda\delta^{-1})^{n-2}} |E \cap B^{n-1}(y_i, \delta/\sqrt{\bar \lambda \bar t})|_\delta \\[3pt] & \le C_2(\bar \lambda \delta^{-1})^{n-2}\delta^{-\epsilon}(\delta/\sqrt{\bar \lambda \bar t})^{n-2+s} |E| \\[3pt] & = C_2\delta^{-\epsilon} \frac{\bar \lambda^{(n-2-s)/2}}{\bar t^{(n-2+s)/2}} \delta^{s}|E| \\[3pt] & \lesssim \delta^{-\epsilon} \frac{\delta^{-(n-2)}}{\bar t^s} \le \delta^{-\epsilon} \frac{\delta^{-(n-2)}}{\|f-g\|_{C^2([0,1]^{n-1})}^s}\end{align*}

which gives (3·22).

The proof is complete.

We are able to show Theorem 1·18.

Proof of Theorem 1·18. Let $2\log (1/\delta_0)\lt \delta_0^{-\epsilon}$ .

For each $f \in F$ and $i \in \mathbb{N}$ , define

$$ F_i(f)\;:\!= \{ g \in F\; :\; \|f-g\|_{C^2([0,1]^{n-1})} \in (2^{-(i+1)},2^{-i}] \} . $$

Note that if $i \ge \log (1/\delta)$ , then $F_i(f) =\{f\}$ due to F being $\delta$ -separated. Since F is $(\delta,t,\delta^{-\epsilon})$ -set, we have $|F_i(f)| \le \delta^{-\epsilon}2^{-it}|F|$ . Using Lemma 1·2, we have

(3·25) \begin{align} \sum_{f \in F}\sum_{g \in F, g \ne f} \mathcal{L}^n(f^\delta \cap g^\delta) &= \sum_{f \in F}\sum_{i =1}^{\log (1/\delta)}\sum_{g \in F_i(f)} \mathcal{L}^n(f^\delta \cap g^\delta) \notag\\ &\lesssim \delta^{-\epsilon} |F| \sum_{f \in F}\sum_{i=1}^{\log (1/\delta)}2^{-it} \frac{\delta^2}{2^{-(i+1)}} \notag \\ & \le 2\delta^{-\epsilon}|F|\delta^2\sum_{f \in F} \sum_{i=1}^{\log (1/\delta)} 2^{(1-t)i} \notag\\ & \le 2\delta^{-\epsilon}|F|^2 \delta^{1+t} \log (1/\delta) \le \delta^{-2\epsilon}|F|^2 \delta^{1+t}, \end{align}

where in the second last inequality we note that $\sum_{i=1}^{\log (1/\delta)} 2^{(1-t)i} \le \sum_{i=1}^{\log (1/\delta)} 2^{(1-t)\log (1/\delta) } = \log (1/\delta) \delta^{t-1} $ .

Notice that

(3·26) \begin{equation} \int_{[0,1]^n} \left(\sum_{f \in F} \chi_{f^\delta}\right)^2 \, dx = \sum_{f \in F}\sum_{g \in F}\int_{[0,1]^n} \chi_{f^\delta}\chi_{g^\delta} \, dx = \sum_{f \in F}\sum_{g \in F} \mathcal{L}^n(f^\delta \cap g^\delta). \end{equation}

Combining (3·25) and (3·26), we have

\begin{align*} \int_{ [0,1]^n} \left(\sum_{f \in F} \chi_{f^\delta}\right)^2 \, dx & = \sum_{f \in F} \mathcal{L}^n(f^\delta ) + \sum_{f \in F}\sum_{g \in F, g \ne f} \mathcal{L}^n(f^\delta \cap g^\delta) \\[4pt] & \lesssim |F| \delta +\delta^{-2\epsilon}|F|^2 \delta^{1+t} \lesssim \delta^{-2\epsilon}|F|^2 \delta^{1+t}, \end{align*}

which gives (1·11) as desired.

4· Furstenberg sets of hypersurfaces

In this section, we focus on the dimension lower bound of Furstenberg sets of hypersurfaces. Precisely, we prove Theorem 1·16 and use Theorem 1·16 to show Theorem 1·13.

First, we make some reductions before showing Theorem 1·13. Let $E \subset [0,1]^{n}$ be a $(s, t)_{n-1}$ -Furstenberg set with parameter set $F \subset C^2([0,1]^{n-1}, [0,1])$ with $\mathrm{dim_H} F \geq t$ . Note that if $0 \lt t \le s \le 1$ , then the desired bound is $n-2+s+t$ . Otherwise, the desired bound is $n-2+2s$ . Noticing that once the bound $n-2+s+t$ is proved, the bound $n-2+2s$ immediately follows, since if $t \ge s$ , any $(n-2+s,t)_{n-1}$ -Furstenberg set contains an $(n-2+s,s)_{n-1}$ -Furstenberg set. It suffices to show the bound is $n-2+s+t$ when $0 \lt t \le s \le 1$ .

