Proof. Note that for any $\tau > 0$, $u_\tau \geq 0$, since $u(0)$ is the maximum value for $u$ in $\overline {\Omega }$. Using the fact that $(\lambda ^1_{s,r},\, u(\tau x))$ is the first eigenpair for fractional $r$-Laplacian on $\Omega _\tau$, we obtain that the following equation holds weakly:

(3.2)\begin{align} (-\Delta)^s_r u_\tau(x) ={-}(-\Delta)^s_r u(\tau x) ={-}\lambda^1_{s,r}u(\tau x)^{r-1} \text{ in }\Omega_\tau, \quad u_\tau = \frac{ u(0)}{\tau^{sr^{\prime}}} \text{ in } \mathbb{R}^d \setminus \Omega_\tau. \end{align}

For each $\tau >0$, using proposition 2.1 and [Reference Brasco and Parini14, Theorem 3.13], we get $u_{\tau } \in {C}(\Omega _{\tau })$. Now we divide our proof into two steps. In the first step, we show that $u_{\tau _n}\to \tilde {u}$ in $C_{\text {loc}}(\mathbb {R}^d)$ as $n\to \infty$. In the second step, we prove $\tilde {u}$ is a weak solution to (3.1).

Step 1: Take a ball $B_R(0)$ such that $\overline {B_{4R}(0)}\subset \Omega _\tau$. We choose $\sigma _1>0$ as follows

\[ \sigma_1:=\frac{{\rm d}\gamma}{s}, \text{ where } \gamma >1. \]

By the nonlocal Harnack inequality (see [Reference Giacomoni, Kumar and Sreenadh24, Theorem 2.2]), there exists $C=C(d,\,s,\,r)$ such that

(3.3)\begin{align} \max_{\substack{B_R(0)}}u_\tau \leq C\left(\min_{\substack{B_{2R}(0)}}u_\tau+{\lVert}{\lambda^1_{s,r}\,u^{r-1}}{\rVert}_{L^{\sigma_1}(B_{2R}(0))}^{\frac{1}{r-1}}\right) = C {\lVert}{\lambda^1_{s,r}\,u^{r-1}}{\rVert}_{L^{\sigma_1}(B_{2R}(0))}^{\frac{1}{r-1}}, \end{align}

In (3.3) the last equality follows from the fact $\min _{\substack {B_{2R}(0)}}u_\tau =0$, because origin is the maximum point of $u$ in $\Omega$. Further, for $r\geq 2$ we immediately get $(r-1) \sigma _1 > 1$, and for $1< r<2$ we choose

\[ \gamma> \left\{\begin{array}{@{}ll} \text{max}\left\{\dfrac{sr}{dr+2(sr-d)},1\right\} , & \text{if } sr< d; \\ \text{max}\left\{\dfrac{s}{d(r-1)},1\right\}, & \text{if} \; sr \ge d, \\ \end{array} \right. \]

to get $(r-1)\sigma _1>1$. Then proposition 2.1-(v) and (3.3) yield

(3.4)\begin{equation} \max_{\substack{B_R(0)}}u_\tau \leq C {\lVert}{\lambda^1_{s,r}\,u^{r-1}}{\rVert}_{L^{\sigma_1}( \mathbb{R}^d )}^{\frac{1}{r-1}} \le C, \end{equation}

where $C=C(d,\, s,\, r,\, \lambda ^1_{s,r},\,{\lVert }{u}{\rVert }_{L^{(r-1)\sigma _1}(\mathbb {R}^d)})$. Next, we define the following exponent

(3.5)\begin{equation} \Theta(d,s,r,\sigma_1) := \min \left\{ \frac{1}{r-1} \left( sr - \frac{d}{\sigma_1} \right), 1 \right\}.\, \end{equation}

Then, applying the regularity estimate [Reference Brasco, Lindgren and Schikorra13, Theorem 1.4] when $r \ge 2$ and [Reference Garain and Lindgren22, Theorem 1.2] when $1< r < 2$, for the problem (3.2) we get the following Hölder regularity estimate of the weak solution $u_{\tau }$ for any $s<\delta <\Theta (d,\,s,\,r,\,\sigma _1)$:

(3.6)\begin{align} & [u_\tau]_{C^{0,\delta}(B_{R/8}(0))} \notag\\ & \quad \leq\frac{C}{R^{\delta}} \left[\max_{\substack{B_R(0)}}u_\tau + \mathrm{Tail}_r(u_\tau;0,R) + \left(R^{sr - \frac{d}{\sigma_1} } \lambda^1_{s,r}{\lVert}{u^{r-1}}{\rVert}_{L^{\sigma_1} (B_R(0))} \right)^{\frac{1}{r-1}} \right] \nonumber\\ & \quad :=\frac{C}{R^\delta}\left[\max_{\substack{B_R(0)}}u_\tau+I_1+I_2^{\frac{1}{r-1}}\right], \end{align}

where $C=C(d,\,s,\,r)$. We now estimate the last two terms $I_1,\,I_2$ of (3.6) as follows: Estimate of $I_2$: Choose $a>0$ such that $a >\gamma r-1$. Then, by the change of variable we have

(3.7)\begin{align} I_2& :=\lambda^1_{s,r} R^{ s(r - \frac{1}{\gamma}) } \left( \int_{B_R(0)} {\mid}{u(y)}{\mid}^{(r - 1) \sigma_1} \, {\rm d}y \right)^{\frac{1}{\sigma_1}}\notag\\ & = \lambda^1_{s,r} \frac{R^{ s(r - \frac{1}{\gamma}) }}{R^{ \frac{s a}{\gamma} }} \left( \int_{B_{R^{ a + 1}}(0)} {\mid}{u(z)}{\mid}^{ (r-1) \sigma_1} \, {\rm d}z \right)^{\frac{1}{\sigma_1}} \nonumber\\ & \le \lambda^1_{s,r} R^{ s(r - \frac{1}{\gamma}) - \frac{s a}{\gamma}} \left( \int_{\mathbb{R}^d} {\mid}{u(z)}{\mid}^{ (r-1) \sigma_1} \, {\rm d}z \right)^{\frac{1}{\sigma_1}}, \end{align}

where we see that $r - \frac {1}{\gamma } < \frac {a}{\gamma }$.

Estimate of $I_1$: Note that

(3.8)\begin{align} I_1& :=\mathrm{Tail}_r(u_\tau;0,R)\leq C(r)\left(\mathrm{Tail}_r(u_{\tau}^+;0,R)+\mathrm{Tail}_r(u_{\tau}^-;0,R)\right)\notag\\ & =C(r)\mathrm{Tail}_r(u_{\tau}^+;0,R), \end{align}

where the last equality follows from the non-negativity of $u_\tau$. To estimate $\mathrm {Tail}_r(u_{\tau }^+;0,\,R)$, let $R_1 = 4R$ and $\ell :=\max _{\substack {B_{R}(0)}}u_\tau$. Take $\phi \in C_c^\infty (B_{R})$ satisfying $0\leq \phi \leq 1$, $\phi =1$ in $B_{\frac {R}{2}}$ and ${\mid }{\nabla \phi }{\mid }\leq \frac {8}{R}$. We use the test function $\eta :=(u_\tau -2\ell )\phi ^p$ in the weak formulation of $u_\tau$ and then proceed similarly as in [Reference T. Kuusi and Palatucci32, Lemma 4.2]) to get a constant $C=C(d,\,s,\,r)$ such that

\begin{align*} C\ell{\mid}{B_{R}}{\mid}R^{{-}sr}\mathrm{Tail}_r(u_{\tau}^+;0,R)^{r-1} & \leq C\ell^{r}{\mid}{B_{R}}{\mid}R^{{-}sr}+\lambda^1_{s,r}\int_{B_{R}}u^{r-1}\eta \, {\rm d}x \\ & \leq C\ell^{r}{\mid}{B_{R}}{\mid}R^{{-}sr}+3\ell\lambda^1_{s,r}\int_{B_{R}}u^{r-1} \, {\rm d}x \\ & \leq C\ell^{r}{\mid}{B_{R}}{\mid}R^{{-}sr}+3\ell\lambda^1_{s,r}{\mid}{B_{R}}{\mid}^{\frac{1}{ \sigma_1'}} {\lVert}{u^{r-1}}{\rVert}_{L^{\sigma_1}(B_R)}, \end{align*}

where in the above estimates we used the fact $|u-2\ell |\leq 3\ell$ in $B_{R}$. This implies that

(3.9)\begin{align} \mathrm{Tail}_r(u_{\tau}^+;0,R) & \leq C\left(\max_{\substack{B_{R}(0)}}u_\tau+\left(R^{sr - \frac{d}{\sigma_1} }\lambda^1_{s,r}{\lVert}{u^{r-1}}{\rVert}_{L^{\sigma_1}(B_R)}\right)^{\frac{1}{r-1}}\right)\nonumber\\ & \le C\left(\max_{\substack{B_{R}(0)}}u_\tau + I_2^{\frac{1}{r-1}}\right). \end{align}

Now, plugging the estimates (3.4), (3.7), (3.8), (3.9) into (3.6), we thus obtain

(3.10)\begin{equation} [u_\tau]_{C^{0, \delta}(B_{R/8}(0))}\leq\frac{C}{R^{\delta+\epsilon}}, \end{equation}

where $C=C(d,\,s,\,r,\,\lambda ^1_{s,r},\, {\lVert }{u}{\rVert }_{L^{ (r-1) \sigma _1}(\mathbb {R}^d)})$ and $\epsilon :=\frac {1}{r-1}(\frac {1+a-r\gamma }{\gamma }) > 0$. Let $K \subset \mathbb {R}^d$ be any compact set. Observe that $\Omega _{\tau }$ becoming $\mathbb {R}^d$ when $\tau$ is sufficiently small. Thus, we can choose $R>1$ and $0<\tau _0<<1$ such that $K \subset B_{\frac {R}{8}}(0) \subset \Omega _{\tau }$ for all $\tau \in (0,\, \tau _0)$. Therefore, we use (3.4) and (3.10) to obtain the following uniform estimate for all $\tau \in (0,\,\tau _0)$:

