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The fractional p-Laplacian emerging from homogenization of the random conductance model with degenerate ergodic weights and unbounded-range jumps


We study a general class of discrete p-Laplace operators in the random conductance model with long-range jumps and ergodic weights. Using a variational formulation of the problem, we show that under the assumption of bounded first moments and a suitable lower moment condition on the weights, the homogenized limit operator is a fractional p-Laplace operator. Under strengthened lower moment conditions, we can apply our insights also to the spectral homogenization of the discrete Laplace operator to the continuous fractional Laplace operator.

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Correspondence to Martin Heida.

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Communicated by J. Jost.

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M. Heida is financed by Deutsche Forschungsgemeinschaft (DFG) through Grant CRC 1114 “Scaling Cascades in Complex Systems”, Project C05 Effective models for materials and interfaces with multiple scales. The authors thank Takashi Kumagai and Andrey Piatnitski for inspiring discussions and the referees for the helpful feedback.

A Proofs of auxiliary results

A Proofs of auxiliary results

A.1 Proof of Lemma 15

Lemma 37

Let \(u\in W^{s,p}(\mathbb {R}^{d})\). Then

$$\begin{aligned} \lim _{h\rightarrow 0}\left\| u(\cdot )-u(\cdot -h)\right\| _{s,p}\rightarrow 0. \end{aligned}$$


It is well known that

$$\begin{aligned} \lim _{h\rightarrow 0}\left\| u(\cdot )-u(\cdot -h)\right\| _{L^{p}(\mathbb {R}^{d})}\rightarrow 0 \end{aligned}$$

and it only remains to show

$$\begin{aligned} \lim _{h\rightarrow 0}\left[ u(\cdot )-u(\cdot -h)\right] _{s,p}\rightarrow 0. \end{aligned}$$

Suppose \(u\in C_{c}^{\infty }(\mathbb {R}^{d})\) and let B be a ball that contains the support of u. We write \(u_{h}(x):=u(x-h)\) as well as \(f(x,y)=u(x)-u(y)\) and similarly \(f_{h}(x,y)\). Since, for small h, \(f(x,y)=f_{h}(x,y)=0\) if both \(x,y\not \in 2B\), we observe that

$$\begin{aligned}&\int \limits _{\mathbb {R}^{d}}\int \limits _{\mathbb {R}^{d}}\frac{\left| f(x,y)-f_{h}(x,y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}\,\mathrm {d}x\,\mathrm {d}y\\&\quad =\int \limits _{2B}\int \limits _{2B}\frac{\left| f(x,y)-f_{h}(x,y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}\,\mathrm {d}x\,\mathrm {d}y+2\int \limits _{2B}\int \limits _{\mathbb {R}^{d}\backslash 2B}\frac{\left| f(x,y)-f_{h}(x,y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}\,\mathrm {d}x\,\mathrm {d}y\\&\quad \le 2\int \limits _{2B}\int \limits _{\mathbb {R}^{d}}\frac{\left| u(x)-u_{h}(x)-u(y)+u_{h}(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}\,\mathrm {d}x\,\mathrm {d}y. \end{aligned}$$

For every \(\delta >0\) the right-hand side can be split into an integral over

$$\begin{aligned} A_{\delta }:=\left\{ (x,y):\;x\in 2B,\,\left| x-y\right| <\xi \right\} \end{aligned}$$

and the complement. We find

$$\begin{aligned} \left[ u(\cdot )-u(\cdot -h)\right] _{s,p}&\le 2^{p+1}\int \limits _{A_{\delta }}\frac{\left| u_{h}(x)-u_{h}(y)\right| ^{p}+\left| u(x)-u(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}\\&\quad +2\int \limits _{\mathbb {R}^{2d}\backslash A_{\delta }}\frac{\left| u(x)-u_{h}(x)-u(y)+u_{h}(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}\,\mathrm {d}x\,\mathrm {d}y. \end{aligned}$$

The first integral can be estimated by

$$\begin{aligned} 2^{p+2}\left\| \nabla u\right\| _{\infty }^{p}\int \limits _{A_{\delta }}\frac{1}{\left| x-y\right| ^{d+sp-p}}=2^{p+2}\left\| \nabla u\right\| _{\infty }^{p}|2B|\,|S^{d-1}|\,\delta ^{p-sp}. \end{aligned}$$

The second integral converges to 0 as \(h\rightarrow 0\) as it is bounded by

$$\begin{aligned} \delta ^{-d-sp}4\left\| u-u_{h}\right\| \rightarrow 0. \end{aligned}$$

Hence, we have shown that \(\lim _{h\rightarrow 0}\left[ u(\cdot )-u(\cdot -h)\right] _{s,p}\le C\delta ^{p-sp}\) for every \(\delta >0\), implying (48). For arbitrary \(u\in W^{s,p}(\mathbb {R}^{d})\) the lemma follows from a standard approximation argument. \(\square \)

Remark 38

Via the triangle inequality, the last lemma implies that \(h\mapsto \left\| u(\cdot )-u(\cdot -h)\right\| _{s,p}\) is continuous:

$$\begin{aligned} \left| \left\| u(\cdot )-u(\cdot -h_{1})\right\| _{s,p}-\left\| u(\cdot )-u(\cdot -h_{2})\right\| _{s,p}\right| \le \left\| u(\cdot -h_{1})-u(\cdot -h_{2})\right\| _{s,p}. \end{aligned}$$

Proof of Lemma 15

First note that it is well known that

$$\begin{aligned} \left\| u*\eta _{k}\right\| _{L^{p}(\mathbb {R}^{d})}\le \left\| u\right\| _{L^{p}(\mathbb {R}^{d})}\qquad \text{ and }\qquad \lim _{k\rightarrow \infty }\left\| u*\eta _{k}-u\right\| _{L^{p}(\mathbb {R}^{d})}=0 \end{aligned}$$

and it only remains to show

$$\begin{aligned} \left[ u*\eta _{k}\right] _{s,p}\le \left[ u\right] _{s,p}\qquad \text{ and }\qquad \lim _{k\rightarrow \infty }\left[ u*\eta _{k}-u\right] _{s,p}=0. \end{aligned}$$

