Proof. Given µ > 0 and $ V \subseteq U $, we define $ X(V, \mu) $ as the set of $ x \in V $ for which there exists a polynomial function Q of degree at most 2 such that $ f(y) \leq Q(y) $ for every $ y \in V $, $ f(x) = Q(x) $, $ \| D Q(x) \| \leq \mu $ and $ \| D^2 Q \| \leq \mu $. If $ D \subseteq U $ is a countable dense subset of U and $ I(c) = \{s \in \mathbf{Q} : B_s(c) \subseteq U\} $ for every $ c\in D $, then we notice that

\begin{equation*} \mathcal{S}^\ast(f) \subseteq \bigcup_{c \in D} \bigcup_{s \in I(c)}\bigcup_{i =1}^\infty X(B_s(c), i). \end{equation*}

Henceforth, it is sufficient to show that $ \mathcal{H}^n(\Phi_f(Z)) =0 $ whenever $ Z \subseteq X(U,\mu) $ with $ \mathcal{L}^n(Z) =0 $, for some µ > 0. Notice that f is pointwise differentiable at each point of the set $ X(V,\mu) $, by lemma 2.17.

Now we prove the following estimates: given $ c \in U $ and $ 0 \lt r \lt 1 $ such that $ \overline{B_{3r}(c)} \subseteq U $, then

(3.1)\begin{equation} \| D f(a) - D f(b) \| \leq c(n)\big( \|\mathbf{D}^2 f\|_{L^n(B_{3r}(c))} + \mu r\big) \end{equation}

(3.2)\begin{equation} |f(a) - f(b) | \leq c(n) r \big( \|\mathbf{D}^2 f\|_{L^n(B_{3r}(c))} + \mu \big) \end{equation}

for every $ a, b \in X(U,\mu) \cap B_{r}(c) $. We fix $ a, b \in X(U,\mu) \cap B_r(c) $, a ≠ b, and we define

\begin{equation*} s = \frac{|a-b|}{2} \quad \text{and} \quad d = \frac{a+b}{2}. \end{equation*}

We notice that $ s \leq r $,

\begin{equation*} B_s(d) \subseteq B_{2s}(a) \cap B_{2s}(b) \quad \text{and} \quad B_{2s}(a) \cup B_{2s}(b) \subseteq B_{3r}(c); \end{equation*}

consequently it follows from (2.16) of lemma 2.17 that

\begin{align*} \| \mathbf{D} f -D f(e) \|_{L^n(B_s(d))} & \leq \| \mathbf{D} f - D f(e) \|_{L^n(B_{2s}(e))}\\ & \leq c(n) s \big( \|\mathbf{D}^2 f\|_{L^n(B_{2s}(e))} + \mu s \big)\\ & \leq c(n) s \big( |\mathbf{D}^2 f\|_{L^n(B_{3r}(c))} + \mu r\big), \end{align*}

for $ e \in \{a,b\} $, whence we infer

\begin{align*} \boldsymbol{\alpha}(n)^{1/n}s \| D f(a) - D f(b) \| & \leq \| \mathbf{D} f -D f(a) \|_{L^n(B_s(d))} + \| \mathbf{D} f - D f(b) |_{L^n(B_s(d))}\\ & \leq c(n) s \big( \|\mathbf{D}^2 f\|_{L^n(B_{3r}(c))} + \mu r\big) \end{align*}

and (3.1) is proved. Moreover, combining (2.17) of lemma 2.17 with (3.1)

\begin{align*} & |f(a) - f(b) | \\ & \qquad \leq \| f - L_a\|_{L^\infty(B_s(d))} + \| f - L_b \|_{L^\infty(B_s(d))} + s \| D f(a) \| + s \| D f(b) \| \\ & \qquad \leq \| f - L_a\|_{L^\infty(B_{2s}(a))} + \| f - L_b \|_{L^\infty(B_{2s}(b))} + 2\mu r \\ & \qquad \leq c(n) r \big( \|\mathbf{D}^2 f\|_{L^n(B_{3r}(c))} + \mu \big). \end{align*}

We consider the function $ \overline{f} \times \nabla f $ mapping $ x \in {\rm Diff}(f)$ into $ (\overline{f}(x), \nabla f(x)) \in \mathbf{R}^{2n+1} $. Then it follows from (3.1) and (3.2) that

(3.3)\begin{equation} {\rm diam}\,\big[(\overline{f} \times \nabla f)\big(B_r(c) \cap X(U,\mu)\big)\big] \leq c(n,\mu) \big( \|\mathbf{D}^2 f\|_{L^n(B_{3r}(c))} + r\big). \end{equation}

Let $ Z \subseteq X(U,\mu) $ bounded and $ \mathcal{L}^n(Z) =0 $. Given ϵ > 0, we choose an open set $ V \subseteq U $ such that $ Z \subseteq V $, and

(3.4)\begin{equation} \mathcal{L}^n(V) \leq \epsilon, \quad \| \mathbf{D}^2 f \|_{L^n(V)}^n \leq \epsilon. \end{equation}

We define $ R : Z \rightarrow \mathbf{R} $ and $ \rho : Z \rightarrow \mathbf{R} $ as

\begin{equation*} R(x) = \inf\bigg\{1, \frac{{\rm dist}(x, \mathbf{R}^n \setminus V)}{4}\bigg\}, \quad \text{for}\ x \in Z , \end{equation*}

\begin{equation*} \rho(x) = {\rm diam}\, ((\overline{f} \times \nabla f)(B_{R(x)}(x) \cap X(U,\mu))), \quad \text{for}\ x \in Z . \end{equation*}

We notice that R is a Lipschitzian function with $ {\rm Lip}(R) \leq \frac{1}{4} $ and, noting that $ B_{3R(x)}(x) \subseteq V $ for every $ x \in Z $ and combining (3.3) and (3.4), we obtain

(3.5)\begin{equation} \rho(x) \leq c(n,\mu) \big( \|\mathbf{D}^2 f\|_{L^n(B_{3R(x)}(x))} + R(x)\big) \leq c(n,\mu) \epsilon^{1/n}, \end{equation}

for $ x \in Z $. We prove now the following claim: there exists $ C \subseteq Z $ countable such that

\begin{equation*} \{B_{R(y)/5}(y): y \in C\}\ \text{is disjointed}, \end{equation*}

\begin{equation*} Z \subseteq \bigcup_{y \in C}B_{R(y)}(y), \end{equation*}

and

\begin{equation*} \mathcal{H}^0\big(\{y \in C: B_{3R(y)}(y) \cap B_{3R(x)}(x) \neq \varnothing\}\big) \leq c(n), \quad \text{for every}\ x \in Z . \end{equation*}

Applying Besicovitch covering theorem [Reference Ambrosio, Fusco and Pallara3, Theorem 2.17] (see also the remark at the beginning of p. 52), there exists a positive constant $ \xi(n) $ depending only on n, and there exist $ Z_1, \ldots , Z_{\xi(n)} \subseteq Z $ such that

\begin{equation*} Z \subseteq \bigcup_{i=1}^{\xi(n)} \bigcup_{x \in Z_i}B_{R(x)/5}(x), \end{equation*}

and $ \{B_{R(x)/5}(x) : x \in Z_i\} $ is disjointed for every $ i = 1, \ldots , \xi(n) $. We now apply [Reference Federer10, Lemma 3.1.12] with $ S = Z_i $, U = Z, $ h = \frac{R}{5}$, $ \lambda = \frac{1}{20} $ and $ \alpha = \beta = 15 $, to infer that

\begin{equation*} \mathcal{H}^0 \big( \{y \in Z_i : B_{3R(y)}(y) \cap B_{3R(x)}(x) \neq \varnothing\}\big) \leq c(n), \quad \text{for every}\ i = 1, \ldots , \xi(n) . \end{equation*}

We define $ Z' = \bigcup_{i=1}^{\xi(n)} Z_i $, and we notice that $ Z \subseteq \bigcup_{x \in Z'} B_{R(x)/5}(x) $ and

\begin{equation*} \mathcal{H}^0\big(\{y \in Z' : B_{3R(y)}(y) \cap B_{3R(x)}(x) \neq \varnothing\}\big) \leq \xi(n)c(n), \quad \text{for every}\ x \in Z . \end{equation*}

Now we apply Vitali covering theorem to find a countable set $ C \subseteq Z' $ such that $ \{B_{R(x)/5}(x): x \in C\} $ is disjointed and

\begin{equation*} Z \subseteq \bigcup_{x \in C}B_{R(x)}(x), \end{equation*}

which proves the claim.

