Proof. Let $x_0\in \Omega \backslash U$. Then $u(x_0)=0$. We test $u$ from below at $x_0$ by

(2.1)\begin{equation} \psi_n(x)=\begin{cases} 0 & \text{if}\ |x-x_0|<\frac1n\\ u(x) & \text{otherwise}.\end{cases} \end{equation}

From the inequality

\[ \mathcal{I}^{-}_{k}\psi_n(x_0)\leq0 \]

we infer that, for any $n\in \mathbb {N}$, there exists an orthonormal frame $\left \{\xi _1(n),\ldots,\xi _k(n)\right \}$ such that

(2.2)\begin{equation} C_s\sum_{i=1}^k\int\limits_{0}^{+\infty}\frac{\left[u(x_0+\tau\xi_i(n))+u(x_0-\tau\xi_i(n))\right]\chi_{\left(\frac1n,+\infty\right)}(\tau)}{\tau^{1+2s}}\,d\tau\leq\frac1n\,, \end{equation}

where $\chi _{(\frac 1n,+\infty )}$ denotes the characteristic function of the interval $(\frac 1n,+\infty )$.

Up to a subsequence we can further assume that

\[ \lim_{n\to+\infty}\xi_i(n)=\bar\xi_i\quad\text{for}\ i=1,\ldots,k \]

and that $\left \{\bar \xi _1,\ldots,\bar \xi _k\right \}\in {\mathcal {V}}_k$.

Using Fatou's lemma in (2.2) we obtain

(2.3)\begin{equation} \sum_{i=1}^k\int\limits_{0}^{+\infty}\liminf_{n\to+\infty}\frac{\left[u(x_0+\tau\xi_i(n))+u(x_0-\tau\xi_i(n))\right]\chi_{\left(\frac1n,+\infty\right)}(\tau)}{\tau^{1+2s}}\,{\rm d}\tau\leq0. \end{equation}

Since for any $\tau >0$ we have $\chi _{(\frac 1n,+\infty )}(\tau )\to \chi _{(0,+\infty )}(\tau )$ as $n\to +\infty$ and

\[ u(x_0\pm\tau\bar\xi_i)\leq\liminf_{n\to+\infty}u(x_0\pm\tau\xi_i(n)), \]

by lower semincontinuity, then we infer from (2.3) that

(2.4)\begin{equation} \sum_{i=1}^k\int\limits_{0}^{+\infty}\frac{u(x_0+\tau\bar\xi_i)+u(x_0-\tau\bar\xi_i)}{\tau^{1+2s}}\,{\rm d}\tau\leq0. \end{equation}

Moreover, $u\geq 0$ in $\mathbb {R}^N$ by assumption, hence by (2.4) we conclude that

\[ u(x_0+\tau\bar\xi_i)=0\quad\forall \tau\in\mathbb{R},\ i=1,\ldots,k \]

that is $x_0+\tau \bar \xi _i\notin U$ for any $\tau \in \mathbb {R}$ and any $i=1,\ldots,k$.

Proof. Let us define $u\in LSC(\mathbb {R}^N)$ by

\[ u(x)=\chi_U(x). \]

It is clear that $U=\left \{x\in \mathbb {R}^N\,:\;u(x)>0\right \}$. Moreover, since $\Omega \backslash U\neq \emptyset$ by definition of $G_{\Omega,U,k}$, it holds that $\min _\Omega u=0$.

To prove that $u$ is a viscosity supersolution, let $x_0\in \Omega$. If $x_0\in \Omega \cap U$, then $u(x_0)=1$. Since $U$ is open, the function $u$ is in fact constant in a neighbourhood of $x_0$. Hence, for any $\xi \in \mathbb {R}^N$, with $|\xi |=1$, $\mathcal {I}_\xi u(x_0)$ is well defined and

\[ \mathcal{I}_\xi u(x_0)=C_s\int \limits_{0}^{+\infty}\frac{u(x_0+\tau\xi)+u(x_0-\tau\xi)-2}{\tau^{1+2s}}\,{\rm d}\tau\leq0 \]

considering that $0\leq u(x)\leq 1$ for any $x\in \mathbb {R}^N$. Then we infer that the inequality

\[ \mathcal{I}^{-}_{k}u(x_0)\leq0 \]

holds in classical, and so in the viscosity, sense at $x_0\in \Omega \cap U$.

Assume now that $x_0\in \Omega \backslash U$, so that $u(x_0)=0$. In this case $u$ can be discontinuous at $x_0$ depending on whether $x_0\in \Omega \cap \partial U$ or not.