The inequality can be easily verified from the fact that

$$\begin{aligned}&\int \limits _{\mathbb {R}^{d}}\int \limits _{\mathbb {R}^{d}}\frac{\left| \int \limits _{\mathbb {R}^{d}}\left( \eta _{k}(z)u(x-z)-\eta _{k}(z)u(y-z)\right) \mathrm {d}z\right| ^{p}}{\left| x-y\right| ^{d+sp}}\,\mathrm {d}x\,\mathrm {d}y\\&\quad \le \left\| \eta _{k}\right\| _{L^{1}(\mathbb {R}^{d})}^{p/p^{*}}\int \limits _{\mathbb {R}^{d}}\eta _{k}(z)\int \limits _{\mathbb {R}^{d}}\int \limits _{\mathbb {R}^{d}}\frac{\left| \left( u(x-z)-u(y-z)\right) \right| ^{p}}{\left| x-y\right| ^{d+sp}}\,\mathrm {d}x\,\mathrm {d}y\,\mathrm {d}z\\&\quad =\int \limits _{\mathbb {R}^{d}}\int \limits _{\mathbb {R}^{d}}\frac{\left| \left( u(x)-u(y)\right) \right| ^{p}}{\left| x-y\right| ^{d+sp}}\,\mathrm {d}x\,\mathrm {d}y. \end{aligned}$$

The limit behavior follows from Lemma 37, Remark 38 and the following calculation:

$$\begin{aligned}&\int \limits _{\mathbb {R}^{d}}\int \limits _{\mathbb {R}^{d}}\frac{\left| \int \limits _{\mathbb {R}^{d}}\left( \eta _{k}(z)\left( u(x-z)-u(x)\right) -\eta _{k}(z)\left( u(y-z)-u(y)\right) \right) \mathrm {d}z\right| ^{p}}{\left| x-y\right| ^{d+sp}}\,\mathrm {d}x\,\mathrm {d}y\\&\quad \le \left\| \eta _{k}\right\| _{L^{1}(\mathbb {R}^{d})}^{p/p^{*}}\int \limits _{\mathbb {R}^{d}}\eta _{k}(z)\int \limits _{\mathbb {R}^{d}}\int \limits _{\mathbb {R}^{d}}\frac{\left| \left( u(x-z)-u(x)-u(y-z)+u(y)\right) \right| ^{p}}{\left| x-y\right| ^{d+sp}}\,\mathrm {d}x\,\mathrm {d}y\,\mathrm {d}z\\&\quad \le \left\| \eta _{k}\right\| _{L^{1}(\mathbb {R}^{d})}^{p/p^{*}}\int \limits _{\mathbb {R}^{d}}\eta _{k}(z)\left[ u(\cdot )-u(\cdot -z)\right] _{s,p}\mathrm {d}z\\&\quad \rightarrow 0\quad \text{ as } k\rightarrow \infty . \end{aligned}$$

\(\square \)

A.2 Proof of Remark 17

The remark is a consequence of the following lemma.

Lemma 39

Let \(\varphi \in C_{c}^{1}(\mathbb {R}^{d})\). Then for every \(\varepsilon >0\), \(p\in (1,\infty )\), \(s\in (0,1)\) and \(u\in W^{s,p}(\mathbb {Z}_{\varepsilon }^{d})\) it holds \(\varphi u\in W^{s,p}(\mathbb {Z}_{\varepsilon }^{d})\) and there exists some \(C>0\) which does not depend on \(\varepsilon \) such that

$$\begin{aligned} \left\| \varphi u\right\| _{s,p,\varepsilon }\le C\left\| u\right\| _{s,p,\varepsilon }\left\| \varphi \right\| _{C_{0}^{1}(\mathbb {R}^{d})}. \end{aligned}$$


Writing \(\delta _{f}\left( x,y\right) :=\left| f(x)-f(y)\right| \), we first observe that

$$\begin{aligned} \underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\delta _{u\varphi }\left( x,y\right) ^{p}}{\left| x-y\right| ^{d+ps}}\le \underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| u(x)\right| \delta _{\varphi }\left( x,y\right) +\delta _{u}\left( x,y\right) \left| \varphi (y)\right| }{\left| x-y\right| ^{d+ps}}\delta _{u\varphi }\left( x,y\right) ^{p-1}. \end{aligned}$$

Let \(B(x):=\left\{ y\in \mathbb {R}^{d}:\;|x-y|<1\right\} \) with complement \(B^{\complement }(x)\). Then, for every \(x\in \mathbb {Z}_{\varepsilon }^{d}\) we find

$$\begin{aligned} \underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\delta _{\varphi }\left( x,y\right) ^{p}}{\left| x-y\right| ^{d+ps}}&\le \underset{{\scriptstyle y\in B(x)\cap \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left\| \nabla \varphi \right\| _{\infty }^{p}}{\left| x-y\right| ^{d+ps-p}}+\underset{{\scriptstyle y\in B^{\complement }(x)\cap \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left\| \varphi \right\| _{\infty }^{p}}{\left| x-y\right| ^{d+ps}}\\&\le C\left( \left\| \nabla \varphi \right\| _{\infty }^{p}+\left\| \varphi \right\| _{\infty }^{p}\right) . \end{aligned}$$

