On the rate of convergence in homogenization of time-fractional Hamilton–Jacobi equations

Abstract

Here, we consider periodic homogenization for time-fractional Hamilton–Jacobi equations. By using the perturbed test function method, we establish the convergence, and give estimates on a rate of convergence. A main difficulty is the incompatibility between the function used in the doubling variable method, and the non-locality of the Caputo derivative. Our approach is to provide a lemma to prove the rate of convergence without the doubling variable method with respect to the time variable, which is a key ingredient.

1 Introduction

Let $T > 0$ and $\alpha \in (0, 1)$ be given constants and $\varepsilon > 0$ be a parameter. We are concerned with time-fractional Hamilton–Jacobi equations:

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \partial _{t}^{\alpha }u^{\varepsilon } + H \left( \frac{x}{\varepsilon }, Du^{\varepsilon } \right) \ = 0 &{} \text{ in } \quad {\mathbb {R}}^N \times (0,T), \\ u^\varepsilon (x,0) = u_{0}(x) &{} \text{ on } \quad {\mathbb {R}}^N. \end{array} \right. \end{aligned}$$

(1.1)

Here, $u^\varepsilon :{\mathbb {R}}^N \times [0,T] \rightarrow {\mathbb {R}}$ is an unknown function. The Hamiltonian $H:{\mathbb {R}}^N \times {\mathbb {R}}^N \rightarrow {\mathbb {R}}$, the initial function $u_0:{\mathbb {R}}^N \rightarrow {\mathbb {R}}$ are given continuous functions, which always satisfy

• (A1) The function H is uniformly coercive in the y-variable, i.e.,

  $$\begin{aligned} \displaystyle \lim _{r \rightarrow \infty } \inf \{ H(y, p) \mid y \in {\mathbb {R}}^N, |p| \ge r \} = \infty . \end{aligned}$$

• (A2) The function $y \mapsto H(y, p)$ is ${\mathbb {Z}}^N$-periodic, i.e.,

  $$\begin{aligned} H(y, p) = H(y+k, p) \quad \text{ for } \, \, y, p \in {\mathbb {R}}^N, k \in {\mathbb {Z}}^N. \end{aligned}$$

• (A3) $u_0 \in \mathrm{Lip\,}({\mathbb {R}}^N) \cap \mathrm{BUC\,}({\mathbb {R}}^N)$, where we denote by $\mathrm{Lip\,}({\mathbb {R}}^N)$ and $\mathrm{BUC\,}({\mathbb {R}}^N)$, respectively, the sets of all Lipschitz continuous functions, and bounded, uniformly continuous functions on ${\mathbb {R}}^N$.

Moreover, we denote by $\partial _{t}^{\alpha }u^\varepsilon$ the Caputo fractional derivative of $u^\varepsilon$ with respect to t, that is,

$$\begin{aligned} \displaystyle \partial _{t}^{\alpha }u^\varepsilon (x,t) \mathrel {\mathop :}=\frac{1}{\Gamma (1 - \alpha )} \int _0^{t} (t-s)^{-\alpha } \partial _s u^\varepsilon (x,s) \,ds \end{aligned}$$

for all $(x, t)\in {\mathbb {R}}^N \times (0,T)$, where $\Gamma$ is the Gamma function. If we formally consider $\alpha =1$, then the Caputo fractional derivative coincides with the normal derivative with respect to t, and thus (1.1) turns out to be the standard Hamilton–Jacobi equation.

Fractional derivatives attracted great interest from both mathematics and applications within the last few decades, and developed in wide fields (see [11, 14, 16] for instance). Studying differential equations with fractional derivatives is motivated by mathematical models that describe diffusion phenomena in complex media like fractals, which is sometimes called anomalous diffusion. It has inspired further research on numerous related topics.

The well-posedness of viscosity solutions to (1.1) was established by [9, 17, 19]. More precisely, a comparison principle, Perron’s method, and stability results have been established in [9, 17, 19]. We also mention that the equivalence of two weak solutions for linear uniformly parabolic equations with Caputo’s time-fractional derivatives was proved in [8] by using the resolvent type approximation introduced by [7]. In [12], the authors studied the regularity of viscosity solutions of (1.1), and got the large-time asymptotic result in some special settings.

In our paper, we are interested in the asymptotic behavior of $u^\varepsilon$ as $\varepsilon \rightarrow 0$. This singular limit problem is called “homogenization problem” with a background of the material science. Lions, Papanicolau and Varadhan in [13] were the first who started to study homogenization for Hamilton–Jacobi type equations, and after the perturbed test function method was introduced by Evans [5, 6], there has been much literature concerning on homogenization for nonlinear partial differential equations. It was proved in [13] that, when $\alpha =1$, under assumptions (A1)–(A3), $u^\varepsilon$ converges to ${\overline{u}}$ locally uniformly on ${\mathbb {R}}^N \times [0,T]$ as $\varepsilon \rightarrow 0$, and ${\overline{u}}$ solves the effective equation (1.4) with $\alpha =1$. The effective Hamiltonian ${\overline{H}} \in C({\mathbb {R}}^N)$ is determined in a nonlinear way by H as following. For each $p \in {\mathbb {R}}^N$, it was shown in [13] that there exists a unique constant ${\overline{H}}(p)\in {\mathbb {R}}$, which is called the effective Hamiltonian, such that the stationary problem has a continuous viscosity solution to

$$\begin{aligned} H(y, p+Dv(y, p)) = {\overline{H}}(p) \ \text{ in } \ {\mathbb {T}}^N. \end{aligned}$$

(1.2)

If needed, we write $v=v(y,p)$ to clearly demonstrate the nonlinear dependence of v on p. We call a problem to find a pair $(v(\cdot , p), {\overline{H}}(p)) \in \mathrm{Lip\,}({\mathbb {T}}^N) \times {\mathbb {R}}$ to satisfy (1.2) the cell problem. It is worth mentioning that in general v(y, p) is not unique even up to additive constants.

Heuristically, owing to the two-scale (i.e., x and $\frac{x}{\varepsilon }$) asymptotic expansion, of solutions $u^\varepsilon$ to (1.1) of the form

$$\begin{aligned} u^\varepsilon (x, t) = u^0\left( x, \frac{x}{\varepsilon }, t\right) + \varepsilon u^1\left( x, \frac{x}{\varepsilon }, t\right) + \varepsilon ^2 u^2\left( x, \frac{x}{\varepsilon }, t\right) + \cdots , \end{aligned}$$

(1.3)

by plugging this into (1.1), and using the coercivity of Hamiltonian, we can naturally expect that

$$\begin{aligned} u^\varepsilon (x, t) = {\overline{u}}(x, t) + \varepsilon v\left( \frac{x}{\varepsilon }\right) + {\mathcal {O}}(\varepsilon ^2). \end{aligned}$$

Plugging this into (1.1) again, letting v be a solution of (1.2) with $p= D_x{\overline{u}}(x, t)$, we can expect that $u^\varepsilon$ converges to the limit function ${\overline{u}}$ as $\varepsilon \rightarrow 0$ which is a solution to

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \partial _{t}^{\alpha }{\overline{u}} + {\overline{H}} ( D{\overline{u}} ) \ = 0 &{} \text{ in } \, \, {\mathbb {R}}^N \times (0,T), \\ {\overline{u}}(x,0) = u_{0}(x) &{} \text{ on } \, \, {\mathbb {R}}^N. \end{array} \right. \end{aligned}$$

(1.4)

Our main goal of this paper is to establish the convergence result of $u^\varepsilon$ in $C({\mathbb {R}}^N\times [0,T])$, and obtain a rate of convergence of $u^\varepsilon$ to u, that is, an estimate for $\Vert u^\varepsilon -u\Vert _{L^\infty ({\mathbb {R}}^N \times [0,T])}$ for any given $T>0$ with resect to $\varepsilon$. We give two main results.

Theorem 1.1

Let $u^\varepsilon$ be the viscosity solution to (1.1). The function $u^\varepsilon$ converges locally uniformly in ${\mathbb {R}}^N \times [0, T]$ to the function ${\overline{u}} \in C({\mathbb {R}}^N \times [0, T])$ as $\varepsilon \rightarrow 0$, where ${\overline{u}}$ is the unique viscosity solution to (1.4).

Theorem 1.2

We additionally assume

• (A4) There exists $C > 0$ such that

  $$\begin{aligned} |H(x, p) - H(y, q)| \le C(|x - y| + |p - q|) \quad \text{ for } \, \, x, y, p, q \in {\mathbb {R}}^N. \end{aligned}$$

• (A5) There exists $C > 0$ such that

  $$\begin{aligned} H(x, p) \ge C^{-1}|p| - C \quad \text{ for } \, \, x, p \in {\mathbb {R}}^N. \end{aligned}$$

Let $u^\varepsilon$ and ${\overline{u}}$ be the viscosity solutions of (1.1) and (1.4), respectively. For each $\nu \in (0, 1)$, there exists a constant $C > 0$ such that

$$\begin{aligned} \displaystyle \sup _{(x, t) \in {\mathbb {R}}^N \times (0, T)} |u^\varepsilon (x, t) - {\overline{u}}(x, t)| \le C\varepsilon ^{\frac{1}{3(2-\nu )}}. \end{aligned}$$

By the standard perturbed test function method introduced by [5], it is not hard to obtain Theorem 1.1. To obtain Theorem 1.2, we use the method introduced by [3] which is a combination of the perturbed test function method and the discount approximation. A main difficulty we face here is the incompatibility between the doubling variable method which is often used in the theory of viscosity solutions and the non-locality of the Caputo derivative. More precisely, setting $\varphi (t,s)=(t-s)^2$, usually we have $\varphi _{t}(t,s)=-\varphi _{s}(t,s)$, which is an elementary, but important trick in the theory of viscosity solutions, however, we have

$$\begin{aligned} \partial _{t}^{\alpha }\varphi (t,s)\not =-\partial _{s}^{\alpha }\varphi (t,s)\quad \text {for all} \ t, s>0. \end{aligned}$$

Our approach is to provide a lemma to prove Theorem 1.2 without using the doubling variable method with respect to time variable (see Lemma 5.2), which is our key ingredient of this paper. It is inspired by [9, Lemma 3.3].

The study of a rate of convergence in homogenization of Hamilton–Jacobi equations was started by Capuzzo-Dolcetta and Ishii [3], and in recent years, there has been much interest on the optimal convergence rate. Mitake et al. [15] was the first who established the optimal convergence rate for convex Hamilton–Jacobi equations with several conditional settings by using the representation formula for $u^\varepsilon$ from optimal control theory, and weak KAM methods. Then, Cooperman [4] obtained a near optimal convergence rate $|u^\varepsilon (x,t)-{\overline{u}}(x,t)|\le C{\varepsilon }\log (C+{\varepsilon ^{-1}}t)$ when $n\ge 3$ for general convex settings by using a theorem of Alexander [1], originally proved in the context of first-passage percolation. Finally, Tran and Yu [21] has established the optimal rate ${\mathcal {O}}(\varepsilon )$ by using an argument of Burago [2], which concludes the study of this whole program in the convex setting. It is worth mentioning that the best known result on a rate of convergence on $u^\varepsilon$ for the stationary Hamilton–Jacobi equation with the general nonconvex Hamiltonian is still ${\mathcal {O}}(\varepsilon ^\frac{1}{3})$ obtained by [3]. The argument in [3] can be easily adapted to (1.1) with $\alpha =1$ (see [20, Theorem 4.37]). We also refer to [10, 18, 22] for other development on this subject. In all works [4, 10, 15, 18, 21, 22], the argument relies on the optimal control formula for $u^\varepsilon$, and therefore it is seemingly rather challenging to obtain the optimal rate of convergence of (1.1) for $0< \alpha < 1$, which remains completely open.

Organization of the paper. The paper is organized as follows. In Sect. 2, we recall the definition of viscosity solutions to (1.1), and give several basic results on time-fractional Hamilton–Jacobi equations in the theory of viscosity solutions. In Sect. 3, we give regularity results of viscosity solutions to (1.1), and (1.4). Sections 4 and 5 are devoted to prove the convergence of $u^\varepsilon$, Theorem 1.1, and a convergence rate of $u^\varepsilon$ to ${\overline{u}}$ in $L^\infty$, Theorem 1.2.

2 Preliminaries

In this section, we recall the definition of viscosity solutions. First, we give several elementary facts of the Caputo derivative. These are well-known, but we give proofs to make the paper self-contained.

Proposition 2.1

Let $f:[0, T] \mapsto {\mathbb {R}}$ be a function such that $f \in C^1((0, T]) \cap C([0, T])$ and ${\partial _t f} \in L^1(0, T)$. Then,

$$\begin{aligned} \partial _t^\alpha f(t)&= \frac{f(t) - f(0)}{t^\alpha \Gamma (1 - \alpha )} + \frac{\alpha }{\Gamma (1 - \alpha )} \int _0^t \frac{f(t) - f(t-\tau )}{\tau ^{\alpha + 1}} d\tau \\&= \frac{\alpha }{\Gamma (1 - \alpha )} \int _{-\infty }^t \frac{f(t) - {\tilde{f}}(\xi )}{|t - \xi |^{\alpha + 1}} d\xi \nonumber \end{aligned}$$

(2.1)

for any $t \in (0, T]$, where ${\tilde{f}}$ is defined by

$$\begin{aligned} {\tilde{f}}(t) \mathrel {\mathop :}=\left\{ \begin{array}{ll} f(t) &{} \text{ in } \, \, (0, T], \\ f(0) &{} \text{ in } \, \, (-\infty , 0]. \end{array} \right. \end{aligned}$$

Proof

For $\varepsilon > 0$, using the integration by parts, we have

$$\begin{aligned} \int _0^{t-\varepsilon } (t-s)^{-\alpha } \partial _s f(s) ds&= \int _\varepsilon ^t \tau ^{-\alpha } \partial _{\tau }(-f(t - \tau )) d\tau \\&= [\tau ^{-\alpha }(-f(t-\tau ))]_\varepsilon ^t - \int _\varepsilon ^t -\alpha \tau ^{-(1+\alpha )} (-f(t - \tau )) d\tau \\&= -\frac{f(0)}{t^\alpha } + \frac{f(t - \varepsilon )}{\varepsilon ^\alpha } + \alpha \int _\varepsilon ^t \frac{-f(t - \tau )}{\tau ^{1+\alpha }} d\tau . \end{aligned}$$

Note that

$$\begin{aligned} \alpha \int _\varepsilon ^t \frac{f(t)}{\tau ^{1+\alpha }} d\tau = - \frac{f(t)}{t^\alpha } + \frac{f(t)}{\varepsilon ^\alpha }. \end{aligned}$$

Thus,

$$\begin{aligned} \int _0^{t-\varepsilon } (t-s)^{-\alpha } \partial _s f(s) ds&= \frac{f(t) - f(0)}{t^\alpha } + \alpha \int _\varepsilon ^t \frac{f(t) - f(t - \tau )}{\tau ^{1+\alpha }} d\tau \\&- \frac{f(t) - f(t-\varepsilon )}{\varepsilon ^\alpha }. \end{aligned}$$

Using the smoothness of f and taking the limit as $\varepsilon \rightarrow 0$ above, we get (2.1).