To check that the inequality $\mathcal {I}^{-}_{k}u(x_0)\leq 0$ holds in the viscosity sense, let $\varphi \in C^2(\overline {B_\delta (x_0)})$, $\delta >0$, be such that

(2.5)\begin{equation} \varphi(x_0)=0\quad\text{and}\quad\varphi (x)\leq u(x)\;\;\forall x\in B_\delta(x_0). \end{equation}

Consider the function

\[ \psi(x)=\begin{cases} \varphi(x) & \text{if}\ x\in B_\delta(x_0)\\ u(x) & \text{otherwise}. \end{cases} \]

Using the hypothesis $G_{\Omega,U,k}$, there exists an orthonormal frame $\left \{\bar \xi _1,\ldots,\bar \xi _k\right \}$, depending on $x_0$, such that

(2.6)\begin{equation} u(x_0+\tau\bar\xi_i)=0\quad\forall \tau\in\mathbb{R},\;i=1,\ldots,k. \end{equation}

Thus, by (2.5)–(2.6), we conclude

\[ \mathcal{I}^{-}_{k}\psi(x_0)\leq\sum_{i=1}^k\mathcal{I}_{\bar\xi_i}\psi(x_0) =C_s \sum_{i=1}^k\int \limits_{0}^{\delta}\frac{\varphi(x_0+\tau\bar\xi_i)+\varphi(x_0-\tau\bar\xi_i)}{\tau^{1+2s}}\,{\rm d}\tau\leq0\,. \]

Proof. We define

\[ u(x)=\text{dist}(x,\mathbb{R}^N\backslash U). \]

Clearly $u$ is bounded and Lipschitz in $\mathbb {R}^N$. Moreover, since $U$ is open, $u(x)>0$ if, and only if, $x\in U$.

To check that $u$ is a viscosity supersolution of (2.7), let $x_0\in \mathbb {R}^N$ and let $\varphi \in C^2(\overline {B_\delta (x_0)})$, $\delta >0$, be such that

(2.8)\begin{equation} u(x_0)-\varphi(x_0)=0\leq u(x)-\varphi(x)\quad\forall x\in B_\delta(x_0). \end{equation}

Setting

\[ \psi(x)=\begin{cases} \varphi(x) & \text{if $x\in B_\delta(x_0)$}\\ u(x) & \text{otherwise}, \end{cases} \]

we have to prove that $\mathcal {I}^{-}_{k}\psi (x_0)\leq 0$.

We choose $y_0\in \mathbb {R}^N\backslash U$, depending on $x_0$, such that

(2.9)\begin{equation} \varphi(x_0)=u(x_0)=|x_0-y_0|\,. \end{equation}

Moreover, by the definition of $u$, we have

(2.10)\begin{equation} \varphi(x)\leq u(x)\leq |x-y|\quad\forall x\in B_\delta(x_0),\;y\in\mathbb{R}^N\backslash U. \end{equation}

Using (2.9)–(2.10) with $x=y+x_0-y_0$ and setting $\phi (y)=\varphi (y+x_0-y_0)$, we then obtain

(2.11)\begin{equation} \phi(y)\leq \phi(y_0)\quad\forall y\in B_\delta(y_0)\cap(\mathbb{R}^N\backslash U). \end{equation}

Since $y_0\in \mathbb {R}^N\backslash U$, by the assumption $G_{\mathbb {R}^N,U,k}$ there exists an orthonormal frame $\left \{\bar \xi _1,\ldots,\bar \xi _k\right \}$ such that

(2.12)\begin{equation} y_0+\tau\bar\xi_i\notin U\quad\forall\tau\in\mathbb{R},\ i=1,\ldots,k. \end{equation}

Then, from (2.11) and (2.12), we have

\[ \phi(y_0+\tau\bar\xi_i)+\phi(y_0-\tau\bar\xi_i)-2\phi(y_0)\leq0\quad\forall\tau\in[0,\delta),\ i=1,\ldots,k. \]

The above inequality implies that

\[ \sum_{i=1}^k\int \limits_{0}^{\delta}\frac{\phi(y_0+\tau\bar\xi_i)+\phi(y_0-\tau\bar\xi_i)-2\phi(y_0)}{\tau^{1+2s}}\,{\rm d}\tau\leq0. \]

Moreover, since $\phi (y_0\pm \tau \bar \xi _i)=\varphi (x_0\pm \tau \bar \xi _i)$ for $\tau \in [0,\delta )$ and $i=1,\ldots,k$, we obtain

(2.13)\begin{equation} \sum_{i=1}^k\int \limits_{0}^{\delta}\frac{\varphi(x_0+\tau\bar\xi_i)+\varphi(x_0-\tau\bar\xi_i)-2\varphi(x_0)}{\tau^{1+2s}}\,d\tau\leq0\,. \end{equation}