In the following of this section, we assume $0 \lt t \le s \le 1$ . To show Theorem 1·9, we also make some reductions. Let $\Omega$ be the $(\delta,s,t,\delta^{-\epsilon})_{n-1}$ -configuration in Theorem 1·9. Write $F \;:\!= \pi_{1}(\Omega) \subset C^2([0,1]^{n-1})$ , and $E(f) = \{x \in \mathbb{R}^{n}\; :\; (f,x) \in \Omega\} \subset \mbox{graph}(f)$ . By replacing F and E(f) by maximal $\delta$ -separated subsets, we may assume that F, E(f), and $\Omega$ are finite and $\delta$ -separated to begin with. Furthermore, F contains a $(\delta,t,\delta^{-2\epsilon})$ -subset $\bar{F} \subset F$ of cardinality $(1/2) \delta^{2\epsilon}\delta^{-t} \le |\bar{F}| \leq \delta^{-t}$ by [Reference Orponen and Shmerkin 18 , lemma 2·7]. Recalling E(f) is a $(\delta,n-2+s,\delta^{-\epsilon})$ -set implies that $M = |E(f)| \ge \delta^{\epsilon-(n-2+s)}$ . If $M\gt \delta^{-(n-2+s)}$ , we can always find a $(\delta,n-2+s,\delta^{-2\epsilon})$ -set $\bar E(f) \subset E$ such that $\bar M \equiv|\bar E(f)| \ge \delta^{2\epsilon-(n-2+s)}$ for all $f\in \bar{F}$ . Then $\bar{\Omega} \;:\!= \{(f,x)\; :\; f\in \bar{F} \text{ and } x \in \bar E(f)\}$ remains a $(\delta,s,t,\delta^{-2\epsilon})_{n-1}$ -configuration with $|\bar E(f)| \equiv \bar M$ . Apparently, it suffices to prove Theorem 1·9 for this sub-configuration, so we may assume that

(4·1) \begin{equation} |F| \in \left[ \frac12\delta^{2\epsilon-t},\delta^{-t}\right] \text{ and } M \in [ \delta^{2\epsilon-(n-2+s)},\delta^{-(n-2+s)}]\end{equation}

to begin with. Finally, F is a cinematic family implies that $\|f\|_{C^2([0,1]^{n-1})} \le K$ for all $f \in F$ .

The above notions and assumptions will be fixed throughout this section. We need several auxiliary lemmas.

Lemma 4·1 Let $A_1, \ldots, A_m \subset \mathbb{R}^n$ be m Lebesgue measurable subsets of finite Lebesgue measure. Then

$$ \mathcal{L}^n\left(\bigcup_{i=1}^m A_i\right) \ge \frac{[\sum_{i=1}^{m} \mathcal{L}^n(A_i) ]^2}{\sum_{i,j=1}^m\mathcal{L}^n( A_i\cap A_j)} . $$

Proof. Applying Cauchy–Schwarz inequality to $\sum_{i=1}^{m}\chi_{ A_i}$ and noting that $\|\sum_{i=1}^{m}\chi_{ A_i}\|_{L^2(\mathbb{R}^n)} =\sum_{i,j=1}^m\mathcal{L}^n( A_i\cap A_j)$ , we have

$$ \|\sum_{i=1}^{m}\chi_{ A_i}\|_{L^1(\mathbb{R}^n)} = \int_{\mathrm{spt} (\sum_{i=1}^{m}\chi_{ A_i})} \left|\sum_{i=1}^{m}\chi_{ A_i}(x)\right| \, d \mathcal{L}^n (x) \le \mathcal{L}^n\left(\bigcup_{i=1}^m A_i\right)^{1/2} \|\sum_{i=1}^{m}\chi_{ A_i}\|_{L^2(\mathbb{R}^n)} $$

as desired.

In the remaining part of this section, for any set $A \subset [0,1]^n$ , let $P_{A}$ be the projection of A to the first $n-1$ coordinates.

Proof. Let $\bar\delta_0^{-\epsilon} \gt 4^{n-2+s}(2K)^{n-1} $ . Note that $P_{E(f)} \subset [0,1]^{n-1} \subset \mathbb{R}^{n-1}$ . We need to check for any ball $B^{n-1}(x,r) \subset \mathbb{R}^{n-1}$ with $r\ge \delta$ ,

$$ |P_{E(f)} \cap B^{n-1}(x,r)|_{\delta} \le \delta^{-3\epsilon} r^{n-2+s}|P_{E(f)}|_\delta. $$

Since $\|f\|_{C^2([0,1]^{n-1})} \le K$ , graph(f) is a K-Lipschitz graph. Thus by E(f) being $\delta$ -separated, we know $P_{E(f)}$ is at least $(2K)^{-1}\delta$ -separated and $|P_{E(f)}|_\delta \ge (2K)^{-(n-1)}|P_{E(f)}|= (2K)^{-(n-1)}|E(f)|=(2K)^{-(n-1)}M$ . Thus it suffices to show

(4·2) \begin{equation} |P_{E(f)} \cap B^{n-1}(x,r)|_{\delta} \le \delta^{-3\epsilon}r^{n-2+s}(2K)^{-(n-1)}M \quad x \in \mathbb{R}^{n-1}, \ r \ge \delta.\end{equation}

In the following, for any $y \in \mathbb{R}^n$ , we write $y=(\bar y, y_n)$ where $\bar y = (y_1, \cdots, y_{n-1})$ . Notice that

(4·3) \begin{equation} P_{E(f)} \cap B^{n-1}(x,r) = \{ \bar y\; :\; y \in E(f), \ \bar y \in B^{n-1}(x,r) \}.\end{equation}

