Proof of Lemma 2.1

(i) Suppose that (2.2) does not hold. Then, by the lower semi-continuity of $ u $ on $ \bar {\Omega } $ , there exists a point $ \bar {x} \in \bar {\Omega } $ such that

$$\begin{align*}u(\bar{x}) = \min_{\bar{\Omega}} u(x) < 0. \end{align*}$$

Since $ u(x) \geq 0 $ for all $ x \in \mathbb {R}^n \setminus \Omega $ , it follows that $ \bar {x} \in \Omega $ ; that is, $ \bar {x} $ is an interior minimum point of $ u $ in $ \Omega $ . Hence, $ \Delta u(\bar {x}) \geq 0 $ . A direct computation then yields

$$\begin{align*}\begin{aligned} (-\Delta)^s u(\bar{x}) - \Delta u(\bar{x}) &\leq C_{n,s}~\text{P.V.} \int_{\mathbb{R}^n} \frac{u(\bar{x}) - u(y)}{|\bar{x} - y|^{n+2s}} \, dy \\ &\leq C_{n,s}~\text{P.V.} \int_{\mathbb{R}^n \setminus \Omega} \frac{u(\bar{x}) - u(y)}{|\bar{x} - y|^{n+2s}} \, dy \\ &< 0, \end{aligned} \end{align*}$$

which contradicts the assumed inequality (2.1). Therefore, (2.2) must hold.

(ii) Based on (2.2), suppose there exists a point $ \bar {x} \in \Omega $ such that

$$\begin{align*}u(\bar{x}) = \min_{\bar{\Omega}} u(x) = 0. \end{align*}$$

Then,

$$\begin{align*}0 \leq (-\Delta)^s u(\bar{x}) - \Delta u(\bar{x}) \leq C_{n,s}~\text{P.V.} \int_{\mathbb{R}^n} \frac{-u(y)}{|\bar{x} - y|^{n+2s}} \, dy, \end{align*}$$

and since $ u \geq 0 $ in $ \mathbb {R}^n $ , we conclude that

$$\begin{align*}u(x) = 0 \quad \text{almost everywhere in } \mathbb{R}^n. \end{align*}$$

(iii) If $ \Omega $ is unbounded, the condition $ \lim _{|x| \to \infty } u(x) \geq 0 $ guarantees that any negative minimum must be attained at some point. One can then repeat the argument in part (i) to derive a contradiction.

This completes the proof of Lemma 2.1.

Proof of Lemma 2.3

Suppose (2.5) does not hold, then

$$ \begin{align*}A:=\underset{\Omega}{\sup}~W_\lambda(x)>0.\end{align*} $$

Thus, there exists a sequence $\{x^k\} \subset \Omega $ such that

(2.6) $$ \begin{align} W_\lambda(x^k)\rightarrow A,~\text{as}~k\rightarrow +\infty. \end{align} $$

Let

$$ \begin{align*}\zeta(x)=\left\{\begin{array}{@{}lll} M e^{\frac{1}{|x|^2-1}},& |x|< 1,\\ 0,& |x|\geq 1, \end{array}\right.\end{align*} $$

where M is a positive constant such that $\underset {\mathbb R^n}{\max }~\zeta (x)=1$ . Obviously, $\zeta (x)$ is radially decreasing with respect to $|x|$ , and its support is $B_1(0)$ . We rescale $\zeta (x)$ by setting

$$ \begin{align*}\Psi_k(x)=\zeta(\frac{x-x^k}{r_k}),\end{align*} $$

where $r_k=\frac {\mathrm{dist}(x^k,T_\lambda )}{2}.$

We claim that $r_k$ is bounded away from zero uniformly, that is, there exists a constant $\hat C$ such that

(2.7) $$ \begin{align} r_k=\frac{\mathrm{dist}(x^k,T_\lambda)}{2}\geq \hat C>0,~k\in \mathbb N. \end{align} $$

Otherwise, $\{x^k\}$ will converge to a point on $T_\lambda $ . Then, by the uniform continuity of $W_\lambda $ , we have

$$ \begin{align*} W_\lambda(x^k)\rightarrow 0, \,\,\text{as}\,\, k\to \infty. \end{align*} $$

This contradicts assumption (2.6). Hence, (2.7) holds.

To proceed with the proof, we choose a sequence $\varepsilon _k>0$ with $\varepsilon _k \to 0$ such that

$$ \begin{align*}W_\lambda(x^k)+\varepsilon_k\Psi_k(x^k)>A\geq W_\lambda(x)+\varepsilon_k\Psi_k(x),~\forall~x\in \Sigma_\lambda \backslash B_{r_k}(x^k).\end{align*} $$

Hence, there exists a point $ \bar x^k \in B_{r_k}(x^k) \subset \Omega $ satisfying

$$ \begin{align*}W_\lambda(\bar x^k)+\varepsilon_k\Psi_k(\bar x^k)=\underset{\Sigma_\lambda}{\max} (W_\lambda( x )+\varepsilon_k\Psi_k( x ))>A.\end{align*} $$

On the other hand, a straightforward calculation shows that

(2.8) $$ \begin{align} \begin{aligned} &(-\Delta)^s (W_\lambda(\bar x^k)+\varepsilon_k\Psi_k(\bar x^k) )-\Delta (W_\lambda(\bar x^k)+\varepsilon_k\Psi_k(\bar x^k))\\ \geq\ &C_{n,s}P.V.\int_{\mathbb R^n} \frac{W_\lambda(\bar x^k)+\varepsilon_k\Psi_k(\bar x^k)-W_\lambda(y)-\varepsilon_k\Psi_k(y)}{|\bar x^k -y|^{n+2s}} dy\\ =\ &C_{n,s}P.V.\int_{\Sigma_\lambda} \frac{W_\lambda(\bar x^k)+\varepsilon_k\Psi_k(\bar x^k)-W_\lambda(y)-\varepsilon_k\Psi_k(y)}{|\bar x^k -y|^{n+2s}} dy\\ &+C_{n,s} \int_{\Sigma_\lambda} \frac{W_\lambda(\bar x^k)+\varepsilon_k\Psi_k(\bar x^k)-W_\lambda(y^\lambda)-\varepsilon_k\Psi_k(y^\lambda)}{|\bar x^k -y^\lambda|^{n+2s}} dy\\ \geq\ &C_{n,s} \int_{\Sigma_\lambda} \frac{2(W_\lambda(\bar x^k)+\varepsilon_k\Psi_k(\bar x^k)) -\varepsilon_k\Psi_k(y^\lambda )}{|\bar x^k -y^\lambda|^{n+2s}} dy\\ \geq\ &C_{n,s} C \int_{\Sigma_\lambda} \frac{2(W_\lambda(\bar x^k)+\varepsilon_k\Psi_k(\bar x^k)) -\varepsilon_k\Psi_k(y )}{| x^k -y^\lambda|^{n+2s}} dy\\ \geq \ &C_{n,s}C\int_{\Sigma_\lambda}\frac{2A-\varepsilon_k}{| x^k -y^\lambda|}^{n+2s} dy\\ \geq\ &\frac{C_{n,s} C(2A-\varepsilon_k)}{r_k^{2s}}\int_{\mathbb R^n \backslash\Sigma_2} \frac{1}{|z|^{n+2s}} dz\\ \geq\ & \tilde C \frac{2A-\varepsilon_k }{r_k^{2s}}, \end{aligned} \end{align} $$

where $z=\frac {x^k-y^\lambda }{r_k},$ and we have used the fact that

$$ \begin{align*}|\bar x^k-y^\lambda|\leq |x^k-y^\lambda|+|\bar x^k-x^k|\leq \frac{3}{2}|x^k-y^\lambda|.\end{align*} $$

On the other hand, recalling that $\underset {\Omega }{\sup }~W_\lambda (x)=\underset {\Sigma _\lambda }{\sup }~W_\lambda (x)=A>0$ and $r_k\geq \hat C>0$ , we can deduce from (2.4) that

(2.9) $$ \begin{align} \begin{aligned} &(-\Delta)^s (W_\lambda(\bar x^k)+\varepsilon_k\Psi_k(\bar x^k) )-\Delta (W_\lambda(\bar x^k)+\varepsilon_k\Psi_k(\bar x^k)) \\ \leq \ & C_0(x)f(\bar x^k_1)W_\lambda(\bar x^k)+ \varepsilon_k(-\Delta)^s \Psi_k(\bar x^k)-\varepsilon_k \Delta \Psi_k(\bar x^k)\\ \leq\ &C_0(x)f(\bar x^k_1) W_\lambda(\bar x^k)+\frac{ c_1\varepsilon_k}{r_k^{2s}}+\frac{ c_2\varepsilon_k}{r_k^{2}}\\ \leq\ & C_0(x)f(\bar x^k_1) A+ \frac{ c_3\varepsilon_k}{r_k^{2s}}. \end{aligned} \end{align} $$

Combining (2.8) with (2.9), we obtain

$$ \begin{align*}\tilde C \frac{2A-\varepsilon_k}{r_k^{2s}}\leq C_0(x)f(\bar x^k_1) A +\frac{c_3\varepsilon_k}{r_k^{2s}} ,\end{align*} $$

therefore,

$$ \begin{align*}0< 2A \tilde C \leq C_0(x)f(\bar x^k_1) A r_k^{2s}+ \tilde C \varepsilon_k +c_3\varepsilon_k \leq C_0(x)A f(\bar x^k_1) |\bar x^k_1|^{2s}+ \tilde C \varepsilon_k +c_3\varepsilon_k.\end{align*} $$

Since $\lambda $ is chosen sufficiently negative (independently of A), letting $\varepsilon _k \to 0$ as $k \to \infty $ and using (2.3), we arrive at a contradiction. Therefore, (2.5) holds, and the proof of Lemma 2.3 is complete.

Proof We first estimate $((-\Delta )^s-\Delta )\varphi (x)$ in $ B_1^c(0),$ that is, we show that

(2.12) $$ \begin{align} \left| ((-\Delta)^s -\Delta)\varphi(x) \right| \leq C\frac{1}{\mathrm{dist}^s(x, \partial B_1(0))},\,\, x\in B_1^c(0). \end{align} $$

According to [Reference Biswas, Modasiya and Sen10, Reference Mou and Zhang39] and [Reference Su, Valdinoci, Wei and Zhang43, Theorem B.1], the eigenfunction satisfies $\varphi (x)\in C^{2,\kappa }_{loc}(B_1(0))\cap C^{1,\kappa }(\overline {B_1(0)})$ for any $\kappa \in (0,\min \{1,2-2s\})$ , and

$$\begin{align*}\varphi(y) \le C\,\mathrm{dist}(y,\partial B_1(0)),\,\, y\in B_1(0). \end{align*}$$

Now, for $x \in \mathbb {R}^n \backslash \overline {B_1(0)}, $ since $\varphi (x)=0$ , we have

(2.13) $$ \begin{align} \begin{aligned} \left |( (-\Delta)^s-\Delta) \varphi(x)\right| &= C_{n, s}\left | PV \int_{\mathbb{R}^n}\frac{0-\varphi(y)}{|x-y|^{n+2s}}dy\right|\\ &=C_{n, s}\left | \int_{B_1(0)}\frac{\varphi(y)}{|x-y|^{n+2s}}dy\right|\\ &\leq C \int_{B_1(0)}\frac{\mathrm{dist}(y, \partial B_1(0))}{|x-y|^{n+2s}}dy\\ &\leq C \int_{B_1(0)}\frac{(\mathrm{dist}(y, \partial B_1(0)))^{s}}{|x-y|^{n+2s}}dy\\ &\leq C \int_{B_1(0)}\frac{1}{|x-y|^{n+s}}dy, \end{aligned} \end{align} $$

where we have used the fact that

$$\begin{align*}\mathrm{dist}(y, \partial B_1(0)) \leq |x - y|, \,\, x\in \mathbb{R}^n \backslash \overline {B_1(0)},\,\, y\in B_1(0). \end{align*}$$

Next, we estimate the last integral in (2.13). Denote $x=(x_1, 0'), \,\, x_1<0,$ and

$$ \begin{align*}D=\{y \mid 0 < y_1< 2,\,\, |y'|<1 \}. \end{align*} $$

We then obtain

$$ \begin{align*} \begin{aligned} \int_{B_1(0)}\frac{1}{|x-y|^{n+s}}dy & =\int_{B_1(1,0') }\frac{1}{|x-(y_1-1,y')|^{n+s}} dy \\ &\leq \int_{D} \frac{1}{|x-(y_1-1,y')|^{n+s}} dy\\ &= C \int_0^2 \int_0^1 \frac{r^{n-2}}{(r^2+(y_1-x_1-1)^2)^{\frac{n+s}{2}}}drdy_1\\ &= C \int_0^2 \frac{1}{|y_1-x_1-1|^{s+1}}dy_1\\ &\leq C \frac{1}{|x_1|^{s}}. \end{aligned} \end{align*} $$

Combining (2.13) with (2.14), we arrive at (2.12).