On the other hand, using the change of variables, we obtain

$$\begin{aligned} \int _{-\infty }^t \frac{f(t) - {\tilde{f}}(\xi )}{|t - \xi |^{\alpha + 1}} d\xi&= \int _{-\infty }^0 \frac{f(t) - f(0)}{|t - \xi |^{\alpha + 1}} d\xi + \int _0^t \frac{f(t) - f(\xi )}{|t - \xi |^{\alpha + 1}} d\xi \\&= \frac{f(t) - f(0)}{\alpha t^\alpha } + \int _0^t \frac{f(t) - f(t-\tau )}{\tau ^{\alpha + 1}} d\tau , \end{aligned}$$

which completes the proof. $\square$

From the observation in Proposition 2.1, we set

$$\begin{aligned}&\displaystyle J[f](t) \mathrel {\mathop :}=\frac{f(t) - f(0)}{t^\alpha \Gamma (1 - \alpha )}, \\&\displaystyle K_{(a, b)}[f](t) \mathrel {\mathop :}=\frac{\alpha }{\Gamma (1 - \alpha )} \int _a^b \frac{f(t) - f(t-\tau )}{\tau ^{\alpha + 1}} d\tau \end{aligned}$$

for $t \in (0, T]$ and $0 \le a < b \le t$.

Proposition 2.2

Let $f \in C^1((0, T]) \cap C([0, T])$. For any $t \in (0, T]$, the function $K_{(0, t)}[f](t)$ exists and is continuous in (0, T].

Proof

For any $s \in (0, T]$, we set

$$\begin{aligned} g_s(\tau ) \mathrel {\mathop :}=\frac{f(s) - f(s-\tau )}{\tau ^{\alpha + 1}} \chi _{(0, s)}(\tau ) \quad \text{ for } \, \, \tau \in (0, {T}), \end{aligned}$$

where $\chi _I$ is the characteristic function, i.e., $\chi _{I}(\tau ) = 1$ if $\tau \in I$ and $\chi _{I}(\tau ) = 0$ if $\tau \notin I$ for an interval I. Fix $t \in (0, T]$. It suffices to prove that $g_s \rightarrow g_t$ in $L^1(0, T)$ as $s \rightarrow t$. Let $\rho \in (0,\frac{t}{2})$, and for $s \in (t - \rho , \min \{t + \rho , T \})$, we have

$$\begin{aligned} |g_s(\tau )|&\le \frac{|f(s) - f(s-\tau )|}{\tau ^{\alpha + 1}} \chi _{(0, \rho )}(\tau ) + \frac{|f(s) - f(s-\tau )|}{\tau ^{\alpha + 1}} \chi _{(\rho , s)}(\tau ) \\&\le \frac{\sup _{\tau \in (t-2\rho , \min \{t + \rho , T \})} |f^\prime (\tau )|}{\tau ^{\alpha }} \chi _{(0, \rho )}(\tau ) \\&\quad + \frac{2\Vert f\Vert _\infty }{\tau ^{\alpha + 1}} \chi _{(\rho , T)}(\tau ) \quad \text{ for } \, \, \tau \in (0, T). \end{aligned}$$

The right hand side is integrable on (0, T). Thus, using the dominated convergence theorem, we get the desired result. $\square$

Now, we define the definition of viscosity solutions to (1.1).

Definition 1

An upper semicontinuous function $u^\varepsilon : {\mathbb {R}}^N \times [0, T] \mapsto {\mathbb {R}}$ is said to be a viscosity subsolution of (1.1) if for any $\varphi \in C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T])$ one has

$$\begin{aligned} \displaystyle J[\varphi ]({\hat{x}}, {\hat{t}}) + K_{(0, {\hat{t}})}[\varphi ]({\hat{x}}, {\hat{t}}) + H \left( \frac{{\hat{x}}}{\varepsilon }, D\varphi ({\hat{x}}, {\hat{t}}) \right) \le 0, \end{aligned}$$

whenever $u^\varepsilon - \varphi$ attains a maximum at $({\hat{x}}, {\hat{t}}) \in {\mathbb {R}}^N \times (0, T]$ over ${\mathbb {R}}^N \times [0, T]$ and $u^\varepsilon (\cdot , 0) \le u_0$ on ${\mathbb {R}}^N$.

Similarly, a lower semicontinuous function $u^\varepsilon : {\mathbb {R}}^N \times [0, T] \mapsto {\mathbb {R}}$ is said to be a viscosity supersolution of (1.1) if for any $\varphi \in C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T])$ one has

$$\begin{aligned} \displaystyle J[\varphi ]({\hat{x}}, {\hat{t}}) + K_{(0, {\hat{t}})}[\varphi ]({\hat{x}}, {\hat{t}}) + H \left( \frac{{\hat{x}}}{\varepsilon }, D\varphi ({\hat{x}}, {\hat{t}}) \right) \ge 0, \end{aligned}$$

whenever $u^\varepsilon - \varphi$ attains a minimum at $({\hat{x}}, {\hat{t}}) \in {\mathbb {R}}^N \times (0, T]$ over ${\mathbb {R}}^N \times [0, T]$ and $u^\varepsilon (\cdot , 0) \ge u_0$ on ${\mathbb {R}}^N$.

Finally, we call $u^\varepsilon \in C({\mathbb {R}}^N \times [0, T])$ a viscosity solution of (1.1) if $u^\varepsilon$ is both a viscosity subsolution and supersolution of (1.1).

We give several equivalent conditions for viscosity subsolutions/supersolutions to (1.1), which are often used in Sects. 3–5.

Proposition 2.3

For any $\varepsilon > 0$, let $u^\varepsilon : {\mathbb {R}}^N \times [0, T] \mapsto {\mathbb {R}}$ be an upper semicontinuous function. Then, the following statements are equivalent.

• (a) $u^\varepsilon$ is a viscosity subsolution of (1.1).

• (b) For any $\varphi \in C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T])$ one has

  $$\begin{aligned} \displaystyle J[u^\varepsilon ]({\hat{x}}, {\hat{t}}) + K_{(0, \rho )}[\varphi ]({\hat{x}}, {\hat{t}}) + K_{(\rho , {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) + H \left( \frac{{\hat{x}}}{\varepsilon }, D\varphi ({\hat{x}}, {\hat{t}}) \right) \le 0 \end{aligned}$$

  for $0< \rho < {\hat{t}}$, whenever $u^\varepsilon - \varphi$ attains a maximum at $({\hat{x}}, {\hat{t}}) \in {\mathbb {R}}^N \times (0, T]$ over ${\mathbb {R}}^N \times [0, T]$ and $u^\varepsilon (\cdot , 0) \le u_0$ on ${\mathbb {R}}^N$.

• (c) $K_{(0, {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}})$ exists and for any $\varphi \in C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T])$ one has

  $$\begin{aligned} \displaystyle J[u^\varepsilon ]({\hat{x}}, {\hat{t}}) + K_{(0, {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) + H \left( \frac{{\hat{x}}}{\varepsilon }, D\varphi ({\hat{x}}, {\hat{t}}) \right) \le 0, \end{aligned}$$

  whenever $u^\varepsilon - \varphi$ attains a local maximum at $({\hat{x}}, {\hat{t}}) \in {\mathbb {R}}^N \times (0, T]$ over ${\mathbb {R}}^N \times [0, T]$ and $u^\varepsilon (\cdot , 0) \le u_0$ on ${\mathbb {R}}^N$.

Similarly, we obtain equivalent conditions for viscosity supersolutions.

Proof

(a) $\Rightarrow$ (b). Take $\varphi \in C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T])$ such that $u^\varepsilon - \varphi$ has a strict maximum at $({\hat{x}}, {\hat{t}}) \in {\mathbb {R}}^N \times (0, T]$. Without loss of generality, we may assume that $(u^\varepsilon - \varphi )({\hat{x}}, {\hat{t}}) = 0$. Fix $0< \rho < {\hat{t}}$. For any $\sigma > 0$, take a function $\varphi _\sigma \in C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T])$ satisfying

• $\varphi _\sigma = \varphi$ in $B(({\hat{x}}, {\hat{t}}), \frac{\rho }{2})$,

• $u^\varepsilon \le \varphi _\sigma \le \varphi$ in $B(({\hat{x}}, {\hat{t}}), \rho )$,

• $u^\varepsilon \le \varphi _\sigma \le u^\varepsilon + \sigma$ in $\big ({\mathbb {R}}^N\times [0,T]\big )\setminus B(({\hat{x}}, {\hat{t}}), \rho )$.

Noting that $\max (u^\varepsilon - \varphi _\sigma ) = (u^\varepsilon - \varphi _\sigma )({\hat{x}}, {\hat{t}})$, by (a), we have

$$\begin{aligned} \displaystyle J[\varphi _\sigma ]({\hat{x}}, {\hat{t}}) + K_{(0, {\hat{t}})}[\varphi _\sigma ]({\hat{x}}, {\hat{t}}) + H \left( \frac{{\hat{x}}}{\varepsilon }, D\varphi _\sigma ({\hat{x}}, {\hat{t}}) \right) \le 0. \end{aligned}$$

(2.2)

It is clear that $\lim _{\sigma \rightarrow 0} J[\varphi _\sigma ]({\hat{x}}, {\hat{t}}) = J[u^\varepsilon ]({\hat{x}}, {\hat{t}})$ and $D\varphi _\sigma ({\hat{x}}, {\hat{t}}) = D\varphi ({\hat{x}}, {\hat{t}})$. Noting that $\varphi ({\hat{x}}, {\hat{t}}) - \varphi _\sigma ({\hat{x}}, {\hat{t}}) \le \varphi ({\hat{x}}, {\hat{t}} - \tau ) - \varphi _\sigma ({\hat{x}}, {\hat{t}} - \tau )$ for $0< \tau < \rho$, we have

$$\begin{aligned} K_{(0, {\hat{t}})}[\varphi _\sigma ]({\hat{x}}, {\hat{t}})&= K_{(0, \rho )}[\varphi _\sigma ]({\hat{x}}, {\hat{t}}) + K_{(\rho , {\hat{t}})}[\varphi _\sigma ]({\hat{x}}, {\hat{t}}) \\&\ge K_{(0, \rho )}[\varphi ]({\hat{x}}, {\hat{t}}) + K_{(\rho , {\hat{t}})}[\varphi _\sigma ]({\hat{x}}, {\hat{t}}). \end{aligned}$$

By the dominated convergence theorem, we have

$$\begin{aligned} \lim _{\sigma \rightarrow 0} K_{(\rho , {\hat{t}})}[\varphi _\sigma ]({\hat{x}}, {\hat{t}}) = K_{(\rho , {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}). \end{aligned}$$

Thus, sending $\sigma \rightarrow 0$ in (2.2), we obtain

$$\begin{aligned} \displaystyle J[u^\varepsilon ]({\hat{x}}, {\hat{t}}) + K_{(0, \rho )}[\varphi ]({\hat{x}}, {\hat{t}}) + K_{(\rho , {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) + H \left( \frac{{\hat{x}}}{\varepsilon }, D\varphi ({\hat{x}}, {\hat{t}}) \right) \le 0. \end{aligned}$$

(b) $\Rightarrow$ (c). Assume that there exists $\varphi \in C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T])$ such that $u^\varepsilon - \varphi$ has a strict maximum at $({\hat{x}}, {\hat{t}}) \in {\mathbb {R}}^N \times (0, T)$. By (b), we have

$$\begin{aligned} \displaystyle J[u^\varepsilon ]({\hat{x}}, {\hat{t}}) + K_{(0, \rho )}[\varphi ]({\hat{x}}, {\hat{t}}) + K_{(\rho , {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) + H \left( \frac{{\hat{x}}}{\varepsilon }, D\varphi ({\hat{x}}, {\hat{t}}) \right) \le 0, \end{aligned}$$

which implies that

$$\begin{aligned} \int _0^\rho \frac{\varphi ({\hat{x}}, {\hat{t}}) - \varphi ({\hat{x}}, {\hat{t}} - \tau )}{\tau ^{\alpha + 1}} d\tau + \int _\rho ^{{\hat{t}}} \frac{u^\varepsilon ({\hat{x}}, {\hat{t}}) - u^\varepsilon ({\hat{x}}, {\hat{t}} - \tau )}{\tau ^{\alpha + 1}} d\tau \le C \end{aligned}$$

for some $C > 0$ which is independent of $\rho > 0$. Noting that $\varphi \in C^1({\mathbb {R}}^N \times (0, T])$, we have

$$\begin{aligned} \left| \int _0^\rho \frac{\varphi ({\hat{x}}, {\hat{t}}) - \varphi ({\hat{x}}, {\hat{t}} - \tau )}{\tau ^{\alpha + 1}} d\tau \right| \le \int _0^\rho \frac{| \varphi _t({\hat{x}}, {\hat{t}})|}{\tau ^{\alpha }} d\tau \rightarrow 0 \quad \text{ as } \,\, \rho \rightarrow 0. \end{aligned}$$

For $r \in {\mathbb {R}}$, we write $r_+ \mathrel {\mathop :}=\max \{r, 0 \}$, $r_- \mathrel {\mathop :}={-\min \{r, 0 \}}$. Then,

$$\begin{aligned}&\int _\rho ^{{\hat{t}}} \frac{u^\varepsilon ({\hat{x}}, {\hat{t}}) - u^\varepsilon ({\hat{x}}, {\hat{t}} - \tau )}{\tau ^{\alpha + 1}} d\tau \nonumber \\&\quad =\, \int _\rho ^{{\hat{t}}} \frac{(u^\varepsilon ({\hat{x}}, {\hat{t}}) - u^\varepsilon ({\hat{x}}, {\hat{t}} - \tau ))_+}{\tau ^{\alpha + 1}} - \frac{(u^\varepsilon ({\hat{x}}, {\hat{t}}) - u^\varepsilon ({\hat{x}}, {\hat{t}} - \tau ))_-}{\tau ^{\alpha + 1}} d\tau . \end{aligned}$$

(2.3)

Noting that $u^\varepsilon ({\hat{x}}, {\hat{t}}) - \varphi ({\hat{x}}, {\hat{t}}) \ge u^\varepsilon ({\hat{x}}, {\hat{t}} - \tau ) - \varphi ({\hat{x}}, {\hat{t}} - \tau )$, we obtain

$$\begin{aligned} C \ge - \int _\rho ^{{\hat{t}}} \frac{(u^\varepsilon ({\hat{x}}, {\hat{t}}) - u^\varepsilon ({\hat{x}}, {\hat{t}} - \tau ))_-}{\tau ^{\alpha + 1}} d\tau&\ge - \int _\rho ^{{\hat{t}}} \frac{(\varphi ({\hat{x}}, {\hat{t}}) - \varphi ({\hat{x}}, {\hat{t}} - \tau ))_-}{\tau ^{\alpha + 1}} d\tau \nonumber \\&\ge - C^\prime \end{aligned}$$

(2.4)

for some $C^\prime > 0$ which is independent of $\rho > 0$. By the monotone convergence theorem, sending $\rho \rightarrow 0$ in (2.3), we obtain

$$\begin{aligned} \displaystyle J[u^\varepsilon ]({\hat{x}}, {\hat{t}}) + K_{(0, {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) + H \left( \frac{{\hat{x}}}{\varepsilon }, D\varphi ({\hat{x}}, {\hat{t}}) \right) \le 0. \end{aligned}$$

(c) $\Rightarrow$ (a). Take $\varphi \in C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T])$ such that $u^\varepsilon - \varphi$ has a strict maximum at $({\hat{x}}, {\hat{t}}) \in {\mathbb {R}}^N \times (0, T]$. Noting that $(u^\varepsilon -\varphi )({\hat{x}}, {\hat{t}}-\tau ) \le (u^\varepsilon -\varphi )({\hat{x}}, {\hat{t}})$ for all $\tau \in [0, {\hat{t}}]$, we have

$$\begin{aligned} J[\varphi ]({\hat{x}}, {\hat{t}}) \le J[u^\varepsilon ]({\hat{x}}, {\hat{t}}), K_{(0, {\hat{t}})}[\varphi ]({\hat{x}}, {\hat{t}}) \le K_{(0, {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}). \end{aligned}$$

This inequality together with (c), we get (a). $\square$

We state basic results for time-fractional Hamilton–Jacobi equations, and we refer to [17] for the proofs.

