hence has the same bound. The proof of step (vi) also works nicely, once we
observe that $\int_{\mathbb{R}^{N}}\eta_{+}(\mathcal{L}^{\prime}\eta)\geq 0$.

In the one-dimensional case the operator $\mathcal{L}^{\prime}$ does not
appear in the limit equation, so that we deal with entropy solutions of scalar
conservation laws. Uniqueness for such problems with a Hölder-continuous
nonlinearity $u^{q}$, $q\in(0,1)$, was obtained in [51, Proposition 1.1]. ∎

### Acknowledgements

J. Endal received funding from the European Union’s Horizon 2020 research and
innovation programme under the Marie Skłodowska-Curie grant agreement no.
839749 “Novel techniques for quantitative behaviour of convection-diffusion
equations (techFRONT)” and from the Research Council of Norway under the MSCA-
TOPP-UT grant agreement no. 312021.

L. I. Ignat received funding from Romanian Ministry of Research, Innovation
and Digitization, CNCS - UEFISCDI, project number PN-III-P1-1.1-TE-2021-1539,
within PNCDI III.

F. Quirós received funding from MCIN/AEI (Spain) through projects
PID2020-116949GB-I00, RED2018-102650-T and the ICMAT-Severo Ochoa grant
CEX2019-000904-S.

##  Appendix A Truncated operators

The aim of this appendix is to prove that the rescaled truncated operators
$\mathcal{L}^{\lambda,>\rho}$ and $\mathcal{L}^{\lambda,\leq\rho}$ “approach”
as $\lambda\to\infty$ respectively the truncated operators
$\mathcal{L}^{\prime,>\rho}$ and $\mathcal{L}^{\prime,\leq\rho}$. This fact
plays an essential role in the proof of Theorem 1.3 in the convection regime
$q<q_{*}(\alpha)$.

######  Lemma A.1.

Assume $\alpha\in(0,2)$, $\beta>1/\alpha$, and ($\textup{A}_{\nu}$), and
consider the operators $\mathcal{L}^{\lambda,>\rho}$ and
$\mathcal{L}^{\prime,>\rho}$ defined respectively through (4.7), (1.9), and
also (1.6). Then, for all $\phi\in L^{1}(\mathbb{R}^{N})$,

$\|\mathcal{L}^{\lambda,>\rho}\phi-\mathcal{L}^{\prime,>\rho}\phi\|_{L^{1}(\mathbb{R}^{N})}\rightarrow
0\quad\text{as $\lambda\rightarrow\infty$.}$

###### Proof.

Using the definition of the two operators and a change of variables in
$x^{\prime}\in\mathbb{R}^{N-1}$ we obtain

$\begin{split}\int_{\mathbb{R}^{N}}|\mathcal{L}^{\lambda,>\rho}\phi-\mathcal{L}^{\prime,>\rho}\phi|\leq\sum_{\pm}\int_{\rho}^{\infty}\int_{\mathbb{S}^{N-1}}\int_{\mathbb{R}^{N}}|\phi(x\pm
r(0,\lambda^{\frac{1}{\alpha}-\beta}\theta_{N}))-\phi(x)|\,{\rm d}x{\rm
d}\mu(\theta)\frac{{\rm d}r}{r^{1+\alpha}}.\end{split}$

Since, for all $(r,\theta)\in(\rho,\infty)\times\mathbb{S}^{N-1}$,

$\int_{\mathbb{R}^{N}}|\phi(x\pm
r(0,\lambda^{-\frac{1}{\alpha}(\alpha\beta-1)}\theta_{N}))-\phi(x)|\,{\rm
d}x\rightarrow 0\quad\text{as }\lambda\rightarrow\infty,$

and this integral is uniformly bounded by the $L^{1}$-norm of $\phi$, we can
apply the dominated convergence theorem to obtain the desired result. ∎

######  Lemma A.2.

Assume $\alpha\in(0,2)$, $\beta>1/\alpha$, and ($\textup{A}_{\nu}$), and
consider the operators $\mathcal{L}^{\lambda,\leq\rho}$ and
$\mathcal{L}^{\prime,\leq\rho}$ defined respectively through (4.7), (1.9), and
also (1.6). Then, for all $\phi\in C_{\textup{c}}^{3}(\mathbb{R}^{N})$,

$\displaystyle\|\mathcal{L}^{\lambda,\leq\rho}\phi-\mathcal{L}^{\prime,\leq\rho}\phi\|_{L^{\infty}(\mathbb{R}^{N})}\lesssim
c(\rho)\lambda^{\frac{1}{\alpha}-\beta}\Big{(}\|D^{2}\phi\|_{L^{\infty}(\mathbb{R}^{N})}+\|\partial_{x_{N}}D^{2}\phi\|_{L^{\infty}(\mathbb{R}^{N})}\Big{)},$
$\displaystyle\|\mathcal{L}^{\lambda,\leq\rho}\phi-\mathcal{L}^{\prime,\leq\rho}\phi\|_{L^{1}(\mathbb{R}^{N})}\lesssim
c(\rho)\lambda^{\frac{1}{\alpha}-\beta}\Big{(}\|D^{2}\phi\|_{L^{1}(\mathbb{R}^{N})}+\|\partial_{x_{N}}D^{2}\phi\|_{L^{1}(\mathbb{R}^{N})}\Big{)},$

where $c(\rho)\eqsim\rho^{2-\alpha}+\rho^{3-\alpha}$, uniformly in
$\lambda>1$.

###### Proof.

We use Taylor’s theorem with integral reminder:

$f(x+z)-f(x)=\int_{0}^{1}z\cdot\nabla f(x+sz)\,{\rm d}s=z\cdot\nabla
f(x)+\int_{0}^{1}(1-s)zD^{2}f(x+sz)z^{t}\,{\rm d}s.$

Hence, by defining $\sigma:=1/\alpha-\beta<0$, we have

$\displaystyle(\mathcal{L}^{\lambda,\leq\rho}\phi)(x)$
$\displaystyle=-\int_{0}^{\rho}\int_{\mathbb{S}^{N-1}}\int_{0}^{1}(1-s)(\theta^{\prime},\lambda^{\sigma}\theta_{N})D^{2}\phi(x\pm
sr(\theta^{\prime},\lambda^{\sigma}\theta_{N}))(\theta^{\prime},\lambda^{\sigma}\theta_{N})^{t}\,{\rm
d}s\frac{{\rm d}r}{r^{\alpha-1}}\,{\rm d}\mu(\theta),$
$\displaystyle(\mathcal{L}^{\lambda,\leq\rho}\phi)(x)$
$\displaystyle=-\int_{0}^{\rho}\int_{\mathbb{S}^{N-1}}\int_{0}^{1}(1-s)(\theta^{\prime},0)D^{2}\phi(x\pm
sr(\theta^{\prime},0))(\theta^{\prime},0)^{t}\,{\rm d}s\frac{{\rm
d}r}{r^{\alpha-1}}{\rm d}\mu(\theta).$

We only consider the part with $+$ sign, since the other is similar. We write

$|(\mathcal{L}^{\lambda,\leq\rho}\phi)(x)-(\mathcal{L}^{\prime,\leq\rho}\phi)(x)|\leq
I_{1,\lambda}(x)+I_{2,\lambda}(x)+I_{3,\lambda}(x),$

where

$\displaystyle I_{1,\lambda}(x)$
$\displaystyle:=\int_{0}^{\rho}\int_{\mathbb{S}^{N-1}}\int_{0}^{1}(1-s)\Big{|}(\theta^{\prime},\lambda^{\sigma}\theta_{N})\big{(}D^{2}\phi(x+sr(\theta^{\prime},\lambda^{\sigma}\theta_{N}))$
$\displaystyle\hskip
170.71652pt-D^{2}\phi(x+sr(\theta^{\prime},0))\big{)}(\theta^{\prime},\lambda^{\sigma}\theta_{N})^{t}\Big{|}\,{\rm
d}s\frac{{\rm d}r{\rm d}\mu(\theta)}{r^{\alpha-1}},$ $\displaystyle
I_{2,\lambda}(x)$
$\displaystyle:=\int_{0}^{\rho}\int_{\mathbb{S}^{N-1}}\int_{0}^{1}(1-s)\Big{|}\big{(}(\theta^{\prime},\lambda^{\sigma}\theta_{N})-(\theta^{\prime},0)\big{)}D^{2}\phi(x+sr(\theta^{\prime},0))(\theta^{\prime},\lambda^{\sigma}\theta_{N})^{t}\Big{|}\,{\rm
d}s\frac{{\rm d}r{\rm d}\mu(\theta)}{r^{\alpha-1}},$ $\displaystyle
I_{3,\lambda}(x)$
$\displaystyle:=\int_{0}^{\rho}\int_{\mathbb{S}^{N-1}}\int_{0}^{1}(1-s)\Big{|}(\theta^{\prime},0)D^{2}\phi(x+sr(\theta^{\prime},0))\big{(}(\theta^{\prime},\lambda^{\sigma}\theta_{N})-(\theta^{\prime},0)\big{)}^{t}\Big{|}\,{\rm
d}s\frac{{\rm d}r{\rm d}\mu(\theta)}{r^{\alpha-1}}.$

By using the identity

$\displaystyle D^{2}\phi(x+sr(\theta^{\prime},\lambda^{\sigma}\theta_{N}))$
$\displaystyle-D^{2}\phi(x+sr(\theta^{\prime},0))$
$\displaystyle=sr(0,\lambda^{\sigma}\theta_{N})\cdot\int_{0}^{1}\nabla(D^{2}\phi)\Big{(}x+sr((\theta^{\prime},0)+t(0,\lambda^{\sigma}\theta_{N}))\Big{)}\,{\rm
d}t$
$\displaystyle=sr\lambda^{\sigma}\theta_{N}\int_{0}^{1}\partial_{x_{N}}D^{2}\phi\Big{(}x+sr(\theta^{\prime},t\lambda^{\sigma}\theta_{N})\Big{)}\,{\rm
d}t,$

and the facts that $\sigma<0$ and
$(\theta^{\prime},\theta_{N})\in\mathbb{S}^{N-1}$, we obtain that

$\|I_{1,\lambda}\|_{L^{1}(\mathbb{R}^{N})}\lesssim\lambda^{\sigma}\|\partial_{x_{N}}D^{2}\phi\|_{L^{1}(\mathbb{R}^{N})}\int_{0}^{\rho}\int_{\mathbb{S}^{N-1}}\frac{{\rm
d}r{\rm
d}\mu(\theta)}{r^{\alpha-2}}\eqsim\lambda^{\sigma}\rho^{3-\alpha}\|\partial_{x_{N}}D^{2}\phi\|_{L^{1}(\mathbb{R}^{N})}.$

The last two terms can be estimated by brute force in a similar manner:

$\|I_{2,\lambda}\|_{L^{1}(\mathbb{R}^{N})}+\|I_{3,\lambda}\|_{L^{1}(\mathbb{R}^{N})}\lesssim\lambda^{\sigma}\|D^{2}\phi\|_{L^{1}(\mathbb{R}^{N})}\int_{0}^{\rho}\int_{\mathbb{S}^{N-1}}\frac{{\rm
d}r{\rm
d}\mu(\theta)}{r^{\alpha-1}}\eqsim\lambda^{\sigma}\rho^{2-\alpha}\|D^{2}\phi\|_{L^{1}(\mathbb{R}^{N})}.$

A similar estimate holds for the $L^{\infty}$-norm of the three terms. ∎

##  Appendix B Basic results for entropy solutions

In this section, we will build the basic theory of (P),
($\textup{P}_{\varepsilon}$), and ($\textup{P}_{\varepsilon}^{\prime}$) from
scratch based on [11, 54, 1, 14, 27, 3, 57]. The operator
$\mathcal{L}^{\prime}$ defined by (1.9) only acts on the first $N-1$ spatial
variables. However, most of the results we are going to deduce only depend on
upper bounds of the operator, therefore we will always write them for
$\mathcal{L}$ and only give suitable comments on the case
$\mathcal{L}^{\prime}$, if needed.