Secondly, we prove (2.10). For simplicity, let $v(z)= \rho (x-z).$ Therefore, we need to show that

$$ \begin{align*} \int_{\mathbb{R}^n} ((-\Delta)^s -\Delta)\varphi(x) v(x)dx =\int_{\mathbb{R}^n} \varphi(x) ((-\Delta)^s-\Delta) v(x)dx. \end{align*} $$

By the definition of the fractional Laplacian, the above equality is equivalent to

(2.14) $$ \begin{align} \begin{aligned} &\int_{\mathbb{R}^n} v(x)\big[\lim_{\epsilon \to 0}\int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dy-\Delta \varphi(x)\big] dydx \\ =&\int_{\mathbb{R}^n} \varphi(x) \big[ \lim_{\epsilon \to 0}\int_{|y-x|\geq \epsilon}\frac{v(x)-v(y)}{|x-y|^{n+2s}}dy-\Delta v(x)\big] dydx. \end{aligned} \end{align} $$

Hence, it suffices to prove (2.14) in this step.

We analyze the two integrals on both sides of (2.14), respectively. We prove that the limit as $\epsilon \rightarrow 0$ can be interchanged with the integral over y.

We first consider the integral on the left-hand side of (2.14) in $B^c_1(0)$ and $B_1(0),$ respectively. If $x \in \mathbb {R}^n \backslash \overline {B_1(0)},$ by (2.12), we derive

$$ \begin{align*}\left| \int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dy-\Delta \varphi(x) \right|=\left| \int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dy \right| \leq \frac{C}{\mathrm{dist}^s(x, \partial B_1(0))}. \end{align*} $$

By Lebesgue’s dominated convergence theorem, we have

(2.15) $$ \begin{align} &\int_{B_1^c(0)} v(x) \big[\lim_{\epsilon \to 0}\int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dy-\Delta \varphi (x) \big]dx \nonumber\\ =&\lim_{\epsilon \to 0} \int_{B_1^c(0)}v(x) \int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dy dx-\int_{B_1^c(0)} v(x)\Delta \varphi(x)dx. \end{align} $$

If $x \in B_{1-\delta }(0),$ for any fixed $0<\delta <1$ , by $\varphi (x)\in C^{2,\kappa }_{loc}(B_1(0))\cap C^{1,\kappa }(\overline {B_1(0)})$ for any $\kappa \in (0,\min \{1,2-2s\})$ , we obtain

$$ \begin{align*}\left| \int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dy-\Delta \varphi(x) \right| \leq C \left(\|\varphi\|_{C^{1,1}(B_{1-\delta/2}(0))}+\|\varphi\|_{L^\infty(B_1(0))}\right). \end{align*} $$

Similarly, by Lebesgue’s dominated convergence theorem, one has

$$ \begin{align*} &\int_{B_{1-\delta}(0)} v(x)\big[\lim_{\epsilon \to 0}\int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dy -\Delta \varphi (x)\big]dx \nonumber\\ =&\lim_{\epsilon \to 0} \int_{B_{1-\delta}(0)}v(x) \int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dy dx-\int_{B_{1-\delta}(0)}v(x) \Delta \varphi(x)dx. \end{align*} $$

Let $\delta \to 0,$ we obtain

(2.16) $$ \begin{align} \begin{aligned} &\int_{B_1(0)} v(x)\big [\lim_{\epsilon \to 0}\int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dy-\Delta \varphi(x) \big]dx \\ =&\lim_{\epsilon \to 0} \int_{B_1(0)}v(x) \int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dy dx-\int_{B_1(0)} v(x)\Delta \varphi(x) dx, \end{aligned} \end{align} $$

where we have used the fact that $v(x)\cdot \int _{|y-x|\geq \epsilon } \frac {\varphi (x)-\varphi (y)}{|x-y|^{n+2s}}dy \in L^1(B_1(0))$ .

Indeed,

$$ \begin{align*} F_\varepsilon(x) &:= \int_{|y-x|\geq \varepsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}} \, dy \\ &= \int_{\varepsilon \leq |y-x| < \frac{\delta}{2}} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}} \, dy + \int_{\frac{\delta}{2} \leq |y-x| < 2} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}} \, dy \\ &\quad + \int_{|y-x| \geq 2} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}} \, dy \\ &=: I_1 + I_2 + I_3. \end{align*} $$

For the term $I_1$ , we introduce the Hardy–Littlewood maximal function

here, we extend $\Delta \varphi =0$ in $B_1^c(0)$ (including the boundary $\partial B_1(0)$ ).

Define the averaged function

Clearly, $E(0) = 0$ . Then

Hence,

$$\begin{align*}|E'(r)| \leq \frac{r}{n} \, \mathrm{HLM}(x), \quad \text{and thus} \quad |E(r)| \leq \frac{r^2}{2n} \, \mathrm{HLM}(x). \end{align*}$$

It follows that

$$\begin{align*}I_1 \leq \left| \int_{B_{\delta/2}(x)} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}} \, dy \right| = \left| \int_0^{\delta/2} \frac{n \omega_n r^{n-1} E(r)}{r^{n+2s}} \, dr \right| \leq C \, \delta^{2-2s} \, \mathrm{HLM}(x). \end{align*}$$

For $I_2$ , note that if $\frac {\delta }{2} \leq |y-x| < 2$ , then $|\varphi (x)-\varphi (y)| \leq C |x-y|$ , and therefore

$$\begin{align*}|I_2| \leq C \int_{\frac{\delta}{2} \leq |y-x| < 2} \frac{|x-y|}{|x-y|^{n+2s}} \, dy \leq \begin{cases} C \, \delta^{1-2s}, & s \neq \frac{1}{2},\\[2mm] C \, \ln \delta, & s = \frac{1}{2}. \end{cases} \end{align*}$$

For $I_3$ , since $\varphi (y) = 0$ in $B_1^c(0)$ , we have

$$\begin{align*}|I_3| \leq \int_{|y-x| \geq 2} \frac{|\varphi(x)-\varphi(y)|}{|x-y|^{n+2s}} \, dy \leq C \, \|\varphi\|_{L^\infty(B_1(0))}. \end{align*}$$

By [Reference Su, Valdinoci, Wei and Zhang43, Theorem 4.1], we know that $\|\varphi \|_{W^{2,p}(B_1(0))} < \infty $ , which implies $\|D^2 \varphi \|_{L^p(B_1(0))} < \infty $ for some $p>1$ . Consequently, $\|\mathrm {HLM}\|_{L^{p,\infty }(B_1(0))} < \infty $ , and hence $\|\mathrm {HLM}\|_{L^1(B_1(0))} < \infty $ .

As a result, $\|F_\varepsilon \|_{L^1(B_1(0))} < \infty $ . Since $v \in L^\infty (\mathbb {R}^n)$ , it follows that $v F_\varepsilon \in L^1(B_1(0))$ . Therefore, (2.16) holds.

Combining (2.15) with (2.16), we obtain that the limit as $\epsilon \rightarrow 0$ can be interchanged with the integral over x on the left-hand side of (2.14), more precisely,

(2.17) $$ \begin{align} \begin{aligned} &\int_{\mathbb{R}^n}v(x) \big[\lim_{\epsilon \to 0}\int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dy-\Delta \varphi(x)\big] dx \\ =&\lim_{\epsilon \to 0} \int_{\mathbb{R}^n}v(x) \int_{|y-x|\geq \epsilon} \frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dydx- \int_{\mathbb{R}^n}v(x)\Delta \varphi(x)dx. \end{aligned} \end{align} $$

Then we consider the integral on the right-hand side of (2.14). By Lebesgue’s dominated convergence theorem, it is easy to see that the limit as $\epsilon \rightarrow 0$ and the integral over x on the right-hand side of (2.14) can be interchanged, that is,

(2.18) $$ \begin{align} \begin{aligned} &\int_{\mathbb{R}^n} \varphi(x) \big[ \lim_{\epsilon \to 0}\int_{|y-x|\geq \epsilon}\frac{v(x)-v(y)}{|x-y|^{n+2s}}dy-\Delta v(x)\big]dx \\ =&\lim_{\epsilon \to 0} \int_{\mathbb{R}^n} \varphi(x) \int_{|y-x|\geq \epsilon}\frac{v(x)-v(y)}{|x-y|^{n+2s}}dydx-\int_{\mathbb R^n}\varphi(x)\Delta v(x)dx. \end{aligned} \end{align} $$

To conclude the proof of (2.14), since $\varphi =0$ in $B_1^c(0)$ , it can be extended by zero to $\mathbb {R}^n$ and belongs to $W^{1,2}(\mathbb {R}^n)$ . Hence, for any sufficiently regular v,

$$\begin{align*}\int_{\mathbb{R}^n} v \, \Delta \varphi \, dx = \int_{\mathbb{R}^n} \varphi \, \Delta v \, dx, \end{align*}$$

holds in the distributional sense. Combining this with (2.17) and (2.18), it remains to show that

$$ \begin{align*}\int_{\mathbb{R}^n} v(x) \int_{|y-x|\geq \epsilon}\frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dydx- \int_{\mathbb{R}^n} \varphi(x) \int_{|y-x|\geq \epsilon}\frac{v(x)-v(y)}{|x-y|^{n+2s}}dydx=0.\end{align*} $$

This follows from Fubini’s Theorem that

$$ \begin{align*} &\int_{\mathbb{R}^n} v(x) \int_{|y-x|\geq \epsilon}\frac{\varphi(x)-\varphi(y)}{|x-y|^{n+2s}}dydx- \int_{\mathbb{R}^n} \varphi(x) \int_{|y-x|\geq \epsilon}\frac{v(x)-v(y)}{|x-y|^{n+2s}}dydx\\ &\quad =\int_{\mathbb{R}^n}\int_ {|y-x|\geq\epsilon}\frac{\varphi(x)v(x)-\varphi(y)v(x)}{\mid x-y\mid^{n+2s}}dydx \\ &\qquad -\int_{\mathbb{R}^n}\int_ {|y-x|\geq\epsilon}\frac{\varphi(x)v(x)-\varphi(x)v(y)}{\mid x-y\mid^{n+2s}}dydx\\ &\quad =\int_{\mathbb{R}^n}\int_ {|y-x|\geq\epsilon}\frac{-\varphi(y)v(x)}{\mid x-y\mid^{n+2s}}dydx-\int_{\mathbb{R}^n} \int_ {|y-x|\geq\epsilon}\frac{-\varphi(x)v(y)}{\mid x-y\mid^{n+2s}}dydx\\ &\quad =-\int_{\mathbb{R}^n}\int_ {|y-x|\geq\epsilon}\frac{\varphi(y)v(x)}{\mid x-y\mid^{n+2s}}dydx+\int_{\mathbb{R}^n} \int_ {|y-x|\geq\epsilon}\frac{\varphi(y)v(x)}{\mid x-y\mid^{n+2s}}dydx\\ &\quad =0. \end{align*} $$

Hence, (2.14) holds and (2.10) is verified.