Since graph(f) is a K-Lipschitz graph, we know that the set $\{y \in E(f)\; :\; \bar y \in B^{n-1}(x,r)\}$ can be covered by K balls $B^n(z_i,4r) \subset \mathbb{R}^n$ with $\bar z_i =x$ for all $i=1,\ldots, K$ . Using that E(f) is a $(\delta,n-2+s,\delta^{-2\epsilon})$ -set and $|E(f)|=M$ , we deduce that

$$ |\{y \in E(f)\; :\; \bar y \in B^{n-1}(x,4r)\}|_{\delta} \le \sum_{i=1}^K |\{y \in E(f)\; :\; y \in B^{n}(z_i,4r)\}|_{\delta} \le \delta^{-2\epsilon} (4r)^{n-2+s}M. $$

Recalling (4·3), we know

$$ |P_{E(f)} \cap B^{n-1}(x,4r)|_\delta \le |\{y \in E(f)\; :\; \bar y \in B^{n-1}(x,4r)\}|_{\delta} \le \delta^{-2\epsilon} (4r)^{n-2+s}M $$

and by $4^{n-2+s}(2K)^{n-1} \lt \delta^{-\epsilon}$ , we conclude (4·2).

Lemma 4·3. For any $\epsilon\gt0$ , let $\bar \delta_0$ be as in Lemma 4·2. For all $\delta \in (0, \bar \delta_{0}]$ , let $E^\delta(f)$ be the $\delta$ -neighbourhood of E(f) for all $f \in F$ . Then

(4·4) \begin{equation} \mathcal{L}^n(E^\delta(f) \cap E^\delta(g)) \lesssim \delta^{-3\epsilon} \frac{\delta^2}{\|f-g\|_{C^2([0,1]^{n-1})}^s}. \end{equation}

Proof. Write $\|f-g\|_{C^2([0,1]^{n-1})} = d$ . Noting that $\delta$ -balls in $\mathbb{R}^n$ has measure $\sim \delta^n$ , it suffices to show,

$$ |E^\delta(f) \cap E^\delta(g)|_{\delta} \lesssim \delta^{-3\epsilon} \frac{\delta^{-(n-2)}}{d^s}. $$

Also, noticing $E^\delta(f) \cap E^\delta(g) \subset f^\delta \cap g^\delta$ has vertical thickness $\lesssim\delta$ , we know

$$ |P_{E^\delta(f) \cap E^\delta(g)}|_{\delta} \sim |E^\delta(f) \cap E^\delta(g)|_{\delta}. $$

Observe that

$$ P_{E^\delta(f) \cap E^\delta(g)} \subset P_{E^\delta(f)} \cap P_{f^\delta \cap g^\delta}, $$

it further reduces to showing

(4·5) \begin{equation} |P_{E(f)} \cap P_{f^\delta \cap g^\delta}|_\delta \lesssim \delta^{-3\epsilon} \frac{\delta^{-(n-2)}}{d^s}. \end{equation}

Applying Lemma 4·2 that $P_{E(f)}$ is a $(\delta,n-2+s,\delta^{-3\epsilon})$ -set, recalling that $M \le \delta^{-(n-2+s)}$ and using Lemma 3·1 with $E=P_{E(f)}$ (where $P_{f^\delta \cap g^\delta}$ is denoted by $P_{f,g}$ in Lemma 3·1), we deduce (4·5) as desired.

We are able to prove Theorem 1·16. The proof is partially motivated by [Reference Héra, Shmerkin and Yavicoli 9 , Appendix A].

Proof of Theorem 1·16. Let C be the implicit constant in (4·4) and $ \bar \delta_0(\epsilon)$ be as in Lemma 4·3. The relationship between $\delta_0=\delta_0(\epsilon)$ and $\epsilon$ in this theorem is

(4·6) \begin{equation} \delta_0(\epsilon) \lt \bar\delta_0 \quad\mbox{ and }\quad 2C\log(1/\delta_0) \lt \frac12\delta_0^{-\epsilon}. \end{equation}

Let $\mathcal{E}^\delta$ be the $\delta$ -neighbourhood of $\mathcal{E}$ . Then it suffices to show $\mathcal{L}^n(\mathcal{E}^\delta) \geq \delta^{12\epsilon +2- t-s}$ . Observe that

$$ \mathcal{E}^\delta = \bigcup_{f \in F} E^\delta(f). $$

Applying Lemma 4·1, we have

$$ \mathcal{L}^n\left(\mathcal{E}^\delta\right) = \mathcal{L}^n\left(\bigcup_{f \in F} E^\delta(f)\right) \ge \frac{\left[\sum_{f \in F}\mathcal{L}^n( E^\delta(f))\right]^2}{\sum_{f, g\in F}\mathcal{L}^n( E^\delta(f)\cap E^\delta(g))}. $$

Recalling (4·1) and noting that $ \mathcal{L}^n(E^\delta(f)) = M \delta^n \ge \delta^{2\epsilon} \delta^{2-s}$ for all $f \in F$ and $|F|\ge (1/2) \delta^{2\epsilon} \delta^{-t} \ge \delta^{3\epsilon} \delta^{-t}$ , we deduce that

$$ \left[\sum_{f \in F}\mathcal{L}^n( E^\delta(f))\right]^2 \ge [ \delta^{5\epsilon +2 -t-s} ]^2. $$