(3.11)\begin{equation} \max_{K} u_{\tau} \le C, \quad \text{and}\quad [u_\tau]_{C^{0, \delta }(K)} \le C, \end{equation}

where $C$ is independent of both $\tau$ and $K$. Next, for a sequence $(\tau _n)$ converging to zero, we consider the corresponding sequence of functions $(u_{\tau _n})$. Using (3.11) we can show that $(u_{\tau _n})$ is equicontinuous and uniformly bounded in $K$. Therefore, applying the Arzela–Ascoli theorem, up to a subsequence, $u_{\tau _n}\to \tilde {u}\text { in }C(K)$. Thus we have

(3.12)\begin{equation} u_{\tau_n}\to\tilde{u}\text{ in }C_{\text{loc}}(\mathbb{R}^d), \text{ as } n \to \infty. \end{equation}

Step 2: Recalling the weak formulation of (3.2) for $\tau >0$ be any,

(3.13)\begin{align} & \iint\limits_{\mathbb{R}^d\times\mathbb{R}^d}\frac{|u_\tau(x)-u_\tau(y)|^{r-2}(u_\tau(x)-u_\tau(y))(\phi(x)-\phi(y))}{|x-y|^{d+sr}} \, {\rm d}x {\rm d}y \nonumber\\ & \quad ={-}\lambda^1_{s,r} \int_{\Omega_\tau} u(\tau x)^{r-1} \phi(x) \, {\rm d}x, \quad \forall \, \phi\in C_c^\infty(\Omega_\tau). \end{align}

Let $v\in C_c^\infty (\mathbb {R}^d)$ and let $\text {supp}(v) := K$. Since $\Omega _{\tau _n}$ is becoming $\mathbb {R}^d$, as $\tau _n\to 0$, there exists $n_0 \in \mathbb {N}$ such that $K \subset \Omega _{\tau _n}$ for all $n\geq n_0.$ Hence, from (3.13) for every $n \ge n_0$, we write

(3.14)\begin{align} & \iint\limits_{\mathbb{R}^d\times\mathbb{R}^d}\frac{|u_{\tau_n}(x)-u_{\tau_n}(y)|^{r-2}(u_{\tau_n}(x)-u_{\tau_n}(y))(v(x)-v(y))}{|x-y|^{d+sr}} \, {\rm d}x {\rm d}y\notag\\ & \quad ={-}\lambda^1_{s,r} \int_{K} u(\tau_n x)^{r-1} v(x) \, {\rm d}x. \end{align}

We pass the limit as $n\to \infty$ in the R.H.S of (3.14), to get

(3.15)\begin{align} \lim\limits_{n\to\infty} \int_{K} u(\tau_n x)^{r-1}v(x) \, {\rm d}x & = \lim\limits_{n\to\infty}\int_{\mathbb{R}^d}u(\tau_n x)^{r-1}v(x)\chi_{K}(x) \, {\rm d}x\notag\\ & = \int_{K} u(0)^{r-1}v(x)\,{\rm d}x, \end{align}

where the last equality in (3.15) follows using the dominated convergence theorem. Again, applying the dominated convergence theorem, we have

\begin{align*} & \text{L.H.S of}\, (3.14)\\ & \quad = \lim\limits_{k\to\infty}\iint\limits_{B_k(0)\times B_k(0)}\frac{|u_{\tau_n}(x)-u_{\tau_n}(y)|^{r-2}(u_{\tau_n}(x)-u_{\tau_n}(y))(v(x)-v(y))}{|x-y|^{d+sr}} \, {\rm d}x {\rm d}y \\ & \quad := \lim\limits_{k\to\infty}\iint\limits_{B_k(0)\times B_k(0)} F_n(x,y)\,{\rm d}x{\rm d}y. \end{align*}

Proof of theorem 1.9.

For simplicity of notation, we denote $u_0 = \phi _{s_1,p}$ and $v_0 = \phi _{s_2, q}$. We argue by contradiction. Suppose $u_0 = cv_0$ for some non-zero $c \in {\mathbb {R}}$. Without loss of any generality, we can assume that $u_0=v_0.$ By proposition 2.1, $u_0$ is uniformly bounded, $u_0>0$ in $\Omega$ and is in $C( \overline { \Omega })$. This guarantees that $u_0$ has a global extremum point. Since the operator $(-\Delta )_p^{s_1}$ is translation invariant, we can assume that the origin is such a point. Now for $\tau >0$, define

(3.19)\begin{equation} u_\tau(x) := \left\{\begin{array}{@{}ll} \dfrac{ u_0(0)-u_0(\tau x)}{\tau^{s_1{p^{\prime}}}} , & \text{for } x\in\Omega_\tau; \\[9pt] \dfrac{ u_0(0)}{\tau^{s_1{p^{\prime}}}}, & \text{for } \; x \in \mathbb{R}^d \setminus \Omega_\tau, \\ \end{array} \right. \end{equation}

where $\Omega _\tau :=\{x\in \mathbb {R}^d:\tau x\in \Omega \}$. Then by Blow-up lemma 3.1, there exists a sequence $\tau _n\to 0$ such that $u_{\tau _n}\to \tilde {u}$ in $C_{\text {loc}}(\mathbb {R}^d)$, where $\tilde {u}$ is a non-negative solution of

(3.20)\begin{equation} (-\Delta)^{s_1}_p v={-}\lambda^1_{s_1,p}u_0(0)^{p-1} \text{ in }\mathbb{R}^d, \end{equation}

and $\tilde {u}(0)=0.$ Again, by the change of variable we deduce

\begin{align*} (-\Delta)_q^{s_2}u_\tau(x) & =\text{P.V.}\int_{\mathbb{R}^d}\frac{|u_\tau(x)-u_\tau(y)|^{q-2}(u_\tau(x)-u_\tau(y))}{|x-y|^{d+s_2q}} \,{\rm d}y \\ & ={-}\frac{1}{\tau^{s_1{p^{\prime}}(q-1)}}\text{P.V.}\int_{\mathbb{R}^d}\frac{|u_0(\tau x)-u_0(\tau y)|^{q-2}(u_0(\tau x)-u_0(\tau y))}{|x-y|^{d+s_2q}} \,{\rm d}y \\ & ={-}\tau^{s_2q-s_1{p^{\prime}}(q-1)}\,\text{P.V.}\int_{\mathbb{R}^d}\frac{|u_0(\tau x)-u_0(y)|^{q-2}(u_0(\tau x)-u_0(y))}{|\tau x-y|^{d+s_2q}} \,{\rm d}y \\ & ={-}\tau^{s_2q-s_1{p^{\prime}}(q-1)}(-\Delta)_q^{s_2}u_0(\tau x)={-}\tau^{s_2q-s_1{p^{\prime}}(q-1)}\lambda^1_{s_2,q}\,u_0(\tau x)^{q-1}. \end{align*}

This implies that for each $\tau >0$, $u_\tau$ given by (3.19) satisfies the following equation weakly

\[ (-\Delta)_q^{s_2}v={-}\tau^{s_2q-s_1{p^{\prime}}(q-1)}\lambda^1_{s_2,q}\,u_0(\tau x)^{q-1} \text{ in }\Omega_\tau. \]

Using $\frac {s_1{p^{\prime }}}{{q^{\prime }}}< s_2$ we again proceed as in Blow-up lemma 3.1, to obtain that $\tilde {u} \ge 0$ is also a weak solution of the following equation:

\[ (-\Delta)_q^{s_2} v=0\,\text{ in }\mathbb{R}^d. \]

Therefore, by the strong maximum principle [Reference Del P. and Quaas17, Theorem 1.4], we conclude $\tilde {u}=0$ a.e. in $\mathbb {R}^d,$ which gives a contradiction to (3.20) as $u_0(0)>0$. Thus, the set $\{u_0,\,v_0\}$ is linearly independent.

