Proof. By Hölder's inequality,

\[ J_{p,\beta}(u) \geq \frac{1}{p}\|\nabla u\|_{L^p(\Omega)}^p - \biggl(\int_{\Omega}|u|^{p^*}\biggl)^{\frac{q}{p^*}}|\Omega|^{\frac{p^*-q}{p^*}}, \mbox{ for all } u\in W_0^{1,p}(\Omega). \]

Since, by [Reference Gilbarg and Trudinger25, proof of theorem 7.10], for each $u\in W_0^{1,p}(\Omega ),$

(3.8)\begin{equation} \|u\|_{L^{p^*}(\Omega)}\leq \frac{\theta}{\sqrt{N}}\|\nabla u\|_{L^{p}(\Omega)}, \mbox{ where } \theta=\frac{p(N-1)}{N-p}, \end{equation}

we have,

\begin{align*} J_{p,\beta}(u) & \geq\frac{1}{p}\|\nabla u\|_{L^{p}(\Omega)}^p - |\Omega|^{\frac{p^*-q}{p^*}}\biggl(\frac{p(N-1)}{\sqrt{N}(N-p)}\biggl)^{q} \|\nabla u\|_{L^{p}(\Omega)}^q\\ & \geq \frac{1}{\overline{p}}\|\nabla u\|_{L^{p}(\Omega)}^p - C\|\nabla u\|_{L^{p}(\Omega)}^q, \mbox{ for all } u\in W_0^{1,p}(\Omega), \end{align*}

where $C=\bigl (\overline {p}(N-1)/\sqrt {N}(N-\overline {p})\bigl )^{q}\max \{1,\,|\Omega |\}.$

Note that

\begin{align*} \frac{r^p}{\overline{p}} - Cr^{q}\geq \frac{r^q}{\overline{p}} \ \text{ if and only if } 0< r\leq\biggl(\frac{1}{\overline{p}C+1}\biggl)^{\frac{1}{q-p}}.\end{align*}

Then, by choosing $r=\biggl (\frac {1}{\overline {p}C+1}\biggl )^{\frac {1}{q-\overline {p}}}$ and $\alpha =r^{q}/\overline {p},$ we conclude that $(i)$ holds.

Now, let $\varphi \in C^{\infty }_0(\Omega )$ be such that $|\{\varphi >\beta \}|>0,$ where $\{\varphi >\beta \}$ denotes the set $\{x\in \Omega : \varphi (x) > \beta \}.$ For each $t\geq 1,$ we get

\begin{align*} J_{p,\beta}(t\varphi) & = \frac{t^p}{p}\| \varphi \|_{W_0^{1,p}(\Omega)}^p- \int_{\Omega}F_\beta(t\varphi)\,{\rm d}x \\ & \leq \frac{t^p}{p}\| \varphi \|_{W_0^{1,p}(\Omega)}^p-\frac{t^q}{q}\int_{\{\varphi>\beta\}}\varphi^q \,{\rm d}x +\frac{\beta^{q}}{q}|\Omega|, \end{align*}

which implies in the existence of $e$ satisfying $(ii).$

Proof. Let $(u_n)\subset W_0^{1,p}(\Omega )$ be a $(PS)_c$ sequence for $J_{p,\beta },$ that is, $J_{p,\beta }(u_n) \to c$ and $\Lambda _{J_{p,\beta }}(u_{n})\rightarrow 0,$ where $\Lambda _{J_{p,\beta }}$ is defined in (2.4). Hence, it follows from (2.4) and (3.7) that there exists $(\mu _{n})\subset \partial J_{p,\beta }(u_{n})$ such that

\[ \|\mu_{n}\|_{*}=\Lambda_{J_{p,\beta}}(u_{n})=o_{n}(1) \mbox{ and } \mu_{n}=Q'_p(u_{n})-\rho_{n}, \]

where $\rho _{n}\in \partial \mathcal {F}_\beta (u_{n}).$ Then,

(3.9)\begin{align} \begin{aligned} c + 1 + \| u_n\|_{W_0^{1,p}(\Omega)} & \geq J_{p,\beta}(u_n) - \frac{1}{q}\left\langle \mu_n, u_n \right\rangle + o_n(1)\\ & = J_{p,\beta}(u_n) - \frac{1}{q}\left\langle Q_p'(u_n)-\rho_n, u_n \right\rangle + o_n(1)\\ & = \left(\frac{1}{p}-\frac{1}{q}\right)\| u_n\|^p_{W_0^{1,p}(\Omega)} +\int_{\Omega}\left(\frac{1}{q}\rho_n u_n - F_\beta(u_n)\right) \,{\rm d}x + o_n(1). \end{aligned} \end{align}

