Nonlocal equations with gradient constraints

Abstract

We prove the existence and $C^{1,\alpha }$ regularity of solutions to nonlocal fully nonlinear elliptic equations with gradient constraints. We do not assume any regularity about the constraints; so the constraints need not be $C^{1}$ or strictly convex. We also obtain $C^{0,1}$ boundary regularity for these problems. Our approach is to show that these nonlocal equations with gradient constraints are related to some nonlocal double obstacle problems. Then we prove the regularity of the double obstacle problems. In this process, we also employ the monotonicity property for the second derivative of obstacles, which we have obtained in a previous work.

Appendices

Appendix A: Proof of Theorem 3 for nonzero exterior data

We do not use the next two propositions directly in the proof of Theorem 3, however, they enhance our understanding of double obstacle problems and equations with gradient constraints. First let us introduce the following terminology for the solutions of the double obstacle problem (1.13). (The notation is motivated by the physical properties of the elastic–plastic torsion problem, in which E stands for the elastic region, and P stands for the plastic region.)

Definition 5

Let

$$\begin{aligned} P^{+}:=\{x\in U:u(x)=\rho (x)\},{} & {} P^{-}:=\{x\in U:u(x)=-{\bar{\rho }}(x)\}. \end{aligned}$$

Then $P:=P^{+}\cup P^{-}$ is called the coincidence set; and

$$\begin{aligned} E:=\{x\in U:-{\bar{\rho }}(x)<u(x)<\rho (x)\} \end{aligned}$$

is called the non-coincidence set. We also define the free boundary to be $\partial E\cap U$.

Proposition 1

Suppose Assumption 1 holds. Let $u\in C^{1}(U)$ be a viscosity solution of the double obstacle problem (1.13). If $x\in P^{+}$, and y is a $\rho$-closest point on $\partial U$ to x such that $[x,y[\subset U$, then we have $[x,y[\subset P^{+}$. Similarly, if $x\in P^{-}$, and y is a ${\bar{\rho }}$-closest point on $\partial U$ to x such that $[x,y[\subset U$, then we have $[x,y[\subset P^{-}$.

Remark

If the strict Lipschitz property (2.9) for $\varphi$ holds, then by Lemma 3 we automatically have $[x,y[\subset U$.

Proof

Suppose $x\in P^{-}$; the other case is similar. We have

$$\begin{aligned} u(x)=-{\bar{\rho }}(x)=-\gamma (y-x)+\varphi (y). \end{aligned}$$

Let ${\tilde{v}}:=u-(-{\bar{\rho }})\ge 0$, and $\xi :=\frac{y-x}{\gamma (y-x)}=-\frac{x-y}{{\bar{\gamma }}(x-y)}$. Then ${\bar{\rho }}$ varies linearly along the segment ]x, y[, since y is a ${\bar{\rho }}$-closest point to the points of the segment. So we have $D_{\xi }(-{\bar{\rho }})=D_{-\xi }{\bar{\rho }}=1$ along the segment. Note that we do not assume the differentiability of ${\bar{\rho }}$; and $D_{-\xi }{\bar{\rho }}$ is just the derivative of the restriction of ${\bar{\rho }}$ to the segment ]x, y[. Now by Lemma 1 we get

$$\begin{aligned} D_{\xi }u=\langle Du,\xi \rangle \le \gamma ^{\circ }(Du)\gamma (\xi )\le 1. \end{aligned}$$

So we have $D_{\xi }{\tilde{v}}\le 0$ along ]x, y[. Thus as ${\tilde{v}}(x)={\tilde{v}}(y)=0$, and ${\tilde{v}}$ is continuous on the closed segment [x, y], we must have ${\tilde{v}}\equiv 0$ on [x, y]. Therefore $u=-{\bar{\rho }}$ along the segment as desired. $\square$

Proposition 2

Suppose Assumption 1 holds. Let $u\in C^{1}(U)$ be a viscosity solution of the double obstacle problem (1.13). Suppose Assumption 2 holds too. Then we have

$$\begin{aligned} R_{\rho ,0}\cap P^{+}=\emptyset ,&\hspace{2cm}&R_{{\bar{\rho }},0}\cap P^{-}=\emptyset . \end{aligned}$$

Proof

Note that due to Assumption 2, the strict Lipschitz property (2.9) for $\varphi$ holds, and $\gamma$ is strictly convex. Let us show that $R_{{\bar{\rho }},0}\cap P^{-}=\emptyset$; the other case is similar. Suppose to the contrary that $x\in R_{{\bar{\rho }},0}\cap P^{-}$. Then there are at least two distinct points $y,z\in \partial U$ such that

$$\begin{aligned} {\bar{\rho }}(x)=\gamma (y-x)-\varphi (y)=\gamma (z-x)-\varphi (z). \end{aligned}$$

Now by Lemma 3 we know that $[x,y[,[x,z[\subset U$; so by Proposition 1 we get $[x,y[,[x,z[\subset P^{-}$.