Proposition 2.4

(Comparison principle, [9, Theorem 3.1]) Let $u, v:{\mathbb {R}}^N \times [0, T] \rightarrow {\mathbb {R}}$ be a viscosity subsolution and a viscosity supersolution of (1.1), respectively. If $u, v \in \mathrm{BUC\,}({\mathbb {R}}^N \times [0, T])$ and $u(\cdot , 0) \le v(\cdot , 0)$ on ${\mathbb {R}}^N$, then $u \le v$ on ${\mathbb {R}}^N \times [0, T]$.

Proposition 2.5

(Existence of a solution, [9, Theorem 4.2]) Let $u_-, u_+:{\mathbb {R}}^N \times [0, T] \rightarrow {\mathbb {R}}$ be a viscosity subsolution and a viscosity supersolution of (1.1), respectively. Suppose that $u_- \le u_+$ in ${\mathbb {R}}^N \times [0, T] \rightarrow {\mathbb {R}}$. There exists a viscosity solution u of (1.1) that satisfies $u_- \le u \le u_+$ in ${\mathbb {R}}^N \times [0, T] \rightarrow {\mathbb {R}}$.

Proposition 2.5 follows from Lemmas 2.6 and 2.7. We denote $S^-$ and $S^+$ by the set of all subsolutions and supersolutions of (1.1), respectively.

Lemma 2.6

(Closedness under supremum/infimum operator, [9, Lemma 4.1]) Let X be a nonempty subset of $S^-$ (resp., $S^+$). Set

$$\begin{aligned} \displaystyle u(x, t) \mathrel {\mathop :}=\sup _{v \in X} v(x, t) \quad (resp., \inf _{v \in X} v(x, t)) \quad \text{ for } \, \, (x, t) \in {\mathbb {R}}^N \times [0, T]. \end{aligned}$$

If $u < \infty$ (resp., $u > -\infty$) on ${\mathbb {R}}^N \times [0, T]$, then u is a viscosity subsolution (resp., supersolution) of (1.1).

Lemma 2.7

Let $u_+:{\mathbb {R}}^N \times [0, T] \rightarrow {\mathbb {R}}$ be a supersolution of (1.1). Define

$$\begin{aligned} X \mathrel {\mathop :}=\{ v \in S^- \mid v \le u_+ \, \text{ in } \, {\mathbb {R}}^N \times [0, T] \}. \end{aligned}$$

If $u \in X$ is not a supersolution of (1.1), then there exists a function $w \in X$ and a point $(y, s) \in {\mathbb {R}}^N \times (0, T]$ such that $u(y, s) < w(y, s)$.

Proof

Since if u is not a supersolution of (1.1), there exists $(({\hat{x}}, {\hat{t}}), \varphi ) \in ({\mathbb {R}}^N \times (0, T]) \times (C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T]))$ such that

$$\begin{aligned} (u - \varphi )(x, t) > (u - \varphi )({\hat{x}}, {\hat{t}}) = 0 \quad \text{ for } \text{ all } \,\, (x, t) \ne ({\hat{x}}, {\hat{t}}), \end{aligned}$$

and

$$\begin{aligned} J[\varphi ]({\hat{x}}, {\hat{t}}) + K_{(0, {\hat{t}})}[\varphi ]({\hat{x}}, {\hat{t}}) + H \left( \frac{{\hat{x}}}{\varepsilon }, D\varphi ({\hat{x}}, {\hat{t}}) \right) < 0. \end{aligned}$$

For $r > 0$ small enough, we have

$$\begin{aligned}{} & {} J[\varphi ](x, t) + K_{(0, t)}[\varphi ](x, t) + H \left( \frac{x}{\varepsilon }, D\varphi (x, t) \right) \nonumber \\{} & {} \quad \le 0 \quad \text{ for } \,\, B(({\hat{x}}, {\hat{t}}), 2r) \cap ({\mathbb {R}}^N \times (0, T]), \end{aligned}$$

(2.5)

where we denote by $B(({\hat{x}}, {\hat{t}}), 2r)$ a cylindrical neighborhood of $({\hat{x}}, {\hat{t}})$.

It is clear to see that $\varphi \le u \le u_+$ in ${\mathbb {R}}^N \times [0, T]$. If $(u_+ - \varphi )({\hat{x}}, {\hat{t}})=0$, then it is clear to see a contradiction since $u_+$ is a supersolution. Thus, we only consider the case where $\lambda \mathrel {\mathop :}=\frac{1}{2}(u_+ - \varphi )({\hat{x}}, {\hat{t}}) > 0$. Since $u_+ - \varphi$ is lower semicontinuous, if $r > 0$ is small enough, we have $\varphi + \lambda \le u_+$ in $B(({\hat{x}}, {\hat{t}}), 2r)$. Since $u > \varphi$ in ${\mathbb {R}}^N \times [0, T] {\setminus } \{({\hat{x}}, {\hat{t}})\}$, there exists $\lambda ^\prime \in (0, \lambda )$ such that $\varphi + 2\lambda ^\prime \le u$ in $\big (B({\hat{x}}, 2r) \times [0, T]\big ) {\setminus } B(({\hat{x}}, {\hat{t}}), r)$. Define the function $w:{\mathbb {R}}^N \times [0, T] \mapsto {\mathbb {R}}$ by

$$\begin{aligned} w \mathrel {\mathop :}=\left\{ \begin{array}{ll} \max \{u, \varphi + \lambda ^\prime \} &{} \text{ in } \, \, B(({\hat{x}}, {\hat{t}}), r), \\ u &{} \text{ in } \, \, \big ({\mathbb {R}}^N \times [0, T]\big )\setminus B(({\hat{x}}, {\hat{t}}), r). \end{array} \right. \end{aligned}$$

We claim that $w \in X$. It is easy to see that $w \le u_+$ in ${\mathbb {R}}^N \times [0, T]$. Take $\psi \in C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T])$ such that $w - \psi$ has a strict maximum at $({\hat{y}}, {\hat{s}}) \in {\mathbb {R}}^N \times (0, T]$ with $(w-\psi )({\hat{y}}, {\hat{s}})=0$. If $w = u$ at $({\hat{y}}, {\hat{s}})$, since u is a subsolution, we are done. We consider the case where $w = \varphi + {\lambda '}$ at $({\hat{y}}, {\hat{s}})$. By definition of w, we have $({\hat{y}}, {\hat{s}}) \in B(({\hat{x}}, {\hat{t}}), r)$ and $w({\hat{y}}, {\hat{s}} - \tau ) \ge \varphi ({\hat{y}}, {\hat{s}} - \tau ) + \lambda ^\prime$ for $\tau \in [0, {\hat{s}}]$. Noting that $0 = (w - \psi )({\hat{y}}, {\hat{s}}) \ge (w - \psi )({\hat{y}}, {\hat{s}} - \tau )$, we have

$$\begin{aligned}&\psi ({\hat{y}}, {\hat{s}}) - \psi ({\hat{y}}, {\hat{s}} - \tau ) \le w({\hat{y}}, {\hat{s}}) - w({\hat{y}}, {\hat{s}} - \tau ) \\&\le \varphi ({\hat{y}}, {\hat{s}}) + \lambda ^\prime - (\varphi ({\hat{y}}, {\hat{s}} - \tau ) + \lambda ^\prime ) = \varphi ({\hat{y}}, {\hat{s}}) - \varphi ({\hat{y}}, {\hat{s}} - \tau ) \end{aligned}$$

for $\tau \in [0, {\hat{s}}]$ which yields $J[\psi ]({\hat{y}}, {\hat{s}}) \le J[\varphi ]({\hat{y}}, {\hat{s}})$ and $K_{(0, {\hat{s}})}[\psi ]$$({\hat{y}}, {\hat{s}}) \le K_{(0, {\hat{s}})}[\varphi ]({\hat{y}}, {\hat{s}})$. Also, since $\varphi {+\lambda '} = \psi$ at $({\hat{y}}, {\hat{s}})$ and $0 > w - \psi \ge \varphi + \lambda ^\prime - \psi$ in $\big ({\mathbb {R}}^N \times [0, T]\big ) {\setminus } \{ ({\hat{y}}, {\hat{s}}) \}$, we have $D\varphi = D\psi$ at $({\hat{y}}, {\hat{s}})$. Combining this with (2.5), we obtain

$$\begin{aligned}&J[\psi ]({\hat{y}}, {\hat{s}}) + K_{(0, {\hat{s}})}[\psi ]({\hat{y}}, {\hat{s}}) + H \left( \frac{{\hat{y}}}{\varepsilon }, D\psi ({\hat{y}}, {\hat{s}}) \right) \\&\le J[\varphi ]({\hat{y}}, {\hat{s}}) + K_{(0, {\hat{s}})}[\varphi ]({\hat{y}}, {\hat{s}}) + H \left( \frac{{\hat{y}}}{\varepsilon }, D\varphi ({\hat{y}}, {\hat{s}}) \right) \le 0, \end{aligned}$$

which implies that $w \in S^-$ and finishes the proof. $\square$

3 Regularity estimates

Proposition 3.1

Let $u^\varepsilon$ be the viscosity solution of (1.1). There exists a constant $M^\prime > 0$ depending only on $u_0, H$ and T such that

$$\begin{aligned} |u^\varepsilon (x, t)| \le M^\prime \quad \text{ for } \quad (x, t) \in {\mathbb {R}}^N \times [0, T]. \end{aligned}$$

(3.1)

Proof

We set $u^+(x, t) \mathrel {\mathop :}=u_0(x) + M t^\alpha$ and $u^-(x, t) \mathrel {\mathop :}=u_0(x) - M t^\alpha$ with

$$\begin{aligned} M \mathrel {\mathop :}=\frac{1}{\Gamma (1 - \alpha )} \max \{ H(\xi , p) \mid \xi \in {\mathbb {R}}^N, |p| \le \mathrm{Lip\,}[u_0] \}. \end{aligned}$$

(3.2)

Then, $u^+$ and $u^-$ are a viscosity supersolution and a viscosity subsolution of (1.1), respectively. By Proposition 2.4, we obtain

$$\begin{aligned} u_0(x) - M t^\alpha \le u^\varepsilon (x, t) \le u_0(x) + M t^\alpha \end{aligned}$$

on ${\mathbb {R}}^N \times [0, T]$. Thus, we choose $M^{\prime } \mathrel {\mathop :}=\Vert u_0\Vert _{\infty } + M T^\alpha$. This completes the proof. $\square$

Proposition 3.2

Let $u^\varepsilon$ be the viscosity solution of (1.1). There exists a constant $M > 0$ depending only on $u_0$ and H such that

$$\begin{aligned} |u^\varepsilon (x, t) - u^\varepsilon (x, s)| \le M |t - s|^\alpha \end{aligned}$$

(3.3)

for $(x, t, s) \in {\mathbb {R}}^N \times [0, T] \times [0, T]$.

Proof

Let $u^+$ be the function defined in the proof of Proposition 3.1, and let

$$\begin{aligned} X \mathrel {\mathop :}=\{ v \in S^- \mid v \le u^+ \, \text{ in } \, {\mathbb {R}}^N \times [0, T], \, v \, \text{ satisfies } \, (3.3) \}, \end{aligned}$$

where we denote $S^-$ by the set of all subsolutions of (1.1) and M is the constant defined by (3.2). Set $u(x, t) \mathrel {\mathop :}=\sup _{v \in X} v(x, t)$ for $(x, t) \in {\mathbb {R}}^N \times [0, T]$. It is clear to see that u satisfies (3.3).

Next, let us prove that u is a viscosity solution of (1.1). Noting that $u_0(x) - Mt^\alpha \in X$, by Lemma 2.6, we see that u is a viscosity subsolution of (1.1). Suppose that u is not a supersolution of (1.1). Then there exists $(({\hat{x}}, {\hat{t}}), \varphi ) \in ({\mathbb {R}}^N \times (0, T]) \times (C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T]))$ such that

$$\begin{aligned} (u - \varphi )(x, t) > (u - \varphi )({\hat{x}}, {\hat{t}}) = 0 \quad \text{ for } \,\, (x, t) \ne ({\hat{x}}, {\hat{t}}), \end{aligned}$$

and

$$\begin{aligned} J[\varphi ]({\hat{x}}, {\hat{t}}) + K_{(0, {\hat{t}})}[\varphi ]({\hat{x}}, {\hat{t}}) + H \left( \frac{{\hat{x}}}{\varepsilon }, D\varphi ({\hat{x}}, {\hat{t}}) \right) < 0. \end{aligned}$$

By Lemma 2.7, there exists a viscosity subsolution $w \in \mathrm{USC\,}({\mathbb {R}}^N \times [0, T])$ of (1.1) satisfying $w \le u^+$ on ${\mathbb {R}}^N \times [0, T]$ and $w({\hat{x}}, {\hat{t}}) > u({\hat{x}}, {\hat{t}})$. By construction of w in the proof of Lemma 2.7, it is clear to see that w satisfies (3.3), and also $w \in X$. This immediately gives a contradiction. By the uniqueness of viscosity solutions of (1.1), we obtain $u = u^\varepsilon$ in ${\mathbb {R}}^N \times [0, T]$, which completes the proof. $\square$

Proposition 3.3

Let $u^\varepsilon$ be the viscosity solution of (1.1). There exists a modulus of continuity $\omega \in C([0, \infty ))$ independent of $\varepsilon$ such that

$$\begin{aligned} |u^\varepsilon (x, t) - u^\varepsilon (y, t)| \le \omega (|x - y|) \quad \text{ for } \quad (x, y, t) \in {\mathbb {R}}^N \times {\mathbb {R}}^N \times [0, T]. \end{aligned}$$

Furthermore, assume (A5). There exists a constant $C > 0$ independent of $\varepsilon$ such that

$$\begin{aligned} |u^\varepsilon (x, t) - u^\varepsilon (y, t)| \le C(1+\big |\log |x-y|\big |)|x - y| \end{aligned}$$

(3.4)

for $(x, y, t) \in {\mathbb {R}}^N \times {\mathbb {R}}^N \times [0, T]$. In particular, for each $\nu \in (0, 1)$, there exists a constant $C > 0$ independent of $\varepsilon$ such that

$$\begin{aligned} |u^\varepsilon (x, t) - u^\varepsilon (y, t)| \le C|x - y|^\nu \quad \text{ for } \quad (x, y, t) \in {\mathbb {R}}^N \times {\mathbb {R}}^N \times [0, T]. \end{aligned}$$

We do not know whether the regularity 3.4 is optimal or not. It comes from the nonlocal operator $K_{(0,t)}[u]$ as seen in the proof below. We follow the proof of [12, Theorem 2.2] with a slight modification.