###  B.1. Concept of entropy solutions

The definition of entropy solutions of (P) was already stated as Definition
1.1, let us also give it for ($\textup{P}_{\varepsilon}$):

######  Definition B.1 (Entropy solution of ($\textup{P}_{\varepsilon}$)).

A function $u$ is an _entropy solution_ of ($\textup{P}_{\varepsilon}$) if:

  1. (a)

(Regularity) $u\in
L^{\infty}((0,\infty);L_{\textup{loc}}^{1}(\mathbb{R}^{N}))\cap
L^{\infty}_{\textup{loc}}((0,\infty);L^{\infty}(\mathbb{R}^{N}))\cap
L_{\textup{loc}}^{2}((0,\infty);H_{\textup{loc}}^{1}(\mathbb{R}^{N}))$.

  2. (b)

(Entropy inequality) For all $k\in\mathbb{R}$, all $\rho>0$, and all
$0\leq\phi\in C_{\textup{c}}^{\infty}(\overline{Q})$,

(B.1) $\begin{split}&\iint_{Q}\Big{(}|u-k|\phi_{t}+\,{\rm
sgn}(u-k)\big{(}F(u)-F(k)\big{)}\cdot\nabla\phi+\varepsilon|u-k|\Delta\phi\Big{)}\\\
&\qquad-\iint_{Q}\Big{(}\,{\rm
sgn}(u-k)\phi(\mathcal{L}^{>\rho}u)+|u-k|(\mathcal{L}^{\leq\rho}\phi)\Big{)}+\int_{\mathbb{R}^{N}}|u_{0}-k|\phi(0)\geq
0.\end{split}$

######  Remark B.2.

In the above definition, $u\in
L^{\infty}_{\textup{loc}}((0,\infty);L^{\infty}(\mathbb{R}^{N}))$ implies that
$u\in L_{\textup{loc}}^{2}(Q)$. Hence, the last regularity assumption is
actually $\nabla u\in L_{\textup{loc}}^{2}(Q)$. Its use is better seen if we
rewrite the term involving the Laplacian:

$\iint_{Q}|u-k|\Delta\phi=\iint_{Q}\,{\rm sgn}(u-k)\nabla u\cdot\nabla\phi.$

###  B.2. Known uniqueness and a priori estimates

We follow the uniqueness argument presented in [3, Section 4]. It starts with
a Kato inequality; see e.g. [54, Theorem 3.9] and [27, Proposition 4.2].

######  Proposition B.3 (Kato inequality).

Assume ($\textup{A}_{q}$), and ($\textup{A}_{\nu}$). Let $u,\bar{u}$ be
entropy solutions of (P) with initial data $u_{0},\bar{u}_{0}\in
L^{\infty}(\mathbb{R}^{N})$, respectively. Then, for all $0\leq\phi\in
C_{\textup{c}}^{\infty}(\overline{Q})$,

(B.2) $\begin{split}&\iint_{Q}\Big{(}|u-\bar{u}|\phi_{t}+\,{\rm
sgn}(u-\bar{u})\big{(}F(u)-F(\bar{u})\big{)}\cdot\nabla\phi-|u-\bar{u}|(\mathcal{L}\phi)\Big{)}+\int_{\mathbb{R}^{N}}|u_{0}-\bar{u}_{0}|\phi(0)\geq
0.\end{split}$

######  Remark B.4.

For entropy solutions $u,\bar{u}$ of ($\textup{P}_{\varepsilon}$), we add the
term $\displaystyle\varepsilon\iint_{Q}|u-\bar{u}|\Delta\phi$ in (B.2).

To proceed we need the auxiliary function

(B.3) $\Phi(x)=\begin{cases}1&\text{if }|x|<1,\\\ |x|^{-N-\alpha}&\text{if
}|x|>2,\end{cases}$

which is nonnegative and belongs to $W^{2,1}\cap
W^{2,\infty}(\mathbb{R}^{N})$. Let us obtain some estimates on it.

######  Lemma B.5 (Lemma 4.9 in [3]).

Assume ($\textup{A}_{\nu}$). There is a constant $\gamma$ (depending on
$\Lambda_{1}$) such that

$|\nabla\Phi|\leq\gamma\Phi,\qquad|\Delta\Phi|\leq\gamma\|\Phi\|_{W^{2,\infty}(\mathbb{R}^{N})},\quad\text{and
}|\mathcal{L}\Phi|\leq\gamma\|\Phi\|_{W^{2,\infty}(\mathbb{R}^{N})}.$

######  Remark B.6.

In [3] the authors use for the isotropic case
$\mathcal{L}=(-\Delta)^{\alpha/2}$ a function $\Phi$ that satisfies
$|(-\Delta)^{\frac{\alpha}{2}}\Phi|\leq\gamma\Phi$ for all $\alpha\in(0,2)$.
The existence of a function $\Phi$ satisfying this property is not clear for a
general anisotropic $\mathcal{L}$. However, the uniqueness proof for
$L^{1}\cap L^{\infty}$-solutions in [3] still works substituting the estimate
(28) in that paper by

$\begin{split}\int_{>\frac{1}{\varepsilon}}|u-v|(\mathcal{L}\Phi)(\varepsilon\cdot)\geq-\gamma\|\Phi\|_{W^{2,\infty}(\mathbb{R}^{N})}\int_{>\frac{1}{\varepsilon}}|u-v|\end{split}$

This allows us to obtain estimate (32) in [3] with $s=1$ and $\sigma=q$. The
proof is then completed without more effort since the integrability of
solutions is already assumed.

######  Theorem B.7 ([3]).

(a) Assume $p\in[1,\infty)$, $u_{0}\in L^{\infty}(\mathbb{R}^{N})$,
($\textup{A}_{q}$), and ($\textup{A}_{\nu}$). Then there is _at least_ one
entropy solution $u\in L^{\infty}(Q)\cap
C([0,\infty);L_{\textup{loc}}^{p}(\mathbb{R}^{N}))$ of (P) with initial data
$u_{0}$. Moreover, let $u,\bar{u}$ be entropy solutions of (P) with initial
data $u_{0},\bar{u}_{0}$, respectively. Then:

  1. (i)

(Comparison principle) If $u_{0}\leq\bar{u}_{0}$ a.e. on $\mathbb{R}^{N}$,
then $u\leq\bar{u}$ a.e. in $Q$.

  2. (ii)

($L^{1}$-contraction) If $u_{0}-\bar{u}_{0}\in L^{1}(\mathbb{R}^{N})$, then,
for a.e. $t>0$,

$\int_{\mathbb{R}^{N}}|u(t)-\bar{u}(t)|\leq\int_{\mathbb{R}^{N}}|u_{0}-\bar{u}_{0}|.$

  3. (iii)

($L^{\infty}$-bound) For a.e. $(t,x)\in Q$,
$\operatorname*{ess\,inf}_{x\in\mathbb{R}^{N}}u_{0}(x)\leq
u(t,x)\leq\operatorname*{ess\,sup}_{x\in\mathbb{R}^{N}}u_{0}(x)$.

  4. (iv)

($L^{1}$-bound) For a.e. $t>0$,
$\|u(t)\|_{L^{1}(\mathbb{R}^{N})}\leq\|u_{0}\|_{L^{1}(\mathbb{R}^{N})}$.

  5. (v)

(Very weak solutions) For all $\phi\in C_{\rm c}^{\infty}(\overline{Q})$,

$\begin{split}\iint_{Q}\Big{(}u\partial_{t}\phi+F(u)\cdot\nabla\phi-u(\mathcal{L}\phi)\Big{)}+\int_{\mathbb{R}^{N}}u_{0}\phi(0)=0.\end{split}$

  6. (vi)

(Mass conservation) For a.e. $t>0$,

$\int_{\mathbb{R}^{N}}u(t)=\int_{\mathbb{R}^{N}}u_{0}.$

(b) Assume, in addition, $u_{0}\in L^{1}(\mathbb{R}^{N})\cap
L^{\infty}(\mathbb{R}^{N})$. Then there is _at most_ one entropy solution $u$
of (P) with initial data $u_{0}$, and moreover, the solution satisfies $u\in
C([0,\infty);L^{1}(\mathbb{R}^{N}))$.

######  Remark B.8.

All of the above properties also hold for entropy solutions of
($\textup{P}_{\varepsilon}$): In item (v), we have the additional term
$\varepsilon\iint_{Q}u\Delta\phi$ on the left-hand side.

###### Proof of Theorem B.7.

Let us comment on the two last items:

(v) By the convex inequality (or Kato inequality)

$\iint_{Q}\,{\rm
sgn}(u-k)\phi(\mathcal{L}^{>\rho}(u-k))\geq\iint_{Q}|u-k|(\mathcal{L}^{>\rho}\phi)$

and the choices $k\geq\pm\|u\|_{L^{\infty}(Q)}$ in (B.1), the desired result
follows.