Finally, we prove (2.11).

By the definitions of the fractional Laplacian and Laplacian, (2.10), and the mollifier, we obtain

$$ \begin{align*} &(( - \Delta ) ^s -\Delta )\varphi_1 (x)\\ =&C_{n,s} PV \int_{\mathbb{R}^n} \frac{\int_{\mathbb{R}^n}\rho (x-z)\varphi(z)dz-\int_{\mathbb{R}^n}\rho (y-z)\varphi(z)dz}{|x-y|^{n+2s}} dy-\Delta \varphi_1(x)\\ =& \int_{\mathbb{R}^n} \varphi(z) \big[(-\Delta)_x^s \rho (x-z)-\Delta_x \rho(x-z)\big]dz\\ =& \int_{\mathbb{R}^n} \varphi(z)\big[ (-\Delta)_z^s \rho (x-z)-\Delta_z\rho(x-z)\big]dz\\ =& \int_{B_1(0)} ((-\Delta)_z^s-\Delta_z) \varphi(z) \rho (x-z)dz+\int_{\mathbb{R}^n \backslash B_1(0)} ((-\Delta)_z^s-\Delta_z) \varphi(z) \rho (x-z)dz\\ &\leq \int_{B_1(0)} ((-\Delta)_z^s-\Delta_z) \varphi(z) \rho (x-z)dz\\ &= \int_{B_1(0)} \lambda_1 \varphi(z) \rho (x-z)dz\\ &= \lambda_1 \varphi_1 (x). \end{align*} $$

In the 6th line of the above inequality, we have used the fact that $\varphi =0$ in $B_1^c(0)$ . Therefore, we obtain (2.11) and complete the proof of Lemma 2.5.

Proof Let

$$ \begin{align*}V(x) = v_\lambda(x) - w_\lambda(x).\end{align*} $$

Clearly, (3.1) is equivalent to

$$\begin{align*}(-\Delta)^s V(x) - \Delta V(x) \geq 0 \quad \text{in } E_\lambda. \end{align*}$$

Denote by $\Gamma _\lambda $ the reflection of the set $E_\lambda $ with respect to the hyperplane $T_\lambda $ . Since $V \equiv 0$ in $E_\lambda $ and $V \leq 0$ in $\Sigma _\lambda \setminus E_\lambda $ , we deduce for any $x \in E_\lambda $ that

$$\begin{align*}-\Delta V(x) = 0, \end{align*}$$

and

$$\begin{align*}\begin{aligned} (-\Delta)^s V(x) &= \int_{\mathbb{R}^n} \frac{-V(y)}{|x - y|^{n + 2s}}\,dy \\ &= \left( \int_{\Sigma_\lambda \setminus E_\lambda} + \int_{\Gamma_\lambda} + \int_{\Sigma_\lambda^c \setminus \Gamma_\lambda} \right) \frac{-V(y)}{|x - y|^{n + 2s}}\,dy \\ &= \int_{\Sigma_\lambda \setminus E_\lambda} \frac{-V(y)}{|x - y|^{n + 2s}}\,dy + \int_{E_\lambda} \frac{-V(y^\lambda)}{|x - y^\lambda|^{n + 2s}}\,dy \\ &\quad + \int_{\Sigma_\lambda^c \setminus E_\lambda} \frac{-V(y^\lambda)}{|x - y^\lambda|^{n + 2s}}\,dy \\ &= \int_{\Sigma_\lambda \setminus E_\lambda} (-V(y)) \left[ \frac{1}{|x - y|^{n + 2s}} - \frac{1}{|x - y^\lambda|^{n + 2s}} \right] dy \\ &\geq 0, \end{aligned} \end{align*}$$

where, in the last inequality, we have used the fact that $|x - y| < |x - y^\lambda |$ for all $x \in E_\lambda $ and $y \in \Sigma _\lambda $ .

Therefore, inequality (3.1) holds, and the proof of Lemma 3.1 is complete.

Proof of Theorem 1.1

Let $u(x)$ be a positive bounded classical solution of equation (1.1), and define

$$\begin{align*}u_\lambda(x) = u(x^\lambda), \quad w_\lambda(x) = u_\lambda(x) - u(x), \quad x \in \mathbb{R}^n. \end{align*}$$

We aim to prove that u is strictly increasing in the $x_1$ -direction, that is,

$$\begin{align*}w_\lambda(x)> 0, \quad \forall\, x \in \Sigma_\lambda,\ \text{for all } \lambda \in \mathbb{R}. \end{align*}$$

The proof is divided into the following three steps.

Step 1. We show that for $\lambda $ sufficiently negative,

(3.2) $$ \begin{align} w_\lambda(x) \geq 0, \quad \forall\, x \in \Sigma_\lambda. \end{align} $$

From assumptions (F1) and (F2), and using equation (1.1), we have

(3.3) $$ \begin{align} \begin{aligned} (-\Delta)^s w_\lambda(x) - \Delta w_\lambda(x) &= f(x_1^\lambda)g(u_\lambda(x)) - f(x_1)g(u(x)) \\ &= \big(f(x_1^\lambda) - f(x_1)\big)g(u_\lambda(x)) + f(x_1)\big(g(u_\lambda(x)) - g(u(x))\big) \\ &\geq f(x_1)M_\lambda(x)w_\lambda(x), \end{aligned} \end{align} $$

where

$$\begin{align*}M_\lambda(x) = \frac{g(u_\lambda(x)) - g(u(x))}{u_\lambda(x) - u(x)}. \end{align*}$$

Since u is bounded and $g(\cdot )$ is locally Lipschitz continuous and nondecreasing, it follows that

(3.4) $$ \begin{align} M_\lambda(x) \text{ is bounded and nonnegative.} \end{align} $$

We proceed by contradiction. Suppose that there exists $\lambda \leq 0$ such that $w_\lambda (x) < 0$ somewhere in $\Sigma _\lambda $ . Then we can define the nonempty open set

$$\begin{align*}E_\lambda = \{x \in \Sigma_\lambda \mid w_\lambda(x) < 0\}, \end{align*}$$

and introduce the auxiliary function

$$ \begin{align*}v_\lambda(x)=w_\lambda(x) \chi_{E_\lambda(x)}\leq 0,\end{align*} $$

where

$$ \begin{align*}\chi_{E_\lambda}(x) =\left\{\begin{array}{@{}ll} 1,& \quad x \in E_\lambda,\\ 0,&\quad x \not \in E_\lambda. \end{array}\right.\end{align*} $$

Obviously, for $x \in \Sigma _\lambda $ , we have $x_1^\lambda> x_1$ . Using the monotonicity of both f and g, as well as the nonnegativity of g, it follows that for $x \in E_\lambda $ , similarly to (3.3),

(3.5) $$ \begin{align} (-\Delta)^s w_\lambda(x) - \Delta w_\lambda(x) \geq f(x_1) M_\lambda(x) w_\lambda(x):= C_0(x) f(x_1) w_\lambda(x), \end{align} $$

where $C_0(x)$ is bounded and nonnegative by (3.4).

Employing Lemma 3.1 and (3.5), we obtain

$$ \begin{align*} (-\Delta)^s v_\lambda(x) - \Delta v_\lambda(x) \geq (-\Delta)^s w_\lambda(x) - \Delta w_\lambda(x) \geq C_0(x) f(x_1) w_\lambda(x), \quad x \in E_\lambda. \end{align*} $$

Since $v_\lambda = 0$ for $x \in \mathbb {R}^n \setminus E_\lambda $ , applying the maximum principle for the mixed local and nonlocal operator in unbounded domains (Lemma 2.3) yields $w_\lambda \geq 0$ in $E_\lambda $ . This contradicts the assumption that $w_\lambda < 0$ somewhere in $\Sigma _\lambda $ , verifying that $E_\lambda $ must be empty. Hence, $w_\lambda \geq 0$ in $\Sigma _\lambda $ for sufficiently negative $\lambda $ , that is, (3.2) holds.

Step 2. The above inequality provides a starting point for applying the method of moving planes. Now we move plane $T_\lambda $ toward the right as long as the inequality continue to hold. Our goal is to show that the plane $T_\lambda $ can be moved all the way to infinity. More precisely, let

$$ \begin{align*}\lambda_0=\sup\{\lambda\in \mathbb R\mid w_\mu\geq 0~\text{in}~\Sigma_\mu~\text{for every}~\mu<\lambda\},\end{align*} $$

and we will prove that

$$ \begin{align*}\lambda_0=+\infty.\end{align*} $$

Assume, for contradiction, that $\lambda _0 < +\infty $ . Then, by the definition of $\lambda _0$ , there exists a sequence $\{\lambda _k\} \searrow \lambda _0$ and a corresponding positive sequence $\{m_k\}$ such that

(3.6) $$ \begin{align} \inf_{\Sigma_{\lambda_k}} w_{\lambda_k}(x) := -m_k < 0. \end{align} $$

We first claim that

(3.7) $$ \begin{align} m_k \rightarrow 0 \quad \text{as } k \rightarrow +\infty. \end{align} $$

Suppose, on the contrary, that there exists a subsequence (still denoted by $\{m_k\}$ ) such that $m_k> \bar C$ for some positive constant $\bar C$ . Then there exists a sequence $\{y^k\} \subset \Sigma _{\lambda _k}$ such that

(3.8) $$ \begin{align} w_{\lambda_k}(y^k) \leq -\bar C < 0. \end{align} $$

Case (i). If $y^k \in \Sigma _{\lambda _k} \setminus \Sigma _{\lambda _0}$ , then since $\lambda _k \rightarrow \lambda _0$ as $k \rightarrow \infty $ , we have

$$\begin{align*}|y^k - (y^k)^{\lambda_k}| = 2|\lambda_k - y^k_1| \rightarrow 0 \quad \text{as } k \rightarrow \infty. \end{align*}$$

By the uniform continuity of u, it follows that

$$\begin{align*}w_{\lambda_k}(y^k) = u((y^k)^{\lambda_k}) - u(y^k) \rightarrow 0 \quad \text{as } k \rightarrow \infty. \end{align*}$$

Case (ii). If $y^k \in \Sigma _{\lambda _0}$ , then using the uniform continuity of u, the fact that $\lambda _k \rightarrow \lambda _0$ , and the definition of $\lambda _0$ , we obtain

$$\begin{align*}\begin{aligned} w_{\lambda_k}(y^k) &= u((y^k)^{\lambda_k}) - u((y^k)^{\lambda_0}) + w_{\lambda_0}(y^k) \\ &\geq u((y^k)^{\lambda_k}) - u((y^k)^{\lambda_0}) \rightarrow 0 \quad \text{as } k \rightarrow \infty. \end{aligned} \end{align*}$$

In both cases, we arrive at a contradiction with (3.8). Therefore, the claim (3.7) must hold.

Secondly, from (3.7), there exists a sequence $\{x^k\}\subset \Sigma _{\lambda _k}$ such that

(3.9) $$ \begin{align} w_{\lambda_k}(x^k)=-m_k+m_k^2<0. \end{align} $$

We will show that

(3.10) $$ \begin{align} d_k=\frac{1}{2}\mathrm{dist}\{x^k,T_{\lambda_k}\}\geq \bar C>0. \end{align} $$

Without loss of generality, we assume $d_k<1$ , and hence $d_k<d_k^s$ .