Furthermore, note that

$$\begin{aligned}&\underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| u(x)\right| \delta _{\varphi }\left( x,y\right) }{\left| x-y\right| ^{d+ps}}\left| \delta _{u\varphi }\left( x,y\right) \right| ^{p-1}\\&\quad \le \underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| u(x)\right| \left( \underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\delta _{\varphi }\left( x,y\right) ^{p}}{\left| x-y\right| ^{d+ps}}\right) ^{\frac{1}{p}}\left( \underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| \delta _{u\varphi }\left( x,y\right) \right| ^{p}}{\left| x-y\right| ^{d+ps}}\right) ^{\frac{p-1}{p}}\\&\quad \le C\left( \left\| \nabla \varphi \right\| _{\infty }^{p}+\left\| \varphi \right\| _{\infty }^{p}\right) \left\| u\right\| _{L^{p}(\mathbb {Z}_{\varepsilon }^{d})}\left( \underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| \delta _{u\varphi }\left( x,y\right) \right| ^{p}}{\left| x-y\right| ^{d+ps}}\right) ^{\frac{p-1}{p}} \end{aligned}$$

as well as

$$\begin{aligned} \underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\delta _{u}\left( x,y\right) \left| \varphi (y)\right| }{\left| x-y\right| ^{d+ps}}\left| \delta _{u\varphi }\left( x,y\right) \right| ^{p-1}&\le \left\| \varphi \right\| _{\infty }^{p}\left[ u\right] _{s,p,\varepsilon }\left( \underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| \delta _{u\varphi }\left( x,y\right) \right| ^{p}}{\left| x-y\right| ^{d+ps}}\right) ^{\frac{p-1}{p}}. \end{aligned}$$

Hence we obtain (49). \(\square \)

A.3 Proof of Theorem 18

We prove Theorem 18 after three auxiliary lemmas. The first lemma is an equivalent to Lemma 6.1 in [9].

Lemma 40

Let \(1\le p<\infty \), \(s\in (0,1)\). There exists C depending only on s, p and d such that

$$\begin{aligned} \underset{{\scriptstyle y\in E^{\complement }\cap \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}\ge C\left| E\right| ^{-sp/d} \end{aligned}$$

for every \(\varepsilon >0\), every \(x\in \mathbb {Z}_{\varepsilon }^{d}\) and every measurable set \(E\subset \mathbb {R}^{d}\) with finite measure.


Let \(\rho :=2d^{\frac{1}{d}}\left( \left| E\right| _{\varepsilon }\right) ^{\frac{1}{d}}\), where \(\left| E\right| _{\varepsilon }=\varepsilon ^{d}\sharp \left\{ E\cap \mathbb {Z}_{\varepsilon }^{d}\right\} \), see the beginning of Sect. 3. Then, for every \(\tilde{\rho }\ge \rho \) we find \(\left| B_{\tilde{\rho }}(x)\right| _{\varepsilon }\ge \left| E\right| _{\varepsilon }\) and hence

$$\begin{aligned} \left| E^{\complement }\cap B_{\tilde{\rho }}(x)\right| _{\varepsilon }&=\left| B_{\tilde{\rho }}(x)\right| _{\varepsilon }-\left| E\cap B_{\tilde{\rho }}(x)\right| _{\varepsilon }\ge \left| E\right| _{\varepsilon }-\left| E\cap B_{\tilde{\rho }}(x)\right| _{\varepsilon }\\&\ge \left| E\cap B_{\tilde{\rho }}^{\complement }(x)\right| _{\varepsilon }. \end{aligned}$$

Hence, we infer that

$$\begin{aligned} \underset{{\scriptstyle y\in E^{\complement }\cap \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}&=\underset{{\scriptstyle y\in E^{\complement }\cap B_{\tilde{\rho }}(x)}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}+\underset{{\scriptstyle y\in E^{\complement }\cap B_{\tilde{\rho }}^{\complement }(x)}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}\\&\ge \frac{\left| E^{\complement }\cap B_{\tilde{\rho }}(x)\right| _{\varepsilon }}{\left| \tilde{\rho }\right| ^{d+sp}}+\underset{{\scriptstyle y\in E^{\complement }\cap B_{\tilde{\rho }}^{\complement }(x)}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}\\&\ge \frac{\left| E\cap B_{\tilde{\rho }}^{\complement }(x)\right| _{\varepsilon }}{\left| \tilde{\rho }\right| ^{d+sp}}+\underset{{\scriptstyle y\in E^{\complement }\cap B_{\tilde{\rho }}^{\complement }(x)}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}\\&\ge \underset{{\scriptstyle y\in E\cap B_{\tilde{\rho }}^{\complement }(x)}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}+\underset{{\scriptstyle y\in E^{\complement }\cap B_{\tilde{\rho }}^{\complement }(x)}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}\\&=\underset{{\scriptstyle y\in B_{\tilde{\rho }}^{\complement }(x)}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}\, . \end{aligned}$$

Next, we consider cells \(C_{\varepsilon }(z):=z+\varepsilon (-\frac{1}{2},\frac{1}{2})\), \(z\in \mathbb {Z}_{\varepsilon }^{d}\backslash \{0\}\). On each of these cells, we want to estimate the ratio between the maximal and the minimal value of the function \(f(y)=|y|^{-d-ps}\). Due to the polynomial decay of this function, the closer one of the cells \(C_\varepsilon (z)\) lies next to 0, the higher will be the ratio in f. The biggest value that f can attain on \(\mathbb {R}^{d}\backslash C_\varepsilon (0)\) is \(\varepsilon ^{-d-ps}\). Furthermore, all neighboring cells to \(C_\varepsilon (0)\) lie within the cube \(\left( -\frac{3}{2}\varepsilon ,\frac{3}{2}\varepsilon \right) \) and the minimal value of f is on this domain is the value of f is \(\left( \frac{3}{2}d^{\frac{1}{d}}\varepsilon \right) ^{-d-sp}\). Hence we obtain