Proof. $\boldsymbol {d>s_1p\,}$: Let $M\geq 0$, define $u_M=\min \{u,\,M\}$. Clearly $u_M$ is non-negative and is in $L^{\infty }(\Omega )$. Since $u\in {W_0^{s_1,p}(\Omega )}$, then $u_M\in {W_0^{s_1,p}(\Omega )}.$ Fixed $\sigma \geq 1$, define $\phi =u_M^{\sigma }$. Then, $\phi \in {W_0^{s_1,p}(\Omega )}.$ Thus taking $\phi$ as a test function in the weak formulation of $u$, we have

(4.1)\begin{align} & \iint\limits_{\mathbb{R}^d\times\mathbb{R}^d}|u(x)-u(y)|^{p-2}(u(x)-u(y))(\phi(x)-\phi(y)) \, {\rm d}\mu_1 \nonumber\\ & \qquad +\iint\limits_{\mathbb{R}^d\times\mathbb{R}^d}|u(x)-u(y)|^{q-2}(u(x)-u(y))(\phi(x)-\phi(y))\, {\rm d}\mu_2 \nonumber\\ & \quad =\alpha\displaystyle{\int_{\Omega}} u(x)^{p-1}\phi(x) \,{\rm d}x+\beta\displaystyle{\int_{\Omega}} u(x)^{q-1}\phi(x)\,{\rm d}x\notag\\ & \quad \leq \alpha\displaystyle{\int_{\Omega}} u(x)^{p+\sigma-1}\,{\rm d}x+\beta\displaystyle{\int_{\Omega}} u(x)^{q+\sigma-1}\,{\rm d}x. \end{align}

Now, using [Reference Brasco, Lindgren and Parini12, Lemma C.2] we estimate

\begin{align*} I_1 & :=\iint\limits_{\mathbb{R}^d\times\mathbb{R}^d}|u(x)-u(y)|^{p-2}(u(x)-u(y))(\phi(x)-\phi(y)) \, {\rm d}\mu_1\\ & \geq\frac{\sigma p^p}{(\sigma+p-1)^p}\iint\limits_{\mathbb{R}^d\times\mathbb{R}^d}\left|u_M(x)^{\frac{\sigma+p-1}{p}}-u_M(y)^{\frac{\sigma+p-1}{p}}\right|^{p}\,{\rm d}\mu_1 \\ & \geq \frac{C(d,s_1,p)\sigma p^p}{(\sigma+p-1)^p} \left(\int_{\mathbb{R}^d}\left(u_M(x)^{\frac{\sigma+p-1}{p}} \right)^{p^*_{s_1}}\,{\rm d}x\right)^{\frac{p}{p^*_{s_1}}}, \end{align*}

where in the last inequality we use ${W_0^{s_1,p}(\Omega )} \hookrightarrow L^{p^*_{s_1}}(\mathbb {R}^d)$. Since $s_2q < d$, using ${W_0^{s_2,q}(\Omega )} \hookrightarrow L^{q^*_{s_2}}(\mathbb {R}^d)$ we estimate $I_2$ as

\begin{align*} I_2 & :=\iint\limits_{\mathbb{R}^d\times\mathbb{R}^d}|u(x)-u(y)|^{q-2}(u(x)-u(y))(\phi(x)-\phi(y))\, {\rm d}\mu_2\\ & \geq \frac{C(d,s_2,q)\sigma q^q}{(\sigma+q-1)^q} \left(\int_{\mathbb{R}^d}\left(u_M(x)^{\frac{\sigma+q-1}{q}}\right)^{q^*_{s_2}}\,{\rm d}x\right)^{\frac{q}{q^*_{s_2}}}. \end{align*}

Plugging the estimates of $I_1$ and $I_2$ into (4.1) we obtain

\begin{align*} & \frac{C(d,s_1,p)\sigma p^p}{(\sigma+p-1)^p} \left(\int_{\mathbb{R}^d}\left(u_M(x)^{\frac{\sigma+p-1}{p}} \right)^{p^*_{s_1}}\,{\rm d}x\right)^{\frac{p}{p^*_{s_1}}} \\ & \qquad +\frac{C(d,s_2,q)\sigma q^q}{(\sigma+q-1)^q} \left(\int_{\mathbb{R}^d}\left(u_M(x)^{\frac{\sigma+q-1}{q}} \right)^{q^*_{s_2}}\,{\rm d}x\right)^{\frac{q}{q^*_{s_2}}} \\ & \quad\leq\alpha\displaystyle{\int_{\Omega}} u(x)^{p+\sigma-1}\,{\rm d}x+\beta\displaystyle{\int_{\Omega}} u(x)^{q+\sigma-1}\,{\rm d}x. \end{align*}

Letting $M\to \infty$ in above, the monotone convergence theorem yields

(4.2)\begin{align} & \frac{C(d,s_1,p)\sigma p^p}{(\sigma+p-1)^p} \left(\int_{\mathbb{R}^d}\left(u(x)^{\frac{\sigma+p-1}{p}}\right)^{p^*_{s_1}}\,{\rm d}x\right)^{\frac{p}{p^*_{s_1}}}\notag\\ & \qquad +\frac{C(d,s_2,q)\sigma q^q}{(\sigma+q-1)^q} \left(\int_{\mathbb{R}^d}\left(u(x)^{\frac{\sigma+q-1}{q}}\right)^{q^*_{s_2}}\,{\rm d}x\right)^{\frac{q}{q^*_{s_2}}} \nonumber\\ & \quad \leq\alpha\displaystyle{\int_{\Omega}} u(x)^{p+\sigma-1}\,{\rm d}x+\beta\displaystyle{\int_{\Omega}} u(x)^{q+\sigma-1}\,{\rm d}x. \end{align}

Proof. $\boldsymbol {\alpha,\, \beta \ge 0}$: Since $u$ is a non-negative supersolution of (EV; α, β), we obtain

\[ \left< A_p(u), v \right>{+} \left< B_q(u), v \right> \ge \alpha \displaystyle{\int_{\Omega}} u^{p-1}v + \beta \displaystyle{\int_{\Omega}} u^{q-1}v \ge 0, \]

for every $v \in {W_0^{s_1,p}(\Omega )}$ with $v \ge 0$. Now we can use [Reference Ambrosio3, (2) of Theorem 1.1] (by taking $c(x)=0$) with modifications (due to the presence of multiple parameters $s_1,\,s_2$) to conclude either $u>0$ a.e. in $\Omega$ or $u \equiv 0$ a.e. in $\mathbb {R}^d$.

$\boldsymbol {\alpha,\, \beta \le 0}$ or $\boldsymbol {\alpha \beta \leq 0}$: Let $x_0 \in \Omega$ and $R>0$ be such that $B_R(x_0) \subset \Omega$. Since $u$ is a non-negative supersolution of (EV; α, β), then we proceed as in [Reference Ambrosio3, Lemma 2.1], for any $R_1 > 0$ satisfying $B_{R_1} = B_{R_1}(x_0) \subset B_{\frac {R}{2}}(x_0)$, and obtain the following logarithmic estimate

(4.12)\begin{align} & \iint\limits_{B_{R_1} \times B_{R_1}}\left| \log\left(\frac{u(x)+\delta}{u(y)+\delta}\right)\right|^q\, {\rm d}\mu_2\notag\\ & \quad \le C \bigg(\delta^{1-q} R_1^d \bigg[ R^{{-}s_1 p}\,\mathrm{Tail}_{p}(u_-;x_0,R)^{p-1} \ + R^{{-}s_2 q}\,\mathrm{Tail}_{q}(u_-;x_0,R)^{q-1} \bigg] R_1^{d-s_1 p}\notag\\ & \qquad + R_1^{d-s_2 q} \left({\lVert}{u}{\rVert}_{L^\infty(\mathbb{R}^d)}+\delta\right)^{p-q} \ + \left(|\alpha|+|\beta|\,{\lVert}{u}{\rVert}_{L^\infty(\mathbb{R}^d)}^{p-q}\right)|B_{ 2 R_1 }(x_0)| \bigg), \end{align}

where $\delta \in (0,\,1)$ and $C= C(d,\,s_1,\,p,\,s_2,\,q)>0$. Now the result follows using (4.12) and the arguments given in [Reference Ambrosio3, Lemma 2.3].

Proof. Let $u_n \rightharpoonup u$ in ${W_0^{s_1,p}(\Omega )}$. Then using the compactness of the embeddings ${W_0^{s_1,p}(\Omega )} \hookrightarrow L^p(\Omega )$, ${W_0^{s_2,q}(\Omega )} \hookrightarrow L^{q}(\Omega )$, and the weak lower semicontinuity of the seminorm, we get

\begin{align*} \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} {I_+(u_n)} & = \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} \frac{[u_n]_{s_1,p}^p}{p} + \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} \frac{[u_n]_{s_2,q}^q}{q} - \alpha \lim_{n \rightarrow \infty} \frac{{\lVert}{u_n^+}{\rVert}^p_p }{p} \\ & \quad - \beta \lim_{n \rightarrow \infty} \frac{{\lVert}{u_n^+}{\rVert}^q_q}{q} \ge I_+(u). \end{align*}

Now we prove the coercivity of $I_+$. Suppose $\alpha \le 0$. Then using ${W_0^{s_1,p}(\Omega )} \hookrightarrow L^q(\Omega )$,

(5.1)\begin{equation} I_+(u) \ge \frac{[u]_{s_1,p}^p}{p} - \beta \frac{{\lVert}{u^+}{\rVert}^q_q}{q} \ge \frac{[u]_{s_1,p}^p}{p} - C \beta \frac{[u]_{s_1,p}^q}{q}, \quad \forall\ u \in {W_0^{s_1,p}(\Omega)} \setminus \{ 0\}. \end{equation}

If $\alpha > 0$, then there exists $a \in (0,\,1)$ such that $\alpha = a \lambda _{s_1,p}^1$. In this case, using ${W_0^{s_1,p}(\Omega )} \hookrightarrow L^q(\Omega )$, we get

(5.2)\begin{equation} I_+(u) \ge \frac{[u]_{s_1,p}^p}{p} - a \lambda_{s_1,p}^1 \frac{{\lVert}{u^+}{\rVert}_p^p}{p} - \beta \frac{{\lVert}{u^+}{\rVert}^q_q}{q} \ge \frac{[u]_{s_1,p}^p}{p} - a \lambda_{s_1,p}^1 \frac{{\lVert}{u^+}{\rVert}_p^p}{p} - C \beta \frac{[u]_{s_1,p}^q}{q}, \end{equation}

for every $u \in {W_0^{s_1,p}(\Omega )} \setminus \{0\}$. From the definition of $\lambda _{s_1,p}^1$, we have $[u]_{s_1,p}^p \ge \lambda _{s_1,p}^1 {\lVert }{u}{\rVert }_p^p \ge \lambda _{s_1,p}^1 {\lVert }{u^+}{\rVert }_p^p$. Therefore, (5.1) yields

(5.3)\begin{equation} I_+(u) \ge \frac{1-a}{p} [u]_{s_1,p}^p - C \beta \frac{[u]_{s_1,p}^q}{q}, \quad \forall\ u \in {W_0^{s_1,p}(\Omega)} \setminus \{ 0 \}. \end{equation}