Definition 2

Let $f: [0, T] \mapsto {\mathbb {R}}$ be a bounded function. For each $\delta > 0$, we define $f^\delta$ and $f_\delta$ by

$$\begin{aligned} f^\delta (t) \mathrel {\mathop :}=\sup _{\xi \in [0, T]} \left\{ f(\xi ) - \frac{|t - \xi |^2}{2\delta } \right\} , \quad f_\delta (t) \mathrel {\mathop :}=\inf _{\xi \in [0, T]} \left\{ f(\xi ) + \frac{|t - \xi |^2}{2\delta } \right\} \end{aligned}$$

for $t \in [0, T]$. We call $f^\delta$ and $f_\delta$ the sup-convolution and the inf-convolution, respectively.

Lemma 3.4

Let $f: [0, T] \mapsto {\mathbb {R}}$ be a bounded function. There exists $C > 0$, for each $\delta > 0$,

• (a) $f(t) \le f^\delta (t) \le \Vert f \Vert _{\infty }$ for $t \in [0, T]$,

• (b) $\displaystyle |f^\delta (t) - f^\delta (s)| \le \frac{C}{\delta } |t-s|$ for $t, s \in [0, T]$.

Furthermore, assume that $f \in C^{0, \alpha }([0, T])$ for $\alpha \in (0, 1]$, then

• (c) $|t - \xi _\delta |^{2-\alpha } \le 2 M \delta$ for $\xi _\delta \in \mathop {\mathrm {arg\,max}}\limits \{ f(\xi ) - \frac{|t - \xi |^2}{2\delta } \mid \xi \in [0, T] \}$,

• (d) $\Vert f^\delta - f \Vert _{\infty } \le (2\,M)^{\frac{2}{2 - \alpha }} \delta ^{\frac{\alpha }{2-\alpha }}$,

• (e) $f^\delta \in C^{0, \alpha }([0, T])$. In particular, there exists $C>0$ which only depends on M (is independent of $\delta >0$) such that $\Vert f^\delta \Vert _{C^{0, \alpha }([0, T])}\le C$.

where M is the Hölder constant of f.

Proof

We omit the proofs of (a) and (b), as they are standard. We only prove (c), (d) and (e). Take $\xi _\delta \in [0, T]$ so that $f^\delta (t) = f(\xi _\delta ) - \frac{|t - \xi _\delta |^2}{2\delta }$. Note that

$$\begin{aligned} \frac{|t - \xi _\delta |^2}{2\delta } = f(\xi _\delta ) - f^\delta (t) \le f(\xi _\delta ) - f(t) \le M|t - \xi _\delta |^\alpha . \end{aligned}$$

(3.5)

Thus, $|t - \xi _\delta |^{2-\alpha } \le 2 M \delta$. By (3.5), we have

$$\begin{aligned} |f^\delta (t) - f(t)|&= \left| f(\xi _\delta ) - \frac{|t - \xi _\delta |^2}{2\delta } - f(t)\right| \\&\le |f(\xi _\delta ) - f(t)| + \frac{|t - \xi _\delta |^2}{2\delta } \le 2 M |t - \xi _\delta |^\alpha \le (2 M)^{\frac{2}{2 - \alpha }} \delta ^{\frac{\alpha }{2-\alpha }} \end{aligned}$$

for $t \in [0, T]$, which proves (d). Finally, let $t, s \in [0, T]$. We consider the case $|t - s|^{2 - \alpha } \ge \delta$. By applying (d), we have

$$\begin{aligned} |f^\delta (t) - f^\delta (s)|&\le |f^\delta (t) - f(t)| + |f(t) - f(s)| + |f(s) - f^\delta (s)| \\&\le 2 (2 M)^{\frac{2}{2 - \alpha }} \delta ^{\frac{\alpha }{2 - \alpha }} + M |t - s|^\alpha \le C |t - s|^\alpha . \end{aligned}$$

We consider the case $|t - s|^{2 - \alpha } \le \delta$. We have

$$\begin{aligned} f^\delta (t) - f^\delta (s)&\le f(\xi _\delta ) - \frac{|t - \xi _\delta |^2}{2\delta } - \left( f(\xi _\delta ) - \frac{|s - \xi _\delta |^2}{2\delta } \right) \\&= \frac{1}{2\delta } (|s - \xi _\delta |^2 - |t - \xi _\delta |^2) = \frac{1}{2\delta } (|s - t + t - \xi _\delta |^2 - |t - \xi _\delta |^2) \\&\le \frac{1}{2\delta } (|t - s|^2 + 2|t - \xi _\delta ||t - s|). \end{aligned}$$

By (c), we obtain

$$\begin{aligned} f^\delta (t) - f^\delta (s)&\le \frac{1}{2\delta } (|t - s|^2 + 2 (2 M \delta )^{\frac{1}{2 - \alpha }} |t - s|) \le C\left( \frac{|t - s|^2}{\delta } + \frac{|t - s|}{\delta ^{\frac{1 - \alpha }{2 - \alpha }}}\right) \\&\le C \left( \frac{|t - s|^2}{|t - s|^{2-\alpha }} + \frac{|t - s|}{|t - s|^{(2-\alpha )\frac{1 - \alpha }{2 - \alpha }}}\right) = 2 C |t - s|^\alpha , \end{aligned}$$

which completes the proof. $\square$

Lemma 3.5

Let u be a bounded viscosity subsolution of (1.1), and for $\delta \in (0, 1)$, we write $u^\delta$ and $u_\delta$ for the sup-convolution and inf-convolution of u, respectively. Then $u^\delta$ and $u_\delta$ are, respectively, a viscosity subsolution and a viscosity supersolution of

$$\begin{aligned}&\partial _t^\alpha u^{\delta } + H \left( \frac{x}{\varepsilon }, Du^{\delta } \right) \le \eta _\delta \quad \text{ in } \,\, {\mathbb {R}}^N \times (M\delta ^{\frac{1}{4}} , T], \\&\partial _t^\alpha u_{\delta } + H \left( \frac{x}{\varepsilon }, Du_{\delta } \right) \ge -\eta _\delta \quad \text{ in } \,\, {\mathbb {R}}^N \times (M\delta ^{\frac{1}{4}} , T], \nonumber \end{aligned}$$

(3.6)

where M is a constant such that $M^2 \ge 2\Vert u \Vert _\infty$ and $\eta _\delta$ is a constant such that $\eta _\delta \rightarrow 0$ as $\delta \rightarrow 0$.

We refer to [19, Lemma 4.2] and [17, Proposition 4.2] for similar results.

Proof

We only prove that $u^\delta$ is a viscosity subsolution of (3.6). Take $(({\hat{x}}, {\hat{t}}), \varphi ) \in ({\mathbb {R}}^N \times (M\delta ^{\frac{1}{4}}, T]) \times (C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T]))$ so that

$$\begin{aligned} \max _{{\mathbb {R}}^N \times [M\delta ^{\frac{1}{4}}, T]} (u^\delta - \varphi ) = (u^\delta - \varphi )({\hat{x}}, {\hat{t}}). \end{aligned}$$

Let ${\hat{t}}_\delta \in [0, T]$ be such that

$$\begin{aligned} u^\delta ({\hat{x}}, {\hat{t}}) = u({\hat{x}}, {\hat{t}}_\delta ) - \frac{ |{\hat{t}} - {\hat{t}}_\delta |^2 }{\delta }. \end{aligned}$$

Then, we see that

$$\begin{aligned} |{\hat{t}} - {\hat{t}}_\delta |^2 = (u({\hat{x}}, {\hat{t}}_\delta ) - u^\delta ({\hat{x}}, {\hat{t}})) \delta \le 2 \Vert u \Vert _\infty \delta . \end{aligned}$$

Thus, we have ${\hat{t}}_\delta \ge {\hat{t}} - (2 \Vert u \Vert _\infty \delta )^\frac{1}{2}> M\delta ^\frac{1}{4} - (2 \Vert u \Vert _\infty \delta )^\frac{1}{2} \ge M\delta ^\frac{1}{4}(1 - \delta ^\frac{1}{4})> 0$. Noting that $(u^\delta - \varphi )(x, t) \le (u^\delta - \varphi )({\hat{x}}, {\hat{t}})$, we see that

$$\begin{aligned} u(&{\hat{x}}, {\hat{t}}_\delta ) - \frac{|{\hat{t}} - {\hat{t}}_\delta |^2}{\delta } - \varphi ({\hat{x}}, {\hat{t}}) = (u^\delta - \varphi )({\hat{x}}, {\hat{t}}) \ge (u^\delta - \varphi )(x, t) \\&= \sup _{\tau \in [0, T]} \left\{ u(x, \tau ) - \frac{|t - \tau |^2}{\delta } \right\} - \varphi (x, t) \ge u(x, t + {\hat{t}}_\delta - {\hat{t}}) {-} \frac{|{\hat{t}} {-} {\hat{t}}_\delta |^2}{\delta } {-} \varphi (x, t) \end{aligned}$$

for (x, t) in a neighborhood of $({\hat{x}}, {\hat{t}})$. Thus, the function $(x, t) \mapsto u(x, t) - \varphi (x, t - {\hat{t}}_\delta + {\hat{t}})$ attains a local maximum at $({\hat{x}}, {\hat{t}}_\delta )$. Therefore, there exists ${\tilde{\varphi }} \in C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T])$ such that ${\tilde{\varphi }}(x, t) = \varphi (x, t - {\hat{t}}_\delta + {\hat{t}})$ for (x, t) in a neighborhood of $({\hat{x}}, {\hat{t}}_\delta )$ and

$$\begin{aligned} \max _{{\mathbb {R}}^N \times [0, T]} (u - {\tilde{\varphi }}) = (u - {\tilde{\varphi }})({\hat{x}}, {\hat{t}}_\delta ). \end{aligned}$$

Since u is a viscosity subsolution to (1.1), by Proposition 2.3, we have

$$\begin{aligned} J[u]({\hat{x}}, {\hat{t}}_\delta ) + K_{(0, {\hat{t}}_\delta )}[u]({\hat{x}}, {\hat{t}}_\delta ) + H \left( \frac{{\hat{x}}}{\varepsilon }, D{\tilde{\varphi }}({\hat{x}}, {\hat{t}}_\delta ) \right) \le 0. \end{aligned}$$

Define

$$\begin{aligned} {\tilde{u}}(x, t) \mathrel {\mathop :}=\left\{ \begin{array}{ll} u(x, t) &{}\quad \text{ in } \, \, {\mathbb {R}}^N \times [0, T], \\ u(x, 0) &{}\quad \text{ in } \, \, {\mathbb {R}}^N \times (-\infty , 0]. \end{array} \right. \end{aligned}$$

Then, we see that

$$\begin{aligned} J[u]({\hat{x}}, {\hat{t}}_\delta ) + K_{(0, {\hat{t}}_\delta )}[u]({\hat{x}}, {\hat{t}}_\delta )&= \frac{\alpha }{\Gamma (1 - \alpha )} \int _{-\infty }^{{\hat{t}}_\delta } \frac{u({\hat{x}}, {\hat{t}}_\delta ) - {\tilde{u}}({\hat{x}}, \tau )}{|{\hat{t}}_\delta - \tau |^{\alpha + 1}} d\tau \\&= \frac{\alpha }{\Gamma (1 - \alpha )} \int _{-\infty }^{{\hat{t}}} \frac{u({\hat{x}}, {\hat{t}}_\delta ) - {\tilde{u}}({\hat{x}}, \xi + {\hat{t}}_\delta - {\hat{t}})}{|{\hat{t}} - \xi |^{\alpha + 1}} d\xi \\&= \frac{\alpha }{\Gamma (1 - \alpha )} \int _{-\infty }^{{\hat{t}}} \frac{u^\delta ({\hat{x}}, {\hat{t}}) + \frac{|{\hat{t}}_\delta - {\hat{t}}|^2}{\delta } {-} {\tilde{u}}({\hat{x}}, \xi {+} {\hat{t}}_\delta {-} {\hat{t}})}{|{\hat{t}} {-} \xi |^{\alpha + 1}} d\xi . \end{aligned}$$

We set $\Psi (\xi ) \mathrel {\mathop :}=u^\delta ({\hat{x}}, {\hat{t}}) + \frac{|{\hat{t}}_\delta - {\hat{t}}|^2}{\delta } - {\tilde{u}}({\hat{x}}, \xi + {\hat{t}}_\delta - {\hat{t}})$ for $\xi \in (-\infty , {\hat{t}}]$. We divide into three cases: (1) $\xi \le -M\delta ^\frac{1}{2}$, (2) $-M\delta ^\frac{1}{2} < \xi \le M\delta ^\frac{1}{2}$, and (3) $M\delta ^\frac{1}{2} < \xi \le {\hat{t}}$. First, we consider case (1). Since $\xi + {\hat{t}}_\delta - {\hat{t}} \le -M\delta ^\frac{1}{2} + M\delta ^\frac{1}{2} = 0$, we have ${\tilde{u}}({\hat{x}}, \xi + {\hat{t}}_\delta - {\hat{t}}) = u({\hat{x}}, 0) \le u^\delta ({\hat{x}}, 0) = {\tilde{u}}^\delta ({\hat{x}}, \xi )$, where ${\tilde{u}}^\delta$ be the sup-convolution of ${\tilde{u}}$. Note that by Lemma 3.4 (c), in case (1), letting $\tau _\delta \in \arg \max \left\{ {\tilde{u}}({\hat{x}},\tau )-\frac{(\xi -\tau )^2}{\delta }\right\}$, we have

$$\begin{aligned} \tau _\delta \le \xi +(2M\delta )^\frac{1}{2-\alpha } \le -M\delta ^\frac{1}{2}+(2M\delta )^\frac{1}{2-\alpha }<0 \end{aligned}$$

if $\delta >0$ is small enough. Thus, we have ${\tilde{u}}^\delta ({\hat{x}}, \xi )=u^\delta ({\hat{x}}, 0)$. We obtain

$$\begin{aligned} \Psi (\xi ) \ge u^\delta ({\hat{x}}, {\hat{t}}) + \frac{(t_\delta - {\hat{t}})^2}{\delta } - u^\delta ({\hat{x}}, 0) \ge u^\delta ({\hat{x}}, {\hat{t}}) - {\tilde{u}}^\delta ({\hat{x}}, \xi ). \end{aligned}$$

Next, in case (2), we have

$$\begin{aligned} \Psi (\xi )&\ge u^\delta ({\hat{x}}, {\hat{t}}) - {\tilde{u}}({\hat{x}}, \xi + {\hat{t}}_\delta - {\hat{t}}) \\&\ge u^\delta ({\hat{x}}, {\hat{t}}) - {\tilde{u}}^\delta ({\hat{x}}, \xi ) + {\tilde{u}}^\delta ({\hat{x}}, \xi ) - {\tilde{u}}({\hat{x}}, \xi + {\hat{t}}_\delta - {\hat{t}}) \\&\ge u^\delta ({\hat{x}}, {\hat{t}}) - {\tilde{u}}^\delta ({\hat{x}}, \xi ) - 2\Vert u \Vert _\infty . \end{aligned}$$