(vi) Consider a standard cut-off function $\mathcal{X}_{R}$ defined in Section
1.7. The idea is then to choose $\phi$ as
$\mathds{1}_{[0,\tau]}(t)\mathcal{X}_{R}(x)$ in the very weak formulation.
Since $\|\mathcal{L}\mathcal{X}_{R}\|_{L^{\infty}}\leq
R^{-\alpha}\|\mathcal{L}\mathcal{X}\|_{L^{\infty}}$ by $\alpha$-homogeneity,
the gradient and nonlocal terms go to zero as $R\to\infty$. ∎

###  B.3. Concept of classical solutions

We define, for all $\phi\in C_{\textup{c}}^{\infty}(\mathbb{R}^{N})$,

$\begin{gathered}(L\phi)(x):=-\varepsilon(\Delta\phi)(x)+(\mathcal{L}\phi)(x),\quad\text{i.e.,}\\\
\widehat{L\phi}(\xi)=a(\xi)\widehat{\phi}(\xi)\quad\text{with
}a(\xi):=\varepsilon|\xi|^{2}+\int_{\mathbb{S}^{N-1}}|\xi\cdot\theta|^{\alpha}\,{\rm
d}\mu(\theta).\end{gathered}$

Now, solutions of ($\textup{P}_{\varepsilon}$) with initial data
$0<\tilde{\varepsilon}\leq u_{0}\in L^{\infty}(\mathbb{R}^{N})$ are mild and
also classical on $Q$:

###### Proof of Lemma 3.1.

By the assumption ($\textup{A}_{\nu}$), we get $\varepsilon|\xi|^{2}\leq
a(\xi)\leq\varepsilon|\xi|^{2}+\Lambda_{2}|\xi|^{\alpha}$. This means that
assumptions (2.8), (2.9), and (2.10) in [7] are fulfilled (the latter with
$a_{0}=\varepsilon$ and $\tilde{\alpha}=\alpha$). By Remark 1.2 in [22], we
are in the setting of Theorem 1.1 in [23]. However, that theorem requires
$F\in C^{\infty}(\mathbb{R})$, but with our assumption on the initial data,
$F(u)\in
C^{\infty}((\tilde{\varepsilon},\|u_{0}\|_{L^{\infty}(\mathbb{R}^{N})}))$. ∎

We continue by providing a uniform bound for $\nabla u$ (cf. Lemma 2.1 and
Corollary 2.2 in [57]).

######  Lemma B.9.

Assume $p\in[2,\infty)$, $0<\tilde{\varepsilon}\leq u_{0}\in
L^{\infty}(\mathbb{R}^{N})$, ($\textup{A}_{q}$), and ($\textup{A}_{\nu}$). Let
$\Phi$ be as in (B.3) and $0\leq\Theta\in
C_{\textup{c}}^{\infty}((0,\infty))$. Then the unique mild solution $u$ of
($\textup{P}_{\varepsilon}$) satisfies

$\displaystyle\varepsilon\frac{4(p-1)}{p^{2}}\iint_{Q}|\nabla
u^{\frac{p}{2}}(t,x)|^{2}\Theta(t)\Phi(x)\,{\rm d}x{\rm d}t\leq
C,\quad\text{where}$ $\displaystyle C:=\max_{\tilde{\varepsilon}\leq
u(x,t)\leq\|u_{0}\|_{L^{\infty}}}\bigg{\\{}\frac{1}{p}u^{p}+\gamma
u^{p-1}|F(u)|+\gamma\frac{1}{p}u^{p}$
$\displaystyle\qquad+\frac{1}{2}\Big{(}\varepsilon^{-\frac{1}{2}}(p-1)^{\frac{1}{2}}u^{\frac{p}{2}-1}|F(u)|+\varepsilon^{\frac{1}{2}}\gamma(p-1)^{-\frac{1}{2}}u^{\frac{p}{2}}\Big{)}^{2}\bigg{\\}}\bigg{(}\iint_{Q}\max\\{|\Theta^{\prime}|,\Theta\\}\big{(}\Phi+|\mathcal{L}\Phi|\big{)}\bigg{)}.$

######  Remark B.10.

Note that indeed when $p=2$ we have, for all compact
$K\subset\textup{supp}\,\Phi\times\mathbb{R}^{N}$, that

$\begin{split}C&\geq\varepsilon\iint_{Q}|\nabla
u(t,x)|^{2}\Theta(t)\Phi(x)\,{\rm d}x{\rm d}t\geq\varepsilon\iint_{K}|\nabla
u(t,x)|^{2}\Theta(t)\Phi(x)\,{\rm d}x{\rm
d}t\geq\varepsilon\min_{K}\\{\Theta\Phi\\}\iint_{K}|\nabla u|^{2}.\end{split}$

###### Proof of Lemma B.9.

Since $u\in C_{\textup{b}}^{\infty}(Q)$ is a classical solution of the PDE in
($\textup{P}_{\varepsilon}$),

(B.4) $\partial_{t}u+\nabla\cdot F(u)+\mathcal{L}u-\varepsilon\Delta
u=0\quad\textup{pointwise in }Q.$

Let $0\leq\phi\in C_{\textup{c}}^{\infty}(Q)$. Using Kato’s inequality and the
identity $\nabla u\cdot\nabla(u^{p-1})=\frac{4(p-1)}{p^{2}}|\nabla
u^{\frac{p}{2}}|^{2}$,

$\begin{gathered}\iint_{Q}(\mathcal{L}u)u^{p-1}\phi\geq\frac{1}{p}\iint_{Q}(\mathcal{L}u^{p})\phi=\iint_{Q}\frac{1}{p}u^{p}(\mathcal{L}\phi),\\\
-\varepsilon\iint_{Q}(\Delta
u)u^{p-1}\phi=\varepsilon\frac{4(p-1)}{p^{2}}\iint_{Q}|\nabla
u^{\frac{p}{2}}|^{2}\phi+\varepsilon\iint_{Q}u^{p-1}\nabla
u\cdot\nabla\phi.\end{gathered}$

Therefore, if we multiply (B.4) by $u^{p-1}\phi$ and integrate over $Q$,

$\begin{split}&\iint_{Q}\Big{(}\frac{1}{p}u^{p}\partial_{t}\phi+u^{p-1}F(u)\cdot\nabla\phi+\phi
F(u)\cdot\nabla u^{p-1}\Big{)}\\\
&\qquad\geq\iint_{Q}\Big{(}\varepsilon\frac{4(p-1)}{p^{2}}|\nabla
u^{\frac{p}{2}}|^{2}\phi+\varepsilon u^{p-1}\nabla
u\cdot\nabla\phi+\frac{1}{p}u^{p}(\mathcal{L}\phi)\big{)}.\end{split}$

Define $I:=\varepsilon\frac{4(p-1)}{p^{2}}\iint_{Q}|\nabla
u^{\frac{p}{2}}|^{2}\Theta\Phi$. By the choice $\phi(x,t):=\Theta(t)\Phi(x)$,
the identities

$\nabla
u^{p-1}=(p-1)^{\frac{1}{2}}u^{\frac{p}{2}-1}\frac{2(p-1)^{\frac{1}{2}}}{p}\nabla
u^{\frac{p}{2}}\quad\textup{and }\quad u^{p-1}\nabla
u=(p-1)^{-\frac{1}{2}}u^{\frac{p}{2}}\frac{2(p-1)^{\frac{1}{2}}}{p}\nabla
u^{\frac{p}{2}},$

and the regularity $\nabla u\in L^{\infty}(Q)$, we get

$\begin{split}I&\leq\iint_{Q}\Big{(}\Phi\frac{1}{p}u^{p}\partial_{t}\Theta+\Theta
u^{p-1}F(u)\cdot\nabla\Phi+\Theta\Phi(p-1)^{\frac{1}{2}}u^{\frac{p}{2}-1}F(u)\cdot\Big{(}\frac{2(p-1)^{\frac{1}{2}}}{p}\nabla
u^{\frac{p}{2}}\Big{)}\\\
&\qquad-\Theta\frac{1}{p}u^{p}\big{(}\mathcal{L}\Phi\big{)}-\varepsilon\Theta(p-1)^{-\frac{1}{2}}u^{\frac{p}{2}}\Big{(}\frac{2(p-1)^{\frac{1}{2}}}{p}\nabla
u^{\frac{p}{2}}\Big{)}\cdot\nabla\Phi\Big{)}\\\
&\leq\iint_{Q}\Big{(}\Phi\frac{1}{p}u^{p}|\Theta^{\prime}|+\Theta
u^{p-1}|F(u)||\nabla\Phi|+\Theta\Phi(p-1)^{\frac{1}{2}}u^{\frac{p}{2}-1}|F(u)|\Big{|}\frac{2(p-1)^{\frac{1}{2}}}{p}\nabla
u^{\frac{p}{2}}\Big{|}\\\
&\qquad+\Theta\frac{1}{p}u^{p}\big{|}\mathcal{L}\Phi\big{|}+\varepsilon\Theta(p-1)^{-\frac{1}{2}}u^{\frac{p}{2}}\Big{|}\frac{2(p-1)^{\frac{1}{2}}}{p}\nabla
u^{\frac{p}{2}}\Big{|}|\nabla\Phi|\Big{)}.\end{split}$

Using now Lemma B.5,

$\begin{split}I&\leq\iint_{Q}\Big{(}\Phi\frac{1}{p}u^{p}|\Theta^{\prime}|+\Theta
u^{p-1}|F(u)|\gamma\Phi+\Theta\Phi(p-1)^{\frac{1}{2}}u^{\frac{p}{2}-1}|F(u)|\frac{2(p-1)^{\frac{1}{2}}}{p}|\nabla
u^{\frac{p}{2}}|\\\
&\qquad+\Theta\frac{1}{p}u^{p}|\mathcal{L}\Phi|+\varepsilon\Theta(p-1)^{-\frac{1}{2}}u^{\frac{p}{2}}\frac{2(p-1)^{\frac{1}{2}}}{p}|\nabla
u^{\frac{p}{2}}|\gamma\Phi\Big{)}\\\
&=\iint_{Q}\bigg{(}\Phi\max\\{|\Theta^{\prime}|,\Theta\\}\Big{(}\frac{1}{p}u^{p}+\gamma
u^{p-1}|F(u)|+\gamma\frac{1}{p}u^{p}\Big{)}\\\
&\qquad+(\varepsilon\Theta\Phi)^{\frac{1}{2}}\frac{2(p-1)^{\frac{1}{2}}}{p}|\nabla
u^{\frac{p}{2}}|\Big{(}\varepsilon^{-\frac{1}{2}}(\Theta\Phi)^{\frac{1}{2}}(p-1)^{\frac{1}{2}}u^{\frac{p}{2}-1}|F(u)|+(\varepsilon\Theta\Phi)^{\frac{1}{2}}\gamma(p-1)^{-\frac{1}{2}}u^{\frac{p}{2}}\Big{)}\bigg{)}\\\
&\qquad+\iint_{Q}\Theta\frac{1}{p}u^{p}|\mathcal{L}\Phi|.\end{split}$

Now, apply Young’s inequality $ab\leq\frac{1}{2}a^{2}+\frac{1}{2}b^{2}$ to
obtain