Now, we use the inequality

\[ u(x)\leq |x-y|\quad\forall x\in\mathbb{R}^N,\ y\in\mathbb{R}^N\backslash U \]

with the particular choice $x=x_0\pm \tau \bar \xi _i$ and $y=y_0\pm \tau \bar \xi _i$ to infer that

\[ u(x_0\pm\tau\bar\xi_i)\leq|x_0-y_0|=u(x_0) \quad\forall \tau\in[0,+\infty),\;i=1,\ldots,k\,. \]

Thus,

(2.14)\begin{equation} \sum_{i=1}^k\int \limits_{\delta}^{+\infty}\frac{u(x_0+\tau\bar\xi_i)+u(x_0-\tau\bar\xi_i)-2u(x_0)}{\tau^{1+2s}}\,{\rm d}\tau\leq0\,. \end{equation}

The conclusion

\[ \mathcal{I}^{-}_{k}\psi(x_0)\leq0 \]

easily follows from (2.13) and (2.14).

Proof. Let $x_0\in \Omega \cap \partial U$. By assumption, and by an orthogonal transformation, we may assume that $x_0=0$ and that for some $r>0$

(2.18)\begin{equation} U\cap B_r=\left\{x\in B_r\,:\;x_N>f(x')\right\}, \end{equation}

where $x=(x',x_N)$ and

(2.19)\begin{equation} f\in C^2(B_r(x')),\quad f(0')=0, \quad Df(0')=0'. \end{equation}

Moreover, the principal curvatures of $\Omega \cap \partial U$ at $x_0=0$ are the eigenvalues of $D^2f(0')$.

Since $G_{\Omega,U,k}$ holds and $x_0=0\in \Omega \backslash U$, then there exists $\left \{\bar \xi _i\right \}_{i=1}^k\in {\mathcal {V}}_k$ such that

(2.20)\begin{equation} \tau\bar\xi_i\notin U\quad\forall\tau\in\mathbb{R},\ i=1,\ldots,k. \end{equation}

We claim that $\left \langle \bar \xi _i,e_N\right \rangle =0$ for any $i=1,\ldots,k$, where $e_N=(0',1)$.

If not, then $\left \langle \bar \xi _i,e_N\right \rangle \neq 0$ for some $i\in \left \{1,\ldots,k\right \}$. Replacing $\bar \xi _i$ with $-\bar \xi _i$ if necessary, we can further suppose that $\left \langle \bar \xi _i,e_N\right \rangle >0$. Since

\[ f\left(\tau\bar\xi'_i\right)=o(\tau)\quad\text{as}\ \tau\to0, \]

we infer that for any $\tau$ positive and small enough

\[ \tau\left\langle \bar\xi_i,e_N\right\rangle> f\left(\tau\,\bar\xi'_i\right). \]

Thus, using (2.18), we have that for any $\tau$ positive and small enough

\[ \tau\bar\xi_i\in U \]

which contradicts (2.20).

Since $\left \langle \bar \xi _i,e_N\right \rangle =0$, we can write

(2.21)\begin{equation} \bar\xi_i=(\bar\xi'_i,0). \end{equation}

Moreover, $\left \{\bar \xi '_1,\ldots,\bar \xi '_k\right \}$ is an orthonormal frame in $\mathbb {R}^{N-1}$.

Consider now, for $i=1,\ldots,k$, the functions

\[ g_i(\tau)=f(\tau\bar\xi'_i)\quad\tau\in({-}r, r). \]

Using (2.18)–(2.19)–(2.20)–(2.21) we obtain that

\[ g_i(\tau)\geq0=g(0)\quad\forall \tau\in({-}r,r). \]

Hence for any $i=1,\ldots,k$

\[ g_i''(0)=\left\langle D^2f(0')\bar\xi'_i,\bar\xi'_i \right\rangle\geq0, \]

from which we conclude that

\begin{align*} \sum_{i=1}^k\kappa_{N-i}(x_0)& =\sup\left\{\sum_{i=1}^k\left\langle D^2f(0')\xi_i,\xi_i\right\rangle\;:\;\;\left\{\xi_1,\ldots,\xi_k\right\} \text{orthonormal set in}\ \mathbb{R}^{N-1}\right\}\\ & \geq\sum_{i=1}^k\left\langle D^2f(0')\bar\xi'_i,\bar\xi'_i\right\rangle\geq0. \end{align*}

Proof. Let $x_0\in \Omega \backslash U$. We proceed as in the proof of theorem 2.1 testing $u$ from below at $x_0$ by the sequence of function $\psi _n$ defined in (2.1). Hence for any $n\in \mathbb {N}$, there exists a $k$-dimensional linear space $V=V(n)$ such that