To this end, we construct a perturbation of $w_{\lambda _k}$ near $x^k$ as follows:

(3.11) $$ \begin{align} V_k(x)=w_{\lambda_k}(x)-m_k^2\varsigma_k(x),~x\in \mathbb R^n, \end{align} $$

where

$$ \begin{align*}\varsigma_k(x)=\varsigma\left(\frac{x-x^k}{d^s_k}\right),\end{align*} $$

and $\varsigma \in C^{\infty }_0(\mathbb R^n)$ is a cut-off smooth function satisfying

$$ \begin{align*}\left\{\begin{array}{@{}ll} 0\leq \varsigma \leq 1,& \text{in}~\mathbb R^n,\\ \varsigma=1,& \text{in}~B_{\frac{1}{2}}(0),\\ \varsigma=0, & \text{in}~\mathbb R^n \backslash B_1(0). \end{array} \right.\end{align*} $$

Combining (3.6), (3.9), and (3.11), we obtain

$$ \begin{align*} \begin{cases} V_{ k}(x^k)= w_{\lambda_k} (x^k) -m_k^2\varsigma_k(x^k)=-m_k, &\\ V_{ k}(x )=w_{\lambda_k} (x ) -m_k^2\varsigma_k(x )\geq -m_k, &\text{in }~ \mathbb R^n \setminus B_{d_k^s}(x^k),\\ V_{ k}(x )= -m_k^2\varsigma_k(x )\geq -m_k^2, &\text{on}~ T_{\lambda_k}. \end{cases} \end{align*} $$

This implies that each $V_k$ must attain its minimum, which is no greater than $-m_k$ , at some point $\bar x^k \in \overline {B_{d^s_k}(x^k) \cap \Sigma _{\lambda _k}} \subset \Sigma _{\lambda _k}$ (see Figure 1), that is to say,

(3.12) $$ \begin{align} \exists~ \{ \bar x^k \} \subset \overline{B_{d^s_k}(x^k)\cap \Sigma_{\lambda_k}} ~s.t. ~-m_k-m^2_k\leq V_k(\bar x^k)=\underset{\Sigma_{\lambda_k} }{\inf}~ V_k(x )\leq -m_k, \end{align} $$

Figure 1:

The domain $ B_{d_k^s}(x^k)\cap \Sigma _{\lambda _k}$ .

which indicates that

(3.13) $$ \begin{align} - m_k\leq {w}_{\lambda_k}(\bar x^k )\leq -m_k+m^2_k<0. \end{align} $$

From (3.12) and the definitions of the operators $(-\Delta )^s$ and $-\Delta $ , together with the anti-symmetry of $w_{\lambda _k}$ in $ \Sigma _{\lambda _k}$ , we deduce

(3.14) $$ \begin{align} -\Delta V_k(\bar x^k ) \leq 0, \end{align} $$

and

(3.15) $$ \begin{align} \begin{aligned} (-\Delta)^s V_k(\bar x^k )&=C_{n,s}P.V.\int_{\mathbb R^n} \frac{V_k(\bar x^k ) -V_k(y)}{|\bar x^k-y|^{n+2s}}dy \\ &=C_{n,s}P.V.\int_{\Sigma_{\lambda_k}} \frac{V_k(\bar x^k ) -V_k(y )}{|\bar x^k-y|^{n+2s}}dy +C_{n,s}\int_{\Sigma_{\lambda_k}} \frac{V_k(\bar x^k ) -V_k(y^{\lambda_k} )}{|\bar x^k-y^{\lambda_k}|^{n+2s}}dy \\ &\leq C_{n,s}\int_{\Sigma_{\lambda_k}} \frac{2V_k(\bar x^k )-V_k(y )-V_k(y^{\lambda_k} )}{|\bar x^k-y^{\lambda_k}|^{n+2s}}dy \\ &= C_{n,s}\int_{\Sigma_{\lambda_k}} \frac{2V_k(\bar x^k )+m_k^2 \varsigma_k(y)+m_k^2 \varsigma_k(y^{\lambda_k})}{|\bar x^k-y^{\lambda_k}|^{n+2s}}dy \\ &\leq 2C_{n,s}(V_k(\bar x^k )+m^2_k)\int_{\Sigma_{\lambda_k}}\frac{1}{|\bar x^k-y^{\lambda_k}|^{n+2s}}dy \\ &\leq 2 C_{n,s}(V_k(\bar x^k )+m^2_k)\int_{\Sigma_{\lambda_k}}\frac{1}{| x^k-y^{\lambda_k}|^{n+2s}}dy \\ &\leq \frac{c_1(-m_k+m^2_k)}{d_k^{2s}}, \end{aligned} \end{align} $$

where the second inequality from the bottom follows from the estimate

$$ \begin{align*}|\bar x^k-y^{\lambda_k}|\leq |\bar x^k-x^k|+|x^k-y^{\lambda_k}|\leq \frac{3}{2}|x^k-y^{\lambda_k}|.\end{align*} $$

Combining (3.3), (3.11), and (3.14) with (3.15), we obtain

(3.16) $$ \begin{align} \begin{aligned} \frac{c_1(-m_k+m^2_k)}{d_k^{2s}}&\geq (-\Delta)^s V_k(\bar x^k )-\Delta V_k(\bar x^k )\\ &=(-\Delta)^s w_{\lambda_k}(\bar x^k )-\Delta w_{\lambda_k}(\bar x^k )-m_k^2(-\Delta)^s\varsigma_k(\bar x^k)+m_k^2 \Delta \varsigma_k(\bar x^k)\\ &\geq f(\bar x^k_1)M_{\lambda_k}(\bar x^k)w_{\lambda_k}(\bar x^k )-\frac{c_2 m_k^2 }{d_k^{2s^2}}- \frac{c_3 m_k^2 }{d_k^{2s}}\\ &\geq f(\lambda_k) M_{\lambda_k}(\bar x^k)w_{\lambda_k}(\bar x^k )-\frac{c_2 m_k^2 }{d_k^{2s^2}}- \frac{c_3 m_k^2 }{d_k^{2s}}, \end{aligned} \end{align} $$

where $\bar x^k_1$ is the first component of $\bar x^k $ and we have used the fact that

$$ \begin{align*}|(-\Delta)^s \varsigma_k(\bar x^k)|\leq \frac{c_2 }{d_k^{2s^2}} \,\, \text{and}\,\, |\Delta \varsigma_k(\bar x^k)|\leq \frac{c_3 }{d_k^{2s}}\end{align*} $$

and

$$ \begin{align*}f(\bar x_1^k) \leq f(\lambda_k)\end{align*} $$

due to $\bar x^k_1 \leq \lambda _k$ . Putting together (3.13) and (3.16), we conclude that

(3.17) $$ \begin{align} c_1(1-m_k ) \leq |f(\lambda_k)|M_{\lambda_k}(\bar x^k) d_k^{2s} +c_2m_k d_k^{2s(1-s)} +c_3m_k. \end{align} $$

If (3.10) is not true, then $d_k\rightarrow 0$ as $k\rightarrow +\infty .$ Noting that $m_k\rightarrow 0$ as $k\rightarrow +\infty ,$ therefore, if we let $k \to \infty $ in (3.17), we derive a contradiction. Hence, (3.10) holds.

Next, we continue to show $\lambda _0=+\infty $ . Accordingly, we perturb $w_{\lambda _k}$ near $x^k$ again as follows:

(3.18) $$ \begin{align} \tilde V_k(x)=w_{\lambda_k}(x)-m_k^2 \tilde \varsigma_k(x),~x\in \mathbb R^n, \end{align} $$

where

$$ \begin{align*}\tilde \varsigma_k(x)=\tilde \varsigma\left(\frac{x-x^k}{d _k}\right)\end{align*} $$

with $d_k=\frac {1}{2}\mathrm{dist}\{x^k,T_{\lambda _k}\}\geq \bar C>0$ and $\tilde \varsigma \in C^{\infty }_0(\mathbb R^n)$ is a cut-off smooth function satisfying

$$ \begin{align*}\left\{\begin{array}{@{}ll} 0\leq \tilde \varsigma \leq 1,& \text{in}~\mathbb R^n,\\ \tilde \varsigma=1,& \text{in}~B_{\frac{1}{2}}(0),\\ \tilde \varsigma=0, & \text{in}~\mathbb R^n \backslash B_1(0). \end{array} \right.\end{align*} $$

It then follows from (3.6), (3.9), and (3.18) that

$$ \begin{align*} \begin{cases} \tilde V_k(x^k)= w_{\lambda_k} (x^k) -m_k^2\tilde\varsigma_k(x^k)=-m_k ,&\\ \tilde V_k(x)=w_{\lambda_k} (x ) -m_k^2\tilde\varsigma_k(x )\geq -m_k ,&~\text{in}~ \mathbb R^n \setminus B_{d_k }(x^k),\\ \tilde V_k(x )= -m_k^2\tilde\varsigma_k(x )\geq -m_k ,&~\text{on}~ T_{\lambda_k}. \end{cases} \end{align*} $$

This implies that $\tilde {V}_k$ must attain its minima, which is at most $-m_k$ , within $\overline {B_{d_k}(x^k)} \subset \Sigma _{\lambda _k}$ (see Figure 2), more precisely,

(3.19) $$ \begin{align} \exists~ \{ \tilde x^k \} \subset \overline{B_{d _k}(x^k) } ~s.t. ~-m_k-m^2_k\leq \tilde V_k(\tilde x^k)=\underset{\Sigma_{\lambda_k} }{\inf}~ \tilde V_k(x )\leq -m_k. \end{align} $$

Figure 2:

The domain $ B_{d_k }(x^k) $ .

Hence, we obtain

(3.20) $$ \begin{align} - m_k\leq {w}_{\lambda_k}(\tilde x^k )\leq -m_k+m^2_k<0. \end{align} $$

From (3.19), using the same calculations as in (3.14) and (3.15), but with $V_k$ and $\bar {x}^k$ replaced by $\tilde {V}_k$ and $\tilde {x}^k$ , respectively, we obtain

(3.21) $$ \begin{align} -\Delta \tilde V_k(\tilde x^k ) \leq 0, \end{align} $$

and

(3.22) $$ \begin{align} (-\Delta)^s \tilde V_k(\tilde x^k ) \leq \frac{c_1(-m_k+m^2_k)}{d_k^{2s}}, \end{align} $$

Therefore, combining (3.18), (3.21), and (3.22) with (3.10), we obtain

(3.23) $$ \begin{align} (-\Delta)^sw_{\lambda_k}(\tilde x^k )-\Delta w_{\lambda_k}(\tilde x^k ) &=[(-\Delta)^s-\Delta]\tilde V_k(\tilde x^k)+m^2_k[(-\Delta)^s-\Delta]\tilde\varsigma_k(\tilde x^k)\nonumber\\ &\leq \frac{c_1(-m_k+m^2_k)}{d_k^{2s}}+\frac{c_2 m_k^2 }{d_k^{2s }}+\frac{c_3 m_k^2 }{d_k^{2}}\nonumber\\ &\leq \frac{c (-m_k+m^2_k)}{d_k^{2s}}, \end{align} $$

where we have used the fact that

$$ \begin{align*}|(-\Delta)^s \tilde\varsigma_k(\tilde x^k)|\leq \frac{c_2 }{d_k^{2s }}\,\, \text{ and } \,\, |\Delta \tilde\varsigma_k(\tilde x^k)|\leq \frac{c_3 }{d_k^{2 }}.\end{align*} $$

Combining (3.23) with (3.3) yields

(3.24) $$ \begin{align} \frac{c(-m_k+m^2_k)}{d_k^{2s}}\geq f( \tilde x^k_1) M_{\lambda_k} (\tilde x^k ) w_{\lambda_k}(\tilde x^k ). \end{align} $$

If $f(\tilde {x}_1^k) \leq 0$ , this leads to a contradiction immediately, and the proof is complete. Hence, we may assume that $f(\tilde {x}_1^k)> 0$ . By (3.20), we arrive at

(3.25) $$ \begin{align} c(1-m_k ) \leq f(\tilde x_1^k)M_{\lambda_k} (\tilde x^k)d_k^{2s}\leq C M_{\lambda_k} (\tilde x^k) f(\tilde x_1^k)|\tilde x_1^k|^{2s}. \end{align} $$

Therefore, by assumption (1.4) and the boundedness of $M_{\lambda _k} (\tilde x^k ) $ , we obtain

$$ \begin{align*}\{\tilde x^k_1\}~\text{is bounded away from}~-\infty\end{align*} $$

which together with $f\in C(\mathbb R^n)$ and $ \tilde x^k\in \overline {B_{d _k}(x^k) }$ with $d_k=\frac {1}{2} \mathrm{dist}\{\tilde x^k,T_{\lambda _k}\}$ gives rise to

$$ \begin{align*}f(\tilde x_1^k)\leq c,~ d_k^{2s}\leq c,\end{align*} $$

for sufficiently large k.