It suffices to show

(4·7) \begin{equation} \sum_{f, g\in F}\mathcal{L}^n( E^\delta(f)\cap E^\delta(g)) \le \delta^{-6\epsilon}\delta^{2-t-s} \end{equation}

Similar to the proof of Theorem 1·18, for each $f \in F$ and $i \in \mathbb{N}$ , define

$$ F_i(f)\;:\!= \{ g \in F\; :\; \|f-g\|_{C^2([0,1]^{n-1})} \in (2^{-(i+1)},2^{-i}] \} . $$

Note that if $i \ge \log (1/\delta)$ , then $F_i(f) =\{f\}$ due to F being $\delta$ -separated. Since F is $(\delta,t,\delta^{-2\epsilon})$ -set, we have $|F_i(f)| \le \delta^{-2\epsilon}2^{-it}|F|$ . Using Lemma 4·3, we have

(4·8) \begin{align} \sum_{f \in F}\sum_{g \in F, f\ne g} \mathcal{L}^n( E^\delta(f)\cap E^\delta(g)) &= \sum_{f \in F}\sum_{i =1}^{\log (1/\delta)}\sum_{g \in F_i(f)} \mathcal{L}^n( E^\delta(f)\cap E^\delta(g)) \notag\\ &\le C\delta^{-5\epsilon} |F| \sum_{f \in F}\sum_{i=1}^{\log (1/\delta)}2^{-it} \frac{\delta^2}{2^{-(i+1)s}} \notag \\ & \le 2C\delta^{-5\epsilon}|F|\delta^2\sum_{f \in F} \sum_{i=1}^{\log (1/\delta)} 2^{(s-t)i} \notag\\ & \le 2C\delta^{-5\epsilon}|F|^2 \delta^2 \log (1/\delta)\delta^{t-s} \le \frac12 \delta^{-6\epsilon}\delta^{2-t-s}, \end{align}

where in the second last inequality we note that $s\ge t$ and $\sum_{i=1}^{\log (1/\delta)} 2^{(s-t)i} \le \log (1/\delta) 2^{(s-t)\log (1/\delta) } = \log (1/\delta) \delta^{t-s} $ , and in the last inequality we recall $|F| \le \delta^{-t}$ and note $\delta^{-5\epsilon}|F|^2 \le \delta^{-5\epsilon-2t}$ and (4·6). Finally, noticing that

$$ \sum_{f, g\in F}\mathcal{L}^n( E^\delta(f)\cap E^\delta(g)) = \sum_{f \in F} \mathcal{L}^n( E^\delta(f)) + \sum_{f \in F}\sum_{g \in F, g \ne f} \mathcal{L}^n(E^\delta(f)\cap E^\delta(g)) $$

and

$$ \sum_{f \in F} \mathcal{L}^n( E^\delta(f)) \le|F| \delta^n M \le \delta^{2-s-t} \le \frac12 \delta^{-6\epsilon}\delta^{2-s-t}, $$

we conclude (4·7) as desired.

We prove Theorem 1·13 using Theorem 1·16.

Proof of Theorem 1·13. Fix $0 \lt t \leq s \leq 1$ and moreover, fix $t' \in [t/2,t]$ and $t' \leq s' \lt s$ . Since $\mathcal{H}^{t'}_{\infty}(F) \gt 0$ , there exists $\alpha = \alpha(F,t') \gt0$ and $F_1 \subset F$ such that $\mathcal{H}^{t'}_\infty(F_1)\gt \alpha$ , where

(4·9) \begin{equation} F_1 \;:\!= \{ f\in F\; :\; \mathcal{H}^{n-2+s'}_\infty(E \cap \mbox{graph}(f))\gt \alpha\}.\end{equation}

This follows from the sub-additivity of Hausdorff content.

Next, let $\epsilon(\bar s,\bar t)\gt0$ be the constant such that Theorem 1·16 holds for $\bar s,\bar t$ . Define $\epsilon\;:\!= \min_{\bar t \in [t',t], \bar s \in [s',s]}\{\epsilon(\bar s,\bar t)\}$ (where we note that $\epsilon(\bar s,\bar t)\gt0$ whenever $\bar s,\bar t\gt0$ ). We further choose $k_0 = k_{0}(\alpha,\epsilon) = k_0(F,t',\epsilon) \in \mathbb{N}$ satisfying

(4·10) \begin{equation} \alpha\gt\sum_{k=k_0}^{\infty} \frac{1}{k^2} \quad \text{and} \quad k_{0}^{2} \leq \frac{2^{\epsilon k_{0}}}{C},\end{equation}

where $C \geq 1$ is a constant to be determined later. Let $\mathcal{U} = \{ D(x_i,r_i)\}_{i \in\mathcal{I}}$ be an arbitrary cover of E by dyadic $r_i$ -cubes with $r_i \le 2^{-k_0}$ and $E \cap D(x_i,r_i) \ne \emptyset$ for all $i \in \mathcal{I}$ . For $k \ge k_0$ , write

$$ \mathcal{I}_k\;:\!= \{i \in \mathcal{I}\; :\; r_i = 2^{-k} \} \quad \mbox{and} \quad E_k\;:\!= \{\cup D(x_i,r_i)\; :\; i \in \mathcal{I}_k \}.$$