In view of (5.1), observe that (5.3) holds for every $\alpha < \lambda _{s_1,p}^1$. For any $\epsilon >0$, applying Young’s inequality with the conjugate pair $(\frac {p}{q},\, \frac {p}{p-q})$ we obtain

\[ [u]_{s_1,p}^q \le \epsilon \frac{q}{p} [u]_{s_1,p}^p + \frac{p-q}{p} \epsilon^{-\frac{p}{p-q}}. \]

Hence from (5.1) we have the following estimate for every $u \in {W_0^{s_1,p}(\Omega )} \setminus \{ 0\}$:

\[ I_+(u) \ge \frac{1-a}{p} [u]_{s_1,p}^p - \epsilon \frac{C \beta}{p} [u]_{s_1 , p}^p - \frac{C \beta(p-q)}{qp} \epsilon^{-\frac{p}{p-q}}. \]

We choose $\epsilon >0$ so that $C \beta \epsilon < \frac {1-a}{2}$. Therefore, from the above estimate, we get

\[ I_+(u) \ge \frac{1-a}{2p} [u]_{s_1,p}^p - \frac{C \beta(p-q)}{qp} \epsilon^{-\frac{p}{p-q}} \quad \forall \, u \in {W_0^{s_1,p}(\Omega)} \setminus \{ 0\}. \]

Thus the functional $I_+$ is coercive on ${W_0^{s_1,p}(\Omega )}$. Next, we prove that $I_+$ is bounded below. Set $M>0$ such that $M^{p-q} \ge p(1+C \beta q^{-1})$. Then using (5.1), we get

\[ I_+(u) \ge [u]_{s_1,p}^q \left( \frac{[u]_{s_1,p}^{p-q}}{p} - \frac{C \beta}{q} \right) \ge M^q, \ \text{ provided } \ [u]_{s_1,p} \ge M. \]

Further, if $[u]_{s_1,p} \le M$, then $I_+(u) \ge - \frac { M^q C \beta }{q}$. Thus, $I_+$ is bounded below on ${W_0^{s_1,p}(\Omega )}$.

Proof. First, we show that the sequence $(u_n)$ is bounded in ${W_0^{s_1,p}(\Omega )}$. On a contrary, assume that $[u_n]_{s_1,p} \rightarrow \infty$, as $n \rightarrow \infty$. Using (i) of lemma 2.2, note that

\[ [u_n^{-}]^p_{s_1,p} \le [u_n^{-}]^p_{s_1,p} + [u_n^{-}]^q_{s_2,q} \le \left| \left< I_+'(u_n), -u_n^{-} \right>\right| \le {\lVert}{I_+'(u_n)}{\rVert} [u_n^{-}]_{s_1,p}. \]

Hence $[u_n^-]_{s_1, p} \rightarrow 0$, as $n \rightarrow \infty$. Set $w_n=u_n [u_n]_{s_1,p}^{-1}$. Up to a subsequence, $w_n \rightharpoonup w$ in ${W_0^{s_1,p}(\Omega )}$ and by the compactness of ${W_0^{s_1,p}(\Omega )} \hookrightarrow L^p(\Omega )$, $w_n \rightarrow w$ in $L^p(\Omega )$. Further, $[w_n^-]_{s_1,p} = [u_n^-]_{s_1,p} [u_n]_{s_1,p}^{-1} \rightarrow 0$, as $n \rightarrow \infty$. Therefore, $w_n^- \rightarrow 0$ in ${W_0^{s_1,p}(\Omega )}$ and hence in $L^p(\Omega )$. This implies that $w_n^+ \rightarrow w$ in $L^p(\Omega )$, which yields $w \ge 0$ a.e. in $\Omega$. We show that $w$ is an eigenfunction of the fractional $p$-Laplacian corresponding to $\alpha$. For any $\phi \in {W_0^{s_1,p}(\Omega )}$, we write

(5.4)\begin{equation} \left< A_p(u_n), \phi\right>{+} \left< B_q(u_n), \phi \right>{-} \alpha \displaystyle{\int_{\Omega}} {\mid}{u_n}{\mid}^{p-2}u_n \phi - \beta \displaystyle{\int_{\Omega}} {\mid}{u_n}{\mid}^{q-2}u_n \phi =\epsilon_n, \end{equation}

where $\epsilon _n \rightarrow 0$ as $n \rightarrow \infty$. From the above inequality, we obtain

(5.5)\begin{align} & \left< A_p(w_n),\phi\right>{+} [u_n]_{s_1,p}^{q-p} \left< B_q(w_n), \phi \right>{-} \alpha \displaystyle{\int_{\Omega}} {\mid}{w_n}{\mid}^{p-2} w_n \phi - \beta [u_n]_{s_1,p}^{q-p} \displaystyle{\int_{\Omega}} {\mid}{w_n}{\mid}^{q-2} w_n \phi\notag\\ & \quad = \frac{\epsilon_n}{[u_n]_{s_1,p}^{p-1}}. \end{align}

Using the Hölder’s inequality with the conjugate pair $(q,\, q')$, the Poincaré inequality ${\lVert }{\phi }{\rVert }_q \le C(\Omega ) [\phi ]_{s_1,p}$, and the boundedness of $(w_n)$ in ${W_0^{s_2,q}(\Omega )}$ we have

\begin{align*} {\mid}{\left< B_q(w_n), \phi \right>}{\mid} & \le [w_n]_{s_2,q}^{q-1} [\phi]_{s_2,q} \le C [\phi]_{s_1,p}, \quad \text{and}\\ \displaystyle{\int_{\Omega}} {\mid}{w_n}{\mid}^{q-1} {\mid}{\phi}{\mid} & \le {\lVert}{w_n}{\rVert}_{q}^{q-1}{\lVert}{\phi}{\rVert}_{q} \le C[\phi]_{s_1,p}, \end{align*}

We choose $\phi =w_n-w$ in (5.5), and take the limit as $n \rightarrow \infty$ to get $\left < A_p(w_n),\, w_n-w\right > \rightarrow 0.$ Further, since $A_p$ is a continuous functional on ${W_0^{s_1,p}(\Omega )}$, we also have $\left < A_p(w),\, w_n-w\right > \rightarrow 0$. Further, using the definition of $A_p$

(5.6)\begin{equation} \left< A_p(w_n) - A_p(w), w_n-w\right> \ge \left( [w_n]^{p-1}_{s_1,p} - [w]^{p-1}_{s_1,p} \right) \left( [w_n]_{s_1,p} - [w]_{s_1,p} \right). \end{equation}

Therefore, $[w_n]_{s_1,p} \rightarrow [w]_{s_1,p}$, and hence the uniform convexity of ${W_0^{s_1,p}(\Omega )}$ ensures that $w_n \rightarrow w$ in ${W_0^{s_1,p}(\Omega )}$. Further, since $[w]_{s_1,p}=1$ we also have $w \neq 0$ in $\Omega$. Now using (5.5), we obtain

\[ \left< A_p(w), \phi \right>{=} \alpha \displaystyle{\int_{\Omega}} |w|^{p-2}w \phi, \quad \forall\ \phi \in {W_0^{s_1,p}(\Omega)}. \]

Thus $w$ is a nonnegative weak solution to the problem

(5.7)\begin{equation} (-\Delta)^{s_1}_pu=\alpha|u|^{p-2}u \ \text{in }\Omega, \ u=0 \ \text{in } \Omega^c. \end{equation}

Now by the strong maximum principle for fractional $p$-Laplacian [Reference Brasco and Parini14, Proposition 2.6], we conclude that $w>0$ a.e. in $\Omega$. Therefore, the uniqueness of $\lambda ^1_{s_1,p}$ (proposition 2.1) yields $\alpha = \lambda ^1_{s_1,p}$, resulting in a contradiction. Thus, the sequence $(u_n)$ is bounded in ${W_0^{s_1,p}(\Omega )}$. By the reflexivity, up to a subsequence, $u_n \rightharpoonup \tilde {u}$ in ${W_0^{s_1,p}(\Omega )}$. By taking $\phi =u_n-\tilde {u}$ in (5.4) and using the compact embeddings of ${W_0^{s_1,p}(\Omega )} \hookrightarrow L^{\gamma }(\Omega )$ with $\gamma \in [1,\,p]$, we get $\left < A_p(u_n),\, u_n-\tilde {u} \right > + \left < B_q(u_n),\, u_n - \tilde {u} \right > \rightarrow 0$. Therefore, $\left < A_p(u_n)- A_p(\tilde {u}),\, u_n-\tilde {u} \right > + \left < B_q(u_n)- B_q(\tilde {u}),\, u_n - \tilde {u} \right > \rightarrow 0$, which implies $\left < A_p(u_n)- A_p(\tilde {u}),\, u_n-\tilde {u} \right >\to 0$. Thus, $[u_n]_{s_1,p} \rightarrow [\tilde {u}]_{s_1,p}$ (by using (5.6)), and from the uniform convexity, $u_n \rightarrow \tilde {u}$ in ${W_0^{s_1,p}(\Omega )}$, as required.