On the other hand, by (3.25), we conclude that

(3.26) $$ \begin{align} f(\tilde x_1^k),~ M_{\lambda_k}(\tilde x^k )\geq C>0, \end{align} $$

for sufficiently large k. Then by the definition of $M_{\lambda _k}$ , the assumption (1.5) on $g,$ and the positivity of u, we obtain

(3.27) $$ \begin{align} u(\tilde x^k )\geq C>0~ \text{and}~g(u(\tilde x^k ))\geq C>0. \end{align} $$

Since $w_{\lambda _k}(\tilde x^k)\rightarrow 0$ as $k\rightarrow \infty $ , by (3.27), we have

(3.28) $$ \begin{align} g( u_{\lambda_k}(\tilde x^k ))\geq C>0, \end{align} $$

for sufficiently large k.

Next, starting from the original equation (3.3), we revise (3.24) as follows:

(3.29) $$ \begin{align} \frac{c(-m_k + m_k^2)}{d_k^{2s}} \geq \left( f((\tilde{x}_1^k)^{\lambda_k}) - f(\tilde{x}_1^k) \right) g(u_{\lambda_k}(\tilde{x}^k)) + f(\tilde{x}_1^k) M_{\lambda_k}(\tilde{x}^k) w_{\lambda_k}(\tilde{x}^k). \end{align} $$

Noting that $|{\tilde {x}_1^k} - \lambda _k| \sim d_k$ , and using the continuity and monotonicity of f along with (3.26), we obtain

(3.30) $$ \begin{align} f((\tilde x_1^k)^{\lambda_k})-f (\tilde x_1^k) \geq C>0. \end{align} $$

Combining (3.13), (3.26), and (3.28)–(3.30), we arrive at a contradiction for sufficiently large k.

Therefore, $\lambda _0=+\infty $ .

Step 3. Based on the already established result,

$$\begin{align*}w_\lambda(x) \geq 0, \quad \forall\, x \in \Sigma_\lambda,\ \forall\, \lambda \in \mathbb{R}, \end{align*}$$

we will now further prove that

(3.31) $$ \begin{align} w_\lambda(x)> 0, \quad \forall\, x \in \Sigma_\lambda,\ \forall\, \lambda \in \mathbb{R}. \end{align} $$

Otherwise, for some fixed $\lambda $ , there exists $\bar {x} \in \Sigma _\lambda $ such that

$$\begin{align*}w_\lambda(\bar{x}) = \min_{x \in \Sigma_\lambda} w_\lambda(x) = 0. \end{align*}$$

At this minimum point, we have

$$\begin{align*}(-\Delta)^s w_\lambda(\bar{x}) - \Delta w_\lambda(\bar{x}) \leq 0. \end{align*}$$

On the other hand, from equation (3.3), we derive

$$ \begin{align*} \begin{aligned} (-\Delta)^s w_\lambda(\bar{x}) - \Delta w_\lambda(\bar{x}) &= \big(f(\bar{x}_1^\lambda) - f(\bar{x}_1)\big) g(u_\lambda(\bar{x})) + f(\bar{x}_1) \big(g(u_\lambda(\bar{x})) - g(u(\bar{x}))\big) \\ &= \big(f(\bar{x}_1^\lambda) - f(\bar{x}_1)\big) g(u_\lambda(\bar{x})) \\ &> 0, \end{aligned} \end{align*} $$

where the last inequality follows from the positivity of u and g, and the strict monotonicity of f with respect to $x_1$ . This leads to a contradiction. Hence, (3.31) is established.

We have thus shown that every positive solution must be strictly increasing in the $x_1$ direction. This completes the proof of Theorem 1.1.

Proof of Theorem 1.3

Since condition (1.6) holds and $f(x_1)$ is strictly increasing in $x_1$ , we may assume that there exists a sufficiently large constant R such that $f(x_1)> 0$ for all $x_1 \in (R - 2, \infty )$ . Let $B_1(x_0)$ denote the unit ball centered at $x_0 = (R, 0, \ldots , 0)$ , and let $\lambda _1$ be the first (nonzero) eigenvalue of the problem

$$ \begin{align*} \left\{ \begin{array}{@{}ll} (-\Delta)^s \varphi(x) - \Delta \varphi(x) = \lambda_1 \varphi(x), & x \in B_1(x_0), \\ \varphi(x)> 0, & x \in B_1(x_0), \\ \varphi(x) = 0, & x \in B_1^c(x_0). \end{array} \right. \end{align*} $$

According to [Reference Su, Valdinoci, Wei and Zhang43], the corresponding eigenfunction $\varphi (x)$ satisfies $\varphi \in C^{2,\kappa }_{\mathrm {loc}}(B_1(x_0)) \cap C^{1,\kappa }(\overline {B_1(x_0)})$ for any $\kappa \in (0,1)$ .

Let $\xi _u = \min _{B_1(x_0)} u$ . For all sufficiently large R, since u is positive and monotone increasing in the $x_1$ -direction, there exists a constant $c>0$ such that $\xi _u \geq c$ and

$$\begin{align*}m_0 := \frac{g(\xi_u)}{\sup_{\mathbb{R}^n} u}> 0. \end{align*}$$

Using the fact that $u(x)$ is monotone increasing in the $x_1$ direction again, we obtain

$$\begin{align*}(-\Delta)^s u(x) - \Delta u(x) \geq f(R-1)\, m_0\, u(x), \quad x \in B_1(x_0). \end{align*}$$

By taking R sufficiently large, and noting that $f(x_1) \to +\infty $ as $x_1 \to +\infty $ , we can ensure that

$$\begin{align*}(-\Delta)^s u(x) - \Delta u(x) \geq \lambda_1 u(x), \quad x \in B_1(x_0). \end{align*}$$

Define

$$\begin{align*}A_1 = \max_{x \in B_1(x_0)} \frac{\varphi(x)}{u(x)}, \quad \text{and} \quad D_1(x) = A_1 u(x). \end{align*}$$

Then $ D_1(x) $ satisfies

$$ \begin{align*} \left\{ \begin{array}{@{}ll} (-\Delta)^s (D_1(x) - \varphi(x)) - \Delta (D_1(x) - \varphi(x)) \geq 0, & x \in B_1(x_0), \\ D_1(x) - \varphi(x)> 0, & x \in \mathbb{R}^n \backslash B_1(x_0). \end{array} \right. \end{align*} $$

Applying the maximum principle for mixed local and nonlocal equations (Lemma 2.1), we obtain

(3.32) $$ \begin{align} D_1(x)> \varphi(x), \quad x \in B_1(x_0). \end{align} $$

On the other hand, by the definition of $A_1$ , there exists a point $\bar {x} \in B_1(x_0)$ such that

$$\begin{align*}D_1(\bar{x}) = \frac{\varphi(\bar{x})}{u(\bar{x})} u(\bar{x}) = \varphi(\bar{x}), \end{align*}$$

which contradicts (3.32).

Therefore, equation (1.1) admits no bounded positive solution. This completes the proof of Theorem 1.3.

Proof of Theorem 1.5

We will carry out the proof in three steps.

Step 1. In the first step, we show that for $\lambda $ sufficiently negative, the following holds:

(4.2) $$ \begin{align} w_\lambda(x, t) \geq 0, \quad (x, t) \in \Sigma_\lambda \times \mathbb{R}. \end{align} $$

We proceed by contradiction and use a perturbation argument. Suppose (4.2) does not hold. Since u is bounded, there exists some $\lambda < -R$ for a constant $R> 0$ to be determined later, and a constant $B> 0$ such that

(4.3) $$ \begin{align} \inf_{(x, t) \in \Sigma_\lambda \times \mathbb{R}} w_\lambda(x, t) = -B < 0. \end{align} $$

If the minimum of $w_\lambda $ is attained at some point, then a contradiction can be derived directly at that point. However, since both x and t vary over unbounded domains, the minimum may not be achieved, and this case requires a more delicate treatment.

From (4.3), we deduce that there exists a sequence of approximate minima $\{(x^k, t_k)\} \subset \Sigma _\lambda \times \mathbb {R}$ and a sequence $\{\delta _k\} \searrow 0$ such that

(4.4) $$ \begin{align} w_\lambda(x^k, t_k) = -B + \delta_k < 0. \end{align} $$

To construct a sequence of functions that attain their minima, we perturb $w_\lambda $ near $(x^k, t_k)$ as follows:

(4.5) $$ \begin{align} W_k(x , t ) =w_\lambda(x,t)-\delta_k\xi_k(x,t),~(x,t)\in \mathbb R^n\times \mathbb R, \end{align} $$

where

$$\begin{align*}\xi_k(x,t) = \xi\left( \frac{x - x^k}{d_k}, \frac{t - t_k}{d_k^{\frac{2s}{\alpha}}} \right), \end{align*}$$

with $ d_k = \frac {1}{2} \operatorname {dist}(x^k, T_\lambda )> 0 $ , and $ \xi \in C_0^\infty (\mathbb {R}^n \times \mathbb {R}) $ is a smooth cutoff function satisfying

$$\begin{align*}\left\{ \begin{array}{@{}ll} 0 \leq \xi \leq 1, & \text{in } \mathbb{R}^n \times \mathbb{R},\\[5pt] \xi = 1, & \text{in } B_{\frac{1}{2}}(0) \times \left[-\tfrac{1}{2}, \tfrac{1}{2} \right],\\[5pt] \xi = 0, & \text{in} \left(\mathbb{R}^n \times \mathbb{R}\right)\backslash \left(B_1(0) \times [-1, 1]\right). \end{array} \right. \end{align*}$$

We now claim that the sequence $ \{x^k\} $ does not approach the hyperplane $ T_\lambda $ , that is, there exists a constant $ \tilde {c}> 0 $ such that

(4.6) $$ \begin{align} d_k = \tfrac{1}{2} \operatorname{dist}(x^k, T_\lambda) \geq \tilde{c}, \quad \text{for all } k \in \mathbb{N}. \end{align} $$

If this is not true, then $ x^k \to T_\lambda $ and for sufficiently large $ k $ , from the uniform continuity of u, we have

$$\begin{align*}W_k(x^k, t_k) = w_\lambda(x^k, t_k) - \delta_k \xi_k(x^k, t_k) = w_\lambda(x^k, t_k) - \delta_k \to 0. \end{align*}$$

On the other hand, from (4.4), we know

$$\begin{align*}W_k(x^k, t_k) = w_\lambda(x^k, t_k) - \delta_k \xi_k(x^k, t_k) = -B < 0, \end{align*}$$

which yields a contradiction, and hence (4.6) holds.