Denoting with ϕ_δ the size δ approximating measure of the n-dimensional Hausdorff measure $ \mathcal{H}^n $ of $ \mathbf{R}^{2n+1} $ (cf. [Reference Federer10, 2.10.1, 2.10.2]), and combining (3.5) with the claim above and with remark 3.4, we have

\begin{align*} &\phi_{c(n, \mu)\epsilon^{1/n}} ((\overline{f} \times \nabla f)(Z)) \leq c(n) \sum_{y \in C} \rho(y)^n\\ & \qquad \leq c(n,\mu) \sum_{y \in C} \Big( \|\mathbf{D}^2 f \|_{L^n(B_{3R(y)}(y))} + R(y)\Big)^n\\ & \qquad \leq c(n,\mu) \sum_{y \in C} R(y)^n + c(n,\mu) \sum_{y \in C}\int_{B_{3R(y)}(y)}\|\mathbf{D}^2 f \|^n\, d\mathcal{L}^n \\ &\qquad \leq c(n, \mu)\mathcal{L}^n(V) + c(n, \mu) \int_V \| \mathbf{D}^2 f \|^n\, d\mathcal{L}^n \\ & \qquad \leq c(n,\mu) \epsilon. \end{align*}

Henceforth, letting ϵ → 0 we deduce that $ \mathcal{H}^n((\overline{f} \times \nabla f)(Z)) =0 $. Since $ \Phi_f = (\bf{1}_{\mathbf{R}^{n+1}} \times \psi) \circ (\overline{f} \times \nabla f) $ (see remark 3.2), we conclude that $ \mathcal{H}^n(\Phi_f(Z)) =0 $.

Proof. We prove the assertion by contradiction. Suppose $ 0 \in U $ and $ f(0) =0 $; hence there exists $ c \in U \setminus \{0\} $ such that

\begin{equation*} B^{n+1}(c,|c|) \cap G = \varnothing \quad \text{and} \quad K : = \overline{B^{n+1}(c,|c|) \cap (\mathbf{R}^n \times \{0\})} \subseteq U. \end{equation*}

Suppose L > 0 such that $ | f(x) - f(y) | \leq L |x-y|^\gamma $ for every $ x, y \in K $, and define

\begin{equation*} h_{\pm}(x) = \pm \sqrt{|c|^2 -|x-c|^2} = \pm \sqrt{2 c \bullet x - | x |^2} \end{equation*}

for $ x \in K $. Henceforth, either $ h_+(x) \leq f(x) $ for every $ x \in K $, or $ f(x) \leq h_-(x) $ for every $ x \in K $. In both cases, replacing x with $ t c $ and $ 0 \lt t \lt 1 $, one obtains

\begin{equation*} t^{1-2\gamma}(2-t)\leq L^2 |c|^{2\gamma-2} \end{equation*}

for $ 0 \lt t \lt 1 $. This is clearly impossible, since $ 1-2\gamma \lt 0$.

Proof. Suppose $(z, \nu) \in N_f$, where $ z = (x, f(x)) $ with $ x \in A $, and s > 0 such that $ B^{n+1}(z+s\nu, s) \cap E_f= \varnothing $. Since $ \nu \notin \mathbf{R}^n \times \{0\} $ by lemma 3.6, we can easily see that there exists an open set $ W \subseteq U \times \mathbf{R} $ with $ z\in W $, an open set $ V \subseteq U $ with $ x \in V $, and a smooth function $ g : V \rightarrow \mathbf{R} $ such that $ f(x) = g(x)$ and

\begin{equation*} W \cap B^{n+1}(z+s\nu, s) = \{(y,u) : y \in V, \; u \gt g(y)\}; \end{equation*}

in particular $ f(y) \leq g(y) $ for every $ y \in V $. It follows that $ x \in \mathcal{S}^\ast(f) $, $ x \in {\rm Diff}(f) $ and $ D f(x) = D g(x) $ by lemma 2.17. Noting that

\begin{equation*} \nu = \frac{(-\nabla g(x), 1)}{\sqrt{1 + | \nabla g(x)|^2}} \end{equation*}

we conclude $(z,\nu) = \Phi_f(x) $ and $ N_f \cap (A \times \mathbf{R} \times \mathbf{R}^{n+1}) \subseteq \Phi_f(\mathcal{S}^\ast(f)\cap A) $.

The opposite inclusion is clear, since for every $ x \in \mathcal{S}^\ast(f) $ there exists a polynomial function P and an open neighbourhood V of x, such that $ P(x) = f(x) $ and $ P(y) \geq f(y) $ for every $ y \in V $.

We prove now the area formula in (3.7). Since $ \Phi_f $ is a $ W^{1,n} $-map (see remark 3.3), we apply lemma 2.19 with f replaced by $ \Phi_f $ and we find countably many $ \mathcal{L}^n $-measurable sets B_i of U such that $ {\rm Lip} \big(\Phi_f|B_i\big) \lt \infty $ for every $ i \geq 1 $ and

\begin{equation*} \mathcal{L}^n\bigg(U \setminus \bigcup_{i=1}^\infty B_i \bigg) =0. \end{equation*}

Then we set

\begin{equation*} A_1 = B_1 \cap \mathcal{S}^\ast(f), \qquad A_i = B_i \cap \mathcal{S}^\ast(f) \setminus {\textstyle\bigcup_{\ell = 1}^{i-1}} B_\ell \quad \text{for}\ i \geq 2 \end{equation*}

and we notice that

(3.9)\begin{equation} \mathcal{L}^n \big( U \setminus \textstyle{\bigcup_{i=1}^\infty} A_i\big)=0 \quad \text{and} \quad \mathcal{H}^n\big(N_f \setminus \textstyle{\bigcup_{i=1}^\infty} \Phi_f(A_i) \big) =0 \end{equation}

by lemma 2.16, lemma 3.5, and (3.6). If $ \beta : U \times \mathbf{R} \times \mathbf{R}^{n+1} \rightarrow \mathbf{R} $ is a $ \mathcal{H}^n $-measurable non-negative function, firstly we notice that (cf. [Reference Federer10, 2.4.8])

\begin{equation*} \int_{U} \beta(\Phi_f(x))\, \operatorname{ap} J_n \Phi_f(x)\, d\mathcal{L}^n(x) = \sum_{i=1}^\infty \int_{A_i}\beta(\Phi_f(x))\,\operatorname{ap} J_n\Phi_f(x)\, d\mathcal{L}^n(x), \end{equation*}

then, recalling [Reference Federer10, 2.10.43], we use the area formula for Lipschitzian maps in [Reference Federer10, 3.2.5] and the injectivity of $ \Phi_f $ to compute

\begin{equation*} \int_{A_i}\beta(\Phi_f(x))\,\operatorname{ap} J_n\Phi_f(x)\, d\mathcal{L}^n(x) =\int_{\Phi_f(A_i)}\beta(y)\, d\mathcal{H}^n(y) \qquad \text{for}\ i \geq 1 \end{equation*}

and we use (3.9) to conclude

\begin{align*} \int_{U} \beta(\Phi_f(x))\, \operatorname{ap} J_n \Phi_f(x)\, d\mathcal{L}^n(x) & = \sum_{i=1}^\infty \int_{\Phi_f(A_i)}\beta(y)\, d\mathcal{H}^n(y)\\ & = \int_{N_f}\beta(y)\, d\mathcal{H}^n(y). \end{align*}