$$\begin{aligned} \inf _{z\in \mathbb {Z}_{\varepsilon }^{d}\backslash \{0\}}\inf _{y\in C_{\varepsilon }(z)}\left| y\right| ^{-d-sp}\left( \sup _{y\in C_{\varepsilon }(z)}\left| y\right| ^{-d-sp}\right) ^{-1}\ge \left( \frac{\varepsilon }{2}\right) ^{d+sp}\left( \frac{3}{2}d^{\frac{1}{d}}\varepsilon \right) ^{-d-sp}=\left( 3d^{\frac{1}{d}}\right) ^{-d-sp}, \end{aligned}$$

and we conclude that

$$\begin{aligned} \underset{{\scriptstyle y\in E^{\complement }\cap \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}\ge \underset{{\scriptstyle y\in B_{\tilde{\rho }}^{\complement }(x)}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}\ge \left( 3d^{\frac{1}{d}}\right) ^{-d-sp}\int \limits _{y\in B_{\tilde{\rho }}^{\complement }(x)}\frac{1}{\left| x-y\right| ^{d+sp}}\mathrm {d}y. \end{aligned}$$

Now the theorem follows from integration using polar coordinates. \(\square \)

Lemma 41

([9, Lemma 6.2]) Let \(s\in (0,1)\) and \(p\in [1,\infty )\) be such that \(sp<d\). Fix \(T>1\) and let \(N\in \mathbb {Z}\), and

$$\begin{aligned} a_{k} \text{ be } \text{ a } \text{ non-increasing } \text{ sequence } \text{ such } \text{ that } a_{k}=0 \text{ for } \text{ every } k\ge N. \end{aligned}$$


$$\begin{aligned} \sum _{k\in \mathbb {Z}}a_{k}^{(d-sp)/d}T^{k}\le C\sum _{k\in \mathbb {Z},\,a_{k}\not =0}a_{k+1}a_{k}^{-sp/d}T^{k}, \end{aligned}$$

for a suitable constant \(C=C(d,s,p,T)\), independent of N.

We are now in the position to prove the following variant of [9], Lemma 6.3.

Lemma 42

Let \(s\in (0,1)\) and \(p\in [1,\infty )\) be such that \(sp<d\). Let \(f\in L^{\infty }(\mathbb {Z}_{\varepsilon }^{d})\) be compactly supported. For any \(k\in \mathbb {Z}\) let

$$\begin{aligned} a_{k}:=\left| \left\{ \left| f\right| >2^{k}\right\} \right| . \end{aligned}$$


$$\begin{aligned} \underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+ps}}\ge C\sum _{a_k \ne 0} 2^{pk}a_{k+1}a_{k}^{-sp/d} \end{aligned}$$

for some suitable constant \(C=C(d,s,p)>0\), which depends not on \(\varepsilon \).


We first emphasize that \(\left| \left| f(x)\right| -\left| f(y)\right| \right| \le \left| f(x)-f(y)\right| \) and hence we only consider \(f\ge 0\), possibly replacing f by \(\left| f\right| \).

We define

$$\begin{aligned} A_{k}&:=\left\{ f>2^{k}\right\} \quad \text{ with }\quad A_{k+1}\subset A_{k}\\ a_{k}&:=\left| \left\{ f>2^{k}\right\} \right| \quad \text{ with }\quad a_{k+1}\le a_{k}\, . \end{aligned}$$

We define

$$\begin{aligned}&D_{k}:=A_{k}\backslash A_{k+1}=\left\{ 2^{k+1}\ge f>2^{k}\right\} \quad \text{ and }\quad d_{k}:=\left| D_{k}\right| \quad \text{ with }\\&d_{k} \text{ and } a_{k} \text{ are } \text{ bounded } \text{ and } \text{ they } \text{ become } \text{ zero } \text{ when } k \text{ is } \text{ large } \text{ enough, } \end{aligned}$$

since f is bounded. We define \(D_{-\infty }=\left\{ f=0\right\} \) and further observe that the sets \(D_{k}\) are mutually disjoint and

$$\begin{aligned} D_{-\infty }\cup \bigcup _{l\in \mathbb {Z},l\le k}D_{l}=A_{k+1}^{\complement },\qquad \bigcup _{l\in \mathbb {Z},l\ge k}D_{l}=A_{k}. \end{aligned}$$

As a consequence, we have

$$\begin{aligned} a_{k}=\sum _{l=k}^{\infty }d_{l},\qquad d_{k}=a_{k}-\sum _{l=k+1}^{\infty }d_{l}. \end{aligned}$$

The first equality implies that the series \(\sum _{l\ge k}d_{l}\) are convergent. For convenience of notation, in the following we write for arbitrary expressions g(y)

$$\begin{aligned} \sum _{j=-\infty }^{i-2}\underset{{\scriptstyle y\in D_{j}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}g(y):=\underset{{\scriptstyle y\in D_{-\infty }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}g(y)+\sum _{\begin{array}{c} l\in \mathbb {Z}\\ l\le i-2 \end{array} }\underset{{\scriptstyle y\in D_{j}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}g(y) \end{aligned}$$

Now, we fix \(i\in \mathbb {Z}\) and \(x\in D_{i}\). For every \(j\in \mathbb {Z}\), \(j\le i-2\) and every \(y\in D_{j}\) we have

$$\begin{aligned} \left| f(x)-f(y)\right| \ge 2^{i}-2^{j+1}\ge 2^{i}-2^{i-1}=2^{i-1} \end{aligned}$$

and hence by the first equality in (50) it holds

$$\begin{aligned} \sum _{j=-\infty }^{i-2}\underset{{\scriptstyle y\in D_{j}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}&\ge 2^{p(i-1)}\sum _{j=-\infty }^{i-2}\underset{{\scriptstyle y\in D_{j}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}\\&\ge 2^{p(i-1)}\underset{{\scriptstyle y\in A_{i-1}^{\complement }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{1}{\left| x-y\right| ^{d+sp}}. \end{aligned}$$