In other words, $u=-{\bar{\rho }}$ on both of these segments. Therefore by Lemma 3, u varies linearly on both of these segments. Hence we get

$$\begin{aligned} \big \langle Du(x),\frac{y-x}{\gamma (y-x)}\big \rangle =1=\big \langle Du(x),\frac{z-x}{\gamma (z-x)}\big \rangle . \end{aligned}$$

However, since $\gamma$ is strictly convex, and by Lemma 1 we know that $\gamma ^{\circ }(Du(x))\le 1$, this equality implies that z, y must be on the same ray emanating from x. But this contradicts the fact that $[x,y[,[x,z[\subset U$. $\square$

Remark

Since here we do not have $C^{1,1}$ regularity for u, we cannot imitate the proof given in the local case to show that $P^{+},P^{-}$ do not intersect $R_{\rho },R_{{\bar{\rho }}}$. In addition, merely knowing that the solution does not touch the obstacles at their singularities is not enough to obtain uniform bounds for $Iu_{k}$ in the proof of Theorem 3, because we need to have a uniform positive distance from the ridges too, due to the nonlocal nature of the operator. As we have seen in the proof of Theorem 3, we overcome these limitations by using some suitable barriers.

Next let us review some well-known facts from convex analysis which are needed in the following proof. Consider a compact convex set K. Let $x\in \partial K$ and $\textrm{v}\in {\mathbb {R}}^{n}-\{0\}$. We say the hyperplane

$$\begin{aligned} \Gamma _{x,\textrm{v}}:=\{y\in {\mathbb {R}}^{n}:\langle y-x,\textrm{v}\rangle =0\} \end{aligned}$$

(A.1)

is a supporting hyperplane of K at x if $K\subset \{y:\langle y-x,\textrm{v}\rangle \le 0\}$. In this case we say $\textrm{v}$ is an outer normal vector of K at x (see Fig. 2). The normal cone of K at x is the closed convex cone

$$\begin{aligned} N(K,x):=\{0\}\cup \{\textrm{v}\in {\mathbb {R}}^{n}-\{0\}:\textrm{v}\text { is an outer normal vector of }K\text { at }x\}. \end{aligned}$$

(A.2)

It is easy to see that when $\partial K$ is $C^{1}$ we have

$$\begin{aligned} N(K,x)=\{tD\gamma (x):t\ge 0\}. \end{aligned}$$

For more details see [44, Sections 1.3 and 2.2].

Proof of Theorem 3for nonzero exterior data As before we approximate $K^{\circ }$ by a sequence $K_{k}^{\circ }$ of compact convex sets, that have smooth boundaries with positive curvature, and

$$\begin{aligned} K_{k+1}^{\circ }\subset \textrm{int}(K_{k}^{\circ }),&\qquad&K^{\circ }={\textstyle \bigcap }K_{k}^{\circ }. \end{aligned}$$

Then $K_{k}$’s are strictly convex compact sets with 0 in their interior, which have smooth boundaries with positive curvature. Furthermore we have $K=(K^{\circ })^{\circ }\supset K_{k+1}\supset K_{k}$. To simplify the notation we use $\gamma _{k},\gamma _{k}^{\circ },\rho _{k},{\bar{\rho }}_{k}$ instead of $\gamma _{K_{k}},\gamma _{K_{k}^{\circ }},\rho _{K_{k},\varphi },{\bar{\rho }}_{K_{k},\varphi }$, respectively. Note that $K_{k},U,\varphi$ satisfy the Assumption 2. In particular we have $\gamma _{k}^{\circ }(D\varphi )<1$, since $D\varphi \in K^{\circ }\subset \textrm{int}(K_{k}^{\circ })$. Hence as we have shown in [40], $\rho _{k},{\bar{\rho }}_{k}$ satisfy the assumptions of Theorem 1 of [41] (see Appendix B). Thus there are viscosity solutions $u_{k}\in C_{\textrm{loc}}^{1,\alpha }(U)$ of the double obstacle problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \max \{\min \{-Iu_{k},\;u_{k}+{\bar{\rho }}_{k}\},u_{k}-\rho _{k}\}=0 &{} \text {in }U,\\ u_{k}=\varphi &{} \text {in }{\mathbb {R}}^{n}-U. \end{array}\right. } \end{aligned}$$

(A.3)

And $\alpha >0$ depends only on $n,\lambda ,\Lambda ,s_{0}$.

As before, we can easily see that $u_{k}$’s and their derivatives are uniformly bounded. Hence a subsequence of them converges uniformly to a continuous function u, which we extend to all of ${\mathbb {R}}^{n}$ by setting it equal to $\varphi$ in ${\mathbb {R}}^{n}-U$. In Part I we consider the barrier

$$\begin{aligned} \rho _{k,B^{c}}:=\rho _{K_{k},\varphi }(\cdot ;{\mathbb {R}}^{n}-{\overline{B}}) =\underset{z\in \partial B}{\min }[\gamma _{k}(\cdot -z)+\varphi (z)], \end{aligned}$$

(A.4)

where $B\subset {\mathbb {R}}^{n}-{\overline{U}}$ is an open ball such that $\partial B\cap \partial U=\{y_{0}\}$, in which $y_{0}\in \partial U$ is a $\rho _{k}$-closest point to $x_{0}$.