Choosing β = 1 we conclude that $ \mathcal{H}^n(N_f) \lt \infty $ and we deduce that $ \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} N_f, (z,\nu)) $ is an n-dimensional plane for $ \mathcal{H}^n $ a.e. $(z, \nu) \in N_f $. Let D_i be the set of $x \in A_i $ such that $\boldsymbol{\Theta}^n(\mathcal{L}^n\operatorname{\llcorner} \mathbf{R}^n \setminus A_i, x) = 0 $, $ \operatorname{ap} D \Phi_f(x) $ is injective and $ \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} N_f, \Phi_f(x)) $ is an n-dimensional plane. Noting [Reference Federer10, 2.10.19] and remark 3.3, we deduce that

\begin{equation*} \mathcal{H}^n \big(A_i \setminus D_i\big) =0. \end{equation*}

Noting that $ \operatorname{Tan}^n(\mathcal{L}^n \operatorname{\llcorner} A_i , x) = \mathbf{R}^n $ for $ x \in D_i $, and noting that $ \Psi|A_i $ is a bi-lipschitz homeomorphism onto $ \Phi_f(A_i) $, we employ [Reference Santilli40, Lemma B.2] to conclude

\begin{align*} \operatorname{ap} D \Phi_f(x)[\mathbf{R}^n] & = D \Psi_i(x)\big[\operatorname{Tan}^n(\mathcal{L}^n \operatorname{\llcorner} A_i, x)\big]\\ & \subseteq \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} \Psi_i(A_i), \Phi_f(x)) \subseteq \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} N_f, \Phi_f(x)) \end{align*}

for every $ x \in D_i $. Since $ \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} N_f, \Phi_f(x)) $ is an n-dimensional plane and $ \operatorname{ap} D \Phi_f(x) $ is injective for every $ x \in D_i $, we conclude that

\begin{equation*} \operatorname{ap} D \Phi_f(x)[\mathbf{R}^n] = \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} N_f, \Phi_f(x)) \quad \text{for every}\ x \in D_i . \end{equation*}

Proof. We identify $ \mathbf{R}^{n+1} \simeq \mathbf{R}^{n} \times \mathbf{R}$ and we consider the orthonormal basis $ \epsilon_1, \ldots , \epsilon_n $ of Rⁿ such that $ (\epsilon_i,0) = e_i $ for $ i =1, \ldots, n $ (cf. (2.3)). We use the notation $ D_1, \ldots D_n $ and $ \operatorname{ap} D_1 , \ldots , \operatorname{ap} D_n $ for the partial derivatives and the approximate partial derivatives with respect to $ \epsilon_1, \ldots , \epsilon_n $, respectively. We notice by [Reference Federer10, 3.1.4] that $ \operatorname{ap} D_i \Phi_f $ is a $ \mathcal{L}^n \operatorname{\llcorner} U $-measurable map for $ i = 1, \ldots , n $. Henceforth, by the classical Lusin theorem (cf. [Reference Federer10, 2.3.5, 2.3.6]) there exists a Borel map $ \xi_i : U \rightarrow \mathbf{R}^{n} \times \mathbf{R} \times \mathbf{R}^{n+1} $ such that ξ_i is $ \mathcal{L}^n \operatorname{\llcorner} U $ almost equal to $ \operatorname{ap} D_i \Phi_f $ for $ i = 1, \ldots , n $.

Since $ \int_U | \xi_1 \wedge \cdots \wedge \xi_n |\, d\mathcal{L}^n \lt \infty $ by remark 3.3, we define

(3.10)\begin{equation} T(\phi) = \int_U \langle \xi_1(x) \wedge \cdots \wedge \xi_n(x), \phi(\Phi_f(x)) \rangle\, d\mathcal{L}^n(x) \end{equation}

for $ \phi \in \mathcal{D}^n(U \times \mathbf{R} \times \mathbf{R}^{n+1}) $ and we notice that $ T \in \mathcal{D}_n(U \times \mathbf{R} \times \mathbf{R}^{n+1}) $. We choose now a sequence $ f_k \in C^\infty(U)\cap W^{2,n}(U) $ such that $f_k \to f$ in $ W^{2,n}(U) $, $ f_k(x) \to f(x) $ and $ \mathbf{D} f_k(x) \to \mathbf{D} f(x) $ for $ \mathcal{L}^n $ a.e. $ x \in U $; see [Reference Gilbarg and Trudinger11, Theorem 7.9]. Since $ \Phi_{f_k} : U \rightarrow U \times \mathbf{R} \times \mathbf{R}^{n+1} $ is a smooth proper map, we define

\begin{equation*} T_k = (\Phi_{f_k})_{\#}\big( (\mathcal{L}^n \operatorname{\llcorner} U) \wedge \epsilon_1 \wedge \cdots \wedge \epsilon_n \big) \in \mathcal{D}_n(U \times \mathbf{R} \times \mathbf{R}^{n+1}) \end{equation*}

and we prove that

(3.11)\begin{equation} T_k \to T \ \text{in}\ \mathcal{D}_n(U \times \mathbf{R} \times \mathbf{R}^{n+1}) . \end{equation}

Noting that

\begin{equation*}T_k(\phi) = \int_{U} \langle D_1 \Phi_{f_k}(x) \wedge \cdots \wedge D_n \Phi_{f_k}(x), \phi(\Phi_{f_k}(x)) \rangle\, d\mathcal{L}^n(x)\end{equation*}

whenever $ \phi \in \mathcal{D}^n(U \times \mathbf{R} \times \mathbf{R}^{n+1}) $, we estimate