Proof. (i) Let $\rho >0$ and $u \in \mathbb {S}_{\rho }$. Then using ${W_0^{s_1,p}(\Omega )} \hookrightarrow L^\gamma (\Omega )$ for $\gamma \in [1,\,p]$,

(5.8)\begin{align} I_+(u) & \ge \frac{[u]_{s_1,p}^p}{p} - \alpha \frac{{\lVert}{u^+}{\rVert}^p_p}{p} - \beta \frac{{\lVert}{u^+}{\rVert}^q_q}{q}\notag\\ & \ge [u]_{s_1,p}^q \left( \frac{[u]_{s_1,p}^{p-q}}{p} - C \alpha \frac{[u]_{s_1,p}^{p-q}}{p} - C\frac{\alpha}{q}\right) = \rho^{q} A(\rho), \end{align}

where $A(\rho )=\frac {\rho ^{p-q}}{p} - C \alpha \frac {\rho ^{p-q}}{p} - C\frac {\alpha }{q}$. Choose $0 < \alpha _1 < \frac {\rho ^{p-q}}{p} ( C\frac {\rho ^{p-q}}{p} + \frac {C}{q} )^{-1}$ and $\delta = \rho ^{q} A(\rho )$ with $\alpha \in (0,\, \alpha _1)$. Therefore, from (5.8), $I_+(u) \ge \delta$ for all $\alpha \in (0,\, \alpha _1)$. (ii) Note that

\[ I_+(t\phi_{s_1,p}) = \frac{t^p}{p} \left([\phi_{s_1,p}]_{s_1,p}^p - \alpha {\lVert}{\phi_{s_1,p}}{\rVert}_p^p \right) + \frac{t^q}{q} \left([\phi_{s_1,p}]_{s_2,q}^q - \beta {\lVert}{\phi_{s_1,p}}{\rVert}_q^q \right). \]

Since $p > q$ and $\alpha > \lambda ^1_{s_1,p}$, we obtain $I_+(t\phi _{s_1,p}) \rightarrow -\infty$, as $t \rightarrow \infty$. Hence there exists $t_1 > \rho [\phi _{s_1,p}]_{s_1,p}^{-1}$ such that $I_+(t\phi _{s_1,p})<0$ for all $t \ge t_1$. Thus, $v=t\phi _{s_1,p}$ with $t>t_1$ is the required function.

Proof. (i) Let $\alpha < \lambda _{s_1,p}^1$ and $\beta < \lambda _{s_2,q}^1$. Suppose $u \in {W_0^{s_1,p}(\Omega )} \setminus \{0\}$ is a solution of (EV; α, β). Then using the definition of $\lambda _{s_1,p}^1$ and $\lambda _{s_2,q}^1$ (proposition 2.1), we get

(6.1)\begin{align} 0 < (\lambda_{s_1,p}^1 - \alpha){\lVert}{u}{\rVert}^p_p \le [u]_{s_1,p} - \alpha {\lVert}{u}{\rVert}^p_p = \beta {\lVert}{u}{\rVert}^q_q - [u]_{s_2,q} \le (\beta - \lambda_{s_2,q}^1){\lVert}{u}{\rVert}^q_q < 0. \end{align}

A contradiction. Therefore, (EV; α, β) does not admit a solution. For other cases, contradiction similarly follows using (6.1). (ii) For $\alpha = \lambda _{s_1,p}^1$ and $\beta = \lambda _{s_2,q}^1$, if $u \in {W_0^{s_1,p}(\Omega )} \setminus \{0\}$ is a solution of (EV; α, β), then the equality occurs in (6.1). As a consequence, $u$ becomes an eigenfunction corresponding to both $\lambda _{s_1,p}^1$ and $\lambda _{s_2,q}^1$, i.e., (LI) violates. Conversely, suppose (LI) does not hold. For $\alpha \le \lambda _{s_1,p}^1$ and $\beta \le \lambda _{s_2,q}^1$, using remark 5.9 we have $I_+(u) \ge 0$ for any $u \in {W_0^{s_1,p}(\Omega )} \setminus \{0\}$. Thus $0$ is the global minimizing point for $I_+$. Further, since $\phi _{s_1,p} = c\phi _{s_2,q}$ for some nonzero $c \in {\mathbb {R}}$, by setting $\tilde {u}=c_1\phi _{s_1,p} = c_2\phi _{s_2,q}$ (where $c_1,\, c_2 \neq 0$) we see that $I_+(\tilde {u})=0$. Therefore, $\tilde {u} \neq 0$ is a solution of (EV; α, β).

Proof of theorem 1.2. (i) $\boldsymbol {\alpha > \lambda _{s_1,p}^1,\,\beta < \lambda _{s_2,q}^1\,}$: Let $\beta > 0$. Then using $\alpha > \lambda _{s_1,p}^1$ and $\beta < \lambda _{s_2,q}^1$, we get

\[ \lambda_{s_1,p}^1 \left(\Omega, \frac{\alpha}{\beta} \right) = \frac{\lambda_{s_1,p}^1 }{\alpha}\beta < \beta < \lambda_{s_2,q}^1 = \lambda_{s_2,q}^1(\Omega,1). \]

Hence $\beta > \min \{ \lambda _{s_1,p}^1 (\Omega,\, \frac {\alpha }{\beta }),\, \lambda _{s_2,q}^1(\Omega,\,1) \}$ and using theorem 6.2 with $m_p = \frac {\alpha }{\beta }$ and $m_q=1$ we obtain that (EV; α, β) admits a positive solution. Let $\beta \le 0$. Then using proposition 5.3 and lemma 5.4, $I_+$ satisfies all the conditions of the Mountain pass theorem (see [Reference Ambrosetti and Rabinowitz1, Theorem 2.1]). Therefore, by the Mountain pass theorem and remark 5.1, (EV; α, β) admits a nonnegative and nontrivial solution $u \in {W_0^{s_1,p}(\Omega )}$. Further, from the strong maximum principle (proposition 4.3), $u >0$ a.e. in $\Omega$.

$\boldsymbol {\alpha < \lambda _{s_1,p}^1,\,\beta > \lambda _{s_2,q}^1\,}$: Let $\alpha >0$. Then using $\alpha < \lambda _{s_1,p}^1$ and $\beta > \lambda _{s_2,q}^1$, we get $\lambda _{s_2,q}^1(\Omega,\, \frac {\beta }{\alpha })< \alpha < \lambda _{s_1,p}^1(\Omega,\,1)$. Therefore, theorem 6.2 with $m_p=1$ and $m_q=\frac {\beta }{\alpha }$ yields a positive solution for (EV; α, β). If $\alpha \le 0$, then from proposition 5.2, we get the existence of a global minimizer $\tilde {u}$ of $I_+$, and hence using remark 5.1, $\tilde {u}$ is a nonnegative solution of (EV; α, β). Next, we show that $\tilde {u} \neq 0$ in $\Omega$. Observe that, for $t>0$, $G_{\beta }(t\phi _{s_2,q}) = t^q G_{\beta }(\phi _{s_2,q})<0$, and using remark 5.9, $H_{\alpha }(t\phi _{s_2,q}) = t^p H_{\alpha }(\phi _{s_2,q})>0$. Now, if $0< t << 1$, then $I_+(t\phi _{s_2,q})<0$, which implies that $I_+(\tilde {u})<0$ and $\tilde {u} \neq 0$. Therefore, by the strong maximum principle (proposition 4.3), $\tilde {u}>0$ a.e. in $\Omega$.

$\boldsymbol {\alpha = \lambda _{s_1,p}^1,\,\beta = \lambda _{s_2,q}^1\,}$: Let (LI) violates. Then using (ii) of proposition 6.1, we see that (EV; α, β) admits a nonnegative solution $u \in {W_0^{s_1,p}(\Omega )}$. Further, using the strong maximum principle (proposition 4.3), $u>0$ a.e. in $\Omega$.

(ii) Suppose, there exists nonzero $c \in {\mathbb {R}}$ such that $\phi _{s_1,p}=c\phi _{s_2,q}$. We also assume that (EV; α, β) admits a solution $u >0$ a.e. in $\Omega$. Using the Picone’s inequality ((i) of lemma 2.3) and proposition 2.1, we get

\begin{align*} & \iint \limits_{\mathbb{R}^d \times \mathbb{R}^d} |u_k(x)-u_k(y)|^{p-2}(u_k(x)-u_k(y)) \left(\frac{\phi_{s_1,p}(x)^{p}}{u_k(x)^{p-1}} - \frac{\phi_{s_1,p}(y)^p}{u_k(y)^{p-1}} \right) \, {\rm d}\mu_1 \nonumber \\ & \le \iint \limits_{\mathbb{R}^d \times \mathbb{R}^d} {\mid}{\phi_{s_1,p}(x)-\phi_{s_1,p}(y)}{\mid}^p \, {\rm d}\mu_1 = \lambda_{s_1,p}^1 \displaystyle{\int_{\Omega}} \phi_{s_1,p}(x)^p \, {\rm d}x. \end{align*}

Since for $x,\,y \in \mathbb {R}^d$, $u_k(x)-u_k(y)=u(x)-u(y)$ the above inequality yields

(6.2)\begin{equation} \left< A_p(u), \frac{ \phi_{s_1,p}^p}{u_k^{p-1}} \right> \le \lambda_{s_1,p}^1 \displaystyle{\int_{\Omega}} \phi_{s_1,p}(x)^p \, {\rm d}x. \end{equation}

We again use the Picone’s inequality ((ii) of lemma 2.3) to obtain

(6.3)\begin{align} & \iint \limits_{\mathbb{R}^d \times \mathbb{R}^d} |u_k(x)-u_k(y)|^{q-2}(u_k(x)-u_k(y))\left(\frac{ \phi_{s_1,p}(x)^p}{u_k(x)^{p-1}}-\frac{ \phi_{s_1,p}(y)^p}{u_k(y)^{p-1}}\right) \, {\rm d}\mu_2 \nonumber\\ & \quad \le \iint \limits_{\mathbb{R}^d \times \mathbb{R}^d} |\phi_{s_1,p}(x)-\phi_{s_1,p}(y)|^{q-2}(\phi_{s_1,p}(x)-\phi_{s_1,p}(y)) \notag\\ & \qquad \times \left(\frac{\phi_{s_1,p}(x)^{p-q+1}}{u_k(x)^{p-q}} - \frac{\phi_{s_1,p}(y)^{p-q+1}}{u_k(y)^{p-q}} \right) \, {\rm d}\mu_2. \end{align}