$\begin{split}\frac{1}{2}I&\leq\iint_{Q}\bigg{(}\Phi\max\\{|\Theta^{\prime}|,\Theta\\}\Big{(}\frac{1}{p}u^{p}+\gamma
u^{p-1}|F(u)|+\gamma\frac{1}{p}u^{p}\Big{)}\\\
&\qquad+\frac{1}{2}\Big{(}\varepsilon^{-\frac{1}{2}}(\Theta\Phi)^{\frac{1}{2}}(p-1)^{\frac{1}{2}}u^{\frac{p}{2}-1}|F(u)|+(\varepsilon\Theta\Phi)^{\frac{1}{2}}\gamma(p-1)^{-\frac{1}{2}}u^{\frac{p}{2}}\Big{)}^{2}\bigg{)}+\iint_{Q}\Theta\frac{1}{p}u^{p}|\mathcal{L}\Phi|\\\
&\leq\iint_{Q}\Phi\max\\{|\Theta^{\prime}|,\Theta\\}\Big{(}\frac{1}{p}u^{p}+\gamma
u^{p-1}|F(u)|+\gamma\frac{1}{p}u^{p}\\\
&\qquad+\frac{1}{2}\Big{(}\varepsilon^{-\frac{1}{2}}(p-1)^{\frac{1}{2}}u^{\frac{p}{2}-1}|F(u)|+\varepsilon^{\frac{1}{2}}\gamma(p-1)^{-\frac{1}{2}}u^{\frac{p}{2}}\Big{)}^{2}\Big{)}+\iint_{Q}\Theta\frac{1}{p}u^{p}|\mathcal{L}\Phi|.\end{split}$

Taking the maximum over the range of $u$ and multiplying by $2$ gives the
final estimate. ∎

To write down our much needed stability result, we define, for all
$\tilde{\varepsilon}>0$,

(B.5) $L^{\infty}(\mathbb{R}^{N})\ni
u_{0,\tilde{\varepsilon}}:=u_{0}+\tilde{\varepsilon}\qquad\textup{with $u_{0}$
satisfying~{}\eqref{u_0as}.}$

Now, since mild solutions are classical solutions and then entropy solutions
of ($\textup{P}_{\varepsilon}$), they enjoy all the properties of Theorem B.7.
We then get the following convergence result.

######  Proposition B.11.

Assume ($\textup{A}_{q}$) and ($\textup{A}_{\nu}$). Let
$u_{\tilde{\varepsilon}}$ be the unique mild solution of
($\textup{P}_{\varepsilon}$) with initial data $u_{0,\tilde{\varepsilon}}$
defined in (B.5). Then there exists a function $u\in L^{\infty}(Q)$ such that

$\displaystyle u_{\tilde{\varepsilon}}\to u\quad\textup{a.e.~{}in }Q\text{ as
}\tilde{\varepsilon}\to 0^{+},$ $\displaystyle\nabla
u_{\tilde{\varepsilon}}^{\frac{p}{2}}\rightharpoonup\nabla
u^{\frac{p}{2}}\quad\textup{in }L_{\textup{loc}}^{2}(Q)\text{ as
}\tilde{\varepsilon}\to 0^{+}\quad\text{for all }p\in[2,\infty).$

Moreover, $u$ is an entropy solution of ($\textup{P}_{\varepsilon}$) with
initial data $u_{0}$ satisfying ($\textup{A}_{u_{0}}$).

###### Proof.

If $0\leq\tilde{\varepsilon}_{1}\leq\tilde{\varepsilon}_{2}$, then $0\leq
u_{0}\leq u_{0,\tilde{\varepsilon}_{1}}\leq u_{0,\tilde{\varepsilon}_{2}}$,
whence the comparison principle (see Theorem B.7) implies that $0\leq
u_{\tilde{\varepsilon}_{1}}\leq u_{\tilde{\varepsilon}_{2}}$. Therefore, the
sequence
$\\{u_{\tilde{\varepsilon}}(t,x)\\}_{\tilde{\varepsilon}>0}\subset\mathbb{R}$
is nonincreasing and uniformly bounded from below by $0$. Thus, there exists a
function $u$ such that $u_{\tilde{\varepsilon}}\to u$ a.e. in $Q$ as
$\tilde{\varepsilon}\to 0^{+}$. Moreover, if we assume
$0<\tilde{\varepsilon}\leq 1$, then $0\leq u\leq
u_{\tilde{\varepsilon}}\leq\|u_{0}\|_{L^{\infty}}+\tilde{\varepsilon}\leq\|u_{0}\|_{L^{\infty}}+1$,
and, by Lemma B.9, $\nabla u_{\tilde{\varepsilon}}^{\frac{p}{2}}\in
L_{\textup{loc}}^{2}(Q)$ uniformly in $\tilde{\varepsilon}$, which yields

$\nabla u_{\tilde{\varepsilon}}^{\frac{p}{2}}\rightharpoonup h\quad\textup{in
}L_{\textup{loc}}^{2}(Q)\text{ as }\tilde{\varepsilon}\to 0^{+}.$

Since weak limits are distributional limits, and distributional limits are
unique, we can identify the limit by considering $\phi\in
C_{\textup{c}}^{\infty}(Q)$ such that

$\iint_{Q}\nabla
u_{\tilde{\varepsilon}}^{\frac{p}{2}}\phi=-\iint_{Q}u_{\tilde{\varepsilon}}^{\frac{p}{2}}\nabla\phi.$

Now, $\phi\in C_{\textup{c}}^{\infty}(Q)\subset L^{2}(Q)$,
$u_{\tilde{\varepsilon}}\to u$ a.e. as ${\tilde{\varepsilon}}\to 0^{+}$, and
$0\leq u_{\tilde{\varepsilon}}(t,x)\leq\|u_{0}\|_{L^{\infty}}+1$ give

$\iint_{Q}h\phi=-\iint_{Q}u^{\frac{p}{2}}\nabla\phi,$

i.e., by the definition of weak derivatives, $h=\nabla u^{\frac{p}{2}}$ in
$L_{\textup{loc}}^{2}(Q)$.

Finally, since $u_{\tilde{\varepsilon}}$ is an entropy solution with initial
data $u_{0,\tilde{\varepsilon}}$ in the sense of Definition B.1, we obtain
that $u$ is an entropy solution with initial data $u_{0}$ in the sense of
Definition B.1 by simply taking the a.e.-limit as $\tilde{\varepsilon}\to
0^{+}$ in the entropy inequality (B.1). ∎

###  B.4. Entropy solutions enjoy an energy estimate

The goal of this section is to show that entropy solutions $u$ of
($\textup{P}_{\varepsilon}$) with initial data $0\leq u_{0}\in
L^{1}(\mathbb{R}^{N})\cap L^{\infty}(\mathbb{R}^{N})$ indeed satisfy the
$L^{p}$-energy inequality with $p\in[2,\infty)$: for a.e. $t>0$,

(B.6)
$\begin{split}&\frac{1}{p}\|u(t)\|^{p}_{L^{p}(\mathbb{R}^{N})}+\Lambda_{1}\frac{4(p-1)}{p^{2}}\int_{0}^{t}\int_{\mathbb{R}^{N}}|(-\Delta)^{\alpha/4}u^{\frac{p}{2}}|^{2}+\varepsilon\frac{4(p-1)}{p^{2}}\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla
u^{\frac{p}{2}}|^{2}\leq\frac{1}{p}\|u_{0}\|_{L^{p}(\mathbb{R}^{N})}.\end{split}$

Hence, entropy solutions $u$ of ($\textup{P}_{\varepsilon}$) belong to
$L^{2}((0,\infty);H^{1}(\mathbb{R}^{N})\cap H^{\alpha/2}(\mathbb{R}^{N}))$;
Lemma 3.5(b) follows.

For the cut-off function $\mathcal{X}_{R}$ defined in Section 1.7, we will
basically choose $u_{\tilde{\varepsilon}}\mathcal{X}_{R}$ as a test function,
and then take the limit as $\tilde{\varepsilon}\to 0^{+}$ in the resulting
estimate.

######  Lemma B.12.

Assume $p\in[2,\infty)$, $u_{0,\tilde{\varepsilon}}$ as in (B.5),
($\textup{A}_{q}$), and ($\textup{A}_{\nu}$). Then the unique mild solution
$u_{\tilde{\varepsilon}}$ of ($\textup{P}_{\varepsilon}$) with initial data
$u_{0,\tilde{\varepsilon}}$ satisfies

(B.7)
$\begin{split}&\frac{1}{p}\int_{\mathbb{R}^{N}}u_{\tilde{\varepsilon}}(t)^{p}\mathcal{X}_{R}+\int_{0}^{t}\mathcal{E}(u_{\tilde{\varepsilon}}(s),u_{\tilde{\varepsilon}}^{p-1}(s)\mathcal{X}_{R})\,{\rm
d}s+\varepsilon\frac{4(p-1)}{p^{2}}\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla
u_{\tilde{\varepsilon}}^{\frac{p}{2}}|^{2}\mathcal{X}_{R}\\\
&\qquad=\int_{0}^{t}\int_{\mathbb{R}^{N}}u_{\tilde{\varepsilon}}^{q+p-1}\partial_{x_{N}}\mathcal{X}_{R}+\frac{1}{p}\int_{0}^{t}\int_{\mathbb{R}^{N}}u_{\tilde{\varepsilon}}^{p}\Delta\mathcal{X}_{R}+\frac{1}{p}\int_{\mathbb{R}^{N}}u_{0,\tilde{\varepsilon}}^{p}\mathcal{X}_{R},\end{split}$

where $\mathcal{E}$ is the bilinear form associated with $\mathcal{L}$ defined
in Section 1.7.