By the pigeonhole principle and (4·10) we deduce that for each $f \in F_1$ , there exists $k(f) \ge k_0$ such that

$$ \mathcal{H}^{n-2+s'}_\infty( E \cap \mbox{graph}(f) \cap E_{k(f)})\gt k(f)^{-2} .$$

Using the pigeonhole principle again we obtain that there exists $k_1 \ge k_0$ such that

(4·11) \begin{equation} \mathcal{H}^{t'}_\infty(F_2)\gt k_1^{-2}, \end{equation}

where $ F_2 \;:\!= \{f \in F_1\; :\; k(f) = k_1\}.$ By the construction of $F_2$ , we have

\begin{align*}\mathcal{H}^{n-2+s'}_\infty(\mbox{graph}(f)\cap F_{k_1})\ge \mathcal{H}^{n-2+s'}_\infty(E \cap \mbox{graph}(f) \cap E_{k_1})\gt k_1^{-2}, \qquad f \in F_{2}.\end{align*}

Write $\delta =2^{-k_1}$ . By (4·11) and [Reference Orponen and Shmerkin 18 , lemma 2·7], we know that there exists a $\delta$ -separated $(\delta,t',Ck_{1}^{2})$ -set $P \subset F_2$ satisfying $(k_{1}^{-2}/C)\delta^{-t'} \leq |P| \leq \delta^{-t'}$ . Since $P \subset F_{2}$ , we have

(4·12) \begin{equation} \mathcal{H}^{n-2+s'}_\infty(\mbox{graph}(f)\cap E_{k_1})\gt k_1^{-2}, \qquad f \in P.\end{equation}

Applying [Reference Orponen and Shmerkin 18 , lemma 2·7] again to $\mbox{graph}(f)\cap E_{k_1}$ , $f \in P$ , we obtain $\delta$ -separated $(\delta,n-2+s',Ck_{1}^{2})$ -sets $E(f) \subset \mbox{graph}(f)\cap E_{k_1}$ such that

\begin{equation*} |E(f)| \equiv M \geq (k_{1}^{-2}/C)\delta^{-(n-2+s')} \stackrel{\eqref{para31}}{\geq} \delta^{\epsilon -(n-2+s')}, \qquad f \in P.\end{equation*}

By (4·10), P is a $(\delta,t',\delta^{-\epsilon})$ -set, and each E(f) is a $(\delta,n-2+s',\delta^{-\epsilon})$ -set. Therefore,

$$ \Omega\;:\!= \{(f,x)\; :\ f \in P \text{ and } x \in E(f)\} $$

is a $(\delta,s',t',\delta^{-\epsilon},M)_{n-1}$ -configuration. Recall that $\epsilon \leq \epsilon(t')$ by the definition of $\epsilon$ . Letting

$$ \mathcal{E}\;:\!= \bigcup_{f \in P} E(f) $$

and applying Theorem 1·16, we deduce that $|\mathcal{E}|_{\delta} \geq \delta^{16\epsilon - (n-2)-s'-t'}$ .

Since $E(f) \subset E_{k_1}$ for each $f \in P$ , we have $\mathcal{E} \subset E_{k_1}$ , which implies

$$ |\mathcal{I}_{k_1}| = |E_{k_1}|_\delta \geq |\mathcal{E}|_{\delta} \geq \delta^{16\epsilon - (n-2)-s'-t'}.$$

Then

$$ \sum_{i \in \mathcal{I}} r_i^{ (n-2)+s'+t'-16\epsilon} \ge \sum_{i \in \mathcal{I}_{k_1}} r_i^{ (n-2)+s'+t'-16\epsilon} = \delta^{ (n-2)+s'+t'-16\epsilon}|\mathcal{I}_{k_1}| \geq 1. $$

As the covering was arbitrary, we infer that $\mathrm{dim_H} E \ge (n-2)+s'+t'-16\epsilon$ . Sending $s' \nearrow s$ , $t' \nearrow t$ , and $\epsilon \searrow 0$ , we arrive at the desired result.

5· Proof of Theorem 1·4

We prove Theorem 1·9 and Theorem 1·4 in this section. We first use Theorem 1·9 to show Theorem 1·4 by verifying Theorem 1·4 is a special case of Theorem 1·9. To this end, we recall the definition of principle curvature.

Definition 5·1 (Principal curvature). Let M be an n-dimensional $C^2$ -manifold and $\Sigma \subset M$ be an m-dimensional $C^2$ -manifold ( $m \lt n$ ). Let $\nu \in C^2(TM)$ be a normal vector field, i.e. for each $x \in \Sigma$ , $\nu(x) \perp T_x\Sigma$ . Then the m real eigenvalues $\kappa_1, \ldots \kappa_m$ of $II_x(\nu)$ are called principal curvatures of $\Sigma$ at x (with respect to $\nu$ ) and the corresponding eigenvectors are called principal directions where $II_x(\nu)$ is the second fundamental form (of $\nu$ ), i.e.

$$ II_x(\xi,\eta)\;:\!= -\langle \nabla_\xi \nu, \eta\rangle \quad \xi,\eta \in T_x\Sigma. $$

We relate the “non-degenerate” condition for surfaces to principal curvatures.

Proof. Fix any $x \in \Sigma$ . Let $\{e_1,\cdots,e_{n-1}\} \subset T_x\Sigma$ be the principal directions and $\gamma_i\; :\; [-1,1] \to \Sigma$ such that $\gamma_i(0)=x$ and $\dot \gamma_i(0) = e_i$ , $i=1,\cdots n-1$ . Thus by definition, the $n-1$ principal curvatures are $\kappa_i = -\langle \nabla_{e_i} \nu, e_i\rangle $ , $i = 1, \cdots, n-1$ .