(3.12)\begin{align} |T_k&(\phi) - T(\phi) |\\ & \leq \int_U \big| \langle D_1 \Phi_{f_k}(x) \wedge \cdots \wedge D_n \Phi_{f_k}(x) - \xi_1(x) \wedge \cdots \wedge \xi_n(x), \phi(\Phi_{f_k}(x)) \rangle\big|\, d\mathcal{L}^n(x)\notag \\ & \quad + \int_U \big|\langle \xi_1(x) \wedge \cdots \wedge \xi_n(x), \phi(\Phi_{f_k}(x)) - \phi(\Phi_f(x)) \rangle\big| \, d\mathcal{L}^n(x)\notag \\ & \leq \| \phi \|_{L^\infty(U)} \int_U \big| D_1 \Phi_{f_k}(x) \wedge \cdots \wedge D_n \Phi_{f_k}(x) - \xi_1(x) \wedge \cdots \wedge \xi_n(x)\big|\, d\mathcal{L}^n(x) \notag\\ & \quad + \int_U \text{ap}\,J_n\Phi_f(x)\, \| \phi(\Phi_f(x)) - \phi(\Phi_{f_k}(x)) \|\, d\mathcal{L}^n(x) \notag \end{align}

for $ \phi \in \mathcal{D}^n(U \times \mathbf{R} \times \mathbf{R}^{n+1}) $. Moreover, noting that $ \operatorname{ap} J_n \Phi_f \in L^1(U) $ (see remark 3.3) and

\begin{equation*} \operatorname{ap} J_n \Phi_f(x) \, \| \phi(\Phi_f(x)) - \phi(\Phi_{f_k}(x)) \| \leq 2 \| \phi \|_{L^\infty(U \times \mathbf{R} \times \mathbf{R}^n)} \operatorname{ap} J_n\Phi_f(x) \end{equation*}

for $ \mathcal{L}^n $ a.e. $ x \in U $ and for every $ k \geq 1 $, it follows from the dominated convergence theorem that

(3.13)\begin{equation} \lim_{k \to \infty}\int_U \operatorname{ap} J_n\Phi_f(x)\, \| \phi(\Phi_f(x)) - \phi(\Phi_{f_k}(x)) \|\, d\mathcal{L}^n(x) =0. \end{equation}

We observe that

\begin{align*} &D_1 \Phi_{f_k} \wedge \cdots \wedge D_n \Phi_{f_k} - \xi_1 \wedge \cdots \wedge \xi_n \\ & \quad = \sum_{i=1}^{n} \xi_1 \wedge \cdots \wedge \xi_{i-1} \wedge \big(D_i \Phi_{f_k}-\xi_i\big) \wedge D_{i+1} \Phi_{f_k} \wedge \cdots \wedge D_n\Phi_{f_k} \end{align*}

and we use the generalized Holder’s inequality to estimate

(3.14)\begin{align} &\int_U \big| D_1 \Phi_{f_k} \wedge \cdots \wedge D_n \Phi_{f_k} - \xi_1 \wedge \cdots \wedge \xi_n\big|\, d\mathcal{L}^n\\ & \quad \leq \sum_{i=1}^n \int_U \big|\xi_1 \wedge \cdots \wedge \xi_{i-1}\big|\cdot \big| \xi_i - D_i \Phi_{f_k}\big|\cdot \big| D_{i+1} \Phi_{f_k} \wedge \cdots \wedge D_n\Phi_{f_k}\big|\,d\mathcal{L}^n \notag \\ & \quad \leq \sum_{i=1}^n \int_U \| \operatorname{ap}D \Phi_f \|^{i-1}\cdot \| \operatorname{ap}D \Phi_f - D \Phi_{f_k}\| \cdot \| D\Phi_{f_k} \|^{n-i}\, d\mathcal{L}^n \notag \\ & \quad \leq \sum_{i=1}^n \bigg(\int_U \| \operatorname{ap}D \Phi_f \|^{n}\, d\mathcal{L}^n \bigg)^{\frac{i-1}{n}}\cdot \bigg( \int_U \| \operatorname{ap}D \Phi_f - D \Phi_{f_k}\|^{n}\, d\mathcal{L}^n \bigg)^{\frac{1}{n}}\notag\\ &\qquad\cdot \bigg(\int_U \| D\Phi_{f_k} \|^{n}\, d\mathcal{L}^n\bigg)^{\frac{n-i}{n}}.\notag \end{align}

Moreover, by (3.2),

\begin{align*} & \int_U \| D \Phi_{f_k} - \operatorname{ap}D \Phi_f \|^n \, d\mathcal{L}^n\\ & \quad\leq c(n)\bigg( \int_U \| D \overline{f}_k - \mathbf{D} \overline{f} \|^n \, d\mathcal{L}^n + \int_U \| D (\psi \circ \nabla f_k) - \mathbf{D} ( \psi \circ \boldsymbol{\nabla}f) \|^n \, d\mathcal{L}^n\bigg)\\ & \quad\leq c(n)\bigg( \int_U \| D \overline{f}_k - \mathbf{D} \overline{f} \|^n \, d\mathcal{L}^n + \int_U \| D \psi (\nabla f_k)\|^n \|D^2 f_k - \mathbf{D}^2 f_k \|^n\, d\mathcal{L}^n \\ & \qquad + \int_U \| D \psi(\nabla f_k) - D \psi(\boldsymbol{\nabla} f) \|^n \| \mathbf{D}^2 f \|^n\, d\mathcal{L}^n\bigg)\\ & \quad\leq c(n) \bigg( \int_U \| D \overline{f}_k - \mathbf{D} \overline{f} \|^n \, d\mathcal{L}^n + \int_U \|D^2 f_k - \mathbf{D}^2 f_k \|^n\, d\mathcal{L}^n \\ & \qquad+ \int_U \| D \psi(\nabla f_k) - D \psi(\boldsymbol{\nabla} f) \|^n \| \mathbf{D}^2 f \|^n\, d\mathcal{L}^n\bigg) \end{align*}

and

\begin{equation*} \lim_{k \to \infty}\int_U \| D \psi(\nabla f_k) - D \psi(\boldsymbol{\nabla} f) \|^n \| \mathbf{D}^2 f \|^n\, d\mathcal{L}^n = 0 \end{equation*}

by dominated convergence theorem. Consequently, $ \| D \Phi_{f_k} - \mathbf{D} \Phi_f \|_{L^n(U)} \to 0 $, and combining (3.12), (3.13), and (3.14) we obtain (3.11).

Since $ \partial T_k =0 $ for every $ k \geq 1 $, we readily infer from (3.11) that $ \partial T =0 $. Define $ G = [\Phi_f|\mathcal{S}^\ast(f)]^{-1} : N_f \rightarrow U $ and notice that G is simply the restriction on N_f of the linear function that maps a point of $ \mathbf{R}^{n+1} \times \mathbf{R}^{n+1} $ onto its first n coordinates; in particular G is a Borel map (recall that N_f is a Borel set). We employ lemma 3.8 to see that

(3.15)\begin{equation} T(\phi) = \int_{N_f} \langle \xi\big[G(z,\nu)\big], \phi(z,\nu) \rangle\, d\mathcal{H}^n(z, \nu) \end{equation}

for every $ \phi \in \mathcal{D}^n(U \times \mathbf{R} \times \mathbf{R}^{n+1}) $, where $ \xi : U \rightarrow \bigwedge_n(\mathbf{R}^{n+1} \times \mathbf{R}^{n+1})$ is the Borel map defined as

\begin{equation*} \xi(x) = \frac{\xi_1(x) \wedge \cdots \wedge \xi_n(x)}{\big|\xi_1(x) \wedge \cdots \wedge \xi_n(x)\big|}. \end{equation*}

Then we define the Borel map

\begin{equation*} \eta : N_f \rightarrow {\textstyle\bigwedge_n}(\mathbf{R}^{n+1} \times \mathbf{R}^{n+1}), \quad \eta = \xi \circ G, \end{equation*}

and we readily infer that $ \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} N_f, (z, \nu)) $ is associated with $ \eta(z, \nu) $ for $ \mathcal{H}^n $ a.e. $(z, \nu) \in N_f $ by lemma 3.8, and $ (\mathcal{H}^n \operatorname{\llcorner} N_f) \wedge \eta $ is a Legendrian cycle by lemma 2.11. Finally, we use remark 3.3 and shuffle formula (see [Reference Federer10, p. 19]) to compute

\begin{align*} \big\langle \big[\textstyle{\bigwedge_n \pi_0}\big]&(\xi(x)) \wedge \psi(\nabla f(x)), E' \big\rangle\\ & = \frac{1}{\operatorname{ap} J_n \Phi_f(x)} \, \langle D_1\overline{f}(x) \wedge \cdots \wedge D_n\overline{f}(x) \wedge \psi(\nabla f(x)), E' \rangle \\ & = \frac{e'_{n+1}(\psi(\nabla f(x)))}{\operatorname{ap} J_n \Phi_f(x)} \gt 0 \end{align*}

for $ \mathcal{L}^n $ a.e. $ x \in U $.