(3.2)\begin{equation} C_s\sup_{\xi\in V,\;|\xi|=1}\int\limits_{0}^{+\infty}\frac{\left[u(x_0+\tau\xi)+u(x_0-\tau\xi)\right]\chi_{\left(\frac1n,+\infty\right)}(\tau)}{\tau^{1+2s}}\,{\rm d}\tau\leq\frac1n. \end{equation}

Choose an orthonormal basis $\left \{\xi _1(n),\ldots,\xi _k(n)\right \}$ of $V$. Passing to a subsequence, we can also assume that

\[ \lim_{n\to+\infty}\xi_i(n)=\bar\xi_i\quad i=1,\ldots,k. \]

Let $\bar V=\text {span}\left \{\bar \xi _1,\ldots,\bar \xi _k\right \}$. We have $\dim \bar V=k$. We claim that $\bar V$ satisfies also condition $(ii)$ in definition 1.2. For this we shall prove that for any fixed unit vector $\xi \in \bar V$, then

(3.3)\begin{equation} x_0+\tau\xi\notin U\quad \forall \tau\in\mathbb{R}. \end{equation}

Let $\xi \in \bar V$ be such that $|\xi |=1$ and let

\[ v(n):=\sum_{i=1}^k\left\langle \xi_i(n),\xi\right\rangle\xi_i(n). \]

Note that $v(n)\in V=V(n)$ and that

(3.4)\begin{equation} \lim_{n\to+\infty}v(n)=\xi. \end{equation}

Hence, for any $n$ sufficiently large, $|v(n)|>0$. Using (3.2) with $\xi (n)=\frac {v(n)}{|v(n)|}\in V$ we have

\[ C_s\int\limits_{0}^{+\infty}\frac{\left[u(x_0+\tau\xi(n))+u(x_0-\tau\xi(n))\right]\chi_{\left(\frac1n,+\infty\right)}(\tau)}{\tau^{1+2s}}\,{\rm d}\tau\leq\frac1n. \]

Passing to the limit in the above inequality as $n\to +\infty$ and using Fatou's Lemma, $u\in LSC(\mathbb {R}^N)$ and (3.4) we obtain

\[ \int\limits_{0}^{+\infty}\frac{u(x_0+\tau\xi)+u(x_0-\tau\xi)}{\tau^{1+2s}}\,{\rm d}\tau\leq0. \]

The assumption $u\geq 0$ in $\mathbb {R}^N$ yields $u(x_0+\tau \xi )=0$ for any $\tau \in \mathbb {R}$. Thus, (3.3) holds.

Proof. Let $U_0$ be a connected component of $U$. Note that $U_0$ is an open subset of $U$ since $U$ is locally connected.

To the contrary that $U_0$ is a convex set, we suppose that $U_0$ is not convex, which implies that there are two distinct points $x,y\in U_0$ such that the line segment $[x,y]:=\{tx+(1-t)y : t\in [0,1]\}$ is not contained in $U_0$. This implies that $[x,y]$ is not contained in $U$. Indeed, if we suppose that $[x,y]\subset U$, then $[x,y]$ is a connected subset of $U$ and intersects $U_0$, and therefore, $U_0\cup [x,y]$ is a connected subset of $U$, which yields a contradiction that $[x,y]\subset U_0$. Thus, we find that $[x,y]$ is not contained in $U$. We can choose $\lambda \in (0,1)$ so that $z:=\lambda x+(1-\lambda )y\not \in U$. By the property ${\mathcal {G}}_{\mathbb {R}^N,U,N-1}$, we can choose $(N-1)$-dimensional linear subspace $V$ of $\mathbb {R}^N$ such that $(z+V)\cap U=\emptyset$. Select $\nu \in \mathbb {R}^N\setminus \{0\}$ so that $z+V=\{p\in \mathbb {R}^N : \langle \nu, p-z\rangle =0\}$ and set

\[ H^+{=}\{p\in\mathbb{R}^N : \langle \nu, p-z\rangle>0\}\ \text{and}\ H^-{=}\{p\in\mathbb{R}^N : \langle \nu, p-z\rangle<0\}. \]

Observe that

(3.5)\begin{equation} U_0=(H^+ \cap U_0)\ \cup\ (H^- \cap U_0), \end{equation}

that the right-hand side of (3.5) is a disjoint union of two open subsets of $U$, and that either $x\in H^+$ and $y\in H^-$, or $x\in H^-$ and $y\in H^+$, which assures that both $H^+\cap U_0$ and $H^-\cap U_0$ are nonempty. Hence, (3.5) contradicts the connectedness of $U_0$, which completes the proof.