Since $\phi _{s_1,p} \in L^{\infty }(\Omega )$ ((v) of proposition 2.1), using lemma 2.4, $u_k^{q-p}\phi _{s_1,p}^{p-q+1} \in {W_0^{s_1,p}(\Omega )}$. Therefore, we have the following identity:

(6.4)\begin{align} & \iint \limits_{\mathbb{R}^d \times \mathbb{R}^d} |\phi_{s_1,p}(x)-\phi_{s_1,p}(y)|^{q-2}(\phi_{s_1,p}(x)-\phi_{s_1,p}(y)) \nonumber\\ & \quad \times \left(\frac{\phi_{s_1,p}(x)^{p-q+1}}{u_k(x)^{p-q}} - \frac{\phi_{s_1,p}(y)^{p-q+1}}{u_k(y)^{p-q}} \right) \, {\rm d}\mu_2 = \lambda_{s_2,q}^1 \displaystyle{\int_{\Omega}} \frac{\phi_{s_1,p}(x)^p}{u_k(x)^{p-q}} \, {\rm d}x. \end{align}

Set $f_k := u_k^{q-p}\phi _{s_1,p}^p$ and $f:=u^{q-p}\phi _{s_1,p}^p$. It is easy to see that $f_k$ is increasing and $f_k \in L^1(\Omega )$. Moreover, for $\gamma \le p$, using $u_k(x)^{\gamma -p} \rightarrow u(x)^{\gamma -p}$ a.e. in $\Omega$, we get $f_k(x) \rightarrow f(x)$ a.e. in $\Omega$. Therefore, the monotone convergence theorem yields $f \in L^1(\Omega )$, and $\int _{\Omega } f_k(x) \, {\rm d}x \rightarrow \int _{\Omega } f(x) \, {\rm d}x$, as $k \rightarrow \infty$. Hence from (6.3) and (6.4), we obtain

(6.5)\begin{equation} \lim_{k \rightarrow \infty} \left< B_q(u), \frac{ \phi_{s_1,p}^p}{u_k^{p-1}} \right> \le \lambda_{s_2,q}^1 \displaystyle{\int_{\Omega}} \frac{\phi_{s_1,p}(x)^p}{u(x)^{p-q}} \, {\rm d}x. \end{equation}

Now since $u$ is a solution of (EV; α, β), taking $u_k^{1-p}\phi _{s_1,p}^{p} \in {W_0^{s_1,p}(\Omega )}$ (by (v) of proposition 2.1, and lemma 2.4) as a test function,

(6.6)\begin{align} & \left< A_p(u), \frac{ \phi_{s_1,p}^p}{u_k^{p-1}} \right>{+} \left< B_q(u), \frac{ \phi_{s_1,p}^p}{u_k^{p-1}} \right>\notag\\ & \quad = \alpha \displaystyle{\int_{\Omega}} \frac{u(x)^{p-1}}{u_k(x)^{p-1}} \phi_{s_1,p}(x)^p \, {\rm d}x + \beta \displaystyle{\int_{\Omega}} \frac{u(x)^{q-1}}{u_k(x)^{p-1}} \phi_{s_1,p}(x)^p \, {\rm d}x. \end{align}

Furthermore, for $\gamma \in (1,\,p]$, the Hölder’s inequality with the conjugate pair $(\gamma,\,\gamma ')$ yields,

\begin{align*} \displaystyle{\int_{\Omega}} \frac{u(x)^{\gamma-1}}{u_k(x)^{p-1}} \phi_{s_1,p}(x)^p \, {\rm d}x \le {\lVert}{u}{\rVert}_{\gamma}^{\gamma-1} {\lVert}{\frac{\phi_{s_1,p}^p}{u_k^{p-1}}}{\rVert}_{\gamma} \le C(\Omega, \gamma) {\lVert}{u}{\rVert}_{p}^{\gamma-1} {\lVert}{\frac{\phi_{s_1,p}^p}{u_k^{p-1}}}{\rVert}_p. \end{align*}

Moreover, $\frac {u^{\gamma -1}}{u_k^{p-1}} \phi _{s_1,p}^p \rightarrow u^{\gamma -p} \phi _{s_1,p}^p$ a.e. in $\Omega$, and the sequence $(u_k^{1-p})$ is increasing. Hence, again applying the monotone convergence theorem

\[ \displaystyle{\int_{\Omega}} \frac{u^{p-1}}{u_k^{p-1}} \phi_{s_1,p}^p \rightarrow \displaystyle{\int_{\Omega}} \phi_{s_1,p}^p \quad \text{and} \quad \displaystyle{\int_{\Omega}} \frac{u^{q-1}}{u_k^{p-1}} \phi_{s_1,p}^p \rightarrow \displaystyle{\int_{\Omega}} \frac{\phi_{s_1,p}^p}{u^{p-q}}, \text{ as } k \rightarrow \infty. \]

Therefore, (6.2), (6.5) and (6.6) yield

\begin{align*} \alpha \displaystyle{\int_{\Omega}} \phi_{s_1,p}^p + \beta \displaystyle{\int_{\Omega}} \frac{\phi_{s_1,p}^p}{u^{p-q}} & = \lim_{k \rightarrow \infty} \left\{ \left< A_p(u), \frac{ \phi_{s_1,p}^p}{u_k^{p-1}} \right>{+} \left< B_q(u), \frac{ \phi_{s_1,p}^p}{u_k^{p-1}} \right> \right\}\\ & \le \lambda_{s_1,p}^1 \displaystyle{\int_{\Omega}} \phi_{s_1,p}^p + \lambda_{s_2,q}^1 \displaystyle{\int_{\Omega}} \frac{\phi_{s_1,p}^p}{u^{p-q}}. \end{align*}

The above inequality infer that, $(\alpha,\, \beta ) \in ((\lambda _{s_1,p}^1,\, \infty ) \times (-\infty,\, \lambda _{s_2,q}^1)) \cup ((-\infty,\, \lambda _{s_1,p}^1) \times (\lambda _{s_2,q}^1,\,\infty )) \cup ( \{\lambda _{s_1,p}^1\} \times \{\lambda _{s_2,q}^1\} )$. This completes our proof.

Proof. Notice that, $G_{\beta }(u) = [u]^q_{s_2,q} - \lambda _{s_2,q}^1 {\lVert }{u^+}{\rVert }_q^q \ge [u]^q_{s_2,q} - \lambda _{s_2,q}^1 {\lVert }{u}{\rVert }_q^q \ge 0$ for $u \in {W_0^{s_1,p}(\Omega )} \setminus \{0\}$. Let $u \in {\mathcal {N}}_{\alpha, \beta }$. If $G_{\beta }(u) = 0$, then we get

\[ \frac{[u]_{s_2,q}^q}{{\lVert}{u}{\rVert}_q^q} \le \lambda_{s_2,q}^1 \le \frac{[u]_{s_2,q}^q}{{\lVert}{u}{\rVert}_q^q}. \]

By the simplicity of $\lambda _{s_2,q}^1$ ((iv) of proposition 2.1), $u=c\phi _{s_2,q}$ for some $c \in {\mathbb {R}}$. Hence

\[ H_{\alpha}(u) > [u]_{s_1,p}^p - \alpha^*_{s_1,p} {\lVert}{u^+}{\rVert}_p^p = C \left([\phi_{s_2,q}]_{s_1,p}^p - \alpha^*_{s_1,p} {\lVert}{\phi_{s_2,q}}{\rVert}_p^p \right)=0. \]

On the other hand, since $u \in {\mathcal {N}}_{\alpha, \beta }$, $H_{\alpha }(u) = - G_{\beta }(u) = 0$, a contradiction. Therefore, we must have $G_{\beta }(u) >0$. Further, since $u \in {\mathcal {N}}_{\alpha, \beta }$, we obtain $H_{\alpha }(u)<0< G_{\beta }(u)$.

Proof. (i) We show that $d := \min \{I_+(u): u \in {\mathcal {N}}_{\alpha, \beta }\}$ is attained. Let $(u_n)$ be the minimizing sequence in ${\mathcal {N}}_{\alpha,\beta }$, i.e., $\big < I_+'(u_n),\,u_n\big >=0$ for all $n \in \mathbb {N}$ and $I_+(u_n) \rightarrow d$ as $n \rightarrow \infty$. From lemma 6.3, $H_{\alpha }(u_n) < 0 < G_{\beta }(u_n)$. Step 1: This step proves the boundedness of $(u_n)$ in ${W_0^{s_1,p}(\Omega )}$. On a contrary, suppose $[u_n]_{s_1,p} \rightarrow \infty$, as $n \rightarrow \infty$. Set $w_n = u_n [u_n]_{s_1,p}^{-1}$. By the reflexivity, $w_n \rightharpoonup w$ in ${W_0^{s_1,p}(\Omega )}$ and $w_n \rightarrow w$ in $L^p(\Omega )$. Since $H_{\alpha }(u_n) < 0$, we have ${\lVert }{w_n}{\rVert }_p^p = {\lVert }{u_n}{\rVert }_p^p [u_n]^{-p}_{s_1,p} > \frac {1}{\alpha }$. This gives ${\lVert }{w}{\rVert }_p^p \ge \frac {1}{\alpha }$, and hence $w \neq 0$. Now using (i) of remark 5.8,