We now define the parabolic cylinder centered at $ (x^k, t_k) $ as

$$\begin{align*}E(x^k, t_k) := B_{d_k}(x^k) \times \left[ t_k - d_k^{\frac{2s}{\alpha}},\, t_k + d_k^{\frac{2s}{\alpha}} \right] \subset \Sigma_\lambda \times \mathbb{R} \end{align*}$$

(see Figure 3).

Figure 3:

The domain $E(x^k,t_k)$ .

Here, for the case of the mixed local and nonlocal parabolic equations with $\alpha = 1$ , the parameter $\alpha $ in the expressions for $\xi _k(x, t)$ and $E(x^k, t_k)$ should also be taken as 1.

It follows from (4.3)–(4.5) that

$$\begin{align*}\left\{ \begin{array}{@{}ll} W_k(x^k, t_k) = -B, & \\ W_k(x, t) = -\delta_k \xi_k(x, t)> -B & \text{on}\,\, T_\lambda \times \mathbb{R},\\ W_k(x, t) = w_\lambda(x, t) \geq -B & \text{in}\,\, (\Sigma_\lambda \times \mathbb{R}) \setminus E(x^k, t_k). \end{array} \right.\end{align*}$$

This implies that $W_k$ must attain its minimum, which is at most $-B$ , at some point $(\tilde {x}^k, \tilde {t}_k) \in \overline {E(x^k, t_k)} \subset \Sigma _\lambda \times \mathbb {R}$ , that is,

(4.7) $$ \begin{align} \exists~ \{(\tilde x^k,\tilde t_k)\} \subset \overline{E(x^k,t_k)}~s.t. ~-B-\delta_k\leq W_k(\tilde x^k,\tilde t_k)=\underset{\Sigma_\lambda \times \mathbb R}{\inf} W_k(x,t)\leq -B. \end{align} $$

As a consequence of (4.3) and (4.5), we have

(4.8) $$ \begin{align} \begin{cases} -B\leq w_\lambda(\tilde x^k,\tilde t_k)\leq -B+\delta_k<0,&\\ \partial _tW_k(\tilde x^k,\tilde t_k)= 0, &\\ \partial^\alpha_tW_k(\tilde x^k,\tilde t_k)=C_\alpha \int_{-\infty}^{\tilde t_k} \frac{W_k(\tilde x^k,\tilde t_k)-W_k(\tilde x^k,\tau)}{(\tilde t_k-\tau)^{1+\alpha}}d\tau \leq 0, &\text{if}~0<\alpha<1. \end{cases}\end{align} $$

From (4.7) and the definition of operators $(-\Delta )^s$ and $-\Delta $ , and using the antisymmetry of $W_\lambda $ with respect to x, we deduce

(4.9) $$ \begin{align} -\Delta W_k(\tilde x^k,\tilde t_k) \leq 0, \end{align} $$

and

(4.10) $$ \begin{align} \begin{aligned} (-\Delta)^s W_k(\tilde x^k,\tilde t_k)=\ &C_{n,s}P.V.\int_{\mathbb R^n} \frac{W_k(\tilde x^k,\tilde t_k) -W_k(y,\tilde t_k)}{|\tilde x^k-y|^{n+2s}}dy\\ =\ &C_{n,s}P.V.\int_{\Sigma_\lambda} \frac{W_k(\tilde x^k,\tilde t_k) -W_k(y,\tilde t_k)}{|\tilde x^k-y|^{n+2s}}dy\\ &+C_{n,s}\int_{\Sigma_\lambda} \frac{W_k(\tilde x^k,\tilde t_k) -W_k(y^\lambda,\tilde t_k)}{|\tilde x^k-y^\lambda|^{n+2s}}dy\\ \leq\ & C_{n,s}\int_{\Sigma_\lambda} \frac{2W_k(\tilde x^k,\tilde t_k)-W_k(y,\tilde t_k)-W_k(y^\lambda,\tilde t_k)}{|\tilde x^k-y^\lambda|^{n+2s}}dy\\ =\ & C_{n,s}\int_{\Sigma_{\lambda}} \frac{2W_k(\tilde x^k,\tilde t_k)+\delta_k \xi_k(y)-\delta_k \xi_k(y^{\lambda})}{|\tilde x^k-y^{\lambda}|^{n+2s}}dy\\ \leq\ & 2 C_{n,s}(W_k(\tilde x^k,\tilde t_k)+\delta_k)\int_{\Sigma_\lambda}\frac{1}{|\tilde x^k-y^\lambda|^{n+2s}}dy\\ \leq\ & 2 C_{n,s}(W_k(\tilde x^k,\tilde t_k)+\delta_k)\int_{\Sigma_\lambda}\frac{1}{| x^k-y^\lambda|^{n+2s}}dy\\ \leq\ &\frac{c_1(-B+\delta_k)}{d_k^{2s}}, \end{aligned} \end{align} $$

where the second inequality from the bottom follows from the fact

$$ \begin{align*}|\tilde x^k-y^\lambda|\leq |\tilde x^k-x^k|+|x^k-y^\lambda|\leq \frac{3}{2}| x^k-y^\lambda|.\end{align*} $$

Therefore, combining (4.5), (4.6), and (4.8)–(4.10), we obtain, for sufficiently small $\delta _k$ ,

(4.11) $$ \begin{align} \begin{aligned} &(\partial^\alpha_t+(-\Delta)^s-\Delta) w_\lambda(\tilde x^k,\tilde t_k)\\ =\ &(\partial^\alpha_t+(-\Delta)^s-\Delta) W_k(\tilde x^k,\tilde t_k)+\delta_k(\partial^\alpha_t+(-\Delta)^s-\Delta)\xi_k(\tilde x^k,\tilde t_k)\\ \leq\ & \frac{c_1(-B+\delta_k)}{d_k^{2s}}+\frac{c_2\delta_k}{d_k^{2s}}+\frac{c_3\delta_k}{d_k^2}\\ \leq\ & \frac{c (-B+\delta_k)}{d_k^{2s}}, \end{aligned} \end{align} $$

where we have used the fact that

$$ \begin{align*}\left|(\partial_t^\alpha+(-\Delta)^s) \xi_k(\tilde x^k,\tilde t_k)\right|\leq \frac{c_2}{d_k^{2s}}\,\, \text{and}\,\, \left|\Delta \xi_k(\tilde x^k,\tilde t_k)\right|\leq \frac{c_3}{d_k^2}.\end{align*} $$

Combining (4.11) with (4.1), one has

(4.12) $$ \begin{align} \frac{c(-B+\delta_k)}{d_k^{2s}}\geq f( \tilde x^k_1) M_\lambda(\tilde x^k, \tilde t_k) w_\lambda(\tilde x^k,\tilde t_k). \end{align} $$

If $f( \tilde x^k_1)\leq 0,$ this directly implies a contradiction and thus we complete the proof. If $f( \tilde x^k_1)>0,$ by (4.3), (4.12) and $M_\lambda (\tilde x^k, \tilde t_k)$ is bounded and nonnegative, we arrive at

$$ \begin{align*}\frac{c(-B+\delta_k)}{-B}\leq c_0f( \tilde x^k_1)d_k^{2s}\leq c_0f( \tilde x^k_1)|\tilde x^k_1|^{2s}.\end{align*} $$

Hence, for sufficiently large k, we obtain

$$ \begin{align*}f( \tilde x^k_1)|\tilde x^k_1|^{2s}\geq \frac{c}{2c_0},\end{align*} $$

which is impossible, since $\lambda $ is sufficiently negative, $x^k_1 < \lambda $ , and

$$ \begin{align*}\underset{x_1\rightarrow-\infty}{\limsup}f(x_1)|x_1|^{2s}\leq0.\end{align*} $$

Therefore, (4.2) holds.

Step 2. We move the plane $T_\lambda $ to the right along the $x_1$ -direction as long as the inequality (4.2) holds, until it reaches its limiting position $T_{\lambda _0}$ , where $\lambda _0$ is defined by

$$\begin{align*}\lambda_0 := \sup \left\{ \lambda \,\middle|\, w_\mu \geq 0 \ \text{in} \ \Sigma_\mu \times \mathbb{R} \ \text{for all} \ \mu \leq \lambda \right\}. \end{align*}$$

We will prove

(4.13) $$ \begin{align} \lambda_0=+\infty. \end{align} $$

If (4.13) does not hold, then $\lambda _0 < +\infty $ . By the definition of $\lambda _0$ , there exists a sequence $\{\lambda _k\} \searrow \lambda _0$ and a positive sequence $\{\gamma _k\}$ such that

$$ \begin{align*} \underset{(x,t)\in \Sigma_{\lambda_k}\times \mathbb R}{\inf} w_{\lambda_k}(x,t):=-\gamma_k<0. \end{align*} $$

We first prove that

(4.14) $$ \begin{align} \gamma_k\rightarrow0 ~\text{as}~k\rightarrow \infty. \end{align} $$

Otherwise, there exists a subsequence, still denoted by $\{\gamma _k\}$ , such that $\gamma _k<-m$ for some positive constant m. Hence, there exists a sequence $\{(y^k,s_k)\}\subset \Sigma _{\lambda _k}\times \mathbb R$ such that

(4.15) $$ \begin{align} w_{\lambda_k}(y^k,s_k)\leq-m<0. \end{align} $$

Case (i). When $y^k \in \Sigma _{\lambda _k} \setminus \Sigma _{\lambda _0}$ , together with the fact that $\lambda _k \to \lambda _0$ as $k \to \infty $ , we obtain

$$\begin{align*}|y^k - (y^k)^{\lambda_k}| = 2|\lambda_k - y^k_1| \to 0 \quad \text{as} \quad k \to \infty. \end{align*}$$

Then, by the uniform continuity of u, it follows that

$$\begin{align*}w_{\lambda_k}(y^k, s_k) = u((y^k)^{\lambda_k}, s_k) - u(y^k, s_k) \to 0 \quad \text{as} \quad k \to \infty. \end{align*}$$

Case (ii). When $y^k \in \Sigma _{\lambda _0}$ , by the uniform continuity of u, the convergence $\lambda _k \to \lambda _0$ , and the definition of $\lambda _0$ , we deduce that

$$ \begin{align*} w_{\lambda_k}(y^k,s_k)&= u((y^k)^{\lambda_k},s_k)-u((y^k)^{\lambda_0},s_k)+w_{\lambda_0}(y^k,s_k)\\ &\geq u((y^k)^{\lambda_k},s_k)-u((y^k)^{\lambda_0},s_k)\rightarrow 0,~\text{as}~ k\rightarrow \infty. \end{align*} $$

The conclusions from the above two cases contradict (4.15). Therefore, (4.14) holds.