Let $z\in \partial B$, and let y be a point on $\partial U\cap [z,x_{0}[$. Then by the Lipschitz property (1.7) for $\varphi$ with respect to $\gamma _{k}$ (note that $\gamma _{k}^{\circ }(D\varphi )<1$) we have

$$\begin{aligned} \gamma _{k}(x_{0}-y_{0})&+\varphi (y_{0})\le \gamma _{k}(x_{0}-y)+\varphi (y)\\&=\gamma _{k}(x_{0}-z)+\varphi (z)-\gamma _{k}(y-z)+\varphi (y) -\varphi (z)\le \gamma _{k}(x_{0}-z)+\varphi (z). \end{aligned}$$

Hence $y_{0}$ is also a $\rho _{k,B^{c}}$-closest point to $x_{0}$ on $\partial B$.

Consequently we have

$$\begin{aligned} \rho _{k,B^{c}}(x_{0})=\gamma _{k}(x_{0}-y_{0})+\varphi (y_{0})=\rho _{k}(x_{0}). \end{aligned}$$

Now consider $x\in U\subset {\mathbb {R}}^{n}-{\overline{B}}$, and let $z\in \partial B$ be a $\rho _{k,B^{c}}$-closest point to x. Let y be a point on $\partial U\cap [z,x[$. Then similarly to the above we can show that

$$\begin{aligned} \rho _{k}(x)\le \gamma _{k}(x-y)+\varphi (y)\le \gamma _{k}(x-z)+\varphi (z)=\rho _{k,B^{c}}(x). \end{aligned}$$

As for $\rho _{k}$, we extend $\rho _{k,B^{c}}$ to all of ${\mathbb {R}}^{n}$ by setting it equal to $\varphi$ on B. Then for $x\notin U$ we either have $\rho _{k}(x)=\varphi (x)=\rho _{k,B^{c}}(x)$ when $x\in B$, or

$$\begin{aligned} \rho _{k}(x)=\varphi (x)\le \gamma _{k}(x-z)+\varphi (z)=\rho _{k,B^{c}}(x), \end{aligned}$$

when $x\notin B$ and $z\in \partial B$ is a $\rho _{k,B^{c}}$-closest point to x. Hence, $\phi$ is also touching $\rho _{k,B^{c}}$ from below at $x_{0}$.

The rest of the proof in Part I, and Part III of the proof, go as before. So we only need to prove the properties of the new barrier $\rho _{k,B^{c}}$, similarly to the Part II of the proof in the case of zero exterior data. First let us show that $\rho _{k,B^{c}}$ is $C^{2}$ on ${\mathbb {R}}^{n}-{\overline{B}}$. Let $x\in {\mathbb {R}}^{n}-{\overline{B}}$. By the results of Sect. 2.2 we need to show that x has only one $\rho _{k,B^{c}}$-closest point on $\partial B$, and $\det Q_{k,B^{c}}(x)\ne 0$ where $Q_{k,B^{c}}$ is given by (A.6).

Suppose to the contrary that $y,{\tilde{y}}\in \partial B$ are two $\rho _{k,B^{c}}$-closest points to x. Let $z=\frac{y+{\tilde{y}}}{2}\in B$. We assume that the radius of B is small enough so that $\varphi$ is convex on a neighborhood of it. Then we have

$$\begin{aligned} \gamma _{k}(x-z)+\varphi (z)\le \frac{1}{2}\big (\gamma _{k}(x-y) +\gamma _{k}(x-{\tilde{y}})+\varphi (y)+\varphi ({\tilde{y}})\big )=\rho _{k,B^{c}}(x). \end{aligned}$$

Let ${\tilde{z}}$ be the point on $\partial B\,\cap \,]z,x[$. Then by the strict Lipschitz property (2.9) for $\varphi$ with respect to $\gamma _{k}$ (note that $\gamma _{k}^{\circ }(D\varphi )<1$) we have

$$\begin{aligned} \gamma _{k}(x-{\tilde{z}})+\varphi ({\tilde{z}})=\gamma _{k}(x-z)+\varphi (z)&-\gamma _{k}({\tilde{z}}-z)+\varphi ({\tilde{z}})-\varphi (z)\\&<\gamma _{k}(x-z)+\varphi (z)\le \rho _{k,B^{c}}(x), \end{aligned}$$

which is a contradiction. So x must have a unique $\rho _{k,B^{c}}$-closest point y on $\partial B$.