Therefore, by Lemma 40, there exists a constant \(c_{0}\) such that for every \(i\in \mathbb {Z}\) and every \(x\in D_{i}\) it holds

$$\begin{aligned} \sum _{j=-\infty }^{i-2}\underset{{\scriptstyle y\in D_{j}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}&\ge c_{0}2^{pi}a_{i-1}^{-sp/d},\\ \text{ and }\qquad \sum _{j=-\infty }^{i-2}\underset{{\scriptstyle y\in D_{j}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle x\in D_{i}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}&\ge c_{0}2^{pi}a_{i-1}^{-sp/d}d_{i}. \end{aligned}$$

Summing up the last inequality over \(i\in \mathbb {Z}\) we have on one side

$$\begin{aligned} \sum _{\begin{array}{c} i\in \mathbb {Z}\\ a_{i-1}\not =0 \end{array} }\sum _{j=-\infty }^{i-2}\underset{{\scriptstyle y\in D_{j}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle x\in D_{i}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}\ge c_{0}\sum _{l\in \mathbb {Z},\,a_{l-1}\not =0}2^{pl}a_{l-1}^{-sp/d}d_{l}=:c_{0}S, \end{aligned}$$

implying S to be bounded, and on the other hand, using (51), we have

$$\begin{aligned}&\sum _{\begin{array}{c} i\in \mathbb {Z}\\ a_{i-1}\not =0 \end{array} }\sum _{j=-\infty }^{i-2}\underset{{\scriptstyle y\in D_{j}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle x\in D_{i}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}\nonumber \\&\quad \ge c_{0}\sum _{\begin{array}{c} i\in \mathbb {Z}\\ a_{i-1}\not =0 \end{array} }\left( 2^{pi}a_{i-1}^{-sp/d}a_{i}-\sum _{l=i+1}^{\infty }2^{pi}a_{i-1}^{-sp/d}d_{l}\right) , \end{aligned}$$

where we estimate the second sum by S through

$$\begin{aligned} \sum _{\begin{array}{c} i\in \mathbb {Z}\\ a_{i-1}\not =0 \end{array} }\sum _{\begin{array}{c} l\in \mathbb {Z}\\ l\ge i+1 \end{array} }2^{pi}a_{i-1}^{-sp/d}d_{l}&=\sum _{\begin{array}{c} i\in \mathbb {Z}\\ a_{i-1}\not =0 \end{array} }\sum _{\begin{array}{c} l\in \mathbb {Z}\\ l\ge i+1\\ a_{i-1}d_{l}\not =0 \end{array} }2^{pi}a_{i-1}^{-sp/d}d_{l}\\&\le \sum _{\begin{array}{c} i\in \mathbb {Z} \end{array} }\sum _{\begin{array}{c} l\in \mathbb {Z}\\ l\ge i+1\\ a_{l-1}\not =0 \end{array} }2^{pi}a_{i-1}^{-sp/d}d_{l}\\&=\sum _{\begin{array}{c} l\in \mathbb {Z}\\ a_{l-1}\not =0 \end{array} }\sum _{\begin{array}{c} i\in \mathbb {Z}\\ i\le l-1 \end{array} }2^{pi}a_{i-1}^{-sp/d}d_{l}\\&\le \sum _{\begin{array}{c} l\in \mathbb {Z}\\ a_{l-1}\not =0 \end{array} }\sum _{\begin{array}{c} i\in \mathbb {Z}\\ i\le l-1 \end{array} }2^{pi}a_{l-1}^{-sp/d}d_{l}\le S. \end{aligned}$$

Using the last estimate in (53), we obtain

$$\begin{aligned} \sum _{\begin{array}{c} i\in \mathbb {Z}\\ a_{i-1}\not =0 \end{array} }\sum _{j=-\infty }^{i-2}\underset{{\scriptstyle y\in D_{j}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle x\in D_{i}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}\ge c_{0}\sum _{\begin{array}{c} i\in \mathbb {Z}\\ a_{i-1}\not =0 \end{array} }2^{pi}a_{i-1}^{-sp/d}a_{i}-c_{0}S \end{aligned}$$

and using estimate (52) we find upon relabeling \(c_{0}\) that

$$\begin{aligned} \sum _{\begin{array}{c} i\in \mathbb {Z}\\ a_{i-1}\not =0 \end{array} }\sum _{j=-\infty }^{i-2}\underset{{\scriptstyle y\in D_{j}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle x\in D_{i}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}\ge c_{0}\sum _{\begin{array}{c} i\in \mathbb {Z}\\ a_{i-1}\not =0 \end{array} }2^{pi}a_{i-1}^{-sp/d}a_{i}. \end{aligned}$$

On the other hand, it clearly holds that

$$\begin{aligned} \underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}}\ge \sum _{\begin{array}{c} i\in \mathbb {Z}\\ a_{i-1}\not =0 \end{array} }\sum _{j=-\infty }^{i-2}\underset{{\scriptstyle y\in D_{j}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle x\in D_{i}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+sp}} \end{aligned}$$

and hence the lemma follows. \(\square \)

We are now in the position to prove the first Sobolev theorem.

Proof of Theorem 18

It suffices to prove the claim for

$$\begin{aligned} \left[ f\right] _{s,p,\varepsilon }^{p}=\underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+ps}}<\infty \end{aligned}$$

and for \(f\in L^{\infty }(\mathbb {Z}_{\varepsilon }^{d})\). Indeed, for arbitrary \(f\in W^{s,p}(\mathbb {Z}_{\varepsilon }^{d})\), with \(f_{N}:=\max \left\{ -N,\min \left\{ N,f\right\} \right\} \) we obtain that

$$\begin{aligned} \lim _{N\rightarrow \infty }\underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f_{N}(x)-f_{N}(y)\right| ^{p}}{\left| x-y\right| ^{d+ps}}=\underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\underset{{\scriptstyle y\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left| f(x)-f(y)\right| ^{p}}{\left| x-y\right| ^{d+ps}} \end{aligned}$$

due to the dominated convergence theorem and pointwise convergence \(f_{N}\rightarrow f\).