Secondly, from (4.14), there exists a sequence $\{(x^k,t_k)\}\subset \Sigma _{\lambda _k}\times \mathbb R$ such that

$$ \begin{align*}w_{\lambda_k}(x^k, t_k)=-\gamma_k+\gamma_k^2<0. \end{align*} $$

We will prove that $x^k$ will not converge to $T_{\lambda _k}$ , namely,

(4.16) $$ \begin{align} d_k=\frac{1}{2}\mathrm{dist}\{x^k,T_{\lambda_k}\}\geq \hat C>0. \end{align} $$

Otherwise, we have $d_k \to 0$ . Then we construct a perturbation of $w_{\lambda _k}$ near the point $(x^k, t_k)$ as follows:

$$ \begin{align*}\hat V_k(x,t)=w_{\lambda_k}(x,t)-\gamma_k^2\hat \xi_k(x,t),~(x,t)\in \mathbb R^n\times \mathbb R,\end{align*} $$

where

$$ \begin{align*}\hat \xi_k =\hat \xi\left(\frac{x-x^k}{d^s_k},\frac{t-t_k}{{d_k}^{\frac{2s}{\alpha}}}\right)\end{align*} $$

with $d_k=\frac {1}{2}\mathrm{dist}\{x^k,T_{\lambda _k}\} $ and $\hat \xi \in C^{\infty }_0(\mathbb R^n\times \mathbb R)$ is a smooth cut off functions satisfying

$$ \begin{align*} \begin{cases} 0\leq \hat\xi \leq 1, & \text{in}~ \mathbb R^n \times \mathbb R,\\ \hat\xi=1, & \text{in}~B_{\frac{1}{2}}(0)\times [-\frac{1}{2},\frac{1}{2}],\\ \hat\xi=0, & \text{in}~ (\mathbb R^n \times \mathbb R)\backslash (B_{1}(0)\times [-1,1]). \end{cases} \end{align*} $$

Set

$$ \begin{align*}\hat E_k(x^k,t_k):=(B_{d^s_k}(x^k)\cap\Sigma_{\lambda_k}) \times [t_k-(d_k)^{\frac{2s}{\alpha}}, t_k+d_k^{\frac{2s}{\alpha}}]\subset \Sigma_{\lambda_k}\times \mathbb R.\end{align*} $$

As a result, $\hat V_k$ must achieve its minimum which is at most $-\gamma _k$ in $\overline {\hat E_k(x^k,t_k)}\subset \Sigma _{\lambda _k}\times \mathbb R $ (see Figure 4), that is,

$$ \begin{align*} \exists~ \{(\hat x^k,\hat t_k)\} \subset \overline{\hat E_k(x^k,t_k)}~s.t. ~-\gamma_k-\gamma^2_k\leq \hat V_k(\hat x^k,\hat t_k)=\underset{\Sigma_{\lambda_k} \times \mathbb R}{\inf}\hat V_k(x,t)\leq -\gamma_k. \end{align*} $$

Figure 4:

The domain $\hat E_k(x^k,t_k)$ .

It follows that

$$ \begin{align*} -\gamma_k\leq {w}_{\lambda_k}(\hat x^k, \hat t_k)\leq -\gamma_k+\gamma^2_k<0. \end{align*} $$

Employing arguments parallel to those in (4.8)–(4.10), we conclude

$$ \begin{align*}\partial_t^\alpha \hat V_k(\hat x^k,\hat t_k)+ (-\Delta)^s\hat V_k(\hat x^k,\hat t_k)-\Delta\hat V_k(\hat x^k,\hat t_k)\leq \frac{\hat c(-\gamma_k+\gamma_k^2)}{d_k^{2s}},\end{align*} $$

which together with (4.1) gives rise to

$$ \begin{align*} \frac{\hat c(-\gamma_k+\gamma_k^2)}{d_k^{2s}}&\geq \partial_t^\alpha\hat V_k(\hat x^k,\hat t_k)+ (-\Delta)^s\hat V_k(\hat x^k,\hat t_k)-\Delta\hat V_k(\hat x^k,\hat t_k)\\ &=\left(\partial_t^\alpha + (-\Delta)^s -\Delta \right) w_{\lambda_k}(\hat x^k,\hat t_k)-\gamma_k^2\left(\partial_t^\alpha + (-\Delta)^s -\Delta \right) \hat \xi _k(\hat x^k,\hat t_k)\\ &\geq f(\hat x_1^k)M_{\lambda_k}(\hat x^k,\hat t_k)w_{\lambda_k}(\hat x^k,\hat t_k) -\frac{c_2\gamma_k^2}{d_k^{2s^2}}-\frac{c_3\gamma_k^2}{d_k^{2s}}, \end{align*} $$

therefore,

$$ \begin{align*} \hat c(1-\gamma_k^2) \leq & f(\hat x_1^k)M_{\lambda_k}(\hat x^k,\hat t_k)d_k^{2s}+c_2\gamma_kd_k^{2s(1-s)}+c_3\gamma_k\\ \leq & f(\lambda_0)M_{\lambda_k}(\hat x^k,\hat t_k)d_k^{2s}+c_2\gamma_kd_k^{2s(1-s)}+c_3\gamma_k. \end{align*} $$

This leads to a contradiction when $d_k \to 0$ as $k \to +\infty $ . Therefore, (4.16) holds.

Next, we continue the proof of $\lambda _0 = +\infty $ . To this end, we perturb $w_{\lambda _k}$ once again near the point $(x^k, t_k)$ as follows:

$$\begin{align*}V_k(x, t) = w_{\lambda_k}(x, t) - \gamma_k^2 \, \xi_k(x, t), \quad (x, t) \in \mathbb{R}^n \times \mathbb{R}, \end{align*}$$

where $\xi _k$ is defined as in Step 1, with

$$\begin{align*}d_k = \frac{1}{2} \operatorname{dist}(x^k, T_{\lambda_k}) \geq \hat{C}> 0. \end{align*}$$

Set

$$ \begin{align*}E_k(x^k,t_k):= B_{d _k}(x^k) \times [t_k-(d_k)^{\frac{2s}{\alpha}}, t_k+d_k^{\frac{2s}{\alpha}}]\subset \Sigma_{\lambda_k}\times \mathbb R.\end{align*} $$

Then $V_k$ must achieve its minimum which is at most $-\gamma _k$ within $\overline {E_k(x^k,t_k)}\subset \Sigma _{\lambda _k}\times \mathbb R $ , that is,

$$ \begin{align*} \exists~ \{(\tilde x^k,\tilde t_k)\} \subset \overline{E_k(x^k,t_k)}~s.t. ~-\gamma_k-\gamma^2_k\leq V_k(\tilde x^k,\tilde t_k)=\underset{\Sigma_{\lambda_k} \times \mathbb R}{\inf} V_k(x,t)\leq -\gamma_k, \end{align*} $$

which implies that

(4.17) $$ \begin{align} -\gamma_k\leq {w}_{\lambda_k}(\tilde x^k, \tilde t_k)\leq -\gamma_k+\gamma^2_k<0. \end{align} $$

Then, by applying an argument similar to that in Step 1, we obtain

(4.18) $$ \begin{align} \frac{c(-\gamma_k + \gamma_k^2)}{d_k^{2s}} \geq f(\tilde{x}_1^k) M_{\lambda_k}(\tilde{x}^k, \tilde{t}_k) w_{\lambda_k}(\tilde{x}^k, \tilde{t}_k). \end{align} $$

If $f(\tilde {x}_1^k) \leq 0$ , this leads to a contradiction, and the proof is complete. Hence, we may assume that $f(\tilde {x}_1^k)> 0$ . By (4.17), we then have

(4.19) $$ \begin{align} c(1-\gamma_k ) \leq f(\tilde x_1^k)M_{\lambda_k}(\tilde x^k,\tilde t_k)d_k^{2s}. \end{align} $$

Therefore, by assumption (1.4) in $(F1)$ and the boundedness of $M_{\lambda _k}(\tilde x^k,\tilde t_k)$ , we obtain

$$ \begin{align*}\{\tilde x^k_1\}~\text{is bounded away from}~-\infty\end{align*} $$

which, together with $f\in C(\mathbb R^n)$ with $(\tilde x^k,\tilde t_k)\in \overline {E_k(x^k,t_k)}$ implies that

$$ \begin{align*}f(\tilde x_1^k)\leq c,~ d_k^{2s}\leq c,\end{align*} $$

for sufficiently large k.

Furthermore, in view of (4.19), we conclude that

(4.20) $$ \begin{align} f(\tilde x_1^k),~ M_{\lambda_k}(\tilde x^k,\tilde t_k)\geq C>0, \end{align} $$

for sufficiently large k. Then, using the inequalities

$$\begin{align*}u_{\lambda_k}(\tilde{x}^k, \tilde{t}_k) < \eta_{\lambda_k}(\tilde{x}^k, \tilde{t}_k) < u(\tilde{x}^k, \tilde{t}_k) \end{align*}$$

and the positivity of u, we obtain

(4.21) $$ \begin{align} u(\tilde{x}^k, \tilde{t}_k) \geq C> 0 \quad \text{and} \quad g\big(u(\tilde{x}^k, \tilde{t}_k)\big) \geq C > 0. \end{align} $$

Since $w_{\lambda _k}(\tilde {x}^k, \tilde {t}_k) \to 0$ as $k \to \infty $ , it follows from (4.21) that

(4.22) $$ \begin{align} g( u_{\lambda_k}(\tilde x^k,\tilde t_k))\geq C>0, \end{align} $$

for sufficiently large k.

Next, starting from the original equation (4.1), we refine the estimate (4.18) and rewrite it as

(4.23) $$ \begin{align} \frac{c(-\gamma_k + \gamma_k^2)}{d_k^{2s}} \geq \big[ f((\tilde{x}_1^k)^{\lambda_k}) - f(\tilde{x}_1^k) \big] u_{\lambda_k}^p(\tilde{x}^k, \tilde{t}_k) + f(\tilde{x}_1^k) M_{\lambda_k}(\tilde{x}^k, \tilde{t}_k) w_{\lambda_k}(\tilde{x}^k, \tilde{t}_k). \end{align} $$

Observing that $|\tilde {x}_1^k - \lambda _k| \sim d_k$ , and using the continuity and monotonicity of f, along with (4.20), we obtain

(4.24) $$ \begin{align} f((\tilde{x}_1^k)^{\lambda_k}) - f(\tilde{x}_1^k) \geq C> 0. \end{align} $$

Combining (4.17), (4.20), and (4.22)–(4.24), we arrive at a contradiction for sufficiently large k.

Therefore, (4.13) is established.

Step 3. We have already established that

$$\begin{align*}w_\lambda(x,t) \geq 0,\quad \forall~(x,t)\in \Sigma_\lambda\times \mathbb{R},~\forall~\lambda\in \mathbb{R}. \end{align*}$$

We now proceed to show that the inequality is in fact strict, that is,

(4.25) $$ \begin{align} w_\lambda(x,t)> 0,\quad \forall~(x,t)\in \Sigma_\lambda \times \mathbb{R},~\forall~\lambda\in \mathbb{R}. \end{align} $$

Assume by contradiction that for some fixed $\lambda $ , there exists a point $(\bar {x}, \bar {t}) \in \Sigma _\lambda \times \mathbb {R}$ such that

(4.26) $$ \begin{align} w_\lambda(\bar{x}, \bar{t}) = \min_{\Sigma_\lambda \times \mathbb{R}} w_\lambda(x,t) = 0. \end{align} $$

On the one hand, since $(\bar {x}, \bar {t})$ is a global minimum point, we have

$$\begin{align*}\partial_t^\alpha w_\lambda(\bar{x}, \bar{t}) \leq 0,\quad (-\Delta)^s w_\lambda(\bar{x}, \bar{t}) \leq 0,\quad \text{and} \quad -\Delta w_\lambda(\bar{x}, \bar{t}) \leq 0. \end{align*}$$

Thus,

(4.27) $$ \begin{align} \partial_t^\alpha w_\lambda(\bar{x}, \bar{t}) + (-\Delta)^s w_\lambda(\bar{x}, \bar{t}) - \Delta w_\lambda(\bar{x}, \bar{t}) \leq 0. \end{align} $$

On the other hand, by the second equation in (4.1), we have

$$ \begin{align*} &\partial_t^\alpha w_\lambda(x,t) + (-\Delta)^s w_\lambda(x,t) - \Delta w_\lambda(x,t)\\ =\ & \big( f(x_1^\lambda) - f(x_1) \big) g(u_\lambda(x,t)) + f(x_1) \big( g(u_\lambda(x,t)) - g(u(x,t)) \big). \end{align*} $$

Evaluating this at $(\bar {x}, \bar {t})$ and noting that $w_\lambda (\bar {x}, \bar {t}) = 0$ , that is, $u_\lambda (\bar {x}, \bar {t}) = u(\bar {x}, \bar {t})$ , we obtain

(4.28) $$ \begin{align} \partial_t^\alpha w_\lambda(\bar{x}, \bar{t}) + (-\Delta)^s w_\lambda(\bar{x}, \bar{t}) - \Delta w_\lambda(\bar{x}, \bar{t}) = \big( f(\bar{x}_1^\lambda) - f(\bar{x}_1) \big) g(u_\lambda(\bar{x}, \bar{t})). \end{align} $$

Since f is strictly increasing in $x_1$ , $\bar {x}_1^\lambda> \bar {x}_1$ , and $u> 0$ , and assuming g satisfies condition $(F2)$ (i.e., $g(u)> 0$ for $u> 0$ ), the right-hand side of (4.28) is strictly positive. However, this contradicts (4.27), which implies the left-hand side is nonpositive.