Proof. Notice that $ \mathbf{R}^{n+1} \setminus \overline{C} \neq \varnothing $ since $ \operatorname{nor}(C) \neq \varnothing $. It follows from the continuity of ψ_C that $ \psi_C^{-1}(W \cap \operatorname{nor}(C)) $ is relatively open in $ \operatorname{Unp}(C) $. Henceforth, we conclude from (4.1) that

\begin{equation*} \mathcal{L}^{n+1}\big(\psi_C^{-1}(W \cap \operatorname{nor}(C))\big) \gt 0. \end{equation*}

We define $ T = \psi_C^{-1}(W \cap \operatorname{nor}(C)) $. Since $ J_1 {\rm dist}(\cdot, C) $ is $ \mathcal{L}^{n+1} $ almost equal to the constant function 1 on $ \mathbf{R}^{n+1} \setminus \overline{C} $, we use coarea formula to compute

\begin{equation*} 0 \lt \mathcal{L}^{n+1}(T) = \int\mathcal{H}^n(T \cap S_t(C))\, dt \end{equation*}

we infer there exists τ > 0 so that $ \mathcal{H}^n(T \cap S_\tau(C)) \gt 0 $, and we use (4.3) to conclude

\begin{equation*} \mathcal{H}^n\big((T \cap S_\tau(C)) \setminus \operatorname{Cut}(C)\big) \gt 0. \end{equation*}

Consequently there exists $ s \gt \tau $ so that

\begin{equation*} \mathcal{H}^n \big(T \cap S_\tau(C) \cap \{\rho_C \geq s/ \tau\} \big) \gt 0. \end{equation*}

Since $ \psi_C | S_\tau(C) \cap \{\rho_C \geq s/ \tau\} $ is a bi-lipschitz homeomorphism by [Reference Hug and Santilli14, Theorem 3.16], we conclude that

\begin{equation*} \mathcal{H}^n \big(\psi_C\big[T \cap S_\tau(C) \cap \{\rho_C \geq s/ \tau\}\big] \big) \gt 0, \end{equation*}

whence we infer that $ \mathcal{H}^n(W \cap \operatorname{nor}(C)) \gt 0 $.

Proof. First, we observe that

\begin{equation*} \psi(\nabla f(x)) \in \operatorname{Nor}(E_f, \overline{f}(x)) \qquad \text{for every}\ x \in {\rm Diff}(f) . \end{equation*}

Since ψ is a diffeomorphism of Rⁿ onto $ \mathbf{S}^n_+ $ and f is continuous, the set $ \{(\overline{f}(x), \psi(\nabla f(x))) : x \in U\} $ is dense in $ G \times \mathbf{S}^n_+ $; consequently we infer that $ \operatorname{nor}(E_f) $ is dense in $ G \times \mathbf{S}^n_+ $ by standard approximation of regular normals, see [Reference Rockafellar and Wets34, Exercise 6.18(a)].

Proof. Suppose $ \{\Sigma_i\}_{i \geq 1} $ is a countable family of C ²-hypersurfaces such that

\begin{equation*} \mathcal{H}^n\big(X \setminus {\textstyle\bigcup_{i=1}^\infty} \Sigma_i\big) =0 \end{equation*}

and $ \eta_i : \Sigma_i \rightarrow \mathbf{S}^n $ is a continuously differentiable unit-normal vector field with $ {\rm Lip}(\eta_i) \lt \infty $ for $ i \geq 1 $. By [Reference Federer10, 2.10.19(4)]

\begin{equation*} \Theta^n(\mathcal{H}^n \operatorname{\llcorner} X \setminus \Sigma_i, a) = \Theta^n(\mathcal{H}^n \operatorname{\llcorner} \Sigma_i \setminus X, a) =0 \quad \text{for}\ \mathcal{H}^n\ \text{a.e.}\ a \in \Sigma_i \cap X , \end{equation*}

whence we infer that $ \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} X, a) = \operatorname{Tan}(\Sigma_i,a) $ for $ \mathcal{H}^n $ a.e. $ a \in \Sigma_i \cap X $. In particular, if we define

\begin{equation*} \Sigma_i^+ =\{a \in \Sigma_i \cap X : \eta_i(a) = \nu(a)\} \quad \text{and} \quad \Sigma_i^- =\{a \in \Sigma_i \cap X : \eta_i(a) = - \nu(a)\} \end{equation*}

we infer that $ \mathcal{H}^n\big(\Sigma_i \cap X \setminus (\Sigma_i^+ \cup \Sigma_i^-)\big) =0 $ for each $ i \geq 1 $. Moreover, employing again [Reference Federer10, 2.10.19(4)] we infer that

\begin{equation*} \Theta^n(\mathcal{H}^n \operatorname{\llcorner} X \setminus \Sigma_i^\pm, a) =0 \quad \text{for}\ \mathcal{H}^n \ \text{a.e.}\ a \in \Sigma_i^\pm, \end{equation*}

whence we deduce that ν is $ \mathcal{H}^n \operatorname{\llcorner} X $ approximately differentiable at a with

\begin{equation*} \operatorname{ap} D \nu(a) = \pm D\eta_i(a) \end{equation*}

for $ \mathcal{H}^n $ a.e. $ a \in \Sigma_i^\pm $. Finally, since $ D \eta_i(a) $ is symmetric, we conclude the proof.

Proof. Suppose $ \Omega = F(\Omega') $, where $ \Omega' $ and F are as in definition 5.3. Clearly, $ F(\partial \Omega') = \partial \Omega $ and $ F(\partial^v_+ \Omega') = \partial^v_+ \Omega $. Therefore, assertion (1) follows from theorem 2.16, theorem 7.6, and remark 7.7.

If $ p \in \partial^v_+ \Omega $, we have $ \operatorname{Tan}(\partial \Omega, p) \subseteq \nu_\Omega(p)^\perp; $ since $ \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} \partial \Omega, p) \subseteq \operatorname{Tan}(\partial \Omega, p) $ for every $ p \in \partial \Omega $ and $ \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} \partial \Omega, p) $ is an n-dimensional plane for $ \mathcal{H}^n $ a.e. $ p \in \partial \Omega $, we obtain (2). Finally it follows from definitions that $ \operatorname{Tan}(\Omega, p) = \{v \in \mathbf{R}^{n+1}: v \bullet \nu_\Omega(p) \leq 0\} $ and $ \operatorname{Tan}^{n+1}(\mathcal{L}^{n+1} \operatorname{\llcorner} \Omega, p) = \{v \in \mathbf{R}^{n+1}: v \bullet \nu_\Omega(p) \leq 0\} $ for every $ p \in \partial^v_+ \Omega $.