Finally, we consider case (3). Since $\xi + {\hat{t}}_\delta - {\hat{t}} > M\delta ^\frac{1}{2} - M\delta ^\frac{1}{2} = 0$, we have ${\tilde{u}}({\hat{x}}, \xi + {\hat{t}}_\delta - {\hat{t}}) = u({\hat{x}}, \xi + {\hat{t}}_\delta - {\hat{t}})$. Therefore, we obtain

$$\begin{aligned} \Psi (\xi )&= u^\delta ({\hat{x}}, {\hat{t}}) + \frac{|{\hat{t}}_\delta - {\hat{t}}|^2}{\delta } - {\tilde{u}}({\hat{x}}, \xi + {\hat{t}}_\delta - {\hat{t}}) = u^\delta ({\hat{x}}, {\hat{t}}) + \frac{|\xi - (\xi + {\hat{t}}_\delta - {\hat{t}})|^2}{\delta }\\&\quad - u({\hat{x}}, \xi + {\hat{t}}_\delta - {\hat{t}}) \\&\ge u^\delta ({\hat{x}}, {\hat{t}}) - u^\delta ({\hat{x}}, \xi ) =u^\delta ({\hat{x}}, {\hat{t}}) - {\tilde{u}}^\delta ({\hat{x}}, \xi ). \end{aligned}$$

Hence, we obtain

$$\begin{aligned}&J[u]({\hat{x}}, {\hat{t}}_\delta ) + K_{(0, {\hat{t}}_\delta )}[u]({\hat{x}}, {\hat{t}}_\delta ) = \frac{\alpha }{\Gamma (1 - \alpha )} \int _{-\infty }^{{\hat{t}}} \frac{\Psi (\xi )}{|{\hat{t}} - \xi |^{\alpha + 1}} d\xi \\&\quad \ge \frac{\alpha }{\Gamma (1 - \alpha )} \left( \int _{-\infty }^{{\hat{t}}} \frac{u^\delta ({\hat{x}}, {\hat{t}}) - {\tilde{u}}^\delta ({\hat{x}}, \xi )}{|{\hat{t}} - \xi |^{\alpha + 1}} d\xi - \int _{-M\delta ^\frac{1}{2}}^{M\delta ^\frac{1}{2}} \frac{2 \Vert u\Vert _\infty }{|{\hat{t}} - \xi |^{\alpha + 1}} d\xi \right) \\&\quad \ge J[u^\delta ]({\hat{x}}, {\hat{t}}) + K_{(0, {\hat{t}})}[u^\delta ]({\hat{x}}, {\hat{t}}_\delta ) - \frac{\alpha M^2}{\Gamma (1 - \alpha )} \int _{-M\delta ^\frac{1}{2}}^{M\delta ^\frac{1}{2}} \frac{1}{|{\hat{t}} - \xi |^{\alpha + 1}} d\xi . \end{aligned}$$

Noting that ${\hat{t}} > M\delta ^\frac{1}{4}$ and $\frac{1}{2} \delta ^{\frac{1}{4}} > \delta ^{\frac{1}{2}}$ for small $0< \delta < 1$, we have

$$\begin{aligned} \int _{-M\delta ^\frac{1}{2}}^{M\delta ^{\frac{1}{2}}} \frac{1}{|{\hat{t}} - \xi |^{\alpha + 1}} d\xi&\le \int _{-M\delta ^{\frac{1}{2}}}^{M\delta ^{\frac{1}{2}}} \frac{1}{(M\delta ^\frac{1}{4} - M\delta ^\frac{1}{2})^{\alpha + 1}} d\xi \\&\le \left( \frac{M\delta ^{\frac{1}{4}}}{2} \right) ^{-(\alpha +1)} 2M\delta ^\frac{1}{2} = 2^{\alpha +2} M^{-\alpha } \delta ^{\frac{1-\alpha }{4}}, \end{aligned}$$

which implies

$$\begin{aligned} J[u]({\hat{x}}, {\hat{t}}_\delta ) + K_{(0, {\hat{t}}_\delta )}[u]({\hat{x}}, {\hat{t}}_\delta ) \ge J[u^\delta ]({\hat{x}}, {\hat{t}}) + K_{(0, {\hat{t}})}[u^\delta ]({\hat{x}}, {\hat{t}}_\delta ) - \eta _\delta , \end{aligned}$$

where

$$\begin{aligned} \eta _\delta \mathrel {\mathop :}=\frac{2^{\alpha +2} \alpha M^{2-\alpha }}{\Gamma (1 - \alpha )} \delta ^{\frac{1-\alpha }{4}}. \end{aligned}$$

This completes the proof. $\square$

Proof of Proposition 3.3

For $\delta > 0$, we denote by $u^{\varepsilon , \delta }$ the sup-convolution of $u^\varepsilon$. For fixed $y \in {\mathbb {R}}^N, t_0 > 0,$ and $\beta , L > 0$, we set

$$\begin{aligned} \Phi (x, t) \mathrel {\mathop :}=u^{\varepsilon , \delta }(x, t) - u^{\varepsilon , \delta }(y, t_0) - L|x - y| - \frac{|t - t_0|^2}{\beta ^2}, \end{aligned}$$

where L will be fixed later. Since $\Phi (x, t) \rightarrow -\infty$ as $|x| \rightarrow \infty$ for any $t \in [0, T]$ and $\Phi$ is bounded from above, $\Phi$ attains a maximum at a point $({\hat{x}}, {\hat{t}}) \in {\mathbb {R}}^N \times [0, T]$. We claim that if L is sufficiently large, then ${\hat{x}} = y$. Suppose that ${\hat{x}} \ne y$. Since $u^{\varepsilon , \delta }$ is the viscosity subsolution of (1.1), by Proposition 2.3, for every $0< \rho < {\hat{t}}$, we have

$$\begin{aligned} J[u^{\varepsilon , \delta }]({\hat{x}}, {\hat{t}}) + K_{(0, \rho )}[u^{\varepsilon , \delta }]({\hat{x}}, {\hat{t}}) + K_{(\rho , {\hat{t}})}[u^{\varepsilon , \delta }]({\hat{x}}, {\hat{t}}) + H \left( \frac{{\hat{x}}}{\varepsilon }, L\frac{{\hat{x}} - y}{|{\hat{x}} - y|} \right) \le \eta _\delta . \nonumber \\ \end{aligned}$$

(3.7)

By Proposition 3.2 and Lemma 3.4 (e), we have

$$\begin{aligned} J[u^{\varepsilon , \delta }]({\hat{x}}, {\hat{t}}) = \frac{u^{\varepsilon , \delta }({\hat{x}}, {\hat{t}}) - u^{\varepsilon , \delta }({\hat{x}}, 0)}{{\hat{t}}^\alpha \Gamma (1 - \alpha )} \ge - \frac{C}{\Gamma (1 - \alpha )}, \end{aligned}$$

and

$$\begin{aligned} K_{(\rho , {\hat{t}})}[u^{\varepsilon , \delta }]({\hat{x}}, {\hat{t}})&= \frac{\alpha }{\Gamma (1 - \alpha )} \int _\rho ^{{\hat{t}}} \frac{u^{\varepsilon , \delta }({\hat{x}}, {\hat{t}}) - u^{\varepsilon , \delta }({\hat{x}}, {\hat{t}} - \tau )}{\tau ^{\alpha + 1}} d\tau \\&\ge \frac{\alpha }{\Gamma (1 - \alpha )} \int _{\rho }^{{\hat{t}}} \frac{- C\tau ^\alpha }{\tau ^{\alpha + 1}} d\tau = - \frac{\alpha C}{\Gamma (1 - \alpha )} \log \frac{{\hat{t}}}{\rho }. \end{aligned}$$

Next, by Lemma 3.4 (b), we have

$$\begin{aligned} K_{(0, \rho )}[u^{\varepsilon , \delta }]({\hat{x}}, {\hat{t}})&= \frac{\alpha }{\Gamma (1 - \alpha )} \int _0^\rho \frac{u^{\varepsilon , \delta }({\hat{x}}, {\hat{t}}) - u^{\varepsilon , \delta }({\hat{x}}, {\hat{t}} - \tau )}{\tau ^{\alpha + 1}} d\tau \\&\ge \frac{\alpha }{\Gamma (1 - \alpha )} \int _0^\rho \frac{-\frac{C}{\delta } \tau }{\tau ^{\alpha + 1}} d\tau = - \frac{\alpha }{\Gamma (1 - \alpha )} \frac{C}{\delta } \frac{\rho ^{1 - \alpha }}{1 - \alpha }. \end{aligned}$$

Combining these with (3.7), we obtain

$$\begin{aligned}&H \left( \frac{{\hat{x}}}{\varepsilon }, L\frac{{\hat{x}} - y}{|{\hat{x}} - y|} \right) \le \eta _\delta + \frac{C}{\Gamma (1 - \alpha )} + \frac{\alpha C}{\Gamma (1 - \alpha )} \log \frac{{\hat{t}}}{\rho } + \frac{\alpha }{\Gamma (1 - \alpha )} \frac{C}{\delta } \frac{\rho ^{1 - \alpha }}{1 - \alpha }\\&\quad \le C \left( 1 + \frac{\rho ^{1-\alpha }}{\delta } + \log \frac{{\hat{t}}}{\rho }\right) . \end{aligned}$$

By setting $\rho = {\hat{t}}\delta ^{\frac{1}{1-\alpha }}$, we obtain

$$\begin{aligned} H \left( \frac{{\hat{x}}}{\varepsilon }, L\frac{{\hat{x}} - y}{|{\hat{x}} - y|} \right) \le C (1 + {\hat{t}}^{1-\alpha } + \frac{1}{1-\alpha } |\log \delta |) \le C (1 + |\log \delta |). \end{aligned}$$

(3.8)

If we take $L = L(\delta )$ large, then, by the coercivity of H, we get a contradiction.

Therefore, we have $\Phi (x, {\hat{t}}) \le \Phi ({\hat{x}}, {\hat{t}}) = \Phi (y, {\hat{t}})$, i.e.,

$$\begin{aligned} u^{\varepsilon , \delta }(x, {\hat{t}}) - u^{\varepsilon , \delta }(y, t_0) - L(\delta )|x - y| - \frac{|{\hat{t}} - t_0|^2}{\beta ^2}\le & {} u^{\varepsilon , \delta }(y, {\hat{t}}) - u^{\varepsilon , \delta }(y, t_0) \\{} & {} - \frac{|{\hat{t}} - t_0|^2}{\beta ^2}. \end{aligned}$$

Taking $\beta \rightarrow 0$, we have

$$\begin{aligned} u^{\varepsilon , \delta }(x, t_0) - u^{\varepsilon , \delta }(y, t_0) \le L(\delta ) |x - y| \quad \text{ for } \text{ all } \,\, x \in {\mathbb {R}}^N. \end{aligned}$$

Therefore, by Lemma 3.4 (d), we obtain

$$\begin{aligned} u^\varepsilon (x, t) - u^\varepsilon (y, t) \le u^{\varepsilon , \delta }(x, t) - u^{\varepsilon , \delta }(y, t) + C \delta ^{\frac{\alpha }{2 - \alpha }} \le L(\delta )|x - y| + C \delta ^{\frac{\alpha }{2 - \alpha }} \end{aligned}$$

for some $C > 0$.

Finally, define

$$\begin{aligned} \omega (r) \mathrel {\mathop :}=\inf _{\delta > 0} \{ L(\delta )r + C \delta ^{\frac{\alpha }{2 - \alpha }} \}. \end{aligned}$$

We claim that $\omega (r) \rightarrow 0$ as $r \rightarrow 0$. Letting $\delta _r > 0$ so that $L(\delta _r) = r^{-\frac{1}{2}}$, we have

$$\begin{aligned} 0 \le \omega (r) \le r^{\frac{1}{2}} + C\delta _r^{\frac{\alpha }{2 - \alpha }}. \end{aligned}$$

Since $\delta _r \rightarrow 0$ as $r \rightarrow 0$, taking $r \rightarrow 0$ as above, this completes the proof of modulus of continuity.

In addition, we assume (A5), i.e., there exists a constant $C > 0$ such that

$$\begin{aligned} H(x, p) \ge C^{-1} |p| - C \quad \text{ for } \, \, x, p \in {\mathbb {R}}^N. \end{aligned}$$

By (3.8), we have

$$\begin{aligned} L \le C(1 + |\log \delta |). \end{aligned}$$

By similar argument to the above, we obtain

$$\begin{aligned} u^\varepsilon (x, t) - u^\varepsilon (y, t) \le C (1 + |\log \delta |) |x - y| + C\delta ^{\frac{\alpha }{2 - \alpha }} \end{aligned}$$

(3.9)

for $x, y \in {\mathbb {R}}^N$ and $t \in (0, T]$. Thus, taking the infimum with respect to $\delta$, we have

$$\begin{aligned} u^\varepsilon (x, t) - u^\varepsilon (y, t)&\le \inf _{\delta > 0} \{ C (1 + |\log \delta |) |x - y| + C\delta ^{\frac{\alpha }{2 - \alpha }} \} \\&\le C (1 + |\log |x - y|^{\frac{2-\alpha }{\alpha }} |) |x - y| + C|x - y|^{\frac{2-\alpha }{\alpha } \cdot \frac{\alpha }{2 - \alpha }} \\&\le C \left( 1 + \left| \log |x - y| \right| \right) |x - y|, \end{aligned}$$

which completes the proof. $\square$

Remark 1

In [9, Lemma 6.1], they show that Lipschitz continuous with respect to the space variable. However, applying this result directly to our problem, we obtain

$$\begin{aligned} |u^\varepsilon (x, t) - u^\varepsilon (y, t)| \le \left( \mathrm{Lip\,}[u_0] + \frac{Ct^\alpha }{\varepsilon }\right) |x - y|. \end{aligned}$$

Therefore, this does not give us a uniform Lipschitz estimate. On the other hand, by [9, Lemma 6.1] we see that the limit problem (1.4) has the unique Lipschitz continuous viscosity solution. Therefore, we can expect a $\varepsilon$-uniform Lipschitz estimate, but this still remains open.

4 Perturbed test function method

Lemma 4.1

For each $\varepsilon > 0$, let $u^\varepsilon$ be the viscosity solution of (1.1). Set

$$\begin{aligned} \displaystyle&u^*(x, t) \mathrel {\mathop :}=\limsup _{\varepsilon \rightarrow 0} {}^* u^\varepsilon (x, t) = \lim _{\varepsilon \rightarrow 0} \sup \{ {u^\gamma } (y, s) \mid |x - y| \le \varepsilon , |t - s| \le \varepsilon , { \gamma } \le \varepsilon \}, \\ \displaystyle&u_*(x, t) \mathrel {\mathop :}=\liminf _{\varepsilon \rightarrow 0} {}_* u^\varepsilon (x, t) = \lim _{\varepsilon \rightarrow 0} \inf \{ {u^\gamma } (y, s) \mid |x - y| \le \varepsilon , |t - s| \le \varepsilon , {\gamma } \le \varepsilon \}. \end{aligned}$$

Then, $u^*$ and $u_*$ are a viscosity subsolution and a viscosity supersolution of (1.4), respectively.

Proof

We only prove that $u^*$ is a viscosity subsolution of (1.4), we can similarly prove that $u_*$ is a viscosity supersolution of (1.4).