Moreover, note that by (3.3) and (3.4), we have

(3.10)\begin{equation} \int_{\Omega}\left(\frac{1}{q}\rho_n u_n - F_{\beta}(u_n)\right) \,{\rm d}x = \frac{\beta}{q}\int_{\{u_n=\beta\}}\rho_n \,{\rm d}x + \frac{\beta^q}{q}|\{u_n > \beta\}| \geq 0. \end{equation}

Hence,

(3.11)\begin{equation} c + 1 + \| u_n\|_{W_0^{1,p}(\Omega)} \geq \left(\frac{1}{p}-\frac{1}{q}\right)\| u_n\|^p_{W_0^{1,p}(\Omega)} + o_n(1),\end{equation}

which implies that the sequence $(u_n)$ is bounded in $W_0^{1,p}(\Omega )$. Thus, by Sobolev embedding theorems, passing to a subsequence if necessary, we obtain

(3.12)\begin{equation} \begin{cases} u_n \rightharpoonup u \text{ in } W_0^{1,p}(\Omega), \ u_n \rightarrow u \ \mbox{ in } \ L^{s}(\Omega), \\ u_n(x) \rightarrow u(x) \ \ \mbox{a.e in} \ \Omega, \\ |u_n(x)| \leq h(x) \ \mbox{ for some } \ h \in L^{s}(\Omega), s\in [1,p^*:=\frac{Np}{N-p}). \end{cases} \end{equation}

Using a similar argument than [Reference Arcoya and Calahorrano13, pg. 1071], we conclude that $J_{p,\beta }$ satisfies the nonsmooth Palais–Smale condition.

Proof. By Young's inequality, we have

\[ \int_{\Omega}|\nabla u|^{p_1}\,{\rm d}x\leq\frac{p_1}{p_2}\int_{\Omega}|\nabla u|^{p_2}\,{\rm d}x+\frac{p_2-p_1}{p_2}|\Omega|, \mbox{ for all } 1< p_1\leq p_2, u\in W_0^{1,p}(\Omega). \]

Hence, $I_{p,\beta }$ is nondecreasing with respect to $p$ and arguing as in [Reference Segura de León and Molino Salas32], we conclude that $(I_{p,\beta }(u_{p,\beta }))_{1< p<\overline {p}}$ is increasing. Hence,

(3.14)\begin{equation} c_{p_1}\leq c_{p_2}\end{equation}

for all $1 < p_1\leq p_2.$ Note also that, by (3.13),

\begin{align*} c_{p,\beta}& =I_{p,\beta}(u_{p,\beta})-\frac{1}{q} \left\langle Q_p'(u_{p,\beta})-\rho_{p,\beta}, u_{p,\beta} \right\rangle \\ & =\left(\frac{1}{p}-\frac{1}{q}\right)\int_{\Omega}| \nabla u_{p,\beta}|^p\,{\rm d}x +\int_{\Omega}\left(\frac{1}{q}\rho_{p,\beta} u_{p,\beta} - F_\beta(u_{p,\beta})\right)\,{\rm d}x . \end{align*}

From (3.10) and (3.14), it follows that

(3.15)\begin{equation} \int_{\Omega}|\nabla u_{p,\beta}|^p \,{\rm d}x\leq C, \mbox{ for all } 1< p<\overline{p},\end{equation}

where $C:=\frac {\overline {p}q}{q-\overline {p}}c_{\overline {p},\,\beta }>0$ is a constant independent of $p\in (1,\,\overline {p}).$