Next note that by (2.18) at $y\in \partial B$ we have

$$\begin{aligned} D^{2}\rho _{k,B^{c}}(y)=(I-X_{k}^{T})\big (D^{2}\varphi (y) +\lambda _{k}(y)D^{2}d_{B^{c}}(y)\big )(I-X_{k}), \end{aligned}$$

(A.5)

where $d_{B^{c}}$ is the Euclidean distance to $\partial B$ on ${\mathbb {R}}^{n}-B$, and $\lambda _{k},X_{k}$ are given by (2.10),(2.12) (using $\gamma _{k}^{\circ }$ instead of $\gamma ^{\circ }$). But the eigenvalues of $D^{2}d$ are minus the principal curvatures of the boundary, and a zero eigenvalue corresponding to the normal direction (see [18, Section 14.6],). So the nonzero eigenvalues of $D^{2}d_{B^{c}}$ are $\frac{1}{r_{0}}$, where $r_{0}$ is the radius of B (note that we are in the exterior of B, so the curvatures are $\frac{-1}{r_{0}}$). Hence $D^{2}d_{B^{c}}$ is a positive semidefinite matrix. Therefore $D^{2}\rho _{k,B^{c}}(y)$ is also a positive semidefinite matrix, since $\varphi$ is convex on a neighborhood of $\partial U$. (Although the convexity of $\varphi$ is not really needed here. Because by using $\gamma ^{\circ }(D\varphi )<1$ we can easily show that $\lambda _{k}$ has a uniform positive lower bound independently of k, B. Then by decreasing the radius $r_{0}$ and using the boundedness of $D^{2}\varphi$ we can get the desired.)

Now let us consider the matrices W, Q for $\rho _{k,B^{c}}$ given by (2.19). The eigenvalues of

$$\begin{aligned} W_{k,B^{c}}(y_{0}):=-D^{2}\gamma _{k}^{\circ }(\mu _{k}(y))D^{2}\rho _{k,B^{c}}(y) \end{aligned}$$

must be nonpositive, because $D^{2}\gamma _{k}^{\circ }(\mu _{k})$ is a positive semidefinite matrix. Therefore the eigenvalues of

$$\begin{aligned} Q_{k,B^{c}}(x):=I-\big (\rho _{k,B^{c}}(x)-\varphi (y)\big )W_{k,B^{c}}(y) =I-\gamma _{k}(x-y)W_{k,B^{c}}(y) \end{aligned}$$

(A.6)

are greater than or equal to 1. Hence $\det Q_{k,B^{c}}(x)>0$, and we can conclude that $\rho _{k,B^{c}}$ is $C^{2}$ on a neighborhood of an arbitrary point $x\in {\mathbb {R}}^{n}-{\overline{B}}$, as desired.

In addition, by Lemma 4 we have

$$\begin{aligned} D^{2}\rho _{k,B^{c}}(x)\le D^{2}\rho _{k,B^{c}}(y)=(I-X_{k}^{T})\big (D^{2}\varphi (y) +\lambda _{k}(y)D^{2}d_{B^{c}}(y)\big )(I-X_{k}). \end{aligned}$$

Let us show that $D^{2}\rho _{k,B^{c}}(y)$ is bounded independently of k, y, B. Since the radius of B is fixed, $D^{2}d_{B^{c}}(y)$ is bounded independently of y, B. So we only need to show that $X_{k},\lambda _{k}$ are uniformly bounded. Note that $\gamma _{k}^{\circ }\ge \gamma _{1}^{\circ }$, since $K_{k}^{\circ }\subset K_{1}^{\circ }$. Thus by (2.10) we have

$$\begin{aligned} \gamma _{1}^{\circ }(D\varphi +\lambda _{k}\nu )\le \gamma _{k}^{\circ } (D\varphi +\lambda _{k}\nu )=1. \end{aligned}$$

Hence by (2.3) applied to $\gamma _{1}^{\circ }$, we have $|D\varphi +\lambda _{k}\nu |\le C$ for some $C>0$. Therefore we get $|\lambda _{k}|=|\lambda _{k}\nu |\le C+|D\varphi |$. Thus $\lambda _{k}$ is bounded independently of k, y, B.

Hence we only need to show that the entries of $X_{k}=\frac{1}{\langle D\gamma _{k}^{\circ }(\mu _{k}),\nu \rangle }D\gamma _{k}^{\circ }(\mu _{k})\otimes \nu$ are bounded. Note that by (2.6) we have $\gamma _{k}(D\gamma _{k}^{\circ }(\nu ))=1$. Thus $\gamma (D\gamma _{k}^{\circ }(\nu ))\le 1$ for every k, since $\gamma \le \gamma _{k}$ due to $K\supset K_{k}$. So $D\gamma _{k}^{\circ }(\nu )$ is bounded independently of k. Therefore it only remains to show that $\langle D\gamma _{k}^{\circ }(\mu _{k}),\nu \rangle$ has a positive lower bound independently of k, y, B.