Therefore, the assumption in (4.26) must be false, and (4.25) holds. Consequently, $u(x, t)$ is strictly increasing in the $x_1$ -direction.

This concludes the proof of Theorem 1.5.

Proof Suppose, on the contrary, that (4.30) is false. Then there exists a first time $ \hat {t} \in (1, T] $ such that

$$\begin{align*}u(\hat{t}) = \inf_{t \in (-\infty, \hat{t}]} u(t) = 0. \end{align*}$$

Using the assumption $ u(t)> 0 $ for $ t \leq 1 $ and the definition of the Caputo fractional derivatives, we have

$$\begin{align*}\begin{aligned} a\, \partial_t^\alpha u(\hat{t}) + b\, \partial_t^\beta u(\hat{t}) - C u(\hat{t}) &\leq a\, \partial_t^\alpha u(\hat{t}) + b\, \partial_t^\beta u(\hat{t}) \\ &= a C_\alpha \int_{-\infty}^{\hat{t}} \frac{u(\hat{t}) - u(\tau)}{(\hat{t} - \tau)^{1+\alpha}}\, d\tau + b C_\beta \int_{-\infty}^{\hat{t}} \frac{u(\hat{t}) - u(\tau)}{(\hat{t} - \tau)^{1+\beta}}\, d\tau \\ &< 0, \end{aligned} \end{align*}$$

since $ u(\hat {t}) = 0 $ and $ u(\tau )> 0 $ for all $ \tau < \hat {t} $ . This contradicts the first inequality in (4.29).

Therefore, the conclusion (4.30) must be true, and the proof is complete.

Proof of Theorem 1.6

Suppose, for the sake of contradiction, that (1.2) admits a global positive bounded solution $ u $ .

Consider the first eigenvalue problem

$$ \begin{align*} \begin{cases} (-\Delta)^s \varphi(x) - \Delta \varphi(x) = \lambda_1 \varphi(x), & x \in B_1(R+2, 0'), \\ \varphi(x) = 0, & x \in \mathbb{R}^n \setminus B_1(R+2, 0'), \end{cases} \end{align*} $$

where the large parameter $ R \geq 0 $ will be determined later.

To perform integration by parts, we mollify $ \varphi (x) $ to obtain $ \varphi _1(x) = \rho \ast \varphi (x) \in C_0^\infty (\mathbb {R}^n) $ , where $ \rho \in C_0^\infty (B_1(0)) $ is the standard mollifier satisfying $ \int _{\mathbb {R}^n} \rho (x)\, dx = 1 $ .

It is clear that the support of $ \varphi _1 $ is contained in $ B_2(R+2, 0') $ . Moreover, by Lemma 2.5, we have

(4.31) $$ \begin{align} (-\Delta)^s \varphi_1(x) - \Delta \varphi_1(x) \leq \lambda_1 \varphi_1(x), \quad \text{for all } x \in \mathbb{R}^n. \end{align} $$

We may assume that

(4.32) $$ \begin{align} \int_{\mathbb R^n} \varphi_1(x)dx=1. \end{align} $$

Denote

(4.33) $$ \begin{align} \psi_R(t):= \int_{\mathbb R^n} u(x,t) \varphi_1(x)dx=\int_{B_2(R +2,0')} u(x, t)\varphi_1 (x)dx. \end{align} $$

Case 1. When $ \alpha =1,$ by Remark 2.6, equation (1.2), conditions (F1) and (F2), $\operatorname {supp} \varphi _1\subset B_2(R +2,0')$ , and (4.31), we obtain

(4.34) $$ \begin{align} \frac{d }{dt}\psi_R(t)&= -\int_{\mathbb{R}^n}((-\Delta)^su(x,t)-\Delta u(x,t))\varphi_1 (x)dx + \int_{\mathbb{R}^n}f( x_1) g(u (x,t)) \varphi_1 (x)dx\nonumber\\ &=-\int_{\mathbb{R}^n}u(x,t)((-\Delta)^s \varphi_1 (x)-\Delta \varphi_1(x))dx \nonumber \\ &\quad + \int_{B_2(R +2,0')} f(x_1) g(u(x,t))\varphi_1 (x)dx\nonumber\\ & \geq - \lambda_1\int_{\mathbb R^n}u(x,t)\varphi_1 (x)dx + \int_{B_2(R +2,0')} f(x_1 ) g(u(x,t))\varphi_1 (x)dx\nonumber\\ & \geq - \lambda_1 \psi_R(t) + \frac{ f_R }{M} \underset{B_2(0)\times[1,T]}{\inf} g(u(x,t))\int_{B_2(R +2,0')} u(x,t)\varphi_1 (x)dx \nonumber\\ &=: (-\lambda_1+\frac{f_R}{M}m_T) \psi_R (t), \end{align} $$

where $~M:=\underset {\mathbb R^n \times \mathbb R}{\sup }u,~m_T=\underset {B_2(0)\times [1,T]}{\inf } g(u(x,t)),~f_R \rightarrow +\infty $ as $R\rightarrow \infty $ and we have used the fact that the strictly increasing of u with respect to $x_1$ . Since $ u(x,t) $ is strictly increasing in the $ x_1 $ -direction by Theorem 1.5, it follows that for any fixed $ t \in \mathbb {R} $ , the function $ \psi _R(t) $ is monotone increasing with respect to $ R $ . Therefore,

(4.35) $$ \begin{align} \psi_R(0)\geq 2C_0:=\psi_0(0). \end{align} $$

If $ t \geq 0 $ satisfies $ \psi _R(t) \geq C_0 $ , then, since

$$\begin{align*}\lim_{x_1 \to +\infty} f(x_1) = +\infty, \end{align*}$$

by choosing $ R $ sufficiently large, we have

$$ \begin{align*}-\lambda_1+\frac{f_R}{M}m_T \geq 1.\end{align*} $$

Then (4.34) implies that

$$ \begin{align*}\frac{d }{dt}\psi_R(t) \geq \psi_R(t). \end{align*} $$

Thus, from (4.35), we deuce

$$ \begin{align*}\psi_R(t) \geq 2C_0 e^t. \end{align*} $$

Next, we verify the condition

(4.36) $$ \begin{align} \psi_R(t) \geq C_0, \,\, \forall ~t \geq 0 \end{align} $$

by contradiction. Suppose that (4.36) does not hold, then there exists $t_0>0$ such that

$$ \begin{align*}\psi_R(t_0)=C_0 \,\, \text{ and }\,\, \psi_R(t)>C_0 \,\, \text{ in }\,\, [0, t_0). \end{align*} $$

It follows that $\psi _R(t) \geq 2C_0 e^t \geq 2C_0$ in $[0, t_0),$ which contradicts $\psi _R(t_0)=C_0.$ Hence, (4.36) must hold.

Therefore, $\psi _R(t)$ is monotone increasing with respect to $t,$ and

$$ \begin{align*}\psi_R(t) \geq 2C_0 e^t,\,\, \forall~ t \geq 0. \end{align*} $$

Consequently,

$$ \begin{align*}\psi_R(t) \to +\infty, \,\, \text{as} \,\, t\to +\infty, \end{align*} $$

which contradicts the boundedness of $u(x, t).$

Case 2. When $0<\alpha <1,$ we define

$$ \begin{align*}T=\left(\frac{M}{c_1}\right)^\theta,\end{align*} $$

where

(4.37) $$ \begin{align} \theta>0, ~M:=\underset{\mathbb R^n \times \mathbb R}{\sup}u ~\text{ and }~ 2c_1:=\underset{B_2(R+2,0')\times [0,1]}{\inf} u. \end{align} $$

By equation (1.2), conditions (F1) and (F2), $u(x,\cdot )\in C^1(\mathbb R),\, \operatorname {supp} \varphi _1\subset B_2(R +2,0')$ , Remark 2.6, and (4.31), similarly to (4.34), we conclude that for any $t\in [1,T],$

(4.38) $$ \begin{align} \partial_t^\alpha \psi_R(t)=& \int_{\mathbb R^n} \partial_t^\alpha u(x,t) \varphi_1(x) dx\nonumber\\ \geq & - \lambda_1 \psi_R(t) + \frac{ f_R }{M} \underset{B_2(0)\times[1,T]}{\inf} g(u(x,t))\int_{B_2(R +2,0')} u(x,t)\varphi_1 (x)dx \nonumber\\ =:&\left(-\lambda_1+\frac{f_R}{M}m_T\right) \psi_R(t), \end{align} $$

where

$$\begin{align*}m_T = \inf_{B_2(0) \times [1, T]} g(u(x,t)), \quad f_R \to +\infty \quad \text{as } R \to \infty, \end{align*}$$

and we have used the fact that $ u $ is strictly increasing with respect to $ x_1 $ .

Next, we construct an increasing sub-solution for the linear operator $ \partial _t^\alpha - \hat {C}_\alpha $ for some constant $ \hat {C}_\alpha> 0 $ .

Set

$$\begin{align*}w(t) = t^\theta, \end{align*}$$

where $ 0 < \theta = \frac {1}{2k+1} < \alpha $ for some integer $ k $ . Direct calculation yields

(4.39) $$ \begin{align} \begin{cases} \partial_t^\alpha w(t) = \hat{C}_\alpha t^{\theta - \alpha} \leq \hat{C}_\alpha w(t), & t \in [1, +\infty), \\ w(t) \leq 0, & t \in (-\infty, 0], \\ w(t) \leq 1, & t \in [0,1]. \end{cases} \end{align} $$

Note that $\lim _{x_1 \to +\infty } f(x_1) = +\infty $ . Hence, we may choose $ R $ sufficiently large so that

(4.40) $$ \begin{align} -\lambda_1 + \frac{f_R}{M} m_T> \hat{C}_\alpha. \end{align} $$

By the monotonicity of $ u(x,t) $ with respect to $ x_1 $ established in Theorem 1.5, and combining (4.32), (4.33), and (4.37), we conclude

(4.41) $$ \begin{align} \psi_R(t) \geq 2 c_1, \quad t \in [0,1]. \end{align} $$

Define

$$\begin{align*}v(t) = \psi_R(t) - c_1 w(t). \end{align*}$$

Then, by (4.38)–(4.41), the function $ v(t) $ satisfies

$$\begin{align*}\begin{cases} \partial_t^\alpha v(t) - \hat{C}_\alpha v(t) \geq 0, & t \in [1, T], \\ v(t)> 0, & t \in (-\infty, 1]. \end{cases} \end{align*}$$

Applying the maximum principle (Lemma 4.1) with $ a=1 $ and $ b=0 $ , it follows that

$$\begin{align*}v(t)> 0, \quad \text{for all } t \in [1, T]. \end{align*}$$

That is,

$$\begin{align*}\psi_R(t)> c_1 t^\theta, \quad t \in [1, T]. \end{align*}$$

Evaluating at $ t = T $ and using (4.37), we obtain

$$\begin{align*}\psi_R(T)> M, \end{align*}$$

which contradicts

$$\begin{align*}\psi_R(t) \leq M, \end{align*}$$

where the last inequality follows from (4.32) and (4.33).

This completes the proof of Theorem 1.6.