###### Proof.

We ease the notation by writing $u,u_{0}$ for
$u_{\tilde{\varepsilon}},u_{0,\tilde{\varepsilon}}$. Since $u\in
C_{\textup{b}}^{\infty}(Q)$ is a classical solution of the PDE in
($\textup{P}_{\varepsilon}$), we simply multiply it by
$u^{p-1}\mathcal{X}_{R}$ and integrate over $(0,t)\times\mathbb{R}^{N}$. Let
us calculate term by term keeping in mind that $\mathcal{X}_{R}\in
C_{\textup{c}}^{\infty}$ and e.g. Proposition 4.1 in [18] to handle the
nonlocal term:

$\displaystyle\int_{0}^{t}\int_{\mathbb{R}^{N}}(\partial_{t}u)u^{p-1}\mathcal{X}_{R}$
$\displaystyle=\frac{1}{p}\int_{\mathbb{R}^{N}}u(t)^{p}\mathcal{X}_{R}-\frac{1}{p}\int_{\mathbb{R}^{N}}u_{0}^{p}\mathcal{X}_{R},$
$\displaystyle\int_{0}^{t}\int_{\mathbb{R}^{N}}(\nabla\cdot
F(u))u^{p-1}\mathcal{X}_{R}$
$\displaystyle=\frac{q}{q+p-1}\int_{0}^{t}\int_{\mathbb{R}^{N}}(\nabla\cdot\mathbf{a}u^{q+p-1})\mathcal{X}_{R}$
$\displaystyle=\frac{q}{q+p-1}\int_{0}^{t}\int_{\mathbb{R}^{N}}\nabla(u^{q+p-1}\mathbf{a}\mathcal{X}_{R})-\frac{q}{q+p-1}\int_{0}^{t}\int_{\mathbb{R}^{N}}u^{q+p-1}\mathbf{a}\cdot\nabla\mathcal{X}_{R}$
$\displaystyle=-\frac{q}{q+p-1}\int_{0}^{t}\int_{\mathbb{R}^{N}}u^{q+p-1}\mathbf{a}\cdot\nabla\mathcal{X}_{R},$
$\displaystyle\int_{0}^{t}\int_{\mathbb{R}^{N}}(\mathcal{L}u)u^{p-1}\mathcal{X}_{R}$
$\displaystyle=\int_{0}^{t}\mathcal{E}(u(s),u^{p-1}(s)\mathcal{X}_{R})\,{\rm
d}s,$ $\displaystyle-\varepsilon\int_{0}^{t}\int_{\mathbb{R}^{N}}(\Delta
u)u^{p-1}\mathcal{X}_{R}$
$\displaystyle=\varepsilon\frac{4(p-1)}{p^{2}}\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla
u^{\frac{p}{2}}|^{2}\mathcal{X}_{R}+\varepsilon\int_{0}^{t}\int_{\mathbb{R}^{N}}\nabla(\frac{1}{p}u^{p})\cdot\nabla\mathcal{X}_{R}$
$\displaystyle=\varepsilon\frac{4(p-1)}{p^{2}}\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla
u^{\frac{p}{2}}|^{2}\mathcal{X}_{R}-\varepsilon\int_{0}^{t}\int_{\mathbb{R}^{N}}\frac{1}{p}u^{p}\Delta\mathcal{X}_{R}.$

Adding all the above terms finishes the proof. ∎

The estimate is indeed transferred to the limit $u$ as $\tilde{\varepsilon}\to
0^{+}$.

######  Lemma B.13.

Assume ($\textup{A}_{u_{0}}$), ($\textup{A}_{q}$), and ($\textup{A}_{\nu}$).
Then the unique entropy solution $u$ of ($\textup{P}_{\varepsilon}$) satisfies

$\begin{split}&\frac{1}{p}\int_{\mathbb{R}^{N}}u^{p}(t)\mathcal{X}_{R}+\int_{0}^{t}\mathcal{E}(u(s),u^{p-1}(s)\mathcal{X}_{R})\,{\rm
d}s+\varepsilon\frac{4(p-1)}{p^{2}}\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla
u^{\frac{p}{2}}|^{2}\mathcal{X}_{R}\\\
&\qquad\leq\int_{0}^{t}\int_{\mathbb{R}^{N}}u^{q+p-1}\partial_{x_{N}}\mathcal{X}_{R}+\frac{1}{p}\int_{0}^{t}\int_{\mathbb{R}^{N}}u^{p}\Delta\mathcal{X}_{R}+\frac{1}{p}\int_{\mathbb{R}^{N}}u_{0}^{p}\mathcal{X}_{R},\end{split}$

and moreover, as $R\to\infty$ in the above estimate, we obtain (B.6).

###### Proof.

We have already proved that the mild solution $u_{\tilde{\varepsilon}}$ of
($\textup{P}_{\varepsilon}$) with initial data $u_{0,\tilde{\varepsilon}}$
satisfies (B.7). By Proposition B.11, there exists an entropy solution $u$
with initial data $u_{0}$ such that, as $\tilde{\varepsilon}\to 0^{+}$,

$u_{\tilde{\varepsilon}}\to u\quad\textup{a.e.~{}in $Q$},\qquad\nabla
u_{\tilde{\varepsilon}}^{\frac{p}{2}}\rightharpoonup\nabla
u^{\frac{p}{2}}\quad\textup{in }L_{\textup{loc}}^{2}(Q).$

The first two terms on the left-hand side of (B.7) directly converge when
applying the Fatou lemma and the a.e.-convergence as $\tilde{\varepsilon}\to
0^{+}$. Since $x\mapsto|x|^{2}$ is convex, we have

$\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla
u_{\tilde{\varepsilon}}^{\frac{p}{2}}|^{2}\mathcal{X}_{R}\geq\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla
u^{\frac{p}{2}}|^{2}\mathcal{X}_{R}+2\int_{0}^{t}\int_{\mathbb{R}^{N}}\nabla
u^{\frac{p}{2}}\cdot\big{(}\nabla u_{\tilde{\varepsilon}}^{\frac{p}{2}}-\nabla
u^{\frac{p}{2}}\big{)}\mathcal{X}_{R},$

where the latter term goes to zero by the weak convergence of the gradients in
$L_{\textup{loc}}^{2}(Q)$ as $\tilde{\varepsilon}\to 0^{+}$. The right-hand
side of (B.7) converge through the use of the Lebesgue dominated convergence
theorem.

Recall that $\mathcal{X}_{R}\to 1$ a.e. in $\mathbb{R}^{N}$ as $R\to\infty$.
Hence, the left-hand side of the resulting inequality at the limit
$\tilde{\varepsilon}\to 0^{+}$ converges as $R\to\infty$ by Fatou’s lemma. To
handle the right-hand side, we use that

$\partial_{x_{N}}\mathcal{X}_{R}=\frac{1}{R}\partial_{x_{N}}\mathcal{X},\qquad\Delta\mathcal{X}_{R}=\frac{1}{R^{2}}\Delta\mathcal{X},$

together with the boundedness and integrability of $u$ and $u_{0}$, to obtain
the limit as $R\to\infty$ through Lebesgue’s dominated convergence theorem.

To rewrite the nonlocal term in a convenient form, we use the general Stroock-
Varopoulos inequality (cf. e.g. [18, Lemma 4.10]) and assumption
($\textup{A}_{\nu}$), to get

$\begin{split}\mathcal{E}(u(t),u^{p-1}(t))\geq\frac{4(p-1)}{p^{2}}\int_{\mathbb{R}^{N}}|(\mathcal{L})^{\frac{1}{2}}u^{\frac{p}{2}}(t)|^{2}\geq\Lambda_{1}\frac{4(p-1)}{p^{2}}\int_{\mathbb{R}^{N}}|(-\Delta)^{\alpha/4}u^{\frac{p}{2}}(t)|^{2}.\end{split}$

∎

######  Remark B.14.

When considering ($\textup{P}_{\varepsilon}^{\prime}$) instead of
($\textup{P}_{\varepsilon}$), i.e., $\mathcal{L}^{\prime}$ instead of
$\mathcal{L}$, we need to be careful with the application of the general
Stroock-Varopoulos inequality. Indeed, we can only get control of the seminorm
associated with the space $L^{2}(\mathbb{R};H^{\alpha/2}(\mathbb{R}^{N-1}))$
defined in Appendix C.

###### Proof of Lemma 3.3(c).

Since $u$ is an entropy solution, it is then also a very weak solution (cf.
Theorem B.7(a)(v)). By the energy estimate (B.6) with $p=2$, for all $\phi\in
C_{\textup{c}}^{\infty}(Q)$,

$\begin{split}\iint_{Q}\Big{(}u\partial_{t}\phi+u^{q}\partial_{x_{N}}\phi-\varepsilon\nabla
u\cdot\nabla\phi\Big{)}+\int_{0}^{\infty}\mathcal{E}(u(s),\phi(s))\,{\rm
d}s=0.\end{split}$

Let us choose a smooth bounded domain $\Omega$ and $\phi$ supported in
$(0,T)\times\Omega$. When $q\geq 1/2$ we have that $u^{q}\in
L^{2}(\mathbb{R}^{N})$ and then the regularity of $\partial_{t}$. When $q<1/2$
we choose a large $p$ such that $p\geq 1/q\geq 2\geq p^{\prime}$ and again we
obtain the desired estimate. ∎

###  B.5. Compactness in
$C([0,\infty);L_{\textnormal{{loc}}}^{p}(\mathbb{R}^{N}))$

The initial data $u_{0}$ is assumed to satisfy

(B.8) $0\leq u_{0}\in
L^{\infty}(\mathbb{R}^{N}),\qquad\int_{\mathbb{R}^{N}}|u_{0}(x+\xi)-u_{0}(x)|\,{\rm
d}x\to 0\quad\textup{as }\xi\to 0^{+}.$

With the previous results in mind, it is known that a combination of the
Arzelà-Ascoli and Fréchet-Kolmogorov-Riesz compactness theorems gives
convergence in $C([0,\infty);L_{\textnormal{{loc}}}^{1}(\mathbb{R}^{N}))$ as
long as a uniform control for the time- and space-translations are given on
compact sets.

Due to the $L^{1}$-contraction and the translation invariance of
($\textup{P}_{\varepsilon}$), space-translations are easily controlled: for
all compact $K\subset\mathbb{R}^{N}$, and a.e. $t>0$,

$\begin{split}\int_{K}|u(t,x+\xi)-u(t,x)|\,{\rm
d}x\leq\int_{\mathbb{R}^{N}}|u_{0}(x+\xi)-u_{0}(x)|\,{\rm d}x.\end{split}$

It turns out that the $L^{1}$-contraction also provides control of the time-
translations:

######  Proposition B.15.

Assume $0<\varepsilon\leq 1$, ($\textup{A}_{q}$), and ($\textup{A}_{\nu}$).
Let $u$ be an entropy solution of ($\textup{P}_{\varepsilon}$) with initial
data $u_{0}$ satisfying (B.8). Then, for all compact $K\subset\mathbb{R}^{N}$
and all $t,s\in[0,T]$,

$\int_{K}|u(t)-u(s)|\leq\lambda(|t-s|^{\frac{1}{3}}),$

where $\lambda$ is a modulus of continuity depending on $K$,
$\|u_{0}\|_{L^{\infty}(\mathbb{R}^{N})}$, and
$\|u_{0}(\cdot+\xi)-u_{0}\|_{L^{1}(\mathbb{R}^{N})}$.