Note that for every k, B, $\langle D\gamma _{k}^{\circ }(\mu _{k}),\nu \rangle$ is a continuous positive function on the compact set $\partial B$, as explained in Sect. 2.2. Hence there is $c_{k,B}>0$ such that $\langle D\gamma _{k}^{\circ }(\mu _{k}),\nu \rangle \ge c_{k,B}$. Suppose to the contrary that there is a sequence of balls $B_{j}$, points $y_{j}\in \partial B_{j}$, and $k_{j}$ (which we simply denote by j) such that

$$\begin{aligned} \langle D\gamma _{j}^{\circ }(\mu _{j}(y_{j})),\nu _{j}\rangle \rightarrow 0, \end{aligned}$$

(A.7)

where $\nu _{j}$ is the unit normal to $\partial B_{j}$ at $y_{j}$. By passing to another subsequence, we can assume that $y_{j}\rightarrow y$, since $y_{j}$’s belong to a compact neighborhood of $\partial U$. Now remember that

$$\begin{aligned} \mu _{j}(y_{j})=D\varphi (y_{j})+\lambda _{j}(y_{j})\nu _{j}, \end{aligned}$$

where $\lambda _{j}>0$. As we have shown in the last paragraph, $\lambda _{j}$ is bounded independently of $j,B_{j}$. Hence by passing to another subsequence, we can assume that $\lambda _{j}\rightarrow \lambda ^{*}\ge 0$. Also, $|\nu _{j}|=1$. Thus by passing to yet another subsequence we can assume that $\nu _{j}\rightarrow \nu ^{*}$, where $|\nu ^{*}|=1$. Therefore we have

$$\begin{aligned} \mu _{j}(y_{j})\rightarrow \mu ^{*}:=D\varphi (y)+\lambda ^{*}\nu ^{*}. \end{aligned}$$

On the other hand we have $\gamma _{j}^{\circ }(\mu _{j}(y_{j}))=1$. Hence $\gamma ^{\circ }(\mu _{j}(y_{j}))\ge 1$, since $\gamma _{j}^{\circ }\le \gamma ^{\circ }$ due to $K^{\circ }\subset K_{j}^{\circ }$. Thus we get $\gamma ^{\circ }(\mu ^{*})\ge 1$. However we cannot have $\gamma ^{\circ }(\mu ^{*})>1$. Because then $\mu ^{*}$ will have a positive distance from $K^{\circ }$, and therefore it will have a positive distance from $K_{j}^{\circ }$ for large enough j. But this contradicts the facts that $\mu _{j}(y_{j})\rightarrow \mu ^{*}$ and $\mu _{j}(y_{j})\in K_{j}^{\circ }$. Thus we must have $\gamma ^{\circ }(\mu ^{*})=1$, i.e. $\mu ^{*}\in \partial K^{\circ }$.

Fig. 2

The relative situation of $\textrm{v},\,\nu ^{*}$ does not allow $\langle \textrm{v},\nu ^{*}\rangle$ to be zero

Now note that $\textrm{v}_{j}:=D\gamma _{j}^{\circ }(\mu _{j}(y_{j}))$ belongs to the normal cone $N(K_{j}^{\circ },\mu _{j}(y_{j}))$. In addition we have $\gamma _{j}(\textrm{v}_{j})=1$ due to (2.6). Hence we have $\textrm{v}_{j}\in K_{j}\subset K$. Thus by passing to yet another subsequence we can assume that $\textrm{v}_{j}\rightarrow \textrm{v}\in K$. We also have $\gamma _{1}(\textrm{v}_{j})\ge 1$, since $\gamma _{j}\le \gamma _{1}$ due to $K_{j}\supset K_{1}$. So we get $\gamma _{1}(\textrm{v})\ge 1$. In particular $\textrm{v}\ne 0$. We claim that $\textrm{v}\in N(K^{\circ },\mu ^{*})$. To see this note that we have

$$\begin{aligned} K^{\circ }\subset K_{j}^{\circ }\subset \{z:\langle z-\mu _{j}(y_{j}),\textrm{v}_{j}\rangle \le 0\}. \end{aligned}$$

Hence for every $z\in K^{\circ }$ we have $\langle z-\mu _{j}(y_{j}),\textrm{v}_{j}\rangle \le 0$. But as $j\rightarrow \infty$ we have $z-\mu _{j}(y_{j})\rightarrow z-\mu ^{*}$. So we get $\langle z-\mu ^{*},\textrm{v}\rangle \le 0$. Therefore

$$\begin{aligned} K^{\circ }\subset \{z:\langle z-\mu ^{*},\textrm{v}\rangle \le 0\}, \end{aligned}$$

as desired.