If $\Sigma$ satisfies Definition 1·8(ii), then we know

$$ \kappa_i\kappa_j=\langle \nabla_{e_i} \nu, e_i\rangle \langle \nabla_{e_i} \nu, e_i\rangle= \langle \nabla_{\dot \gamma_i(0)} \nu, \dot \gamma_i(0)\rangle \langle \nabla_{\dot \gamma_j(0)} \nu, \dot \gamma_j(0)\rangle \gt0 \quad 1 \le i, j \le n-1. $$

Hence all $\kappa_i$ are not vanishing and have same sign.

Conversely, let $\gamma_k\; :\; [-1,1] \to \Sigma$ such that $\gamma_k(0)=x$ and $\dot \gamma_k(t) \ne 0$ for all $t \in [-1,1]$ , $k=1,2$ . Write $\dot \gamma_k(t)= \sum_{i=1}^{n-1} a_k^i(t) e_i $ , $k=1,2$ . We have

\begin{align*} \langle \nabla_{\dot \gamma_1(0)} \nu, \dot \gamma_1(0)\rangle \langle \nabla_{\dot \gamma_2(0)} \nu, \dot \gamma_2(0)\rangle & = \left[\sum_{i=1}^{n-1}(a_1^i)^2\langle \nabla_{e_i} \nu, e_i\rangle \right] \left[\sum_{i=1}^{n-1}(a_2^i)^2\langle \nabla_{e_i} \nu, e_j\rangle \right]\\[3pt] & =\sum_{i,j=1}^{n-1}(a_1^i)^2(a_2^j)^2\kappa_i\kappa_j \gt0, \end{align*}

That is, $\Sigma$ satisfies Definition 1·8(ii). The proof is complete.

We are able to show Theorem 1·4.

Proof of Theorem 1·4. Since $\Sigma \subset S^n$ , for all $x \in \Sigma$ , we know there exists a unique $\nu(x)\subset T_x S^n$ (up to $\pm 1$ ) such that $\nu(x) \perp x$ and $\nu(x) \perp T_x \Sigma$ . Also noting that $x \perp T_x S^n$ , we know $\mathrm{span}\{x,T_x \Sigma, \nu(x) \}=\mathbb{R}^{n+1}$ . Thus $\Sigma$ satisfies Definition 1·8(i). By Remark 1·6, we deduce that $\Sigma$ also satisfies Definition 1·8(ii). Thus Theorem 1·9 implies Theorem 1·4.

We are left to show Theorem 1·9. We make some further reductions.

Remark 5·3. By Lemma 5·2, if $\Sigma$ satisfies Definition 1·8(ii), we can without loss of generality assume all the principal curvatures $\kappa_i(x)$ are positive for all $x \in \Sigma$ , $i=1, \ldots,n-1$ . Finally, by the $C^2$ -regularity of $\Sigma$ , we know the $n-1$ principal curvatures $\kappa_i \in C^0(\Sigma, \mathbb{R_+})$ . Thus there exists $\kappa \ge 1$ such that

(5·1) \begin{equation} 0\lt\kappa^{-1} \le \min_{x \in \Sigma}\{\kappa_1(x), \ldots, \kappa_{n-1}(x)\} \le \max_{x \in \Sigma}\{\kappa_1(x), \ldots, \kappa_{n-1}(x)\} \le \kappa \lt \infty. \end{equation}

Proof. First we show (1) in Definition 1·10. For any $x \in \Sigma$ ,

(5·2) \begin{equation} \sup_{f_{z} \in F }\|f_{z}\|_{C^2([0,1]^{n-1})} \le \|\Sigma\|_{C^2([0,1]^{n-1})} \max_{z \in Z}|z| \lesssim 1,\end{equation}

where the implicit constant only depends on $\Sigma$ . Thus (1) holds. Also for (4), recalling Remark 1·11, the modulus of continuity $\alpha$ of F can be chosen as the one of $|\nabla^2 \Sigma|$ . Thus the family $\{\nabla_\xi \nabla_\xi f\}_{f \in F, \xi \in S^{n-2}}$ is equicontinuous.

Fix $y,z \in Z$ . We show (3) in Definition 1·10. Similar as (5·2), we have

(5·3) \begin{equation} \|f_y-f_{z}\|_{C^2([0,1]^{n-1})} \lesssim |y-z|.\end{equation}

Write $h\;:\!= f_y-f_{z}$ . We prove (3) by showing

(5·4) \begin{equation} \inf_{x \in [0,1]^{n-1}, \xi \in S^{n-2}} \{ |h(x)| + |\nabla h(x)| + |\nabla_\xi \nabla_\xi h (x)| \} \gtrsim |y-z|.\end{equation}

To apply the curvature condition (5·1), define $\tilde h \in C^2(\Sigma)$ by $\tilde h(w)\;:\!= (y-z) \cdot w$ , $w \in \Sigma$ . Thus $h=\tilde h \circ \Sigma$ . For each $\xi \in S^{n-2}$ , let $\xi_\ast\;:\!= \nabla \Sigma(x) \cdot \xi \in T_{\Sigma(x)} \Sigma$ . We know for all $x \in [0,1]^{n-1}$ and $\xi \in S^{n-2}$ ,