Let $\varphi \in C^1({\mathbb {R}}^N \times (0, T]) \cap C({\mathbb {R}}^N \times [0, T])$ such that $u^* - \varphi$ has a strict maximum at $({\hat{x}}, {\hat{t}}) \in {\mathbb {R}}^N \times (0, T]$. Let $P = D\varphi ({\hat{x}}, {\hat{t}}) \in {\mathbb {R}}^N$, and let $v \in$ Lip $({\mathbb {T}}^N)$ be a viscosity solution of (1.2). Choose a sequence $\varepsilon _m \rightarrow 0$ $(m \in {\mathbb {N}})$ such that

$$\begin{aligned} \displaystyle u^*({\hat{x}}, {\hat{t}}) = \lim _{m \rightarrow \infty } u^{\varepsilon _m}(x_m, t_m) \quad \text{ for } \, (x_m, t_m) \in {\mathbb {R}}^N \times (0, T). \end{aligned}$$

We set

$$\begin{aligned} \Phi ^{m, a}(x, y, t) \mathrel {\mathop :}=u^{\varepsilon _m}(x, t) - \varphi (x, t) - \varepsilon _m v \left( \frac{y}{\varepsilon _m} \right) - \frac{|x - y|^2}{2a} \quad \text{ for } \, a > 0. \end{aligned}$$

For every $m \in {\mathbb {N}}$ and $a > 0$, there exists $(x_{m, a}, y_{m, a}, t_{m, a}) \in {\mathbb {R}}^N \times {\mathbb {R}}^N \times (0, T)$ such that

$$\begin{aligned} \max _{{\mathbb {R}}^N \times {\mathbb {R}}^N \times [0, T]} \Phi ^{m, a}(x, y, t) = \Phi ^{m, a} (x_{m, a}, y_{m, a}, t_{m, a}) \end{aligned}$$

and up to passing some subsequences

$$\begin{aligned} \left\{ \begin{array}{ll} (x_{m, a}, y_{m, a}, t_{m, a}) \rightarrow (x_m, x_m, t_m) &{} \text{ as } \quad a \rightarrow 0, \\ (x_m, t_m) \rightarrow ({\hat{x}}, {\hat{t}}) &{} \text{ as } \quad m \rightarrow \infty , \\ \displaystyle \lim _{m \rightarrow \infty } \lim _{a \rightarrow \infty } u^{\varepsilon _m}(x_{m, a}, t_{m, a}) = u^*({\hat{x}}, {\hat{t}}). \end{array} \right. \end{aligned}$$

Notice that the function $(x, t) \mapsto \Phi ^{m, a}(x, y_{m, a}, t)$ has a maximum at $(x_{m, a}, t_{m, a})$. Since $u^{\varepsilon _m}$ is a viscosity subsolution of (1.1), we have

$$\begin{aligned}{} & {} \displaystyle J[\varphi ](x_{m, a}, t_{m, a}) + K_{(0, t_{m, a})}[\varphi ](x_{m, a}, t_{m, a}) \nonumber \\{} & {} \qquad H \left( \frac{x_{m, a}}{\varepsilon _m}, D\varphi (x_{m, a}, t_{m, a}) + \frac{x_{m, a} - y_{m, a}}{a} \right) \ \le 0. \end{aligned}$$

(4.1)

On the other hand, notice that the function $y \mapsto -\frac{1}{\varepsilon _m} \Phi ^{m, a}(x_{m, a}, \varepsilon _m y, t_{m, a})$ has a minimum at $\frac{y_{m, a}}{\varepsilon _m}$. Since v is a viscosity supersolution of (1.2), we have

$$\begin{aligned} \displaystyle H \left( \frac{y_{m, a}}{\varepsilon _m}, P + \frac{x_{m, a} - y_{m, a}}{a} \right) \ \ge {\overline{H}}(P). \end{aligned}$$

(4.2)

Since $\Phi ^{m, a}(x_{m, a}, y_{m, a}, t_{m, a}) \ge \Phi ^{m, a}(x_{m, a}, x_{m, a}, t_{m, a})$, we have

$$\begin{aligned} \displaystyle \frac{|x_{m, a} - y_{m, a}|^2}{2a} \le \varepsilon _m \left( v\left( \frac{x_{m, a}}{\varepsilon _m}\right) - v\left( \frac{y_{m, a}}{\varepsilon _m}\right) \right) \le \varepsilon _m L \frac{|x_{m, a} - y_{m, a}|}{\varepsilon _m} = L |x_{m, a} - y_{m, a}|, \end{aligned}$$

where L is a Lipschitz constant of v. Thus, we have $\frac{|x_{m, a} - y_{m, a}|}{a} \le 2\,L$. If necessary, taking a subsequence, we can assume that

$$\begin{aligned} \frac{x_{m, a} - y_{m, a}}{a} \rightarrow Q_m \in {\mathbb {R}}^N \quad \text{ as } \, \, a \rightarrow 0.\nonumber \\ \end{aligned}$$

By Proposition 2.2 and taking the limit $a \rightarrow 0$ in (4.1) and (4.2) to get

$$\begin{aligned} \displaystyle J[\varphi ](x_{m}, t_{m}) + K_{(0, t_{m})}[\varphi ](x_{m}, t_{m}) + H \left( \frac{x_m}{\varepsilon _m}, D\varphi (x_m, t_m) + Q_m \right) \ \le 0, \\ \displaystyle H \left( \frac{x_m}{\varepsilon _m}, P + Q_m \right) \ge {\overline{H}}(P). \end{aligned}$$

Combining the above two inequality, and noting that $|Q_m|$ is bounded, we obtain

$$\begin{aligned} \displaystyle&J[\varphi ](x_{m}, t_{m}) + K_{(0, t_{m})}[\varphi ](x_{m}, t_{m}) + {\overline{H}}(P) \\&\quad \le H \left( \frac{x_m}{\varepsilon _m}, P + Q_m \right) - H \left( \frac{x_m}{\varepsilon _m}, D\varphi (x_m, t_m) + Q_m \right) \\&\quad \le \omega _H (|P - D\varphi (x_m, t_m)|) \end{aligned}$$

for a modulus of continuity of H. Taking the limit $m \rightarrow \infty$ to conclude that

$$\begin{aligned} J[\varphi ]({\hat{x}}, {\hat{t}}) + K_{(0, {\hat{t}})}[\varphi ]({\hat{x}}, {\hat{t}}) + {\overline{H}} ( D\varphi ({\hat{x}}, {\hat{t}}) ) \ \le 0. \end{aligned}$$

$\square$

Lemma 4.2

Let $u^*$ and $u_*$ be the functions defined by Lemma 4.1. Then, $u^* = u_* = {\overline{u}}$ in ${\mathbb {R}}^N \times [0, T]$, where ${\overline{u}}$ is the unique viscosity solution to (1.4).

Proof

By definition of half relaxed limit, we have $u_* \le u^*$ in ${\mathbb {R}}^N \times [0, T]$. On the other hand, by the comparison principle for (1.4), we have $u^* \le u_*$ in ${\mathbb {R}}^N \times [0, T]$, which implies the conclusion. $\square$

Theorem 1.1 is a straightforward result of Lemmas 4.1 and 4.2

5 Rate of convergence

Let us consider the discounted approximation for cell problem (1.2):

$$\begin{aligned} \lambda v^\lambda + H(y, P + Dv^\lambda ) = 0 \quad \text{ in } \, \, {\mathbb {R}}^N. \end{aligned}$$

(5.1)

We recall some results of $v^\lambda$, and refer to [3, Lemma 2.3] and [20, Lemma 4.38] for proofs.

Lemma 5.1

• (a) There exists a constant $C > 0$ independent of $\lambda > 0$ such that

  $$\begin{aligned} \lambda |v^\lambda (y, p) - v^\lambda (y, q)| \le C|p - q| \quad \text{ for } \, \, y, p, q \in {\mathbb {R}}^N. \end{aligned}$$

  In particular, $|{\overline{H}}(p) - {\overline{H}}(q)| \le C|p - q|$ for $p, q \in {\mathbb {R}}^N$.

• (b) For any $p \in {\mathbb {R}}^N$, there exists a constant $C > 0$ independent of $\lambda > 0$ and p such that

  $$\begin{aligned} |\lambda v^\lambda (y, p) + {\overline{H}}(p)| \le C (1 + |p|) \lambda \quad \text{ for } \, \, y \in {\mathbb {R}}^N. \end{aligned}$$

The next lemma is a key ingredient to prove Theorem 1.2.

Lemma 5.2

For $\varepsilon , \lambda > 0$, let $u^\varepsilon$, ${\overline{u}}$ and $v^\lambda$ be the viscosity solutions of (1.1), (1.4) and (5.1), respectively. Let $\Phi : {\mathbb {R}}^{2N}\times [0, T] \mapsto {\mathbb {R}}$ be the function defined by

$$\begin{aligned} \Phi (x, y, t) \mathrel {\mathop :}=u^\varepsilon (x, t) - {\overline{u}}(y, t) - \varepsilon v^{\lambda } \left( \frac{x}{\varepsilon }, \frac{x - y}{\varepsilon ^\beta } \right) - \frac{|x-y|^2}{2\varepsilon ^\beta } - \frac{\delta |y|^2}{2} - \frac{Rt^\alpha }{\Gamma (1 + \alpha )}, \end{aligned}$$

where $\lambda = \varepsilon ^\theta$ and $\theta , \beta , \delta \in (0, 1),$ and $R > 0$. Assume that $\Phi (x, y, t)$ attains a maximum at $({\hat{x}}, {\hat{y}}, {\hat{t}}) \in {\mathbb {R}}^{2N} \times (0, T]$ over ${\mathbb {R}}^{2N} \times [0, T]$ and $\delta ^{\frac{1}{2}} < \lambda$. There exists $\varepsilon _0 \in (0, 1)$ such that for each $\nu \in (0, 1)$, there exists $C > 0$ which is independent of $\varepsilon$ such that

$$\begin{aligned} J[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - J[{\overline{u}}]({\hat{y}}, {\hat{t}}) + K_{(0, {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - K_{(0, {\hat{t}})}[{\overline{u}}]({\hat{y}}, {\hat{t}}) \le C (\varepsilon ^{\theta - \frac{\beta (1-\nu )}{2-\nu }} + \varepsilon ^{1-\theta -\beta }) \end{aligned}$$

for all $\varepsilon \in (0, \varepsilon _0)$.

Proof

STEP 1. We claim that if $1 - \theta - \beta > 0$, then there exists $C > 0$ such that

$$\begin{aligned} |{\hat{y}}| \le C \delta ^{-\frac{1}{2}}. \end{aligned}$$

(5.2)

Additionally, if $\delta < \frac{1}{2}$, then

$$\begin{aligned} |{\hat{x}} - {\hat{y}}| \le C \varepsilon ^{\frac{\beta }{2 - \nu }}. \end{aligned}$$

(5.3)

Noting that $\Phi (0, 0, {\hat{t}}) \le \Phi ({\hat{x}}, {\hat{y}}, {\hat{t}})$, i.e.,

$$\begin{aligned} u^\varepsilon (0, {\hat{t}}) - {\overline{u}}(0, {\hat{t}}) - \varepsilon v^{\lambda }(0, 0)\le & {} u^\varepsilon ({\hat{x}}, {\hat{t}}) - {\overline{u}}({\hat{y}}, {\hat{t}}) - \varepsilon v^{\lambda } \left( \frac{{\hat{x}}}{\varepsilon }, \frac{{\hat{x}} - {\hat{y}}}{\varepsilon ^\beta } \right) \\{} & {} - \frac{|{\hat{x}}-{\hat{y}}|^2}{2\varepsilon ^\beta } - \frac{\delta |{\hat{y}}|^2}{2}, \end{aligned}$$

we have

$$\begin{aligned} \frac{\delta |{\hat{y}}|^2}{2} + \frac{|{\hat{x}}-{\hat{y}}|^2}{2\varepsilon ^\beta }&\le u^\varepsilon ({\hat{x}}, {\hat{t}}) - u^\varepsilon (0, {\hat{t}}) - ({\overline{u}}({\hat{y}}, {\hat{t}}) - {\overline{u}}(0, {\hat{t}}) ) \\&\quad - \varepsilon \left\{ v^{\lambda } \left( \frac{{\hat{x}}}{\varepsilon }, \frac{{\hat{x}} - {\hat{y}}}{\varepsilon ^\beta } \right) - v^{\lambda }(0, 0) \right\} . \end{aligned}$$

Noting that, by Lemma 5.1,

$$\begin{aligned} \left| v^{\lambda } \left( \frac{{\hat{x}}}{\varepsilon }, \frac{{\hat{x}} - {\hat{y}}}{\varepsilon ^\beta } \right) - v^{\lambda }(0, 0) \right|&\le \left| v^{\lambda } \left( \frac{{\hat{x}}}{\varepsilon }, \frac{{\hat{x}} - {\hat{y}}}{\varepsilon ^\beta } \right) - v^{\lambda } \left( \frac{{\hat{x}}}{\varepsilon }, 0 \right) \right| \\&\quad + \left| v^{\lambda } \left( \frac{{\hat{x}}}{\varepsilon }, 0 \right) \right| + |v^{\lambda }(0, 0)| \\&\le \frac{C}{\lambda } \frac{|{\hat{x}} - {\hat{y}}|}{\varepsilon ^\beta } + \frac{2C}{\lambda } (|{\overline{H}}(0)| + \lambda ), \end{aligned}$$

we obtain

$$\begin{aligned} \frac{\delta |{\hat{y}}|^2}{2} + \frac{|{\hat{x}} - {\hat{y}}|^2}{2\varepsilon ^\beta }&\le 2 \Vert u^\varepsilon \Vert _\infty + 2 \Vert {\overline{u}} \Vert _\infty + \varepsilon \left| v^{\lambda } \left( \frac{{\hat{x}}}{\varepsilon }, \frac{{\hat{x}} - {\hat{y}}}{\varepsilon ^\beta } \right) - v^{\lambda }(0, 0) \right| \\&\le C + \frac{C \varepsilon }{\lambda } \left( \frac{|{\hat{x}} - {\hat{y}}|}{\varepsilon ^\beta } + 1 \right) = C + \frac{C \varepsilon ^{1 - \frac{\beta }{2}} |{\hat{x}} - {\hat{y}}|}{\lambda \varepsilon ^{\frac{\beta }{2}}} + \frac{C \varepsilon }{\lambda }. \end{aligned}$$

With $\lambda = \varepsilon ^\theta$, by using the Young inequality, we obtain

$$\begin{aligned} \frac{\delta |{\hat{y}}|^2}{2} + \frac{|{\hat{x}}-{\hat{y}}|^2}{2\varepsilon ^\beta } \le C + C \varepsilon ^{2(1 - \theta - \frac{\beta }{2})} + \frac{|{\hat{x}}-{\hat{y}}|^2}{2\varepsilon ^\beta } + C \varepsilon ^{1-\theta }, \end{aligned}$$

which implies (5.2).