On the other hand, by (A.7) we obtain

$$\begin{aligned} \langle \textrm{v},\nu ^{*}\rangle =\lim \langle \textrm{v}_{j}, \nu _{j}\rangle =0. \end{aligned}$$

(A.8)

Now note that $D\varphi =\mu ^{*}-\lambda ^{*}\nu ^{*}$ belongs to the ray passing through $\mu ^{*}\in \partial K^{\circ }$ in the direction $-\nu ^{*}$. However, we know that $D\varphi$ is in the interior of $K^{\circ }$, since $\gamma ^{\circ }(D\varphi )<1$. Hence we must have $\lambda ^{*}>0$ (since $\gamma ^{\circ }(\mu ^{*})=1$). And thus the ray $t\mapsto \mu ^{*}-t\nu ^{*}$ for $t>0$ passes through the interior of $K^{\circ }$. Therefore this ray and $K^{\circ }$ must lie on the same side of the supporting hyperplane $\Gamma _{\mu ^{*},\textrm{v}}$. In addition, the ray cannot lie on the hyperplane, since it intersects the interior of $K^{\circ }$. Hence we must have $\langle \textrm{v},\nu ^{*}\rangle =-\langle \textrm{v},-\nu ^{*}\rangle >0$, which contradicts (A.8). See Fig. 2 for a geometric representation of this argument.

Thus $\langle D\gamma _{k}^{\circ }(\mu _{k}),\nu \rangle$ must have a positive lower bound independently of k, y, B, as desired. Therefore $D^{2}\rho _{k,B^{c}}(x)$ has a uniform upper bound, independently of k, x, B. Then as before it follows that

$$\begin{aligned} \delta \rho _{k,B^{c}}(x,h)=\rho _{k,B^{c}}(x+h) +\rho _{k,B^{c}}(x-h)-2\rho _{k,B^{c}}(x)\le C|h|^{2} \end{aligned}$$

for some constant C independent of k, x, B, provided that the segment $[x-h,x+h]$ does not intersect B. Now, as in the case of zero exterior data, we truncate $\rho _{k,B^{c}}$ outside of a compact neighborhood of ${\overline{U}}$ to make it bounded. Note that, after the truncation, we can choose the bound for $\rho _{k,B^{c}}$ uniformly, since

$$\begin{aligned} \rho _{k,B^{c}}(x)=\gamma _{k}(x-y)+\varphi (y)\le \gamma _{1}(x-y)+\varphi (y)\le C+\Vert \varphi \Vert _{L^{\infty }}, \end{aligned}$$

where C is an upper bound for $\gamma _{1}$ on the compact set of differences $x-y$ for x, y in the chosen compact neighborhood of ${\overline{U}}$. Also note that we can still make sure that $\rho _{k}\le \rho _{k,B^{c}}$ after truncation, since $\rho _{k}=\varphi$ outside U, and $\varphi$ is bounded. And finally we can show that $I\rho _{k,B^{c}}(x)\le C_{0}$ for $x\in V$, similarly to the case of zero exterior data. $\square$

Appendix B: Nonlocal Double Obstacle Problems

In this appendix, for reader’s convenience, we provide a sketch of proof of the following result from [41], which we relied on in the proof of Theorem 3.

Theorem

Suppose I satisfies Assumption 1, and $\partial U$ is $C^{2}$. Let $f:{\mathbb {R}}^{n}\rightarrow {\mathbb {R}}$ be a continuous function. In addition, suppose $\psi ^{\pm },\varphi :{\mathbb {R}}^{n}\rightarrow {\mathbb {R}}$ are bounded Lipschitz functions with Lipschitz constant $C_{1}$ which satisfy

• (a) $\psi ^{\pm }=\varphi$ on ${\mathbb {R}}^{n}-U$, and for all $x\in U$ we have

  $$\begin{aligned} 0<\psi ^{+}(x)-\psi ^{-}(x)\le 2C_{1}d(x), \end{aligned}$$

  (B.1)

  where d is the Euclidean distance to $\partial U$.

• (b) For every $x\in U$ and $|y|\le d(x)-\epsilon$ we have

  $$\begin{aligned} \pm \delta \psi ^{\pm }(x,y)\le C|y|^{2}, \end{aligned}$$

  (B.2)

  where the constant C depends only on $\epsilon$. In other words, $\psi ^{+},\psi ^{-}$ are respectively semi-concave and semi-convex on compact subsets of U.

Then the double obstacle problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \max \{\min \{-Iu-f,u-\psi ^{-}\},u-\psi ^{+}\}=0 &{} \text {in } \ \ U,\\ u=\varphi &{} \text {in } \ \ {\mathbb {R}}^{n}-U, \end{array}\right. } \end{aligned}$$

(B.3)

has a viscosity solution $u\in C_{\textrm{loc}}^{1,\alpha }(U)$ for some $\alpha >0$ depending only on $n,\lambda ,\Lambda ,s_{0}$. And for an open subset $V\subset \subset U$ we have

$$\begin{aligned} \Vert u\Vert _{C^{1,\alpha }({\overline{V}})}\le C(\Vert u\Vert _{L^{\infty }({\mathbb {R}}^{n})}+\Vert f\Vert _{L^{\infty }(U)}+C_{0}), \end{aligned}$$

where C depends only on $n,\lambda ,\Lambda ,s_{0}$, and $d(V,\partial U)$; and $C_{0}$ depends only on these constants together with $\Vert \psi ^{\pm }\Vert _{L^{\infty }({\mathbb {R}}^{n})}$ and the semi-concavity constants of $\pm \psi ^{\pm }$ on V.