###### Proof.

Assume $s<t$. Define $u_{\delta}(t,x):=u(t,\cdot)\ast\omega_{\delta}(x)$,
where $\omega_{\delta}$ is a standard mollifier. By the triangle inequality
and the $L^{1}$-contraction,

$\begin{split}\|u(t)-u(s)\|_{L^{1}(K)}&\leq\|u(t)-u_{\delta}(t)\|_{L^{1}(K)}+\|u_{\delta}(t)-u_{\delta}(s)\|_{L^{1}(K)}+\|u_{\delta}(s)-u(s)\|_{L^{1}(K)}\\\
&\leq
2\sup_{|\xi|\leq\delta}\|u_{0}(\cdot+\xi)-u_{0}\|_{L^{1}(\mathbb{R}^{N})}+\|u_{\delta}(t)-u_{\delta}(s)\|_{L^{1}(K)}.\end{split}$

Now, it is standard to choose
$(\tau,y)\mapsto\mathds{1}_{[s,t]}(\tau)\omega_{\delta}(x-y)$ as a test
function in the very weak formulation of ($\textup{P}_{\varepsilon}$) to
obtain

$\begin{split}\|u_{\delta}(t)-u_{\delta}(s)\|_{L^{1}(K)}&\leq\int_{K}\int_{s}^{t}\big{|}u(\tau)^{q}\ast\partial_{x_{N}}\omega_{\delta}-u(\tau)\ast(\mathcal{L}\omega_{\delta})+\varepsilon
u(\tau)\ast\Delta\omega_{\delta}\big{|}\,{\rm d}\tau\\\
&\leq|t-s||K|\big{(}\|u_{0}\|_{L^{\infty}}^{q}\|\partial_{x_{N}}\omega_{\delta}\|_{L^{1}}+\|u_{0}\|_{L^{\infty}}\|\mathcal{L}\omega_{\delta}\|_{L^{1}}+\varepsilon\|u_{0}\|_{L^{\infty}}\|\Delta\omega_{\delta}\|_{L^{1}}\big{)}.\end{split}$

Since

$\|\partial_{x_{N}}\omega_{\delta}\|_{L^{1}}\lesssim\delta^{-1},\quad\|\mathcal{L}\omega_{\delta}\|_{L^{1}}\lesssim\delta^{-\alpha},\quad\textup{and
}\|\Delta\omega_{\delta}\|_{L^{1}}\lesssim\delta^{-2},$

we can choose $\delta^{2}=|t-s|^{\frac{2}{3}}$ to obtain the desired result. ∎

In view of the previous discussion on the application of Arzelà-Ascoli and
Fréchet-Kolmogorov-Riesz compactness theorems, we obtain Lemma 3.4.

###  B.6. Compactness in $C([0,\infty);L^{p}(\mathbb{R}^{N}))$

This follows from a tail estimate; see e.g. [19]. Hence we start by proving
Lemma 3.3(d).

###### Proof of Lemma 3.3(d).

We take (after a standard approximation procedure)
$\mathds{1}_{[0,\tau]}(t)\rho_{R}(x)$ as a test function in the very weak
formulation of ($\textup{P}_{\varepsilon}$), where $\rho_{R}$ is the tail-
control function defined in Section 1.7. Then,

$\begin{split}\int_{\mathbb{R}^{N}}u(\tau)\rho_{R}=\int_{\mathbb{R}^{N}}u_{0}\rho_{R}+\int_{0}^{\tau}\int_{\mathbb{R}^{N}}\big{(}F(u)\cdot\nabla\mathcal{\rho}_{R}-u(\mathcal{L}\mathcal{\rho}_{R})+\varepsilon
u\Delta\mathcal{\rho}_{R}\big{)}.\end{split}$

Using that $\mathds{1}_{|x|>2R}\leq\rho_{R}(x)\leq\mathds{1}_{|x|>R}$ and
$u,u_{0}\geq 0$, we get

$\begin{split}\int_{|x|>2R}u(t)\leq\int_{|x|>R}u_{0}+\int_{0}^{\tau}\int_{\mathbb{R}^{N}}\big{(}u^{q}|\partial_{x_{N}}\rho_{R}|+u|\mathcal{L}\rho_{R}|+\varepsilon
u|\Delta\rho_{R}|\big{)}.\end{split}$

Since $\|\Delta\rho_{R}\|_{L^{\infty}(\mathbb{R}^{N})}\lesssim R^{-2}$, the
result follows. ∎

This preliminary tail control provided by Lemma 3.3(d) is not enough for our
purposes, since it is not uniform in time. In order to improve it we use the
$\alpha$-homogeneity of the operator $\mathcal{L}$, that yields
$\|\mathcal{L}\rho_{R}\|_{L^{\infty}(\mathbb{R}^{N})}\lesssim R^{-\alpha}$,
and observe also that

(B.9)
$\int_{0}^{\tau}\int_{\mathbb{R}^{N}}u^{q}|\partial_{x_{N}}\rho_{R}|\lesssim\begin{cases}\tau\|u_{0}\|_{L^{\infty}(\mathbb{R}^{N})}^{q-1}\|u_{0}\|_{L^{1}(\mathbb{R}^{N})}R^{-1}&\textup{when
}q>1,\\\ \tau\|u_{0}\|_{L^{1}(\mathbb{R}^{N})}^{q}R^{-(1-N(1-q))}&\textup{when
}1-\frac{1}{N}<q<1.\end{cases}$

In the second estimate we used Hölder’s inequality. Tail control is thus
established uniformly in $\varepsilon$ in the interval $[0,\tau]$ for all
finite $\tau>0$, which is enough to prove Lemmas 4.3 and 4.5.

######  Remark B.16.

Since the sequences above are always uniformly bounded, the mentioned
convergence results are actually in
$C([0,\infty);L_{\textup{loc}}^{p}(\mathbb{R}^{N}))$ and
$C([0,\infty);L^{p}(\mathbb{R}^{N}))$, respectively, for all $p\in[1,\infty)$.

###  B.7. Primitives of entropy solutions solve parabolic problems

We prove now Lemma 3.7, needed for the proofs of Theorem 1.5 and Lemma 3.8.

###### Proof of Lemma 3.7.

Since $u$ is an entropy solution of ($\textup{P}_{\varepsilon}$),

$\iint_{Q}\Big{(}u\partial_{t}\phi+u^{q}\partial_{x_{N}}\phi-u(\mathcal{L}\phi+\varepsilon\Delta\phi)\Big{)}+\int_{\mathbb{R}^{N}}u_{0}\phi(0)=0\quad\text{for
all }\phi\in C_{\textup{c}}^{\infty}(\overline{Q}).$

We then choose
$\phi_{R}(t,x^{\prime},x_{N})=\psi(t,x^{\prime})\mathcal{X}_{R}(x_{N})$ with
$\psi\in C_{\textup{c}}^{\infty}([0,\infty)\times\mathbb{R}^{N-1})$ and
$\mathcal{X}_{R}$ the usual cut-off function. Since $u\in
C([0,\infty);L^{1}(\mathbb{R}^{N}))\cap L^{\infty}(Q)$, the term involving
$u^{q}$ can be estimated similarly as in (B.9), and we may apply the dominated
convergence theorem to obtain

$\displaystyle\lim_{R\to\infty}\Big{(}\iint_{Q}\big{(}u\partial_{t}\phi_{R}+u^{q}\partial_{x_{N}}\phi_{R}-u(\mathcal{L}\phi_{R}+\varepsilon\Delta\phi_{R})\big{)}+\int_{\mathbb{R}^{N}}u_{0}\phi_{R}(0)\Big{)}$
$\displaystyle\qquad=\int_{0}^{T}\int_{\mathbb{R}^{N-1}}\big{(}v\partial_{t}\psi-v(\widetilde{\mathcal{L}}\psi+\varepsilon\Delta_{x^{\prime}}\psi)\big{)}+\int_{\mathbb{R}^{N-1}}v_{0}\psi(0),$

since (see Remark 2.3)

$\lim_{R\rightarrow\infty}\mathcal{L}(\psi\mathcal{X}_{R})(x^{\prime},x_{N})=\mathcal{L}(\psi(\cdot^{\prime})\mathds{1}_{\mathbb{R}}(\cdot_{N}))(x^{\prime},x_{N})=(\widetilde{\mathcal{L}}\psi)(x^{\prime})\mathds{1}_{\mathbb{R}}(x_{N})\quad\text{for
any }\psi\in C_{\textup{c}}^{\infty}(\mathbb{R}^{N-1}).$

∎

###  B.8. Parabolic regularization when $q>1$

We include a variant of Lemma 3.1, which will be useful in the proof of
Theorem 1.5.

######  Proposition B.17 (Parabolic regularization).

Assume ($\textup{A}_{u_{0}}$), ($\textup{A}_{q}$) with $q>1$, and
($\textup{A}_{\nu}$). Let $u$ be the entropy solution of
($\textup{P}_{\varepsilon}$)/($\textup{P}_{\varepsilon}^{\prime}$) with
initial data $u_{0}$. Then $u\in C^{1+\delta,2+2\delta}(Q)$ for some
$\delta>0$, i.e., $u$ is a classical solution of the PDE in
($\textup{P}_{\varepsilon}$)/($\textup{P}_{\varepsilon}^{\prime}$).

###### Proof.

In the first step we prove that $u\in
L^{p}_{\textup{loc}}((0,\infty);H^{2,p}(\mathbb{R}^{N}))$ and
$\partial_{t}u\in L^{p}_{\textup{loc}}((0,\infty);L^{p}(\mathbb{R}^{N}))$ for
all $p\in(1,\infty)$. Let us first observe that $u\in L^{\infty}(Q)\cap
C([0,\infty);L^{1}(\mathbb{R}^{N}))$ is a distributional solution of

$\partial_{t}u-\varepsilon\Delta u=g$

with $g=\mathcal{L}u+\partial_{x_{N}}(f(u))$. In particular, if $f(u)=u^{q}$
we have that $f(u)\in L^{p}((T^{\prime},T);L^{p}(\mathbb{R}^{N}))$ for all
$p>\max\\{1/q,1\\}$. This implies that $\partial_{x_{N}}(f(u))\in
H^{-1,p}(\mathbb{R}^{N})$. Also, in view of (2.8) and (2.9) we have
$\mathcal{L}u,\mathcal{L}^{\prime}u\in
L^{p}((T^{\prime},T);H^{-\alpha,p}(\mathbb{R}^{N}))$, hence $g\in
L^{p}((T^{\prime},T);H^{-\max\\{\alpha,1\\},p}(\mathbb{R}^{N}))$. Regularity
results for the heat equation [46] give us that $u\in
L^{p}((T^{\prime},T);H^{2-\max\\{\alpha,1\\},p}(\mathbb{R}^{N}))$.