Applying once more Young's inequality, we obtain

\begin{align*} \|u_{p,\beta}\| & \leq \frac{1}{p}\int_\Omega|\nabla u_{p,\beta}|^{p}\,{\rm d}x + \frac{p-1}{p}|\Omega|\\ & \leq \overline{C}+ |\Omega|, \end{align*}

for some constant $\overline {C}>0,$ independent of $p.$

Proof. Here to simplify the notation we put $u=u_{p,\beta }$ and $\rho _{p,\beta }=\rho.$ To obtain the $L^\infty$-estimate we will use the Moser's iteration [Reference Moser31] and a careful analysis of some constants. For each $L>0$, we define

\begin{align*} u_L(x)& := \left\{ \begin{array}{@{}rcl} u(x), & \mbox{if} & u(x)\leq L\\ L, & \mbox{if} & u(x)> L, \end{array} \right.\\ z_{L,n}(x)& :=(u_L^{p(\gamma-1)}u)(x)\ \mbox{ and }\ w_L(x):=(uu_L^{\gamma-1})(x), \end{align*}

with $\gamma >1$ to be determined later. Choosing $\varphi =z_{L,n}$ in (3.2), we get

\[ \int_{\Omega}u_L^{p(\gamma-1)}|\nabla u|^{p}\,{\rm d}x\displaystyle={-}p(\gamma-1)\int_{\Omega}u_L^{p\gamma-p-1}u|\nabla u|^{p-2}\nabla u\nabla u_L \,{\rm d}x+\int_{\Omega}\rho u u_L^{p(\gamma-1)}\,{\rm d}x. \]

Since

\[ p(\gamma-1)\int_{\Omega}u_L^{p\gamma-p-1}u|\nabla u|^{p-2}\nabla u\nabla u_L \,{\rm d}x=p(\gamma-1)\int_{\{u\leq L\}}u_L^{p(\gamma-1)}|\nabla u|^p\,{\rm d}x\geq0 \]

and $0\leq \rho (x)\leq |u(x)|^{q-1}$ for almost every $x\in \Omega,$ see (3.3), we obtain

(3.17)\begin{equation} \int_{\Omega}u_L^{p(\gamma-1)}|\nabla u|^{p}\,{\rm d}x\leq \int_{\Omega}u^qu_L^{p(\gamma-1)}\,{\rm d}x. \end{equation}

On the other hand, by (3.8) it follows that

\[ |w_L|_{L^{p^*}(\Omega)}^{p}\leq c_{p,\beta}\int_{\Omega}|\nabla w_L|^{p}\,{\rm d}x =c_{p,\beta}\int_{\Omega} |\nabla(uu_L^{\gamma-1})|^{p}\,{\rm d}x, \]

where $c_{p,\beta }=\bigl (p(N-1)/(N-p)\bigl )^{p}.$ Thus,

\[ |w_L|_{L^{p^*}(\Omega)}^{p}\leq 2^pc_{p,\beta}\int_{\Omega}u_L^{p(\gamma-1)}|\nabla u|^p\,{\rm d}x+2^pc_{p,\beta}(\gamma-1)^{p}\int_{\Omega}u_L^{p(\gamma-2)}u^p|\nabla u_L|^p \,{\rm d}x, \]

hence, we get

(3.18)\begin{equation} |w_L|_{L^{p^*}(\Omega)}^p\leq 2^pc_{p,\beta}\gamma^p\int_{\Omega}u_L^{p(\gamma-1)}|\nabla u|^p\,{\rm d}x. \end{equation}

Combining (3.17) and (3.18), we obtain

\[ |w_L|_{L^{p^*}(\Omega)}^p\leq 2^pc_{p,\beta}\gamma^p\int_{\Omega}u^{q-p}(uu_L^{\gamma-1})^p\,{\rm d}x, \]

and so,

\[ |w_L|_{L^{p^*}(\Omega)}^{p}\leq 2^pc_{p,\beta} \gamma^p\int_{\Omega}u^{q-p}w_L^p\,{\rm d}x. \]