Proof. Suppose $ \Omega = F(\Omega') $, where $ \Omega' $ and F are as in definition 5.3. We recall the definition of $ \Psi_F $ from (2.13) and notice that

(5.1)\begin{equation} \Psi_F(\operatorname{nor}(\Omega')) = \operatorname{nor}(\Omega) \end{equation}

by [Reference Santilli41, Lemma 2.1]. Since $ F(\partial^v_+ \Omega') = \partial^v_+ \Omega $, we readily infer from (5.1) that

\begin{equation*} \Psi_F(x, \nu_{\Omega'}(x)) = (F(x), \nu_\Omega(F(x))) \quad \text{for every}\ x \in \partial^v_+\Omega' \end{equation*}

and

(5.2)\begin{equation} \Psi_F\big(\overline{\nu_{\Omega'}}(F^{-1}(S))\big) = \overline{\nu_\Omega}(S) \quad \text{for every}\ S \subseteq \partial^v_+ \Omega . \end{equation}

To prove the assertions in (1) and (2) we notice, firstly, that they are true for $ \Omega' $ as a consequence of lemma 3.5 and (3.6) of lemma 3.8; then we apply (5.1) and (5.2).

To prove (3) we first employ lemma 5.2 to find a countable family $ X_i \subseteq \partial^v_+ \Omega $ such that $ \mathcal{H}^n\big(\partial \Omega \setminus \bigcup_{i=1}^\infty X_i\big) =0 $ and $ {\rm Lip}(\nu_\Omega|X_i) \lt \infty $ for every $ i \geq 1 $; then we define Y_i to be the set of $ x \in X_i $ such that $ \nu_\Omega $ is $ \mathcal{H}^n\operatorname{\llcorner} \partial \Omega $ approximately differentiable at x, $ \operatorname{Tan}^n(\mathcal{H}^n\operatorname{\llcorner} \partial \Omega, x) $ and $ \operatorname{Tan}^n(\mathcal{H}^n\operatorname{\llcorner} \operatorname{nor}(\Omega), \overline{\nu_\Omega}(x)) $ are n-dimensional planes, and $ \Theta^n(\mathcal{H}^n\operatorname{\llcorner} \partial \Omega \setminus X_i, x) =0 $. We notice that $ \overline{\nu_\Omega}| X_i $ is bi-lipschitz and, since $\operatorname{Tan}^n(\mathcal{H}^n\operatorname{\llcorner} \operatorname{nor}(\Omega), (x,u)) $ is an n-dimensional plane for $ \mathcal{H}^n $ a.e. $ (x,u) \in \operatorname{nor}(\Omega) $, we conclude that

\begin{equation*} \mathcal{H}^n(X_i \setminus Y_i) =0 \quad \text{for every}\ i \geq 1 . \end{equation*}

It follows from (1) and (2) that

(5.3)\begin{equation} \mathcal{H}^n\big(\operatorname{nor}(\Omega) \setminus {\textstyle\bigcup_{i=1}^\infty \overline{\nu_\Omega}(Y_i)}\big) =0. \end{equation}

We fix now $ x \in Y_i $. Then there exists a map $ g : \mathbf{R}^{n+1} \rightarrow \mathbf{R}^{n+1} \times \mathbf{R}^{n+1} $ pointwise differentiable at x such that $ \Theta^n(\mathcal{H}^n \operatorname{\llcorner} \partial \Omega \setminus \{g = \overline{\nu_\Omega}\}, x) =0 $ and $ \operatorname{ap} D \overline{\nu_\Omega}(x) = D g(x) | \operatorname{Tan}^n(\mathcal{H}^n\operatorname{\llcorner} \partial \Omega, x) $; noting that $ \operatorname{ap} D \overline{\nu_\Omega}(x) $ is injective, $ g | X_i \cap \{g = \overline{\nu_\Omega}\} $ is bi-lipschitz and

\begin{equation*} \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} \partial \Omega, x) = \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} X_i \cap \{g = \overline{\nu_\Omega}\}, x), \end{equation*}

we readily infer by [Reference Santilli40, Lemma B.2] that

\begin{equation*} \operatorname{ap}D \overline{\nu_\Omega}(x)[\operatorname{Tan}^n(\mathcal{H}^n\operatorname{\llcorner} \partial \Omega, x)] = \operatorname{Tan}^n(\mathcal{H}^n\operatorname{\llcorner} \operatorname{nor}(\Omega), \overline{\nu_\Omega}(x)). \end{equation*}

Henceforth, if $ \tau_1, \ldots , \tau_n $ is an orthonormal basis of $ \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} \partial \Omega, x) $ with $ \operatorname{ap} D \nu_\Omega(x)(\tau_i) = {\chi}_{\Omega, i}(x) \tau_i $ for $ i = 1, \ldots , n $, we conclude that

\begin{equation*} \Bigg\{\bigg(\frac{1}{\sqrt{1 + {\chi}_{\Omega, i}(x)^2}}\tau_i, \frac{{\chi}_{\Omega, i}(x)}{\sqrt{1 + {\chi}_{\Omega, i}(x)^2}}\tau_i\bigg): i = 1, \ldots, n\Bigg\} \end{equation*}

is an orthonormal basis of $ \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} \operatorname{nor}(\Omega), \overline{\nu_\Omega}(x)) $. Since x is arbitrarily chosen in Y_i, thanks to (5.3), we deduce from the uniqueness stated in lemma 2.11 that

\begin{equation*} \kappa_{\Omega, i}(x,u) = {\chi}_{\Omega,i}(x) \quad \text{for}\ \mathcal{H}^n \ \text{a.e.}\ (x,u) \in \operatorname{nor}(\Omega). \end{equation*}

Finally, we prove (4). By lemma 2.11 we can choose maps $ \tau_1, \ldots , \tau_n $ defined $ \mathcal{H}^n $ a.e. on $ \operatorname{nor}(\Omega') $ such that $ \tau_{1}(x,u), \ldots , \tau_n(x,u), u $ is an orthonormal basis of $ \mathbf{R}^{n+1} $,

(5.4)\begin{equation} \tau_1(x,u) \wedge \cdots \tau_n(x,u) \wedge u = e_1 \wedge \cdots \wedge e_{n+1} \quad \text{for}\ \mathcal{H}^n \text{a.e.}\ (x,u)\in \operatorname{nor}(\Omega') \end{equation}

and the vectors

\begin{equation*} \zeta_i'(x,u) = \bigg(\frac{1}{\sqrt{1 + \kappa_{\Omega', i}(x,u)^2}} \tau_i(x,u), \frac{\kappa_{\Omega',i}(x,u)}{\sqrt{1 + \kappa_{\Omega',i}(x,u)^2}}\tau_i(x,u) \bigg), \quad i = 1, \ldots, n, \end{equation*}

form an orthonormal basis of $ \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} \operatorname{nor}(\Omega'), (x,u)) $ for $ \mathcal{H}^n $ a.e. $(x,u)\in \operatorname{nor}(\Omega') $. Then we define

\begin{equation*} \eta' = \zeta_1' \wedge \cdots \wedge \zeta_n' \end{equation*}

and notice that

\begin{equation*} | \eta'(x, u)| =1, \quad \eta'(x, u)\ \text{is simple}, \end{equation*}

\begin{equation*} \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} \operatorname{nor}(\Omega'), (x, u))\ \text{is associated with}\ \eta'(x, u) \end{equation*}

and (see (2.2) and (2.5))