(5·5) \begin{equation} \langle\nabla h (x), \xi \rangle = \langle\nabla \tilde h (\Sigma(x)), \xi_\ast \rangle\end{equation}

and

(5·6) \begin{equation} \nabla_\xi \nabla_\xi h (x) = \nabla_{\xi_\ast} \nabla_{\xi_\ast}\tilde h (\Sigma(x)) + \langle\nabla\tilde h (\Sigma(x)), \nabla^2\Sigma(x) (\xi,\xi) \rangle. \end{equation}

where $\nabla^2 \Sigma(x)$ is a linear map from $T_x[0,1]^{n-1} \times T_x[0,1]^{n-1} $ to $T_w \Sigma$ . In particular, noting that $|\nabla \Sigma|$ is nowhere vanishing (since $\Sigma$ is a diffeomorphism), (5·5) implies that

(5·7) \begin{equation} |\nabla h (x)| \sim |\nabla \tilde h (\Sigma(x))|, \quad x \in [0,1]^{n-1}, \end{equation}

where the implicit constant depends on $\max_{x \in [0,1]^{n-1}}\{|\nabla \Sigma(x)|\}$ and $\min_{x \in [0,1]^{n-1}}\{|\nabla \Sigma(x)|\}$ .

Fix $x \in [0,1]^{n-1}$ and $\xi \in S^{n-2}$ in the following and let $w \;:\!=\Sigma(x)$ . Also, let $ \{e_1, \cdots, e_{n-1} \}$ $\subset T \Sigma= \Sigma \times \mathbb{R}^{n-1}$ be an orthonormal basis on each point in $\Sigma$ and $e_1(w) = \xi_\ast$ . Noting that $\tilde h (w)= (y-z) \cdot w$ , we have $\nabla_{e_i} \tilde h (w)= (y-z) \cdot \gamma_i '(0)$ where $\gamma_i :[-\epsilon,\epsilon] \to \Sigma$ is a $C^2$ -curve parametrized by arc-length such that $\gamma_i(0)=w$ and $\gamma_i'(0)=e_i$ . Thus

(5·8) \begin{equation} \nabla \tilde h (w)= \sum_{i=1}^{n-1} [(y-z) \cdot \gamma_i '(0)]e_i = \sum_{i=1}^{n-1} [(y-z) \cdot e_i]e_i.\end{equation}

Since $\Sigma$ satisfies Definition 1·8(i), we know there exists a $C^2$ vector field $\nu$ such that

(5·9) \begin{equation} \nu \perp w, \ \nu \perp T_{w} \Sigma \mbox { and } \mathrm{span} \{w, T_{w} \Sigma, \nu\} = \mathbb{R}^{n+1}.\end{equation}

Furthermore, we note that (5·10) implies $w \ne 0$ for all $w \in \Sigma$ , i.e. $\Sigma$ does not contain the origin. Denoting by $e_0\;:\!= {w}/{|w|}= {\Sigma(x)}/{|\Sigma(x)|}$ , we have

(5·10) \begin{equation} \nu \perp e_0, \ \nu \perp T_{w} \Sigma \mbox { and } \mathrm{span} \{e_0, T_{w} \Sigma, \nu\} = \mathbb{R}^{n+1}.\end{equation}

Thus $y-z$ can be uniquely written as

\begin{equation*} y-z = \sum_{i=0}^{n-1}[(y-z) \cdot e_i] e_i + [(y-z) \cdot \nu] \nu.\end{equation*}

Let $I_\Sigma$ be the first fundamental form of $\Sigma$ and $|I_\Sigma| = \sum_{i,j,k=1}^{n-1} |\Gamma_{ij}^k|$ where $\Gamma_{ij}^k= \langle \nabla_{e_j}e_i,e_k \rangle$ are the Christoffel symbols, $1 \le i \le n-1$ . Define

(5·11) \begin{equation} m\;:\!= \max_{p \in \Sigma} \{\langle e_0,\nabla_{e_1}e_1 \rangle|_{p}\}.\end{equation}

Recall $\kappa$ in (5·1) in Remark 5·3 and denote

$$ M \;:\!= \frac{\min\{\kappa^{-1}, \min_{p \in \Sigma}\{|p|\},1\}}{10n \max\{|I_\Sigma|,m, \|\Sigma\|_{C^2([0,1]^{n-1},\mathbb{R}^{n+1})},1 \}} \in \left(0,\frac{1}{10n}\right], $$

where the fact $M\gt0$ is because (5·10) implies $\min_{p \in \Sigma}\{|p|\} \gt0$ . In addition, we notice that $|p| \ge M$ for all $p \in \Sigma$ .

We consider the following three cases separately.

Case 1. If $|(y-z) \cdot e_0| \ge M |y-z|$ , then by $\tilde h (w)= (y-z) \cdot (|w|e_0)$ , we deduce that

$$ |h(x)| = |\tilde h (w)| = |w| |(y-z) \cdot e_0| \ge |w|M|y-z| \ge M^2|y-z| \sim |y-z|.$$

Thus (5·4) holds.