Next, by $\Phi ({\hat{x}}, {\hat{x}}, {\hat{t}}) \le \Phi ({\hat{x}}, {\hat{y}}, {\hat{t}})$, i.e.,

$$\begin{aligned} - {\overline{u}}({\hat{x}}, {\hat{t}}) - \varepsilon v^{\lambda }\left( \frac{{\hat{x}}}{\varepsilon }, 0\right) - \frac{\delta |{\hat{x}}|^2}{2} \le - {\overline{u}}({\hat{y}}, {\hat{t}}) - \varepsilon v^{\lambda }\left( \frac{{\hat{x}}}{\varepsilon }, \frac{{\hat{x}} - {\hat{y}}}{\varepsilon ^\beta }\right) - \frac{|{\hat{x}} - {\hat{y}}|^2}{2 \varepsilon ^\beta } - \frac{\delta |{\hat{y}}|^2}{2}, \end{aligned}$$

and Proposition 3.3, Lemma 5.1, for $\nu \in (0, 1)$, we obtain

$$\begin{aligned} \frac{|{\hat{x}} - {\hat{y}}|^2}{2 \varepsilon ^\beta }&\le {\overline{u}}({\hat{x}}, {\hat{t}}) - {\overline{u}}({\hat{y}}, {\hat{t}}) + \varepsilon \left\{ v^{\lambda }\left( \frac{{\hat{x}}}{\varepsilon }, 0\right) - v^{\lambda }\left( \frac{{\hat{x}}}{\varepsilon }, \frac{{\hat{x}} - {\hat{y}}}{\varepsilon ^\beta }\right) \right\} + \delta \frac{|{\hat{x}}|^2 - |{\hat{y}}|^2}{2} \nonumber \\&\le C|{\hat{x}} - {\hat{y}}|^\nu + \varepsilon \frac{C}{\lambda } \frac{|{\hat{x}} - {\hat{y}}|}{\varepsilon ^\beta } + \frac{\delta }{2} (|{\hat{x}} - {\hat{y}}|^2 + 2 |{\hat{x}} - {\hat{y}}| |{\hat{y}}|). \end{aligned}$$

(5.4)

We notice that $\nu$ can be an arbitrary fixed number in (0, 1) in light of Proposition 3.3. It is sufficient to consider case $|{\hat{x}} - {\hat{y}}| < 1$ if $\varepsilon$ is small enough. Combining (5.4) with (5.2), we have

$$\begin{aligned} (1 - \delta \varepsilon ^\beta ) \frac{|{\hat{x}} - {\hat{y}}|^2}{2 \varepsilon ^\beta }&\le C ( 1 + \varepsilon ^{1 - \theta - \beta }|{\hat{x}} - {\hat{y}}|^{1-\nu } + \delta ^{\frac{1}{2}}|{\hat{x}} - {\hat{y}}|^{1-\nu } ) |{\hat{x}} - {\hat{y}}|^\nu \\&\le C ( 1 + \varepsilon ^{1 - \theta - \beta } + \delta ^{\frac{1}{2}}) |{\hat{x}} - {\hat{y}}|^\nu . \end{aligned}$$

Since $\delta < \frac{1}{2}$ and $0 < 1 - \theta - \beta$, we get (5.3).

STEP 2. For $\gamma > 0$, we consider the auxiliary function $\Psi :{\mathbb {R}}^{4N}\times [0, T]^2 \mapsto {\mathbb {R}}$ defined by

$$\begin{aligned} \begin{aligned} \Psi (x, y, z, \xi , t, s) \mathrel {\mathop :}=\,&u^\varepsilon (x, t) - {\overline{u}}(y, s) - \varepsilon v^\lambda \left( \xi , \frac{z}{\varepsilon ^\beta } \right) \\&- \frac{|x-y|^2}{2\varepsilon ^\beta } - \frac{\delta |y|^2}{2} - \frac{R t^\alpha }{\Gamma (1 + \alpha )} \\&- \frac{|x - \varepsilon \xi |^2 + |x-y-z|^2 + |t - s|^2}{2\gamma } \\&- \frac{|x - {\hat{x}}|^2 + |y - {\hat{y}}|^2 + |t - {\hat{t}}|^2}{2}. \end{aligned} \end{aligned}$$

Since $\Psi (x, y, z, \xi , t, s) \rightarrow -\infty$ as $|x|, |y|, |z|, |\xi | \rightarrow \infty$ for any $(t, s) \in [0, T]^2$ and $\Psi$ is bounded from above, $\Psi$ attains a maximum at a point $(x_\gamma , y_\gamma , z_\gamma , \xi _\gamma , t_\gamma , s_\gamma ) \in {\mathbb {R}}^{4N}\times [0, T]^2$. By the standard argument of the theory of viscosity solutions, we obtain

$$\begin{aligned} (x_\gamma , y_\gamma , z_\gamma , \xi _\gamma , t_\gamma , s_\gamma ) \rightarrow \left( {\hat{x}}, {\hat{y}}, {\hat{x}} - {\hat{y}}, \frac{{\hat{x}}}{\varepsilon }, {\hat{t}}, {\hat{t}} \right) \quad \text{ as } \,\, \gamma \rightarrow 0. \end{aligned}$$

If necessary, taking $\gamma$ sufficiently small, we can assume that $t_\gamma , s_\gamma > 0$ because we assume ${\hat{t}}>0$. Our goal in STEP 2 is to obtain

$$\begin{aligned}{} & {} J[u^\varepsilon ](x_\gamma , t_\gamma ) - J[{\overline{u}}](y_\gamma , s_\gamma ) + K_{(0, t_\gamma )}[u^\varepsilon ] (x_\gamma , t_\gamma ) - K_{(0, s_\gamma )}[{\overline{u}}](y_\gamma , s_\gamma ) \nonumber \\{} & {} \quad \le C (|z_\gamma |\varepsilon ^{\theta -\beta } + \varepsilon ^\theta + \varepsilon ^{1-\theta -\beta }) + E_\gamma , \end{aligned}$$

(5.5)

where $E_\gamma$ is a constant which satisfies $E_\gamma \rightarrow 0$ as $\gamma \rightarrow 0$, and will be defined later.

STEP 2-1. We claim that there exists $C > 0$ such that

$$\begin{aligned} J[u^\varepsilon ](x_\gamma , t_\gamma ) + K_{(0, t_\gamma )}[u^\varepsilon ](x_\gamma , t_\gamma ) + {\overline{H}} \left( \frac{z_\gamma }{\varepsilon ^\beta } \right) \le C ( |z_\gamma |\varepsilon ^{\theta -\beta } {+} \varepsilon ^\theta {+} \varepsilon ^{1 {-} \theta {-} \beta }) {+} E_{1, \gamma },\nonumber \\ \end{aligned}$$

(5.6)

for some $E_{1, \gamma } > 0$ satisfying $E_{1, \gamma } \rightarrow 0$ as $\gamma \rightarrow 0$. Notice that the function $(x, t) \mapsto \Psi (x, y_\gamma , z_\gamma , \xi _\gamma , t, s_\gamma )$ has a maximum at $(x_\gamma , t_\gamma )$. Since $u^\varepsilon$ is a viscosity subsolution of (1.1), by Proposition 2.3, we have

$$\begin{aligned} \begin{aligned} J[u^\varepsilon ](x_\gamma , t_\gamma )&+ K_{(0, t_\gamma )}[u^\varepsilon ](x_\gamma , t_\gamma ) \\&+ H \left( \frac{x_\gamma }{\varepsilon }, \frac{x_\gamma - y_\gamma }{\varepsilon ^\beta } + \frac{x_\gamma - \varepsilon \xi _\gamma + x_\gamma - y_\gamma - z_\gamma }{\gamma } + x_\gamma - {\hat{x}} \right) \le 0. \end{aligned}\nonumber \\ \end{aligned}$$

(5.7)

On the other hand, notice that the function $\xi \mapsto - \frac{1}{\varepsilon } \Psi (x_\gamma , y_\gamma , z_\gamma , \xi , t_\gamma , s_\gamma )$ has a minimum at $\xi _\gamma$. Since $v^\lambda$ is a viscosity supersolution of (5.1), we obtain

$$\begin{aligned} \lambda v^\lambda \left( \xi _\gamma , \frac{z_\gamma }{\varepsilon ^\beta } \right) + H \left( \xi _\gamma , \frac{z_\gamma }{\varepsilon ^\beta } + \frac{x_\gamma - \varepsilon \xi _\gamma }{\gamma } \right) \ge 0. \end{aligned}$$

(5.8)

By $\Psi (x_\gamma , y_\gamma , x_\gamma - y_\gamma , \xi _\gamma , t_\gamma , s_\gamma ) \le \Psi (x_\gamma , y_\gamma , z_\gamma , \xi _\gamma , t_\gamma , s_\gamma )$ and Lemma 5.1, we see that

$$\begin{aligned} \frac{|x_\gamma - y_\gamma - z_\gamma |^2}{2\gamma }&\le \varepsilon \left\{ v^\lambda \left( \xi _\gamma , \frac{x_\gamma - y_\gamma }{\varepsilon ^\beta } \right) - v^\lambda \left( \xi _\gamma , \frac{z_\gamma }{\varepsilon ^\beta } \right) \right\} \\&\le \varepsilon C \frac{1}{\lambda } \frac{|x_\gamma - y_\gamma - z_\gamma |}{\varepsilon ^\beta } = C \varepsilon ^{1 - \theta - \beta } |x_\gamma - y_\gamma - z_\gamma |, \end{aligned}$$

which implies

$$\begin{aligned} \frac{|x_\gamma - y_\gamma - z_\gamma |}{\gamma } \le C \varepsilon ^{1 - \theta - \beta }. \end{aligned}$$

(5.9)

Combining (5.7) with (5.8) and (5.9), by Lemma 5.1, we obtain

$$\begin{aligned}&J[u^\varepsilon ](x_\gamma , t_\gamma ) + K_{(0, t_\gamma )}[u^\varepsilon ](x_\gamma , t_\gamma ) + {\overline{H}} \left( \frac{z_\gamma }{\varepsilon ^\beta } \right) \\&\quad \le \, \lambda v^\lambda \left( \xi _\gamma , \frac{z_\gamma }{\varepsilon ^\beta }\right) + {\overline{H}} \left( \frac{z_\gamma }{\varepsilon ^\beta } \right) + H \left( \xi _\gamma , \frac{z_\gamma }{\varepsilon ^\beta } + \frac{x_\gamma - \varepsilon \xi _\gamma }{\gamma } \right) \\&\qquad - H \left( \frac{x_\gamma }{\varepsilon }, \frac{x_\gamma - y_\gamma }{\varepsilon ^\beta } + \frac{x_\gamma - \varepsilon \xi _\gamma + x_\gamma - y_\gamma - z_\gamma }{\gamma } + x_\gamma - {\hat{x}} \right) \\&\quad \le \, C \left( 1 + \left| \frac{z_\gamma }{\varepsilon ^\beta } \right| \right) \lambda + C \left( \left| \frac{x_\gamma }{\varepsilon } - \xi _\gamma \right| + \frac{|x_\gamma - y_\gamma - z_\gamma |}{\varepsilon ^\beta } + \frac{|x_\gamma - y_\gamma - z_\gamma |}{\gamma } + |x_\gamma - {\hat{x}}| \right) \\&\quad \le \, C( |z_\gamma |\varepsilon ^{\theta -\beta } + \varepsilon ^\theta + \varepsilon ^{1 - \theta - \beta }) + C \left( \left| \frac{x_\gamma }{\varepsilon } - \xi _\gamma \right| + \frac{|x_\gamma - y_\gamma - z_\gamma |}{\varepsilon ^\beta } + |x_\gamma - {\hat{x}}| \right) \\&\quad =\mathrel {\mathop :}\, C( |z_\gamma |\varepsilon ^{\theta -\beta } + \varepsilon ^\theta + \varepsilon ^{1 - \theta - \beta }) + E_{1, \gamma }. \end{aligned}$$

STEP 2-2. Next, we claim that there exists $C > 0$ such that

$$\begin{aligned} J[{\overline{u}}](y_\gamma , s_\gamma ) + K_{(0, s_\gamma )}[{\overline{u}}](y_\gamma , s_\gamma ) + {\overline{H}} \left( \frac{z_\gamma }{\varepsilon ^\beta } \right) \ge - C ( \varepsilon ^{1 - \theta - \beta } + \varepsilon ^\theta )- E_{2, \gamma }\nonumber \\ \end{aligned}$$

(5.10)

for some $E_{2, \gamma } > 0$ satisfying $E_{2, \gamma } \rightarrow 0$ as $\gamma \rightarrow 0$. Notice that $(y, s) \mapsto - \Psi (x_\gamma , y, z_\gamma , \xi _\gamma , t_\gamma , s)$ has a minimum at $(y_\gamma , s_\gamma )$. By Proposition 2.3, we have

$$\begin{aligned} J[{\overline{u}}](y_\gamma , s_\gamma ) + K_{(0, s_\gamma )}[{\overline{u}}](y_\gamma , s_\gamma ) + {\overline{H}} \left( \frac{x_\gamma - y_\gamma }{\varepsilon ^\beta } - \delta y_\gamma + \frac{x_\gamma - y_\gamma - z_\gamma }{\gamma } - (y_\gamma - {\hat{y}}) \right) \ge 0. \end{aligned}$$

Thus, by (5.2), (5.9) and Lemma 5.1, we obtain

$$\begin{aligned}&J[{\overline{u}}](y_\gamma , s_\gamma ) + K_{(0, s_\gamma )}[{\overline{u}}](y_\gamma , s_\gamma ) + {\overline{H}} \left( \frac{z_\gamma }{\varepsilon ^\beta } \right) \\&\quad \ge {\overline{H}} \left( \frac{z_\gamma }{\varepsilon ^\beta } \right) - {\overline{H}} \left( \frac{x_\gamma - y_\gamma }{\varepsilon ^\beta } - \delta y_\gamma + \frac{x_\gamma - y_\gamma - z_\gamma }{\gamma } - (y_\gamma - {\hat{y}}) \right) \\&\quad \ge - C \left\{ \frac{|x_\gamma - y_\gamma - z_\gamma |}{\varepsilon ^\beta } + \frac{|x_\gamma - y_\gamma - z_\gamma |}{\gamma } + (1 + \delta )|y_\gamma - {\hat{y}}| + \delta |{\hat{y}}| \right\} \\&\quad \ge - C ( \varepsilon ^{1 - \theta - \beta } + \delta ^{\frac{1}{2}} ) - \left\{ \frac{|x_\gamma - y_\gamma - z_\gamma |}{\varepsilon ^\beta } + (1 + \delta )|y_\gamma - {\hat{y}}| \right\} \\&\quad =\mathrel {\mathop :}- C (\varepsilon ^{1 - \theta - \beta } + \delta ^{\frac{1}{2}} ) - E_{2, \gamma }. \end{aligned}$$

Take $0< \delta < \frac{1}{2}$ satisfying $\delta ^{\frac{1}{2}} < \lambda = \varepsilon ^\theta$ to get (5.10). Combining (5.6) with (5.10), we get

$$\begin{aligned} \begin{aligned} J[u^\varepsilon ](x_\gamma , t_\gamma ) - J[{\overline{u}}](y_\gamma , s_\gamma ) + K_{(0, t_\gamma )}[u^\varepsilon ]&(x_\gamma , t_\gamma ) - K_{(0, s_\gamma )}[{\overline{u}}](y_\gamma , s_\gamma ) \\&\le C (|z_\gamma |\varepsilon ^{\theta -\beta } + \varepsilon ^\theta + \varepsilon ^{1-\theta -\beta }) + E_{1, \gamma } + E_{2, \gamma }. \end{aligned} \end{aligned}$$

We set $E_\gamma \mathrel {\mathop :}=E_{1, \gamma } + E_{2, \gamma }$. This completes STEP 2.