Assume that $u\in L^{p}((T^{\prime},T);H^{\sigma,p}(\mathbb{R}^{N}))$ for some
$\sigma\in(0,1]$. The fractional chain rule [63, Proposition 5.1, p. 112]
gives us that

$\|f(u)\|_{H^{\sigma,p}(\mathbb{R}^{N})}\lesssim\|f^{\prime}(u)\|_{L^{\infty}(\mathbb{R}^{N})}\|u\|_{H^{\sigma,p}(\mathbb{R}^{N})}.$

Since $u\in L^{\infty}((0,\infty);L^{\infty}(\mathbb{R}^{N}))$ and
$f(u)=u^{q}$ with $q>1$, we obtain that

$f(u)\in
L^{p}((T^{\prime},T);H^{\sigma,p}(\mathbb{R}^{N})),\qquad\partial_{x_{N}}(f(u))\in
L^{p}((T^{\prime},T);H^{\sigma-1,p}(\mathbb{R}^{N}))$

and then the right-hand side term $g\in
L^{p}((T^{\prime},T);H^{\sigma-\max\\{\alpha,1\\},p}(\mathbb{R}^{N}))$. We set
$\sigma_{0}=2-\max\\{\alpha,1\\}>0$. As long as $\sigma_{n}\leq 1$ we define
$\sigma_{n+1}=2+\sigma_{n}-\max\\{1,\alpha\\}$ and thus $u\in
L^{p}((T^{\prime},T);H^{\sigma_{n+1},p}(\mathbb{R}^{N}))$. Since the sequence
$(\sigma_{n})_{n\geq 0}$ is increasing we have that at some point
$\sigma_{n}>1$, hence $u\in L^{p}((T^{\prime},T);H^{1,p}(\mathbb{R}^{N}))$.

If $\alpha\in(0,1]$ we repeat the above argument with $\sigma=1$ and we obtain
that $u\in L^{p}((T^{\prime},T);H^{2,p}(\mathbb{R}^{N}))$ and then
$\partial_{t}u\in L^{p}_{\textup{loc}}((T^{\prime},T);L^{p}(\mathbb{R}^{N}))$.

Let us consider now the case when $\alpha\in(1,2)$. We know that $u\in
L^{p}((T^{\prime},T);H^{\sigma,p}(\mathbb{R}^{N}))$ for some $\sigma\geq 1$.
It implies that $\partial_{x_{N}}(f(u))\in
L^{p}((T^{\prime},T);L^{p}(\mathbb{R}^{N}))$, $\mathcal{L}u\in
L^{p}((T^{\prime},T);H^{\sigma-\alpha,p}(\mathbb{R}^{N}))$ and $g$ belongs to
$L^{p}((T^{\prime},T);H^{\min\\{\sigma-\alpha,0\\},p}(\mathbb{R}^{N}))$.
Defining $\sigma_{n+1}=2+\min\\{\sigma_{n},\alpha,0\\}$ we obtain that $u\in
L^{p}((T^{\prime},T);H^{\sigma_{n+1},p}(\mathbb{R}^{N}))$. We remark that at
some moment $\sigma_{n}>\alpha$ which implies $\sigma_{n+1}=2$ and we obtain
the desired property.

In the second step we will show that for any $0<T^{\prime}<T$, the function
$g$ is belongs to $C^{\delta,2\delta}((T^{\prime},T)\times\mathbb{R}^{N})$ for
some $\delta>0$, where $C^{\delta,2\delta}$ are the parabolic Hölder spaces
[50, Section 8.5, p. 117]. Regularity results for the inhomogeneous heat
equation [50, Theorem 8.7.3, p. 123] applied to $\tilde{u}=\chi(t)u(t)$ where
$\chi(t)$ is a smooth cut-off function shows that $u\in
C^{1+\delta,2+2\delta}((T^{\prime},T)\times\mathbb{R}^{N})$, so $u$ is a
classical solution.

In the following we will show that $g=\mathcal{L}u+\partial_{x_{N}}(f(u))$
belongs to $C^{\delta,2\delta}((T^{\prime},T)\times\mathbb{R}^{N})$ for some
positive $\delta$. Let us recall that $u\in
L^{p}((T^{\prime},T);H^{2,p}(\mathbb{R}^{N}))$ with $\partial_{t}u\in
L^{p}((T^{\prime},T);L^{p}(\mathbb{R}^{N}))$ for all $1<p<\infty$, i.e.

$u\in L^{p}((T^{\prime},T);H^{2,p}(\mathbb{R}^{N}))\cap
W^{1,p}((T^{\prime},T);L^{p}(\mathbb{R}^{N})).$

Let us show that the nonlocal term has the required regularity. Denoting
$v=\mathcal{L}u$ or $v=\mathcal{L}^{\prime}u$ we have that $v\in
L^{p}((T^{\prime},T);H^{2-\alpha,p}(\mathbb{R}^{N}))\cap
W^{1,p}((T^{\prime},T);H^{-\alpha,p}(\mathbb{R}^{N}))$. Therefore,

$(I-\Delta)^{-\alpha/2}v\in L^{p}((T^{\prime},T);H^{2,p}(\mathbb{R}^{N}))\cap
W^{1,p}((T^{\prime},T);L^{p}(\mathbb{R}^{N})).$

Using the interpolation results for Sobolev spaces in [56, Proposition 3.2] to
$(I-\Delta)^{-\alpha/2}v$, we get

$(I-\Delta)^{-\alpha/2}v\in
W^{s,p}((T^{\prime},T);H^{2(1-s),p}(\mathbb{R}^{N}))\quad\text{for all
}s\in(0,1)$

and then $v\in W^{s,p}((T^{\prime},T);H^{2(1-s)-\alpha,p}(\mathbb{R}^{N}))$.
We emphasize that we have to apply the cited results to
$(I-\Delta)^{-\alpha/2}v$ and not directly to $v$ since the results in [56]
hold for nonnegative index Bessel potential spaces. The classical Morrey
estimates for Bessel potential and Sobolev spaces give us that $v\in
C^{\delta}((T^{\prime},T);C^{2\delta}(\mathbb{R}^{N}))\subset
C^{\delta,2\delta}((T^{\prime},T)\times\mathbb{R}^{N})$ provided
$s-1/p\geq\delta$ and $2(1-s)-\alpha-N/p\geq 2\delta$. Choosing
$s=(2-\alpha)/4$ and $p$ large enough the two inequalities hold for any
$\delta<(2-\alpha)/4$.

Let us analyze the nonlinear term
$v=\partial_{x_{N}}(f(u))=\frac{1}{q}u^{q-1}\partial_{x_{N}}u$. When $q\geq 2$
it is easy to differentiate once in time and space and using again
interpolation and Morrey’s inequality, we have the required regularity for
$v$. Let us consider now the case $q\in(1,2)$. The regularity of $u$ and the
space interpolation give us that for any $s\in(0,1)$,

$u\in W^{s,p}((T^{\prime},T);H^{2(1-s),p}(\mathbb{R}^{N}))\subset
C^{s-1/N}((T^{\prime},T);C^{2(1-s)-N/p}(\mathbb{R}^{N})).$

Choosing $s=1/2$ we obtain that $u\in
C^{\delta,2\delta}((T^{\prime},T)\times\mathbb{R}^{N})$ for all $\delta<1/2$.
This implies that

$u^{q-1}\in
C^{\delta(q-1),2\delta(q-1)}((T^{\prime},T)\times\mathbb{R}^{N})\quad\text{for
all }\delta<1/2.$

Regularity of $u$ implies that

$\displaystyle\partial_{x_{N}}u$ $\displaystyle\in
L^{p}((T^{\prime},T);H^{1,p}(\mathbb{R}^{N})\cap
W^{1,p}((T^{\prime},T);H^{-1,p}(\mathbb{R}^{N}))$ $\displaystyle\subset
W^{s,p}((T^{\prime},T);H^{-s+(1-s),p}(\mathbb{R}^{N})\subset
C^{s-p/N}((T^{\prime},T);C^{1-2s-N/p}(\mathbb{R}^{N})).$

Choosing $s=1/4$ we obtain that $\partial_{x_{N}}u\in
C^{\delta,2\delta}((T^{\prime},T)\times\mathbb{R}^{N})$ for all $\delta<1/4$.
This implies that $u^{q-1}\partial_{x_{N}}u\in
C^{\delta(q-1),2\delta(q-1)}((T^{\prime},T)\times\mathbb{R}^{N})$ for all
$\delta<1/4$. ∎

##  Appendix C Compact embedding of mixed spaces

We gather here two compactness results that are used in the proof of our
convergence result.

######  Theorem C.1 (Aubin-Lions-Simon, [62, Theorem 5]).

Consider the three Banach spaces

$X\hookrightarrow B\hookrightarrow Y,$

where $X\hookrightarrow B$ is compact. Assume $1\leq p\leq\infty$ and:

  1. (i)

$\mathcal{F}\subset L^{p}((0,T);X)$ is bounded.

  2. (ii)

$\textstyle{\int_{0}^{T-h}\|f(t+h)-f(t)\|_{Y}\,{\rm d}t\to 0}$ as $h\to 0^{+}$
uniformly for $f\in\mathcal{F}$.

Then $\mathcal{F}$ is relatively compact in $L^{p}((0,T);B)$ (and in
$C([0,T];B)$ if $p=\infty$).

Assume that $\alpha\in(0,2)$ and
$x=(x^{\prime},x_{N})\in\mathbb{R}^{N-1}\times\mathbb{R}$, and define the
Hilbert space

$\displaystyle L^{2}(\mathbb{R};W^{\alpha/2,2}(\mathbb{R}^{N-1}))$
$\displaystyle:=\\{\phi\in
L^{2}(\mathbb{R}^{N})\,:\,|\phi|_{L^{2}(\mathbb{R};W^{\alpha/2,2}(\mathbb{R}^{N-1}))}<\infty\\},\quad\text{where
}$ $\displaystyle|\phi|_{L^{2}(\mathbb{R};W^{\alpha/2,2}(\mathbb{R}^{N-1}))}$
$\displaystyle:=\bigg{(}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N-1}}\frac{|\phi(x^{\prime}+z^{\prime},x_{N})-\phi(x^{\prime},x_{N})|^{2}}{|z^{\prime}|^{(N-1)+\alpha}}\,{\rm
d}z^{\prime}{\rm d}(x^{\prime},x_{N})\bigg{)}^{\frac{1}{2}},$

and moreover,
$\|\phi\|_{L^{2}(\mathbb{R};W^{\alpha/2,2}(\mathbb{R}^{N-1}))}=\|\phi\|_{L^{2}(\mathbb{R}^{N})}+|\phi|_{L^{2}(\mathbb{R};W^{\alpha/2,2}(\mathbb{R}^{N-1}))}$.
It is classical that the above spaces are the same as the Bessel potential
spaces, i.e.
$W^{\alpha/2,2}(\mathbb{R}^{N-1})=H^{\alpha/2}(\mathbb{R}^{N-1})$.

######  Lemma C.2.

Assume $\alpha\in(0,2)$, $\beta>0$, $R^{\prime},R>0$ and a set $\mathcal{F}$
such that:

  1. (i)

$\|\phi\|_{L^{2}(\mathbb{R};W^{\alpha/2,2}(\mathbb{R}^{N-1}))}+\|\phi\|_{L^{\infty}(\mathbb{R}^{N})}\leq
C$ for some $C$ independent of $\phi\in\mathcal{F}$.