We recall the definitions

$$\begin{aligned} A_{k}&:=\left\{ |f|>2^{k}\right\} \quad \text{ with }\quad A_{k+1}\subset A_{k}\\ a_{k}&:=\left| \left\{ |f|>2^{k}\right\} \right| \quad \text{ with }\quad a_{k+1}\le a_{k} \end{aligned}$$

from the proof of Lemma 42 and obtain

$$\begin{aligned} \left\| f\right\| _{L^{p^{*}}(\mathbb {Z}_{\varepsilon }^{d})}^{p^{\star }}=\sum _{k\in \mathbb {Z}}\underset{{\scriptstyle x\in A_{k}\backslash A_{k+1}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| f(x)\right| ^{p^{\star }}\le \sum _{k\in \mathbb {Z}}\underset{{\scriptstyle x\in A_{k}\backslash A_{k+1}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| 2^{k+1}\right| ^{p^{\star }}\le \sum _{k\in \mathbb {Z}}2^{(k+1)p^{\star }}a_{k}. \end{aligned}$$

Using \(p/p^{\star }=(d-sp)/d=1-sp/d<1\) we can conclude with Lemma 41 that

$$\begin{aligned} \left\| f\right\| _{L^{p^{*}}(\mathbb {Z}_{\varepsilon }^{d})}^{p}&\le 2^{p}\left( \sum _{k\in \mathbb {Z}}2^{kp^{\star }}a_{k}\right) ^{\frac{p}{p^{\star }}}\le 2^{p}\,\sum _{k\in \mathbb {Z}}2^{kp}a_{k}^{(d-sp)/d}\\&\le C\sum _{\begin{array}{c} k\in \mathbb {Z}\\ a_{k}\not =0 \end{array} }2^{kp}a_{k+1}a_{k}^{\frac{-sp}{d}}. \end{aligned}$$

It only remains to apply Lemma 42 and relabeling the constant C to find (23) in case \(q=p^{\star }\). In case \(q=\theta p+(1-\theta )p^{\star }\), \(\theta \in (0,1)\), we obtain from Hölder’s inequality and the case \(q=p^{\star }\) that

$$\begin{aligned} \underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| f(x)\right| ^{q}&=\underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| f(x)\right| ^{\theta p}\left| f(x)\right| ^{(1-\theta )p^{\star }}\le \left( \underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| f(x)\right| ^{p}\right) ^{\theta }\left( \underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| f(x)\right| ^{p^{\star }}\right) ^{1-\theta }\\&=\left\| f\right\| _{L^{p}(\mathbb {Z}_{\varepsilon }^{d})}^{p\theta }\left\| f\right\| _{L^{p^{\star }}(\mathbb {Z}_{\varepsilon }^{d})}^{\left( 1-\theta \right) p^{\star }}\le \left\| f\right\| _{L^{p}(\mathbb {Z}_{\varepsilon }^{d})}^{p\theta }\left[ f\right] _{s,p,\varepsilon }^{\left( 1-\theta \right) p^{\star }}\\&\le \left\| f\right\| _{s,p,\varepsilon }^{p\theta }\left\| f\right\| _{s,p,\varepsilon }^{\left( 1-\theta \right) p^{\star }}=\left\| f\right\| _{s,p,\varepsilon }^{q}. \end{aligned}$$

\(\square \)

A.4 Proof of Theorem 20


Since \(\varvec{Q}\) is a uniform extension domain, the family \(\mathcal {R}_{\varepsilon }^{*}u^{\varepsilon }\) is precompact if and only if \(\mathcal {R}_{\varepsilon }^{*}\mathcal {E}_{\varepsilon }u^{\varepsilon }\) is compact, where we recall the operator \(\mathcal {E}_{\varepsilon }\) from Definition 16. We will apply the Frechet–Kolmogorov(–Riesz) theorem to prove compactness of \(\mathcal {R}_{\varepsilon }^{*}\mathcal {E}_{\varepsilon }u^{\varepsilon }\). More precisely, it suffices to verify the following three properties:

$$\begin{aligned}&\sup _{\varepsilon>0}\left\| \mathcal {R}_{\varepsilon }^{*}\mathcal {E}_{\varepsilon }u^{\varepsilon }\right\| _{L^{q}(\mathbb {R}^{d})}<\infty ,\qquad \lim _{R\rightarrow \infty }\sup _{\varepsilon >0}\left\| \mathcal {R}_{\varepsilon }^{*}\mathcal {E}_{\varepsilon }u^{\varepsilon }\right\| _{L^{q}(\mathbb {R}^{d}\backslash B_{R}(0))}=0, \end{aligned}$$
$$\begin{aligned}&\lim _{|h|\rightarrow 0}\sup _{\varepsilon >0}\left\| \mathcal {R}_{\varepsilon }^{*}\mathcal {E}_{\varepsilon }u^{\varepsilon }(\cdot )-\mathcal {R}_{\varepsilon }^{*}\mathcal {E}_{\varepsilon }u^{\varepsilon }(\cdot +h)\right\| _{L^{q}(\mathbb {R}^{d})}\rightarrow 0. \end{aligned}$$

Note that the conditions in (54) are satisfied due to Theorem 18 and Remark 17. Thus, it only remains to show (55).