Sketch of Proof Let $\eta _{\varepsilon }$ be a standard mollifier whose support is $\overline{B_{\varepsilon }(0)}$, and consider the mollified functions $\psi _{\varepsilon }^{\pm }:=\eta _{\varepsilon }*\psi ^{\pm }$ and $\varphi _{\varepsilon }:=\eta _{\varepsilon }*\varphi$. Let

$$\begin{aligned} U_{\varepsilon }:=\{x\in {\mathbb {R}}^{n}:d(x,{\overline{U}})<\varepsilon \}. \end{aligned}$$

Then it is easy to see that for $x\in U_{\varepsilon }$ we have $\psi _{\varepsilon }^{-}(x)<\psi _{\varepsilon }^{+}(x)$, and for $x\in {\mathbb {R}}^{n}-U_{\varepsilon }$ we have $\psi _{\varepsilon }^{\pm }(x)=\varphi _{\varepsilon }(x)$. In addition, $\partial U_{\varepsilon }$ is $C^{2}$. Also, similarly to Part II of the proof of Theorem 3, we can show that for every bounded open set $V\subset \subset U$ and small enough $\varepsilon$ there is a constant $C_{0}$ such that $\pm I\psi _{\varepsilon }^{\pm }\le C_{0}$.

Now we consider the double obstacle problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \max \{\min \{-Iu_{\varepsilon }-f,u_{\varepsilon }-\psi _{\varepsilon }^{-}\}, u_{\varepsilon }-\psi _{\varepsilon }^{+}\}=0 &{} \text {in }U_{\varepsilon },\\ u_{\varepsilon }=\varphi _{\varepsilon } &{} \text {in }{\mathbb {R}}^{n}-U_{\varepsilon }. \end{array}\right. } \end{aligned}$$

(B.4)

We show that it has a viscosity solution $u_{\varepsilon }\in C_{\textrm{loc}}^{1,\alpha }(U_{\varepsilon })$. To this end, let $\beta$ be a smooth increasing function that vanishes on $(-\infty ,0]$ and equals $\frac{1}{\theta }t$ for $t\ge \theta$. Then the equation

$$\begin{aligned} {\left\{ \begin{array}{ll} -I{\tilde{u}}-f-\beta (\psi _{\varepsilon }^{-}-{\tilde{u}}) +\beta ({\tilde{u}}-\psi _{\varepsilon }^{+})=0 &{} \text {in } \ \ U_{\varepsilon },\\ {\tilde{u}}=\varphi _{\varepsilon } &{} \text {in} \ \ {\mathbb {R}}^{n}-U_{\varepsilon }, \end{array}\right. } \end{aligned}$$

(B.5)

has a viscosity solution ${\tilde{u}}={\tilde{u}}_{\theta }$ (see Theorem 5.6 of [27]).

Now let us show that

$$\begin{aligned} \Vert \beta (\pm ({\tilde{u}}-\psi _{\varepsilon }^{\pm }))\Vert _{L^{\infty }(U_{\varepsilon })}\le C_{2}, \end{aligned}$$

where $C_{2}$ is independent of $\theta$. Note that $\beta (\pm ({\tilde{u}}-\psi _{\varepsilon }^{\pm }))$ is zero on ${\mathbb {R}}^{n}-U_{\varepsilon }$. So assume that $\beta (\pm ({\tilde{u}}-\psi _{\varepsilon }^{\pm }))$ attains its positive maximum at $x_{0}\in U_{\varepsilon }$. Since $\beta$ is increasing, ${\tilde{u}}-\psi _{\varepsilon }^{+}$ has a positive maximum at $x_{0}$ too. Therefore by the definition of viscosity solution we have

$$\begin{aligned} -I\psi _{\varepsilon }^{+}(x_{0})-f(x_{0}) -\beta (\psi _{\varepsilon }^{-}(x_{0})-{\tilde{u}}(x_{0})) +\beta ({\tilde{u}}(x_{0})-\psi _{\varepsilon }^{+}(x_{0}))\le 0. \end{aligned}$$

So at $x_{0}$ we have

$$\begin{aligned} -I\psi _{\varepsilon }^{+}(x_{0})-f(x_{0})\le \beta (\psi _{\varepsilon }^{-} -{\tilde{u}})-\beta ({\tilde{u}}-\psi _{\varepsilon }^{+}) =-\beta ({\tilde{u}}-\psi _{\varepsilon }^{+}), \end{aligned}$$

since by our assumption $\psi _{\varepsilon }^{-}(x_{0})<\psi _{\varepsilon }^{+}(x_{0})<{\tilde{u}}(x_{0})$. Thus $\beta ({\tilde{u}}-\psi _{\varepsilon }^{+})\le I\psi _{\varepsilon }^{+}+f$ at $x_{0}$. Therefore $\beta ({\tilde{u}}-\psi _{\varepsilon }^{+})$ is bounded independently of $\theta$, because $I\psi _{\varepsilon }^{+},f$ are continuous functions.