STEP 3. Finally, we claim that taking the limit infimum $\gamma \rightarrow 0$ in (5.5) yields that

$$\begin{aligned} J[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - J[{\overline{u}}]({\hat{y}}, {\hat{t}}) + K_{(0, {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - K_{(0, {\hat{t}})}[{\overline{u}}]({\hat{y}}, {\hat{t}}) \le C (\varepsilon ^{\theta - \frac{\beta (1-\nu )}{2-\nu }} + \varepsilon ^\theta + \varepsilon ^{1-\theta -\beta }). \end{aligned}$$

We set

$$\begin{aligned} I_1&\mathrel {\mathop :}=J[u^\varepsilon ](x_\gamma , t_\gamma ) - J[{\overline{u}}](y_\gamma , s_\gamma ) \\&= \frac{u^\varepsilon (x_\gamma , t_\gamma ) - u^\varepsilon (x_\gamma , 0)}{t_\gamma ^\alpha \Gamma (1 - \alpha )} - \frac{{\overline{u}}(y_\gamma , s_\gamma ) - {\overline{u}}(y_\gamma , 0)}{s_\gamma ^\alpha \Gamma (1 - \alpha )}, \\ I_2&\mathrel {\mathop :}=\frac{\Gamma (1 - \alpha )}{\alpha } \{ K_{(0, r)}[u^\varepsilon ] (x_\gamma , t_\gamma ) - K_{(0, r)}[{\overline{u}}](y_\gamma , s_\gamma ) \}\\&= \int _0^r \frac{u^\varepsilon (x_\gamma , t_\gamma ) - u^\varepsilon (x_\gamma , t_\gamma - \tau )}{\tau ^{\alpha + 1}} d\tau - \int _0^r \frac{{\overline{u}}(y_\gamma , s_\gamma ) - {\overline{u}}(y_\gamma , s_\gamma - \tau )}{\tau ^{\alpha + 1}} d\tau , \\ I_3&\mathrel {\mathop :}=\frac{\Gamma (1 - \alpha )}{\alpha } \{ K_{(r, t_\gamma )}[u^\varepsilon ] (x_\gamma , t_\gamma ) - K_{(r, s_\gamma )}[{\overline{u}}](y_\gamma , s_\gamma ) \} \\&= \int _r^{t_\gamma } \frac{u^\varepsilon (x_\gamma , t_\gamma ) - u^\varepsilon (x_\gamma , t_\gamma - \tau )}{\tau ^{\alpha + 1}} d\tau - \int _r^{s_\gamma } \frac{{\overline{u}}(y_\gamma , s_\gamma ) - {\overline{u}}(y_\gamma , s_\gamma - \tau )}{\tau ^{\alpha + 1}} d\tau \end{aligned}$$

for $0< r < \min \{t_\gamma , s_\gamma \}$. Notice that $I_1 + \frac{\alpha }{\Gamma (1 - \alpha )}(I_2 + I_3) = J[u^\varepsilon ](x_\gamma , t_\gamma ) - J[{\overline{u}}](y_\gamma , s_\gamma ) + K_{(0, t_\gamma )}[u^\varepsilon ] (x_\gamma , t_\gamma ) - K_{(0, s_\gamma )}[{\overline{u}}](y_\gamma , s_\gamma )$.

First, by the continuity of $u^\varepsilon$ and ${\overline{u}}$, we have

$$\begin{aligned} \lim _{\gamma \rightarrow 0} I_1 = J[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - J[{\overline{u}}]({\hat{y}}, {\hat{t}}). \end{aligned}$$

(5.11)

Next, by $\Psi (x_\gamma , y_\gamma , z_\gamma , \xi _\gamma , t_\gamma , s_\gamma ) \ge \Psi (x_\gamma , y_\gamma , z_\gamma , \xi _\gamma , t_\gamma - \tau , s_\gamma - \tau )$ for all $\tau \in [0, r]$, i.e.,

$$\begin{aligned} \begin{aligned} u^\varepsilon (x_\gamma , t_\gamma ) - {\overline{u}}(y_\gamma , s_\gamma ) -&\frac{R t_\gamma ^\alpha }{\Gamma (1 + \alpha )} - \frac{|t_\gamma - {\hat{t}}|^2}{2} \\&\ge u^\varepsilon (x_\gamma , t_\gamma - \tau ) - {\overline{u}}(y_\gamma , s_\gamma - \tau ) - \frac{R (t_\gamma - \tau )^\alpha }{\Gamma (1 + \alpha )} - \frac{|t_\gamma - \tau - {\hat{t}}|^2}{2}, \end{aligned} \end{aligned}$$

we have

$$\begin{aligned} I_2&\ge \int _0^r \left\{ R\frac{t_\gamma ^\alpha - (t_\gamma - \tau )^\alpha }{\Gamma (1 + \alpha )} + \frac{|t_\gamma - {\hat{t}}|^2 - |t_\gamma - \tau - {\hat{t}}|^2}{2} \right\} \frac{d\tau }{\tau ^{\alpha + 1}} \nonumber \\&\ge \int _0^r \frac{|t_\gamma - {\hat{t}}|^2 - |t_\gamma - \tau - {\hat{t}}|^2}{2} \frac{d\tau }{\tau ^{\alpha + 1}} = \int _0^r \frac{2(t_\gamma - {\hat{t}})\tau - \tau ^2}{2} \frac{d\tau }{\tau ^{\alpha + 1}} \nonumber \\&= \frac{(t_\gamma - {\hat{t}}) r^{1 - \alpha }}{1 - \alpha } - \frac{r^{2 - \alpha }}{2 (2 - \alpha )} \rightarrow - \frac{r^{2 - \alpha }}{2 (2 - \alpha )} =\mathrel {\mathop :}- C_r \quad \text{ as } \,\, \gamma \rightarrow 0, \end{aligned}$$

(5.12)

Finally, we see that

$$\begin{aligned} I_3&= \int _r^T \left( \frac{u^\varepsilon (x_\gamma , t_\gamma ) - u^\varepsilon (x_\gamma , t_\gamma - \tau )}{\tau ^{\alpha + 1}} \chi _{(r, t_\gamma )}(\tau ) - \frac{{\overline{u}}(y_\gamma , s_\gamma ) - {\overline{u}}(y_\gamma , s_\gamma - \tau )}{\tau ^{\alpha + 1}} \chi _{(r, s_\gamma )}(\tau ) \right) d\tau , \end{aligned}$$

where $\chi _I$ is the characteristic function. Notice that by Proposition 3.2

$$\begin{aligned} \begin{aligned} \frac{u^\varepsilon (x_\gamma , t_\gamma ) - u^\varepsilon (x_\gamma , t_\gamma - \tau )}{\tau ^{\alpha + 1}} \chi _{(r, t_\gamma )}(\tau ) - \frac{{\overline{u}}(y_\gamma , s_\gamma ) - {\overline{u}}(y_\gamma , s_\gamma - \tau )}{\tau ^{\alpha + 1}} \chi _{(r, s_\gamma )}(\tau ) \\ \ge - \frac{2C}{\tau } \chi _{(r, T)}(\tau ). \end{aligned} \end{aligned}$$

Since the right-hand side is integrable on [0, T], by Fatou’s lemma, we obtain

$$\begin{aligned} \begin{aligned} \liminf _{\gamma \rightarrow 0} I_3&\ge \int _r^{{\hat{t}}} \frac{u^\varepsilon ({\hat{x}}, {\hat{t}}) - u^\varepsilon ({\hat{x}}, {\hat{t}} - \tau )}{\tau ^{\alpha + 1}} d\tau - \int _r^{{\hat{t}}} \frac{{\overline{u}}({\hat{y}}, {\hat{t}}) - {\overline{u}}({\hat{y}}, {\hat{t}} - \tau )}{\tau ^{\alpha + 1}} d\tau \\&= \frac{\Gamma (1 - \alpha )}{\alpha } ( K_{(r, {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - K_{(r, {\hat{t}})}[{\overline{u}}]({\hat{y}}, {\hat{t}}) ). \end{aligned} \end{aligned}$$

(5.13)

Combining (5.11), (5.12) with (5.13), and sending $\gamma \rightarrow 0$ in (5.5), by (5.3) we obtain

$$\begin{aligned}&J[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - J[{\overline{u}}]({\hat{y}}, {\hat{t}}) - \frac{\alpha C_r}{\Gamma (1 - \alpha )} + K_{(r, {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - K_{(r, {\hat{t}})}[{\overline{u}}]({\hat{y}}, {\hat{t}}) \\&\quad \le C (|{\hat{x}} - {\hat{y}}|\varepsilon ^{\theta - \beta } + \varepsilon ^\theta + \varepsilon ^{1-\theta -\beta }) \le C(\varepsilon ^{\theta -\frac{\beta (1-\nu )}{2-\nu }}+\varepsilon ^\theta +\varepsilon ^{1-\theta -\beta }). \end{aligned}$$

Note that as in the proof of (2.4) in Proposition 2.3, we can prove

$$\begin{aligned} K_{(r, {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - K_{(r, {\hat{t}})}[{\overline{u}}]({\hat{y}}, {\hat{t}}) \ge -C \end{aligned}$$

for some $C>0$ independent of r. Since $C_r \rightarrow 0$ as $r \rightarrow 0$ by (5.12), we obtain the desired result. $\square$

Proof of Theorem 1.2.

Let $v^\lambda$ be the viscosity solution of (5.1). We consider the auxiliary function $\Phi : {\mathbb {R}}^{2N}\times [0, T] \mapsto {\mathbb {R}}$ defined in Lemma 5.2, i.e.,

$$\begin{aligned} \Phi (x, y, t) \mathrel {\mathop :}=u^\varepsilon (x, t) - {\overline{u}}(y, t) - \varepsilon v^{\lambda } \left( \frac{x}{\varepsilon }, \frac{x - y}{\varepsilon ^\beta } \right) - \frac{|x-y|^2}{2\varepsilon ^\beta } - \frac{\delta |y|^2}{2} - \frac{R t^\alpha }{\Gamma (1 + \alpha )}, \end{aligned}$$

where $\lambda = \varepsilon ^\theta , \theta , \beta \in (0, 1), 0< \delta < \lambda ^2$ and $R > 0$. Noting that $\Phi$ is a bounded from above and $\Phi (x, y, t) \rightarrow -\infty$ as $|x|, |y| \rightarrow \infty$ for any $t \in [0, T]$, we see that $\Phi$ attains a maximum at $({\hat{x}}, {\hat{y}}, {\hat{t}}) \in {\mathbb {R}}^{2N}\times [0, T]$.

First, we set $R:= R^\prime (\varepsilon ^{\theta - \frac{\beta (1-\nu )}{2-\nu }}+ \varepsilon ^{1-\theta -\beta })$. We claim that if $R^\prime$ is sufficiently large, then ${\hat{t}} = 0$. Suppose that ${\hat{t}} > 0$. By Lemma 5.2, we have

$$\begin{aligned} J[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - J[{\overline{u}}]({\hat{y}}, {\hat{t}}) + K_{(0, {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - K_{(0, {\hat{t}})}[{\overline{u}}]({\hat{y}}, {\hat{t}}) \le C (\varepsilon ^{\theta - \frac{\beta (1-\nu )}{2-\nu }}+ \varepsilon ^{1-\theta -\beta }).\nonumber \\ \end{aligned}$$

(5.14)

By $\Phi ({\hat{x}}, {\hat{y}}, 0) \le \Phi ({\hat{x}}, {\hat{y}}, {\hat{t}})$ and $\Phi ({\hat{x}}, {\hat{y}}, {\hat{t}} - \tau ) \le \Phi ({\hat{x}}, {\hat{y}}, {\hat{t}})$ for all $\tau \in [0, {\hat{t}}]$, we obtain

$$\begin{aligned}&J[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - J[{\overline{u}}]({\hat{y}}, {\hat{t}})\ge \frac{R}{\Gamma (1 - \alpha ) \Gamma (1 + \alpha )}, \\&K_{(0, {\hat{t}})}[u^\varepsilon ]({\hat{x}}, {\hat{t}}) - K_{(0, {\hat{t}})}[{\overline{u}}]({\hat{y}}, {\hat{t}}) \ge \frac{\alpha R}{\Gamma (1 - \alpha ) \Gamma (1 + \alpha )} \int _0^{{\hat{t}}} \frac{{\hat{t}}^\alpha - ({\hat{t}} - \tau )^\alpha }{\tau ^{\alpha + 1}} d\tau \ge 0. \end{aligned}$$

Combining two inequalities above with (5.14), we have

$$\begin{aligned} R \le C \Gamma (1 - \alpha ) \Gamma (1 + \alpha ) (\varepsilon ^{\theta - \frac{\beta (1-\nu )}{2-\nu }}+ \varepsilon ^{1-\theta -\beta }). \end{aligned}$$

Thus, we set $R^\prime > C \Gamma (1 - \alpha ) \Gamma (1 + \alpha )$, which implies a contradiction.

By a similar argument to the proof of Lemma 5.2, we have $|{\hat{x}} - {\hat{y}}| \le C \varepsilon ^\frac{\beta }{2-\nu }$ if $\delta < \frac{1}{2}$ and $0< \theta < 1 - \beta$.

Next, by $\Phi (x, x, t) \le \Phi ({\hat{x}}, {\hat{y}}, {\hat{t}}) = \Phi ({\hat{x}}, {\hat{y}}, 0)$ for all $(x, t) \in {\mathbb {R}}^N \times (0, T)$, we obtain, by Lemma 5.1,

$$\begin{aligned}&u^\varepsilon (x, t) - {\overline{u}}(x, t) \\&\quad \le u^\varepsilon ({\hat{x}}, 0) - {\overline{u}}({\hat{y}}, 0) + \varepsilon \left\{ v^{\lambda } \left( \frac{x}{\varepsilon }, 0 \right) - v^{\lambda } \left( \frac{{\hat{x}}}{\varepsilon }, \frac{{\hat{x}} - {\hat{y}}}{\varepsilon ^\beta } \right) \right\} + \frac{\delta |x|^2}{2} + \frac{R t^\alpha }{\Gamma (1 + \alpha )} \\&\quad \le \mathrm{Lip\,}[u_0] |{\hat{x}} - {\hat{y}}| + \frac{C\varepsilon }{\lambda } \left( \frac{|{\hat{x}} - {\hat{y}}|}{\varepsilon ^\beta } + 1 \right) + \frac{\delta |x|^2}{2} + (\varepsilon ^{\theta - \frac{\beta (1-\nu )}{2-\nu }} + \varepsilon ^{1 - \theta - \beta }) \frac{R^\prime T^\alpha }{\Gamma (1 + \alpha )} \\&\quad \le C ({\varepsilon ^\frac{\beta }{2-\nu } + \varepsilon ^{1-\theta +\frac{\beta }{2-\nu }-\beta } } + \varepsilon ^{\theta - \frac{\beta (1-\nu )}{2-\nu }}+ \varepsilon ^{1 - \theta - \beta }) + \frac{\delta |x|^2}{2} \end{aligned}$$

for all $(x, t) \in {\mathbb {R}}^N \times (0, T)$. Therefore, by the optimal choice of parameters $\theta = \beta = \frac{1}{3}$, and sending $\delta \rightarrow 0$, we obtain

$$\begin{aligned} u^\varepsilon (x, t) - {\overline{u}}(x, t) \le C \varepsilon ^{\frac{1}{3(2-\nu )}}. \end{aligned}$$

By a symmetric argument, we get the desired result. $\square$