(5.5)\begin{equation} \langle \big[ {\textstyle\bigwedge_n} \pi_0\big](\eta'(x,u)) \wedge u, E' \rangle \gt 0 \quad \text{(by 5.4)} \end{equation}

for $ \mathcal{H}^n $ a.e. $(x,u) \in \operatorname{nor}(\Omega') $. If $ p \in \partial \Omega' $, ϵ > 0, $ \nu \in \mathbf{S}^n $, $ U \subseteq \nu^\perp $ is a bounded open set with $ 0 \in U $ and $ f \in W^{2,n}(U)$ is a continuous function with $ f(0) =0 $ such that

\begin{equation*} \{p + b + \tau \nu : b \in U, \, -\epsilon \lt \tau \leq f(b)\} = \overline{\Omega'} \cap C_{U,\epsilon}, \end{equation*}

where $ C_{U, t} = \{p + b + \tau \nu: b \in U, \, -t \lt \tau \lt t \} $ for each $ 0 \lt t \leq \infty $, then we observe that

\begin{equation*} N_f = \operatorname{nor}(\Omega') \cap (C_{U, \epsilon} \times \mathbf{S}^n),\end{equation*}

where $ N_f = \operatorname{nor}(E_f) \cap (C_{U, \infty} \times \mathbf{S}^n) $ and $ E_f = \{p + b + \tau \nu : b \in U, \, -\infty \lt \tau \leq f(b)\} $. It follows from (5.5) and theorem 3.9 that $ \eta'| \big[ \operatorname{nor}(\Omega') \cap (C_{U,\epsilon} \times \mathbf{S}^n)\big] $ is $ \mathcal{H}^n $ almost equal to a Borel n-vectorfield defined over $ \operatorname{nor}(\Omega') \cap (C_{U,\epsilon} \times \mathbf{S}^n) $ and $ (\mathcal{H}^n \operatorname{\llcorner} \big[\operatorname{nor}(\Omega') \cap (C_{U,\epsilon} \times \mathbf{S}^n)\big]) \wedge \eta' $ is an n-dimensional Legendrian cycle of $ C_{U, \epsilon} $. Henceforth, we define the integer multiplicity locally rectifiable n-current

\begin{equation*} T' = \big(\mathcal{H}^n \operatorname{\llcorner} \operatorname{nor}(\Omega')\big) \wedge \eta' \end{equation*}

and we conclude by lemma 2.5 that T ^ʹ is a Legendrian cycle of $ \mathbf{R}^{n+1} $.

We define now $ \psi = \Psi_F | \operatorname{nor}(\Omega') $ and recalling (5.1) and noting that

(5.6)\begin{equation} \operatorname{ap} D \psi(\psi^{-1}(y,v)) = D \Psi_F(\psi^{-1}(y,v)) \end{equation}

for $ \mathcal{H}^n $ a.e. $ (y,v) \in \operatorname{nor}(\Omega) $, we define

\begin{equation*} \eta(y,v) = \frac{\big[{\textstyle \bigwedge_n \operatorname{ap} D \psi(\psi^{-1}(y,v))}\big] \eta'(\psi^{-1}(y,v))}{J^{\operatorname{nor}(\Omega')}_n \psi(\psi^{-1}(y,v))} \end{equation*}

for $ \mathcal{H}^n $ a.e. $ (y,v) \in \operatorname{nor}(\Omega) $. Since $ \Psi_F $ is a diffeomorphism we have that $ \eta(y,v) \neq 0 $ for $ \mathcal{H}^n $ a.e. $ (y,v) \in \operatorname{nor}(\Omega) $. We now apply [Reference Federer10, 4.1.30] with U, K, W, ξ, G and g replaced by $ \mathbf{R}^{n+1} \times \mathbf{R}^{n+1} $, $ \partial \Omega' \times \mathbf{S}^n $, $ \operatorname{nor}(\Omega') $, $ \Psi_F $ and ψ respectively. We infer that

\begin{equation*} (\Psi_F)_{\#}\big[(\mathcal{H}^n \operatorname{\llcorner} \operatorname{nor}(\Omega')) \wedge \eta' \big] = \big(\mathcal{H}^n \operatorname{\llcorner} \operatorname{nor}(\Omega)\big) \wedge \eta \end{equation*}

and that $ | \eta(y,v) | = 1 $ and $ \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} \operatorname{nor}(\Omega), (y,v)) $ is associated with $ \eta(y,v) $ for $ \mathcal{H}^n $ a.e. $ (y,v) \in \operatorname{nor}(\Omega) $. Clearly, $ \big(\mathcal{H}^n \operatorname{\llcorner} \operatorname{nor}(\Omega)\big) \wedge \eta $ is a cycle, and $\big[ \big(\mathcal{H}^n \operatorname{\llcorner} \operatorname{nor}(\Omega)\big) \wedge \eta\big] \operatorname{\llcorner} \alpha =0 $ by lemma 2.11. Finally, if $ \boldsymbol{\ast} $ is the Hodge-star operator with respect to E (cf. remark 5.8), since $ \tau_1(x,u) \wedge \cdots \wedge \tau_n(x,u) = (-1)^n\, (\boldsymbol{\ast}u) $ and

\begin{align*} & \big[{\textstyle \bigwedge_n} \pi_0 \big](\eta(\Psi_F(x,u))) \\ & \quad = \frac{1}{J_n^{\operatorname{nor}(\Omega')}\psi(x,u)}\, \Bigg(\prod_{i=1}^{n}\frac{1}{\sqrt{1 + \kappa_{\Omega', i}(x,u)^2}}\Bigg)\\ &\qquad\big[ D F(x)(\tau_1(x,u)) \wedge \cdots \wedge D F(x)(\tau_n(x,u))\big] \\ & \quad = \frac{(-1)^n}{J_n^{\operatorname{nor}(\Omega')}\psi(x,u)}\, \Bigg(\prod_{i=1}^{n}\frac{1}{\sqrt{1 + \kappa_{\Omega', i}(x,u)^2}}\Bigg)\, \big[ {\textstyle \bigwedge_n} D F(x)\big](\boldsymbol{\ast} u) \end{align*}

for $ \mathcal{H}^n $ a.e. $(x,u) \in \operatorname{nor}(\Omega') $, it follows by remark 5.8 below and (5.1) that either

\begin{equation*} \langle \big[ {\textstyle\bigwedge_n} \pi_0\big](\eta(y,v)) \wedge v, E' \rangle \gt 0 \quad \text{for}\ \mathcal{H}^n\ \text{a.e.}\ (y,v) \in \operatorname{nor}(\Omega) \end{equation*}

or

\begin{equation*} \langle \big[ {\textstyle\bigwedge_n} \pi_0\big](\eta(y,v)) \wedge v, E' \rangle \lt 0 \quad \text{for}\ \mathcal{H}^n \text{a.e.}\ (y,v) \in \operatorname{nor}(\Omega) . \end{equation*}

This settles the existence part in statement (4). Uniqueness easily follows from the defining conditions of T and the representation of η follows from lemma 2.11.