The bound $\beta (\pm ({\tilde{u}}-\psi _{\varepsilon }^{\pm }))\le C_{2}$ and the definition of $\beta$ imply that

$$\begin{aligned} {\tilde{u}}-\psi _{\varepsilon }^{+}\le \theta (C_{2}+1), \qquad \qquad \psi _{\varepsilon }^{-}-{\tilde{u}}\le \theta (C_{2}+1). \end{aligned}$$

(B.6)

This also shows that ${\tilde{u}}$ is uniformly bounded independently of $\theta$. Then from the equation (B.5) we conclude that $-3C_{2}\le I{\tilde{u}}\le 3C_{2}$ in the viscosity sense. Thus by Theorem 4.1 of [25] we obtain

$$\begin{aligned} \Vert {\tilde{u}}\Vert _{C^{1,\alpha }({\overline{V}})}\le C(\Vert {\tilde{u}}\Vert _{L^{\infty }({\mathbb {R}}^{n})}+3C_{2}), \end{aligned}$$

where $V\subset \subset U_{\varepsilon }$. Therefore ${\tilde{u}}$ is bounded in $C^{1,\alpha }({\overline{V}})$ independently of $\theta$, due to its uniform boundedness in $L^{\infty }$. So, similarly to Part III of the proof of Theorem 3, we can show that as $\theta \rightarrow 0$, a subsequence of the ${\tilde{u}}$’s converges to a solution of (B.4) in V. Then by expanding V and a diagonalization argument we can construct a viscosity solution $u_{\varepsilon }$ of (B.4) in $U_{\varepsilon }$.

Next, similarly to Part I of the proof of Theorem 3, we can show that for every bounded open set $V\subset \subset U$ and every small enough $\varepsilon$ we have

$$\begin{aligned} -C_{0}-\Vert f\Vert _{L^{\infty }(U)}\le Iu_{\varepsilon }\le \Vert f\Vert _{L^{\infty }(U)}+C_{0}, \end{aligned}$$

(B.7)

where the constant $C_{0}$ satisfies $\pm I\psi _{\varepsilon }^{\pm }\le C_{0}$. Heuristically, on the contact set $\{u_{\varepsilon }=\psi _{\varepsilon }^{+}\}$, although a priori we do not have a lower bound for the second derivative of $\psi _{\varepsilon }^{+}$, and hence an upper bound for $-I\psi _{\varepsilon }^{+}$, we can obtain the desired bound for $-Iu_{\varepsilon }$ from the equation.

Thus by Theorem 4.1 of [25], we can show that there is $\alpha$ depending only on $n,\lambda ,\Lambda ,s_{0}$, such that for an open subset $V\subset \subset U$ we have

$$\begin{aligned} \Vert u_{\varepsilon }\Vert _{C^{1,\alpha }({\overline{V}})}\le C(\Vert u_{\varepsilon }\Vert _{L^{\infty }({\mathbb {R}}^{n})} +\Vert f\Vert _{L^{\infty }(U)}+C_{0}), \end{aligned}$$

where C depends only on $n,\lambda ,\Lambda ,s_{0}$, and $d(V,\partial U)$. Therefore $u_{\varepsilon }$ is bounded in $C^{1,\alpha }({\overline{V}})$ independently of $\varepsilon$, because $\Vert u_{\varepsilon }\Vert _{L^{\infty }}$ is bounded by $\Vert \psi _{\varepsilon }^{\pm }\Vert _{L^{\infty }}$, and $\Vert \psi _{\varepsilon }^{\pm }\Vert _{L^{\infty }}$ are uniformly bounded by $\Vert \psi ^{\pm }\Vert _{L^{\infty }}$. Now, again by using a diagonalization argument, we can construct a function u in $C_{\text {loc}}^{1,\alpha }(U)$. We furthermore extend u to all of ${\mathbb {R}}^{n}$ by setting it equal to $\varphi$ on ${\mathbb {R}}^{n}-U$. Then $u_{\varepsilon }$ converges uniformly to u on ${\mathbb {R}}^{n}$. To see this note that since $u,u_{\varepsilon }$ are between their corresponding obstacles we get

$$\begin{aligned} |u_{\varepsilon }-u|\le \max \{|\psi _{\varepsilon }^{+}-\psi ^{-}|, |\psi _{\varepsilon }^{-}-\psi ^{+}|\}\;\,\\ \le |\psi ^{+}-\psi ^{-}|+\max \{|\psi _{\varepsilon }^{+}-\psi ^{+}|,&|\psi _{\varepsilon }^{-}-\psi ^{-}|\}\le {\left\{ \begin{array}{ll} 2C_{1}d(\cdot )+C_{1}\varepsilon &{} \text {in }U,\\ C_{1}\varepsilon &{} \text {in }{\mathbb {R}}^{n}-U, \end{array}\right. } \end{aligned}$$

because $|\psi ^{+}-\psi ^{-}|=0$ outside of U. Finally, due to the stability of viscosity solutions, u must satisfy the double obstacle problem (B.3). $\square$