Case 2. If $|(y-z) \cdot e_i| \ge M |y-z|$ for some $i=1,\cdots, n-1$ , then by (5·7) and (5·8),

$$ |\nabla h(x)| \sim |\nabla \tilde h(w)| \ge |(y-z) \cdot e_i| \ge M |y-z| \gtrsim |y-z|,$$

which gives (5·4).

Case 3. $|(y-z) \cdot e_i| \lt M |y-z| \le (10n)^{-1}|y-z| $ for all $i=0,\cdots, n-1$ , which implies $|(y-z) \cdot \nu| \ge (1/2) |y-z|$ . By direct computation, we have

\begin{align*} \nabla_{e_i}\nabla_{e_i} \tilde h(w) & = \nabla_{e_i}[(y-z)\cdot \gamma'_i(0)] = (y-z)\cdot \nabla_{e_i}e_i\\ & = \sum_{j=0}^{n-1}[(y-z) \cdot e_j]\langle e_j,\nabla_{e_i}e_i \rangle + [(y-z) \cdot \nu]\langle \nu,\nabla_{e_i}e_i \rangle.\end{align*}

Recalling that (5·11) and noting that $M \le {\kappa^{-1}}/{10n m}$ , we know

(5·12) \begin{equation} |\langle e_0,\nabla_{e_1}e_1 \rangle|M \le m \frac{\kappa^{-1}}{10n m} \le \frac{\kappa^{-1}}{10n}.\end{equation}

Moreover, for $j=1,\cdots, n-1$ , noting that $M \le {\kappa^{-1}}/{10n |I_\Sigma|}$ , we have

(5·13) \begin{equation} |\langle e_j,\nabla_{e_i}e_i \rangle| M = |\Gamma_{ji}^i| M \le \frac{\kappa^{-1}}{10n}.\end{equation}

By the definition of of principal curvature of $\Sigma$ as well as recalling (5·1),

(5·14) \begin{equation} \langle \nu, \nabla_{e_1}e_1 \rangle = -\langle \nabla_{e_1}\nu,e_1 \rangle \ge \kappa^{-1} .\end{equation}

As a result, recalling $\xi_\ast =e_1(w)$ and combining (5·12), (5·13) and (5·14), we obtain

(5·15) \begin{align} |\nabla_{\xi_\ast}\nabla_{\xi_\ast} \tilde h (w)| & = \left|\sum_{j=0}^{n-1}[(y-z) \cdot e_j]\langle e_j,\nabla_{\xi_\ast}\xi_\ast \rangle + [(y-z) \cdot \nu]\langle \nu,\nabla_{\xi_\ast}\xi_\ast \rangle \right| \notag\\ & \ge|[(y-z) \cdot \nu]\langle \nu,\nabla_{\xi_\ast}\xi_\ast \rangle| - \left|\sum_{j=0}^{n-1}[(y-z) \cdot e_j]\langle e_j,\nabla_{\xi_\ast}\xi_\ast \rangle \right| \notag\\ & \ge \frac12|y-z|\kappa^{-1} - |y-z|M \sum_{j=0}^{n-1} |\langle e_j,\nabla_{\xi_\ast}\xi_\ast \rangle| \notag\\ & \ge \frac12|y-z| \kappa^{-1} -\frac1{10}|y-z|\kappa^{-1} \notag\\ & \ge \frac2{5\kappa} |y-z|.\end{align}

Recalling that $\nabla^2 \Sigma(x)$ is a linear map from $T_x[0,1]^{n-1} \times T_x[0,1]^{n-1} $ to $T_w \Sigma$ , we know $\nabla^2\Sigma(x) (\xi,\xi) \in T_w \Sigma$ and thus $\nabla^2\Sigma(x) (\xi,\xi) \in$ $\mathrm{span}\{e_1, \ldots,e_{n-1}\}$ . Combining (5·8) and the standing assumption in Case 3, we deduce that

\begin{align*} |\langle\nabla\tilde h (\Sigma(x)), \nabla^2\Sigma(x) (\xi,\xi) \rangle | & \le |\nabla^2\Sigma(x)| \sum_{i=1}^{n-1}|(y-z) \cdot e_i| \\ & \lt \|\Sigma\|_{C^2([0,1]^{n-1},\mathbb{R}^{n+1})} nM |y-z| \\ & \le \frac{1}{10\kappa}|y-z|.\end{align*}

Consequently, using (5·6), we arrive at

$$ |\nabla_\xi \nabla_\xi h (x) | \ge |\nabla_{\xi_\ast}\nabla_{\xi_\ast} \tilde h (w)| - |\langle\nabla\tilde h (\Sigma(x)), \nabla^2\Sigma(x) (\xi,\xi) \rangle | \gtrsim |y-z|, $$

which gives (5·4). Combining Cases 1-3, (3) in Definition 1·10 holds.

Finally, we show (2) in Definition 1·10. Indeed, combining (5·3) and (5·4), we deduce that

\begin{equation*} |y-z| \lesssim \inf_{x \in [0,1]^{n-1}, \xi \in S^{n-2}} \{ |h(x)| + |\nabla h(x)| + |\nabla_\xi \nabla_\xi h (x)| \} \le \|f_y-f_{z}\|_{C^2([0,1]^{n-1})} \lesssim |y-z|,\end{equation*}

which implies that (2) in Definition 1·10 holds and F is a doubling space with the same doubling constant as $\mathbb{R}^{n+1}$ . In particular, F is biLipschitz homeomorphic to Z.

The proof is complete.