(6.8)\begin{equation} \frac{p-q}{pq} G_{\beta}(w_n) = \frac{I_+(u_n)}{[u_n]^q_{s_1,p}} \rightarrow 0, \text{ as } n \rightarrow \infty. \end{equation}

Using (6.8) we obtain

\begin{align*} 0 \le G_{\beta}(w) \le \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} G_{\beta}(w_n)=0. \end{align*}

Therefore, $w = c\phi _{s_2,q}$ for some $c \in {\mathbb {R}}$. Further, using (i) of remark 5.8, and (6.8),

\[ [\phi_{s_2,q}]_{s_1,p}^p - \alpha {\lVert}{\phi_{s_2,q}}{\rVert}_p^p = H_{\alpha}(w) \le \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} H_{\alpha}(w_n) ={-} \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} \displaystyle\frac{G_{\beta}(w_n)}{[u_n]_{s_1,p}^{p-q}} =0. \]

The above inequality yields $\alpha ^*_{s_1,p} \le \alpha$, a contradiction. Therefore, $(u_n)$ must be bounded in ${W_0^{s_1,p}(\Omega )}$. Step 2: By the reflexivity, $u_n \rightharpoonup \tilde {u}$ in ${W_0^{s_1,p}(\Omega )}$. In this step, we show $(u_n)$ converges to $\tilde {u}$ in ${W_0^{s_1,p}(\Omega )}$. On a contrary, suppose $[u_n]_{s_1,p} \not \rightarrow [\tilde {u}]_{s_1,p}$. If $\lim _{n \rightarrow \infty } [u_n]_{s_1,p} < [\tilde {u}]_{s_1,p}$, then $\mathop {\operatorname {\underline {lim}}}_{n \rightarrow \infty } [u_n]_{s_1,p} < [\tilde {u}]_{s_1,p}$ contradicts the weak lower semicontinuity of $[\cdot ]_{s_1,p}$. Henceforth, assume that $[\tilde {u}]_{s_1,p} < \lim _{n \rightarrow \infty } [u_n]_{s_1,p}$. Using this inequality we get $[\tilde {u}]_{s_1,p} < \mathop {\operatorname {\underline {lim}}}_{n \rightarrow \infty } [u_n]_{s_1,p}$ and $H_{\alpha }(\tilde {u}) < \mathop {\operatorname {\underline {lim}}}_{n \rightarrow \infty } H_{\alpha }(u_n) \le 0$. This implies that $\tilde {u}$ is nonzero. Now, $G_{\beta }(\tilde {u}) \ge 0$, and if $G_{\beta }(\tilde {u})=0$, then $\tilde {u}=c\phi _{s_2,q}$ for some $c \in {\mathbb {R}}$. Hence $H_{\alpha }(\phi _{s_2,q}) <0$ which implies that $\alpha > \alpha ^*_{s_1,p}$, a contradiction. Therefore, $H_{\alpha }(\tilde {u}) < 0 < G_{\beta }(\tilde {u})$. Now applying proposition 5.7 there exists a unique $t_{\alpha,\beta } \in {\mathbb {R}}^+$ such that $t_{\alpha,\beta }\tilde {u} \in {\mathcal {N}}_{\alpha, \beta }$ and $0< I_+(t_{\alpha,\beta } \tilde {u})= \max _{t \in {\mathbb {R}}^+} \, I_+(t\tilde {u})$. Moreover, from (ii) of remark 5.8, $I_+(u_n)= \max _{t \in {\mathbb {R}}^+} \, I_+(tu_n)$. Therefore,

\[ {\rm d}\le I_+(t_{\alpha, \beta} \tilde{u}) < \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} I_+(t_{\alpha,\beta} u_n) \le \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} I_+(u_n) = d, \]

a contradiction. Thus, $[u_n]_{s_1,p} \rightarrow [\tilde {u}]_{s_1,p}$ in ${\mathbb {R}}^+$. Hence from the uniform convexity of ${W_0^{s_1,p}(\Omega )}$, $u_n \rightarrow \tilde {u}$ in ${W_0^{s_1,p}(\Omega )}$. Step 3: In this step we prove that $\tilde {u}$ is a positive solution of (EV; α, β). Since $u_n \rightarrow \tilde {u}$ in ${W_0^{s_1,p}(\Omega )}$, we obtain $d = I_+(\tilde {u})$ and $\big < I_+'(\tilde {u}),\, \tilde {u} \big >=0$. Using the continuity of $H_{\alpha }$ and $G_{\beta }$, $H_{\alpha }(\tilde {u}) \le 0 \le G_{\beta }(\tilde {u})$. Next, we show $\tilde {u}$ is nonzero. Set $w_n = u_n [u_n]_{s_1,p}^{-1}$. Then $w_n \rightharpoonup w$ in ${W_0^{s_1,p}(\Omega )}$. Since $H_{\alpha }(u_n)<0$, from the same arguments as in previous steps, $w \neq 0$ and $G_{\beta }(w)>0$. Next, suppose $[u_n]_{s_1,p} \rightarrow 0$ as $n \rightarrow \infty$. Using $G_{\beta }(w_n) \ge 0$ we get

\[ [w]^p_{s_1,p}-\alpha{\lVert}{w}{\rVert}^p_p \le H_{\alpha}(w) \le \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} H_{\alpha}(w_n) ={-} \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} \frac{G_{\beta}(w_n)}{[u_n]_{s_1,p}^{p-q}} ={-}\infty. \]

A contradiction, as $w \in L^p(\Omega )$. Thus $\inf _{n \in \mathbb {N}} [u_n]_{s_1,p} >0$ and $\alpha {\lVert }{\tilde {u}}{\rVert }_p^p \ge \lim _{n \rightarrow \infty } [u_n]_{s_1,p}^p >0$, which implies that $\tilde {u}$ is nonzero in $\Omega$, and hence $\tilde {u} \in {\mathcal {N}}_{\alpha, \beta }$. Moreover, from lemma 6.3, $H_{\alpha }(\tilde {u}) < 0 < G_{\beta }(\tilde {u})$. Now, using proposition 5.6 and remark 5.1, we conclude $\tilde {u}$ is a nonnegative solution of (EV; α, β). Furthermore, by proposition 4.3, $\tilde {u}>0$ a.e. in $\Omega$.

(ii) Our proof uses the method of contradiction. Let $u \in {W_0^{s_1,p}(\Omega )}$ and $u>0$ a.e. in $\Omega$. From (v) of proposition 2.1 and Lemma 2.4, $u_k^{q-p}\phi _{s_2,q}^{p-q+1},\, u_k^{1-p}\phi _{s_2,q}^p \in {W_0^{s_1,p}(\Omega )}$. Applying the discrete Picone’s inequality ((ii) of lemma 2.3),

(6.9)\begin{align} & \iint \limits_{\mathbb{R}^d \times \mathbb{R}^d} |u_k(x)-u_k(y)|^{q-2}(u_k(x)-u_k(y))\left(\frac{ \phi_{s_2,q}(x)^p}{u_k(x)^{p-1}}-\frac{ \phi_{s_2,q}(y)^p}{u_k(y)^{p-1}}\right) \, {\rm d}\mu_2 \nonumber\\ & \quad \le \iint \limits_{\mathbb{R}^d \times \mathbb{R}^d} |\phi_{s_2,q}(x)-\phi_{s_2,q}(y)|^{q-2}(\phi_{s_2,q}(x)-\phi_{s_2,q}(y)) \notag\\ & \qquad \times \left(\frac{\phi_{s_2,q}(x)^{p-q+1}}{u_k(x)^{p-q}} - \frac{\phi_{s_2,q}(y)^{p-q+1}}{u_k(y)^{p-q}} \right) \, {\rm d}\mu_2 = \lambda_{s_2,q}^1 \displaystyle{\int_{\Omega}} \frac{\phi_{s_2,q}(x)^p}{u_k(x)^{p-q}} \, {\rm d}x. \end{align}

The monotone convergence theorem yields $u^{q-p}\phi _{s_2,q}^p \in L^1(\Omega )$ and $\int _{\Omega } u_k^{q-p}\phi _{s_2,q}^p \rightarrow \int _{\Omega } u^{q-p}\phi _{s_2,q}^p$, as $k \rightarrow \infty$. Next, we again use the Picone’s inequality ((i) of lemma 2.3), to get

(6.10)\begin{align} & \iint \limits_{\mathbb{R}^d \times \mathbb{R}^d} |u_k(x)-u_k(y)|^{p-2}(u_k(x)-u_k(y)) \left(\frac{\phi_{s_2,q}(x)^{p}}{u_k(x)^{p-1}} - \frac{\phi_{s_2,q}(y)^p}{u_k(y)^{p-1}} \right) \, {\rm d}\mu_1 \nonumber\\ & \quad \le \iint \limits_{\mathbb{R}^d \times \mathbb{R}^d} {\mid}{\phi_{s_2,q}(x)-\phi_{s_2,q}(y)}{\mid}^p \, {\rm d}\mu_1 = \alpha^*_{s_1,p} \displaystyle{\int_{\Omega}} \phi_{s_2,q}(x)^p \, {\rm d}x. \end{align}

If $u$ is a solution of (EV; α, β), then taking $u_k^{1-p}\phi _{s_2,q}^p$ as a test function we write

(6.11)\begin{align} & \left< A_p(u), \frac{\phi_{s_2,q}^{p}}{u_k^{p-1}} \right>{+} \left< B_q(u), \frac{\phi_{s_2,q}^{p}}{u_k^{p-1}} \right> \nonumber\\ & \quad = \alpha \displaystyle{\int_{\Omega}} \frac{u(x)^{p-1}}{u_k(x)^{p-1}} \phi_{s_2,q}(x)^p \, {\rm d}x + \lambda_{s_2,q}^1 \displaystyle{\int_{\Omega}} \frac{u(x)^{q-1}}{u_k(x)^{p-1}} \phi_{s_2,q}(x)^p \, {\rm d}x. \end{align}