Proof. We know by theorem 5.7(4) that $ N_{\Omega} = (\mathcal{H}^n \operatorname{\llcorner} \operatorname{nor}(\Omega)) \wedge (\zeta_1\wedge \ldots \wedge \zeta_n) $. Noting that

\begin{equation*} J_n^{\operatorname{nor}(\Omega)}\pi_0(x,u) = \prod_{i=1}^n \frac{1}{\sqrt{1 + \kappa_{\Omega,i}(x,u)^2}} \quad \text{for}\ \mathcal{H}^n\ \text{a.e.}\ (x,u) \in \operatorname{nor}(\Omega) , \end{equation*}

we employ theorem 5.7(3) to compute

\begin{align*} [N_\Omega \operatorname{\llcorner} \varphi_{n-k}](\phi) = {n \choose k} \int_{\operatorname{nor}(\Omega)} J_n^{\operatorname{nor}(\Omega)}\pi_0(x,u)\,\phi(x,u)\,H_{\Omega, k}(x)\, d\mathcal{H}^n(x,u), \end{align*}

whence we conclude using area formula in combination with theorem 5.7(2) and lemma 5.5(1).

Proof. Combining remark 5.10 and lemma 5.13 we obtain

\begin{equation*} \big[(\Psi_{F_t})_{\#}N_\Omega \big](\varphi_{n-k+1}) = N_{F_t(\Omega)}(\varphi_{n-k+1}) = {n \choose k-1} \mathcal{A}_{k-1}(F_t(\Omega)) \end{equation*}

for $ k = 1 , \ldots, n+1 $. Hence we use lemma 2.8 and again lemma 5.13 to compute

\begin{equation*} \frac{d}{dt}\big[(\Psi_{F_t})_{\#}N_\Omega \big](\varphi_{n-k+1}) \Big|_{t=0} = k\,{n \choose k}\int_{\partial \Omega} (V(x) \bullet \nu_\Omega(x))\, H_{\Omega, k}(x)\, d\mathcal{H}^n(x) \end{equation*}

for $ k = 1, \ldots , n $ and

\begin{equation*} \frac{d}{dt}\big[(\Psi_{F_t})_{\#}N_\Omega \big](\varphi_{0}) \Big|_{t=0} = 0. \end{equation*}

Proof. We consider the local variation $ F_t(x) = e^t\, x $ for $ (x,t) \in \mathbf{R}^n \times \mathbf{R} $ and we notice that

\begin{equation*} \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} \partial \Omega, x) = \operatorname{Tan}^n(\mathcal{H}^n \operatorname{\llcorner} F_t(\partial \Omega), F_t(x)) \end{equation*}

\begin{equation*} \nu_{F_t(\Omega)}(F_t(x)) = \nu_\Omega(x) \quad \textrm{and} \quad {\chi}_{F_t(\Omega),i}(F_t(x)) = e^{-t} {\chi}_{\Omega,i}(x) \end{equation*}

for $ \mathcal{H}^n $ a.e. $ x \in \partial \Omega $ and $ i = 1, \ldots , n $. Henceforth, we compute by area formula

\begin{align*} \mathcal{A}_{r-1}(F_t(\Omega)) & = \int_{\partial F_t(\Omega)} H_{F_t(\Omega), r-1}\, d\mathcal{H}^n \\ & = e^{-(r-1)t} \int_{F_t(\partial \Omega)} H_{\Omega, r-1}(F_t^{-1}(y))\, d\mathcal{H}^n(y)\\ & = e^{(n-r+1)t} \int_{\partial \Omega} H_{\Omega, r-1}(x)\, d\mathcal{H}^n(x) \end{align*}

and we apply theorem 5.15.

Proof. Suppose $ H_{\Omega,k}(z) =0 $ for $ \mathcal{H}^n $ a.e. $ z \in \partial \Omega $. Then we can employ corollary 5.17 (with r = k) and use (5.8) (for $ i = k-1 $) to infer that $ H_{\Omega, k-1}(z) =0 $ for $ \mathcal{H}^n $ a.e. $ z \in \partial \Omega $. Now we repeat this argument with $ r = k-1 $ and $ i = k-2 $ to infer that $ H_{\Omega, k-2}(z) =0 $ for $ \mathcal{H}^n $ a.e. $ z \in \partial \Omega $, and we continue until we obtain that $ H_{\Omega, 0}(z) =0 $ for $ \mathcal{H}^n $ a.e. $ z \in \partial \Omega $, which means $ \mathcal{H}^n(\partial \Omega) =0 $. Since the latter is clearly impossible, we have proved the assertion.

Proof. We define $ \Omega' = \mathbf{R}^{n+1} \setminus \overline{\Omega} $ and notice that $ \Omega' $ is a $ W^{2,n} $-domain. Since $ \partial^v_+ \Omega' = \partial^v_+ \Omega $ and $ \nu_{\Omega'} = -\nu_\Omega $, it follows from theorem 5.7 that

\begin{equation*} \mathcal{H}^n\big(\operatorname{nor}(\Omega') \setminus \{(z, -\nu_\Omega(z)): z \in \partial^v_+\Omega\}\big) =0, \end{equation*}

and

\begin{equation*} -{\chi}_{\Omega,i}(z)= {\chi}_{\Omega',i}(z) = \kappa_{\Omega',i}(z, -\nu_\Omega(z)) \quad \text{for}\ \mathcal{H}^n\ \text{a.e.}\ z \in \partial^v_+ \Omega . \end{equation*}

Henceforth,

(6.1)\begin{equation} \sum_{i=1}^n \kappa_{\Omega',i}(z,u) = - n\, H_{\Omega, 1}(z) \leq 0 \quad \text{for}\ \mathcal{H}^n\ \text{a.e.}\ (z,u) \in \operatorname{nor}(\Omega') \end{equation}

and we infer from theorem 2.13 and area formula [Reference Federer10, 3.2.20] that

(6.2)\begin{align} (n+1)\mathcal{L}^{n+1}(\Omega) & \leq \int_{\operatorname{nor}(\Omega')}J_n^{\operatorname{nor}(\Omega')}\pi_0(z,u)\,\frac{n}{|\sum_{i=1}^n \kappa_{\Omega',i}(z,u)|}\, d\mathcal{H}^n(z,u) \notag \\ & = \int_{\partial \Omega}\frac{1}{H_{\Omega,1}(z)}\, d\mathcal{H}^n(z). \end{align}

We assume now that $ H_{\Omega, 1}(z) \geq \frac{\mathcal{H}^n(\partial \Omega)}{(n+1)\mathcal{L}^{n+1}(\Omega)} $ for $ \mathcal{H}^n $ a.e. $ z \in \partial \Omega $. Then, we observe that

\begin{equation*} \mathcal{H}^n\bigg( \bigg\{z \in \partial \Omega: H_{\Omega, 1}(z) \geq (1+\epsilon)\frac{\mathcal{H}^n(\partial \Omega)}{(n+1)\mathcal{L}^{n+1}(\Omega)} \bigg\}\bigg) =0 \quad \text{for every}\ \epsilon \gt 0 , \end{equation*}

otherwise we would obtain a contradiction with the inequality (6.2) (cf. proof of [Reference Hug and Santilli14, Corollary 5.16]). This implies that

\begin{equation*} H_{\Omega, 1}(z) = \frac{\mathcal{H}^n(\partial \Omega)}{(n+1)\mathcal{L}^{n+1}(\Omega)} \quad \text{for}\ \mathcal{H}^n\ \text{a.e.}\ z \in \partial \Omega , \end{equation*}

whence we infer that (6.2) holds with equality. Recalling (6.1) we deduce from theorem 2.13 that Ω must be a round ball.

Proof. Combining theorem 5.17 and divergence theorem for sets of finite perimeter, (it is clear by lemma 5.5 that Ω is a set of finite perimeter whose reduced boundary is $\mathcal{H}^n$ almost equal to the topological boundary) we obtain