  2. (ii)

For all $|h_{N}|\leq R$ and some $\beta>0$,

$\int_{|x^{\prime}|\leq R^{\prime},|x_{N}|\leq
R}|\phi(x^{\prime},x_{N}+h_{N})-\phi(x^{\prime},x_{N})|^{2}\,{\rm d}x\leq
C|h_{N}|^{\beta}$

for some constant C independent of $\phi\in\mathcal{F}$.

Then $\mathcal{F}$ is compact in $L^{2}(B_{R^{\prime}}\times(-R,R))$.

######  Remark C.3.

For any $R,R^{\prime}>0$ the space $X_{R,R^{\prime}}\subset
L^{\infty}(\mathbb{R}^{N})\cap
L^{2}(\mathbb{R};W^{\alpha/2,2}(\mathbb{R}^{N-1}))$ endowed with the norm

$\|\psi\|_{X_{R,R^{\prime}}}=\|\phi\|_{L^{2}(\mathbb{R};W^{\alpha/2,2}(\mathbb{R}^{N-1}))}+\|\phi\|_{L^{\infty}(\mathbb{R}^{N})}+\sup_{|h_{N}|<R}\int_{|x^{\prime}|\leq
R^{\prime},|x_{N}|\leq
R}\frac{|\phi(x^{\prime},x_{N}+h_{N})-\phi(x^{\prime},x_{N})|}{|h_{N}|}\,{\rm
d}x$

is compactly embedded in $L^{2}(B_{R^{\prime}}\times(-R,R))$.

###### Proof.

We will employ the Fréchet-Kolmogorov-Riesz compactness theorem (see e.g.
[35]), and hence, we need uniform estimates on translations in $L^{2}$.

By the triangle inequality,

$\begin{split}|\phi(x+h)-\phi(x)|^{2}&=|\phi(x+(h^{\prime},h_{N}))-\phi(x)|^{2}\\\
&\leq|\phi(x+(h^{\prime},h_{N}))-\phi(x+(0,h_{N}))|^{2}+|\phi(x+(0,h_{N}))-\phi(x)|^{2}.\end{split}$

Integrating the above inequality and using the change of variables
$x+(0,h_{N})\mapsto x$ gives

(C.1) $\begin{gathered}\int_{|x^{\prime}|\leq R^{\prime},|x_{N}|\leq
R}|\phi(x+h)-\phi(x)|^{2}\,{\rm d}x\\\
\leq\int_{\mathbb{R}^{N}}|\phi(x+(h^{\prime},0))-\phi(x)|^{2}\,{\rm
d}x+\int_{|x^{\prime}|\leq R^{\prime},|x_{N}|\leq
R}|\phi(x+(0,h_{N}))-\phi(x)|^{2}\,{\rm d}x.\end{gathered}$

The second term can already be estimated by assumption (ii), while the first
term needs further consideration. Using Plancherel’s identity and the
inequality

$|e^{\xi^{\prime}h^{\prime}}-1|\leq\min\\{2,|\xi^{\prime}h^{\prime}|\\}\leq
C_{\alpha}|\xi^{\prime}h^{\prime}|^{\alpha/2}\leq
C_{\alpha}|\xi^{\prime}|^{\alpha/2}|h^{\prime}|^{\alpha/2}$

we get

$\displaystyle\int_{\mathbb{R}^{N}}|\phi(x+(h^{\prime},0))-\phi(x)|^{2}\,{\rm
d}x$
$\displaystyle=\int_{\mathbb{R}^{N}}|e^{\xi^{\prime}h^{\prime}}-1|^{2}|\widehat{\phi}(\xi)|^{2}\,{\rm
d}\xi\leq
C_{\alpha}|h^{\prime}|^{\alpha}\int_{\mathbb{R}^{N}}|\xi^{\prime}|^{\alpha}|\widehat{\phi}(\xi)|^{2}\,{\rm
d}\xi$
$\displaystyle=C_{\alpha}|h^{\prime}|^{\alpha}|\phi|_{L^{2}(\mathbb{R};W^{\alpha/2,2}(\mathbb{R}^{N-1}))}^{2}.$

Going back to (C.1), we therefore get

(C.2) $\int_{|x^{\prime}|\leq R^{\prime},|x_{N}|\leq
R}|\phi(x^{\prime}+h^{\prime},x_{N}+h_{N})-\phi(x^{\prime},x_{N})|^{2}\,{\rm
d}(x^{\prime},x_{N})\leq
C\big{(}|h^{\prime}|^{\alpha}+|h_{N}|^{\beta}\big{)},$

where $C$ is again independent of $\phi$ by assumption (ii).

Now, define

$\psi(x):=\phi(x)\mathds{1}_{B_{R^{\prime}}\times(-R,R)}(x)$

In order to apply the Fréchet-Kolmogorov-Riesz compactness theorem, we need to
check that the family $\\{\psi\,:\,\psi\in L^{2}(\mathbb{R}^{N})\\}\subset
L^{2}(\mathbb{R}^{N})$ is equicontinuous and equitight. The latter is
automatically satisfied in
$\mathbb{R}^{N}\setminus(B_{R^{\prime}}\times(-R,R))$, and we are left with
estimating the translations, for $h\in\mathbb{R}^{N-1}\times(-R,R)$:

$\begin{split}\big{\|}\psi(\cdot+h)-\psi\big{\|}_{L^{2}(\mathbb{R}^{N})}&\leq\big{\|}\phi(\cdot+h)\big{(}\mathds{1}_{B_{R^{\prime}}\times(-R,R)}(\cdot+h)-\mathds{1}_{B_{R^{\prime}}\times(-R,R)}\big{)}\big{\|}_{L^{2}(\mathbb{R}^{N})}\\\
&\quad+\big{\|}\mathds{1}_{B_{R^{\prime}}\times(-R,R)}\big{(}\phi(\cdot+h)-\phi\big{)}\big{\|}_{L^{2}(\mathbb{R}^{N})}\\\
&\leq\big{\|}\phi\big{\|}_{L^{\infty}(\mathbb{R}^{N})}\big{\|}\mathds{1}_{B_{R^{\prime}}\times(-R,R)}(\cdot+h)-\mathds{1}_{B_{R^{\prime}}\times(-R,R)}\big{\|}_{L^{2}(\mathbb{R}^{N})}\\\
&\quad+\big{\|}\phi(\cdot+h)-\phi\big{\|}_{L^{2}(B_{R^{\prime}}\times(-R,R))}\end{split}$

By assumption (i), $\|\phi\|_{L^{\infty}(\mathbb{R}^{N})}$ is uniformly
bounded, and hence, the first term goes to zero as $h\to 0^{+}$ uniformly.
Moreover, the second term does so too since (C.2) holds. Hence, there is a
convergent subsequence in $L^{2}(B_{R^{\prime}}\times(-R,R))$. ∎

##  Appendix D An auxiliary lemma

This appendix is devoted to extend to the case of integrable and locally
bounded functions a result that was already proved in [29, Lemma 1.2] for
smooth functions and in [53, Lemma A.1], [51, Lemma 2.7] for continuous and
integrable ones. This extension is used in the course of the proof of Lemma
3.8 to obtain the hyperbolic estimates when $q\in(0,1)$.

######  Lemma D.1.

Consider a nonnegative function $f\in L^{1}(\mathbb{R})\cap L^{\infty}_{\rm
loc}(\mathbb{R})$ satisfying

(D.1) $(f^{k})^{\prime}\leq
C\quad\text{or}\quad(f^{k})^{\prime}\geq-C\quad\text{in
}\mathcal{D}^{\prime}(\mathbb{R})$

for some real numbers $k>0$ and $C>0$. Then

$0\leq
f(x)\leq\Big{(}\frac{C(k+1)}{k}\|f\|_{L^{1}(\mathbb{R})}\Big{)}^{1/(k+1)}\
a.e.\ x\in\mathbb{R}.$

###### Proof.

It is enough to prove the first case; the second one can be reduced to it by
taking $\tilde{f}(x)=f(-x)$.

Let us first prove that a function $g\in L_{\rm loc}^{1}(\mathbb{R})$ such
that $g^{\prime}\leq C$ in $\mathcal{D}^{\prime}(\mathbb{R})$ satisfies

(D.2) $g(y)-g(x)\leq C(y-x)\quad\text{a.e. }x<y\in\mathbb{R}.$

By density, in the inequality

$-\int_{\mathbb{R}}g\varphi^{\prime}\leq
C\int_{\mathbb{R}}\varphi\quad\text{for all }\varphi\in C^{1}_{\rm
c}(\mathbb{R}),\ \varphi\geq 0,$

we can use nonnegative and compactly supported test functions $\varphi$ which
are only _piecewise_ $C^{1}$. Fix two points $x<y$ and a function
$\varphi_{\varepsilon}$ such that $\varphi_{\varepsilon}=1$ in
$(x+\varepsilon,y-\varepsilon)$, linear on
$(x-\varepsilon,x+\varepsilon)\cup(y-\varepsilon,y+\varepsilon)$ and vanishing
identically outside the interval $(x-\varepsilon,y+\varepsilon)$. Then

$-\frac{1}{2\varepsilon}\int_{x-\varepsilon}^{x+\varepsilon}g+\frac{1}{2\varepsilon}\int_{y-\varepsilon}^{y+\varepsilon}g\leq
C\int_{x-\varepsilon}^{y+\varepsilon}\varphi_{\varepsilon}=C(y-x).$

Since the function $g$ is locally integrable, Lebesgue differentiation theorem
guarantees that for a.e. $x$ and $y$ we can let $\varepsilon\rightarrow 0$ in
the above inequality to obtain (D.2).

Let us now go back to function $f$. Since $f\in L^{\infty}_{\rm
loc}(\mathbb{R})$, then $f^{k}\in L^{1}_{\rm loc}(\mathbb{R})$. Hence, we may
apply the preliminary result to $g=f^{k}$ to obtain, using the inequality
(D.1),

$f^{k}(y)-f^{k}(x)\leq C(y-x)\quad\text{a.e. }x<y\in\mathbb{R}.$

Let us fix a point $y$ for which the above inequality holds. Then

$f^{k}(x)\geq C(x-\bar{y})\quad\text{a.e. }x<y,\quad\text{where
}\bar{y}:=y-\frac{f^{k}(y)}{C}.$

We use this inequality on the interval $(\bar{y},y)$ to get, since $f$ is
nonnegative,

$\displaystyle\|f\|_{L^{1}(\mathbb{R})}$
$\displaystyle=\int_{\mathbb{R}}f\geq\int_{\bar{y}}^{y}f\geq
C^{1/k}\int_{\bar{y}}^{y}(x-\bar{y})^{1/k}\,{\rm
d}x=C^{1/k}\frac{k}{k+1}(y-\bar{y})^{1+1/k}=\frac{k}{C(k+1)}f^{k+1}(y),$

from where the result follows immediately. ∎

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