Now we use the Hölder's inequality (with exponents $p^*/(q-p)$ and $p^*/(p^*-(q-p))$ to get that

\[ |w_L|_{L^{p^*}(\Omega)}^p\leq 2^pc_{p,\beta}\gamma^p\Biggl(\int_{\Omega}u^{p^*}\,{\rm d}x\Biggl)^{\frac{q-p}{p^*}}\Biggl(\int_{\Omega}w_L^{\frac{pp^*}{p^*-(q-p)}}\,{\rm d}x\Biggl)^{\frac{p^*-(q-p)}{p^*}}, \]

where $p<\frac {pp^*}{p^*-(q-p)}< p^*.$

The previous inequality, (3.8) and (3.15) imply that

(3.19)\begin{equation} |w_L|_{L^{p^*}(\Omega)}^{p}\leq (2\theta_p)^p\gamma^p\Biggl(\int_{\Omega}w_L^{\alpha^*}\,{\rm d}x\Biggl)^{\frac{p}{\alpha^*}}, \end{equation}

where

(3.20)\begin{equation} \alpha^*:=\frac{pp^*}{p^*-(q-p)}\ \mbox{ and } \ \theta_p:=\biggl(p\frac{(N-1)}{(N-p)}\biggl)^{\frac{q}{p}}C^{\frac{q-p}{p}}, \end{equation}

with the constant $C$ given in (3.15).

Using that $0\leq w_L=(uu_L^{\gamma -1})\leq u^{\gamma }$ on the right-hand side of (3.19) and then letting $L\rightarrow \infty$ on the left-hand side, as a consequence of Fatou's Lemma on the variable $L$, we have

\[ \Biggl(\int_{\Omega}u^{p^*\gamma}\,{\rm d}x\Biggl)^{\frac{p}{p^*}}\leq (2\theta_p)^p\gamma^p\Biggl(\int_{\Omega}u^{\gamma\alpha^*}\,{\rm d}x\Biggl)^{\frac{p}{\alpha^*}}, \]

from which we get that

(3.21)\begin{equation} |u|_{L^{p^*\gamma}(\Omega)}\leq(2\theta_p)^{\frac{1}{\gamma}}\gamma^{\frac{1}{\gamma}}|u|_{L^{\gamma\alpha^*}(\Omega)}. \end{equation}

Let us define $\sigma :=p^*/\alpha ^*.$ When $\gamma =\sigma$ in (3.21), since $\gamma \alpha ^*=p^*$ we have $u\in L^{p^*\sigma }(\Omega )$ and

(3.22)\begin{equation} |u|_{L^{p^*\sigma}(\Omega)}\leq (2\theta_p)^{\frac{1}{\sigma}}\sigma^{\frac{1}{\sigma}} |u|_{L^{p^*}(\Omega)}. \end{equation}

Now, choosing $\gamma =\sigma ^2$ in (3.21), since $\gamma \alpha ^*=p^*\sigma$ and $p^*\gamma =p^*\sigma ^2,$ we obtain

(3.23)\begin{equation} |u|_{L^{p^*\sigma^2}(\Omega)}\leq (2\theta_p)^{\frac{1}{\sigma^2}}\sigma^{\frac{2}{\sigma^2}} |u|_{L^{p^*\sigma}(\Omega)}, \end{equation}

by using (3.22) and (3.23), we have

\[ |u|_{L^{p^*\sigma^2}(\Omega)}\leq(2\theta_p)^{\frac{1}{\sigma^2}+\frac{1}{\sigma}} \sigma^{\frac{2}{\sigma^2}+{\frac{1}{\sigma}}} |u|_{L^{p^*}(\Omega)}. \]

For $n\geq 1,$ we define $\sigma _n$ inductively so that $\sigma _n=\sigma ^n.$ Then, from (3.21), it follows that

(3.24)\begin{equation} |u|_{L^{p^*\sigma^n}(\Omega)}\leq (2\theta_p)^{\frac{1}{\sigma^n}+\cdots+\frac{1}{\sigma^2}+\frac{1}{\sigma}} \sigma^{\frac{n}{\sigma^n}+\cdots+\frac{2}{\sigma^2}+{\frac{1}{\sigma}}} |u|_{L^{p^*}(\Omega)}. \end{equation}

Note that

\[ \displaystyle\sum_{i=1}^{\infty}\frac{1}{\sigma^i}=\frac{1}{\sigma-1} \ \mbox{ and } \displaystyle\sum_{i=1}^{\infty}\frac{i}{\sigma^i}=\frac{\sigma}{(\sigma-1)^2}. \]