For \(h\in \mathbb {R}^{d}\) we write \(\tau _{h}u(x):=u(x+h)\), whenever this is well defined. Moreover, for every \(\varepsilon >0\) we define

$$\begin{aligned} \left\| u\right\| _{p,\varepsilon }:= \left( \underset{{\scriptstyle x\in \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| u(x)\right| ^{p} \right) ^{1/p}\, . \end{aligned}$$

We first prove the theorem in case \(q=p\). Let \(h\in \mathbb {Z}_{\varepsilon }^{d}\) and \(\eta :=10h\). We define \(B_{\eta ,\varepsilon }:=\left\{ y\in \mathbb {Z}_{\varepsilon }^{d}:\;|y|\le |\eta |\right\} \) and \(B_{\eta }:=\left\{ y\in \mathbb {R}^{d}:\;|y|<\left| \eta \right| \right\} \). Since \(h\in \mathbb {Z}_{\varepsilon }^{d}\) we always have \(\eta \ge 10\varepsilon \) and hence we have

$$\begin{aligned} C_{B}:=\sup _{\varepsilon ,\eta }\left( \frac{\left| B_{\eta ,\varepsilon }\right| _{\varepsilon }}{\left| B_{\eta }\right| }+\frac{\left| B_{\eta }\right| }{\left| B_{\eta ,\varepsilon }\right| _{\varepsilon }}\right) <+\infty \end{aligned}$$


$$\begin{aligned} \tilde{C}_{B}:=\sup _{\varepsilon ,\eta }\left( \underset{{\scriptstyle y\in B_{\eta ,\varepsilon }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| y\right| ^{\left( d+ps\right) /\left( p-1\right) }\right) /\left( \int _{B_{\eta }}\left| y\right| ^{\left( d+ps\right) /\left( p-1\right) }\mathrm {d}y\right) <+\infty \end{aligned}$$

We find

$$\begin{aligned} \left\| u-\tau _{h}u\right\| _{p,\varepsilon }\le \left| B_{\eta ,\varepsilon }\right| _{\varepsilon }^{-1}\underset{{\scriptstyle y\in B_{\eta ,\varepsilon }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left( \left\| u-\tau _{y}u\right\| _{p,\varepsilon }+\left\| \tau _{h}u-\tau _{y}u\right\| _{p,\varepsilon }\right) . \end{aligned}$$

In order to estimate the right-hand side, we apply Hölder’s inequality and obtain

$$\begin{aligned} \underset{{\scriptstyle y\in B_{\eta ,\varepsilon }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left\| u-\tau _{y}u\right\| _{p,\varepsilon }&\le \left( \underset{{\scriptstyle y\in B_{\eta ,\varepsilon }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left\| u-\tau _{y}u\right\| _{p,\varepsilon }^{p}}{\left| y\right| ^{d+ps}}\right) ^{\frac{1}{p}}\left( \underset{{\scriptstyle y\in B_{\eta ,\varepsilon }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| y\right| ^{\left( d+ps\right) /\left( p-1\right) }\right) ^{\frac{p-1}{p}}\\&\le \left( \underset{{\scriptstyle y\in B_{\eta ,\varepsilon }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\,\underset{{\scriptstyle x\in \cap \mathbb {Z}_{\varepsilon }^{d}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left( u(x)-u(x+y)\right) ^{p}}{\left| y\right| ^{d+ps}}\right) ^{\frac{1}{p}}\tilde{C}_{B}^{\frac{p-1}{p}}\left( \int \limits _{B_{\eta }}\left| y\right| ^{\left( d+ps\right) /\left( p-1\right) }\mathrm {d}y\right) ^{\frac{p-1}{p}}\\&=C\left\| u\right\| _{W^{s,p}(\mathbb {Z}_{\varepsilon }^{d})}\left| B_{\eta }\right| \left| \eta \right| ^{s}. \end{aligned}$$

Also with \(B_{2\eta ,\varepsilon }(h):=\left\{ y\in \mathbb {Z}_{\varepsilon }^{d}:\;|y-h|\le 2|\eta |\right\} \) we get from Hölder’s inequality

$$\begin{aligned} \underset{{\scriptstyle y\in B_{\eta ,\varepsilon }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left\| \tau _{h}u-\tau _{y}u\right\| _{p,\varepsilon }&\le \left( \underset{{\scriptstyle y\in B_{\eta ,\varepsilon }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\frac{\left\| \tau _{h}u-\tau _{y}u\right\| _{p,\varepsilon }^{p}}{\left| y-h\right| ^{d+ps}}\right) ^{\frac{1}{p}}\left( \underset{{\scriptstyle y\in B_{\eta ,\varepsilon }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| y-h\right| ^{\left( d+ps\right) /\left( p-1\right) }\right) ^{\frac{p-1}{p}}\\&\le C\left\| u\right\| _{W^{s,p}(\mathbb {Z}_{\varepsilon }^{d})}\left( \underset{{\scriptstyle y\in B_{2\eta ,\varepsilon (h)}}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| y-h\right| ^{\left( d+ps\right) /\left( p-1\right) }\right) ^{\frac{p-1}{p}}\\&=2^{d+ps}C\left\| u\right\| _{W^{s,p}(\mathbb {Z}_{\varepsilon }^{d})}\left| B_{\eta }\right| \left| \eta \right| ^{s}. \end{aligned}$$

This implies

$$\begin{aligned} \left\| u-\tau _{h}u\right\| _{p,\varepsilon }\le C\left\| u\right\| _{W^{s,p}(\mathbb {Z}_{\varepsilon }^{d})}\left| h\right| ^{s}. \end{aligned}$$