Further, applying the monotone convergence theorem

\[ \displaystyle{\int_{\Omega}} \frac{u^{p-1}}{u_k^{p-1}} \phi_{s_2,q}^p \rightarrow \displaystyle{\int_{\Omega}} \phi_{s_2,q}^p; \quad \displaystyle{\int_{\Omega}} \frac{u^{q-1}}{u_k^{p-1}} \phi_{s_2,q}^p \rightarrow \displaystyle{\int_{\Omega}} \frac{\phi_{s_2,q}^p}{u^{p-q}}, \text{ as } k \rightarrow \infty. \]

Therefore, from (6.9), (6.10), and (6.11), we conclude

\begin{align*} \alpha \displaystyle{\int_{\Omega}} \phi_{s_2,q}^p + \lambda_{s_2,q}^1 \displaystyle{\int_{\Omega}} \frac{\phi_{s_2,q}^p}{u^{p-q}} & = \lim_{k \rightarrow \infty} \left\{ \left< A_p(u), \frac{ \phi_{s_2,q}^p}{u_k^{p-1}} \right>{+} \left< B_q(u), \frac{ \phi_{s_2,q}^p}{u_k^{p-1}} \right> \right\}\\ & \le \alpha^*_{s_1,p} \displaystyle{\int_{\Omega}} \phi_{s_2,q}^p + \lambda_{s_2,q}^1 \displaystyle{\int_{\Omega}} \frac{\phi_{s_2,q}^p}{u^{p-q}}. \end{align*}

The above inequality yields $\alpha \le \alpha ^*_{s_1,p}$, which is a contradiction. Thus there does not exist any positive solution for $\alpha > \alpha ^*_{s_1,p}$.

Proof. Due to the homogeneity,

\begin{align*} \beta^{{\star}}(\alpha) & = \inf \left\{[u]_{s_2,q}^q : u \in \mathcal{M} \right\}, \text{ where }\\ \mathcal{M} & := \left\{ u \in {W_0^{s_1,p}(\Omega)}, {\lVert}{u}{\rVert}_q^q =1, \text{ and } H_{\alpha}(u) \le 0 \right\}. \end{align*}

Let $(u_n)$ be a minimizing sequence for $\beta ^{\star }(\alpha )$ in $\mathcal {M}$. Suppose $[u_n]_{s_1,p} \rightarrow \infty$. Then $H_{\alpha }(u_n) \le 0$ implies $\alpha ^{\frac {1}{p}} {\lVert }{u_n}{\rVert }_p \ge [u_n]_{s_1,p} \rightarrow \infty$. Set $w_n=u_n {\lVert }{u_n}{\rVert }_p^{-1}$. Then $[w_n]_{s_1,p} = [u_n]_{s_1,p} {\lVert }{u_n}{\rVert }_p^{-1} \le \alpha ^{\frac {1}{p}}$, and $w_n \rightharpoonup w$ in ${W_0^{s_1,p}(\Omega )}$. Using ${\lVert }{u_n}{\rVert }_q=1$, we get ${\lVert }{w_n}{\rVert }_q \rightarrow 0$ in ${\mathbb {R}}^+$. On the other hand, ${\lVert }{w_n}{\rVert }_p=1$. Now the compact embeddings of ${W_0^{s_1,p}(\Omega )} \hookrightarrow L^{\gamma }(\Omega ); \gamma \in [1,\,p]$ yield:

\[ (a) \, {\lVert}{w}{\rVert}_q = 0 \text{ which implies } w=0 \text{ a.e. in } \Omega; \quad (b) \, {\lVert}{w}{\rVert}_p = 1. \]

Clearly, $(a)$ and $(b)$ contradict each other. Therefore, the sequence $(u_n)$ is bounded in ${W_0^{s_1,p}(\Omega )}$. By the reflexivity, $u_n \rightharpoonup \tilde {u}$ in ${W_0^{s_1,p}(\Omega )}$. Further, $\tilde {u} \in \mathcal {M}$ follows from the compact embedding of ${W_0^{s_1,p}(\Omega )}$ and weak lower semicontinuity of $H_{\alpha }$. Therefore,

\[ \beta^{{\star}}(\alpha) \le [\tilde{u}]_{s_2, q}^q \le \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} [u_n]_{s_2, q}^q = \beta^{{\star}}(\alpha). \]

Thus $\beta ^{\star }(\alpha )$ is attained. Clearly, $\beta ^{\star }(\alpha ) \ge \lambda _{s_2,q}^1$. If $\beta ^{\star }(\alpha ) = \lambda _{s_2,q}^1$, then by the simplicity of $\lambda _{s_2,q}^1$ ((iv) of proposition 2.1), $\tilde {u} =c\phi _{s_2,q}$ for some $c\in {\mathbb {R}}$. Further, since $\alpha < \alpha ^*_{s_1,p}$, we get $H_{\alpha }(\tilde {u})= CH_{\alpha }(\phi _{s_2,q})>0$, a contradiction to $\tilde {u} \in \mathcal {M}$. Thus $\beta ^{\star }(\alpha ) > \lambda _{s_2,q}^1$.

Proof. We adapt the arguments as given in [Reference Bobkov and Tanaka9, Theorem 2.5]. As before, we will show that $d := \min \{I_+(u): u \in {\mathcal {N}}_{\alpha, \beta }\}$ is attained. Since $\beta > \lambda _{s_2,q}^1$ and $\alpha <\alpha ^*_{s_1,p}$, we have $G_{\beta }(\phi _{s_2,q})<0< H_{\alpha }(\phi _{s_2,q})$. Then by proposition 5.7, there exists a unique $t_{\alpha,\beta } \in {\mathbb {R}}^+$ such that $0>I_+(t_{\alpha,\beta }\phi _{s_2,q})=\min _{t \in {\mathbb {R}}^+} I_+(t\phi _{s_2,q})$. We also have $t_{\alpha,\beta }\phi _{s_2,q} \in {\mathcal {N}}_{\alpha,\beta }$. Therefore, $d<0$. Let $(u_n)$ be the minimizing sequence in ${\mathcal {N}}_{\alpha,\beta }$ for $d$. Then there exists $n_0 \in \mathbb {N}$ such that $I_+(u_n)<0$ for $n \ge n_0$. Since $u_n \in {\mathcal {N}}_{\alpha,\beta }$, using (i) of remark 5.8, $G_{\beta }(u_n)< 0 < H_{\alpha }(u_n)$ for $n \ge n_0$. Step 1: In this step, we show that $(u_n)$ is a bounded sequence in ${W_0^{s_1,p}(\Omega )}$. As before, to prove this we argue by contradiction. Suppose $[u_n]_{s_1,p} \rightarrow \infty$, as $n \rightarrow \infty$, and set $w_n = u_n [u_n]_{s_1,p}^{-1}$. Then $w_n \rightharpoonup w$ in ${W_0^{s_1,p}(\Omega )}$. Hence

\[ H_{\alpha}(w) \le \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} H_{\alpha}(w_n) ={-} \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} \frac{G_{\beta}(w_n)}{[u_n]_{s_1,p}^{p-q}} = 0, \text{ and } 1- {\lVert}{w}{\rVert}_p^p= \mathop{\operatorname{\underline{lim}}}_{n \rightarrow \infty} H_{\alpha}(w_n) = 0, \]

which implies that $w \neq 0$. Therefore, from the definition, $\beta ^{\star }(\alpha ) \le [w]_{s_2,q}^q {\lVert }{w}{\rVert }_q^{-q}$. Using this inequality along with $\beta < \beta ^{\star }(\alpha )$, we get $G_{\beta }(w) >0$. On the other hand, $G_{\beta }(w) \le \mathop {\operatorname {\underline {lim}}}_{n \rightarrow \infty } G_{\beta }(w_n) \le 0$, a contradiction. Step 2: Let $u_n \rightharpoonup \tilde {u}$ in ${W_0^{s_1,p}(\Omega )}$. This step shows that $\tilde {u}$ is a positive solution of (EV; α, β). First, we claim ${H_{\alpha }(\tilde {u})>0}$. On a contrary, assume $H_{\alpha }(\tilde {u}) \le 0$. Since $I_+(\tilde {u}) \le \mathop {\operatorname {\underline {lim}}}_{n \rightarrow \infty } I_+(u_n) \le d <0$, we get $\tilde {u}\neq 0.$ Hence $\beta ^{\star }(\alpha ) \le [\tilde {u}]_{s_2,q}^q {\lVert }{\tilde {u}}{\rVert }_q^{-q}$ and $\beta < \beta ^{\star }(\alpha )$ imply $G_{\beta }(\tilde {u}) > 0$. On the other hand, $G_{\beta }(\tilde {u}) \le \mathop {\operatorname {\underline {lim}}}_{n \rightarrow \infty } G_{\beta }(u_n) \le 0$, a contradiction. Therefore, $H_{\alpha }(\tilde {u})>0$. Further, $H_{\alpha }(\tilde {u})+ G_{\beta }(\tilde {u}) = I_+(\tilde {u}) \le \mathop {\operatorname {\underline {lim}}}_{n \rightarrow \infty } I_+(u_n) \le 0$ yields $G_{\beta }(\tilde {u})<0$. Now we can use proposition 5.7, to get a unique $t_{\alpha,\beta } \in {\mathbb {R}}^+$ that minimizes $I_+(t \tilde {u})$ over ${\mathbb {R}}^+$, and $t_{\alpha,\beta } \tilde {u} \in {\mathcal {N}}_{\alpha,\beta }$. Hence