Now, let \(C_{\varepsilon }:=[-\varepsilon ,\varepsilon ]^{d}\) be the cube of size \(\varepsilon \) and let \(h\in \mathbb {R}^{d}\backslash C_{\varepsilon }\). Further, let \(\mathbb {Z}_{\varepsilon ,h}^{d}:=\left\{ z\in \mathbb {Z}_{\varepsilon }^{d}:\;\left( z+C_{\varepsilon }\right) \cap \left( h+C_{\varepsilon }\right) \not =\emptyset \right\} \) and for every \(z\in \mathbb {Z}_{\varepsilon ,h}^{d}\) let \(V(z,h)=\left| \left( z+C_{\varepsilon }\right) \cap \left( h+C_{\varepsilon }\right) \right| \). Then we find

$$\begin{aligned} \left\| \mathcal {R}_{\varepsilon }^{*}u-\tau _{h}\mathcal {R}_{\varepsilon }^{*}u\right\| _{L^{p}(\mathbb {R}^{d})}&\le \sum _{z\in \mathbb {Z}_{\varepsilon ,h}^{d}}V(z,h)\left\| \mathcal {R}_{\varepsilon }^{*}u-\tau _{z}\mathcal {R}_{\varepsilon }^{*}u\right\| _{L^{p}(\mathbb {R}^{d})}\\&=\sum _{z\in \mathbb {Z}_{\varepsilon ,h}^{d}}V(z,h)\left\| u-\tau _{z}u\right\| _{L^{p}(\mathbb {Z}_{\varepsilon }^{d})}\\&{\mathop {\le }\limits ^{(56)}}C\sum _{z\in \mathbb {Z}_{\varepsilon ,h}^{d}}V(z,h)\left\| u\right\| _{W^{s,p}(\mathbb {Z}_{\varepsilon }^{d})}\left| z\right| ^{s}\\&\le C\sum _{z\in \mathbb {Z}_{\varepsilon ,h}^{d}}V(z,h)\left\| u\right\| _{W^{s,p}(\mathbb {Z}_{\varepsilon }^{d})}\left| 2h\right| ^{s}\\&\le C\left\| u\right\| _{W^{s,p}(\mathbb {Z}_{\varepsilon }^{d})}\left| h\right| ^{s}. \end{aligned}$$

Now, let \(h\in C_{\varepsilon }\). Like above, we obtain

$$\begin{aligned} \left\| \mathcal {R}_{\varepsilon }^{*}u-\tau _{h}\mathcal {R}_{\varepsilon }^{*}u\right\| _{L^{p}(\mathbb {R}^{d})}\le C\sum _{z\in \mathbb {Z}_{\varepsilon ,h}^{d}}V(z,h)\left\| u\right\| _{W^{s,p}(\mathbb {Z}_{\varepsilon }^{d})}\left| z\right| ^{s}. \end{aligned}$$

However, this time we find \(V(z,h)\rightarrow 0\) uniformly and linearly in \(\left| h\right| \rightarrow 0\). Hence, we have

$$\begin{aligned} \left\| \mathcal {R}_{\varepsilon }^{*}u-\tau _{h}\mathcal {R}_{\varepsilon }^{*}u\right\| _{L^{p}(\mathbb {R}^{d})}\le C{\left\{ \begin{array}{ll} \left| h\right| ^{s} &{} \text{ if } h\in \mathbb {R}^{d}\backslash C_{\varepsilon }\\ |h| &{} \text{ if } h\in C_{\varepsilon } \end{array}\right. }. \end{aligned}$$

Since C does not depend on \(\varepsilon \), we infer

$$\begin{aligned} \left\| \mathcal {R}_{\varepsilon }^{*}u-\tau _{h}\mathcal {R}_{\varepsilon }^{*}u\right\| _{L^{p}(\mathbb {R}^{d})}\le C{\left\{ \begin{array}{ll} \left| h\right| ^{s} &{} \text{ if } \left| h\right| >1\\ |h| &{} \text{ if } \left| h\right| \le 1 \end{array}\right. }. \end{aligned}$$

This implies (55) in case \(p=q\).

In case \(q<p\), we use Remark 17 and let \(\tilde{\varvec{Q}}\) denote the common support of \(\mathcal {E}_{\varepsilon }u^{\varepsilon }\). We then obtain by Hölder’s inequality

$$\begin{aligned} \left\| \mathcal {R}_{\varepsilon }^{*}\mathcal {E}_{\varepsilon }u^{\varepsilon }-\tau _{h}\mathcal {R}_{\varepsilon }^{*}\mathcal {E}_{\varepsilon }u^{\varepsilon }\right\| _{L^{q}(\mathbb {R}^{d})}\le \left| \tilde{\varvec{Q}}\right| ^{\frac{p-q}{p}}\left\| \mathcal {R}_{\varepsilon }^{*}\mathcal {E}_{\varepsilon }u^{\varepsilon }-\tau _{h}\mathcal {R}_{\varepsilon }^{*}\mathcal {E}_{\varepsilon }u^{\varepsilon }\right\| _{L^{p}(\mathbb {R}^{d})}^{\frac{q}{p}}, \end{aligned}$$

and hence compactness by (57).

In case \(q\in (p,p^{\star })\) we use the same trick as in the proof of Theorem 18: we have for \(f=u-\tau _{h}u\) and for \(q=\theta p+(1-\theta )p^{\star }\) that

$$\begin{aligned} \underset{{\scriptstyle x\in Q^{\varepsilon }}}{\,\sum \,{\scriptstyle \varepsilon }\;\;}\left| \left( u-\tau _{h}u\right) (x)\right| ^{q}&\le \left\| u-\tau _{h}u\right\| _{L^{p}(Q^{\varepsilon })}^{p\theta }\left\| u-\tau _{h}u\right\| _{L^{p^{\star }}(Q^{\varepsilon })}^{\left( 1-\theta \right) p^{\star }}, \end{aligned}$$

and hence (55) follows from Theorem 18 and (57). \(\square \)

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Flegel, F., Heida, M. The fractional p-Laplacian emerging from homogenization of the random conductance model with degenerate ergodic weights and unbounded-range jumps. Calc. Var. 59, 8 (2020).

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Mathematics Subject Classification

  • 80M40
  • 60H25
  • 60K37
  • 35B27
  • 35R60
  • 47B80
  • 47A75