Thus, since $\sigma >1,$ passing to the limit as $n\rightarrow \infty$ in (3.24) we conclude that $u\in L^{\infty }(\Omega )$ and

(3.25)\begin{equation} |u|_{L^\infty(\Omega)}\leq (2\theta_p)^{\frac{1}{\sigma-1}}\sigma^{\frac{\sigma}{(\sigma-1)^2}}|u|_{L^{p^*}(\Omega)}. \end{equation}

Finally, since $\sigma =\frac {N}{N-p}-\frac {q}{p}+1$ and $1< p< q<1^*< p^*,$ using once more (3.8), the expression of $\theta _p$ (see (3.20)) and (3.25) we conclude the proof of the lemma.

Proof. By lemma 3.4, it follows that $(\rho _{p,\beta })_{1 < p < \overline {p}}$ is bounded in $L^{\frac {q}{q-1}}(\Omega )$. Hence, there exists $\rho _{\beta } \in L^{\frac {q}{q-1}}(\Omega )$, such that (3.28) holds. Moreover, if $E \subset \Omega$ is a measurable set, then

(3.30)\begin{equation} \int_E \rho_{p,\beta} \,{\rm d}x = \int_\Omega \rho_{p,\beta} . \chi_E \,{\rm d}x \to \int_\Omega \rho_{\beta} . \chi_E \,{\rm d}x= \int_E \rho_{\beta} \,{\rm d}x, \mbox{ as } p \to 1^+. \end{equation}

Now, let us show that $\rho _\beta$ satisfies (3.29). First of all, note that

(3.31)\begin{equation} 0 \leq \rho_\beta(x), \quad \mbox{a.e. in } \Omega. \end{equation}

Indeed, otherwise, a measurable set $E \subset \Omega$ would exist, such that $\rho _\beta (x) < 0$, in $E$. Then,

\[ \int_E \rho_\beta \,{\rm d}x < 0. \]

Hence, from (3.30), we have a contradiction with the fact that $\rho _{p,\beta }(x) \geq 0$, for all $p > 1$.

Now let us show that

(3.32)\begin{equation} \rho_\beta(x) = 0, \quad \mbox{if}\ u_\beta(x) < \beta. \end{equation}

Let $E = [u_\beta < \beta ]$ and note that

(3.33)\begin{align} 0 & \leq \int_E \rho_{p,\beta} \,{\rm d}x \\ & =\int_{E \cap [u_{p,\beta} \geq \beta]}\rho_{p,\beta} \,{\rm d}x\nonumber\\ & \leq \int_{E \cap [u_{p,\beta} \geq \beta]}u_{p,\beta}^{q-1} \,{\rm d}x \nonumber \end{align}

Claim 1: $\displaystyle \int _{E \cap [u_{p,\beta } \geq \beta ]}u_{p,\beta }^{q-1} \,{\rm d}x = o_p(1)$.

Assuming for a while that claim 1 holds true, then from (3.30), (3.31) and (3.33), it follows that (3.32) holds.

Now, let us show that the claim holds. First of all, let us show that

(3.34)\begin{equation} \chi_{[u_{p,\beta} \geq \beta]} \to 0, \quad \mbox{a.e. in}\ E,\ \mbox{as}\ p \to 1^+. \end{equation}

Indeed, let us suppose by contradiction that there exists $E^* \subset E$ with positive measure such that, for every fixed $x \in E^*$, there exists $(p_{n,x})_{n \in \mathbb {N}}$, such that $p_{n,x} \to 1^+$, as $n \to +\infty$ and

\[ \chi_{[u_{p_{n,x},\beta} \geq \beta]}(x) = 1, \quad \mbox{for all}\ n \in \mathbb{N}. \]

This, in turn, is equivalent to

(3.35)\begin{equation} u_{p_{n,x}}(x) \geq \beta, \quad \mbox{for all}\ n \in \mathbb{N}.\end{equation}

By doing $n \to +\infty$ in (3.35), since $p_{n,x} \to 1^+$, we have that

\[ u_\beta(x) \geq \beta, \quad \mbox{for all}\ x \in E^*. \]

But this contradicts the fact that $E^* \subset E$. Hence (3.34) holds.

Therefore, from (3.34) and the Lebesgue Convergence Theorem, it follows that claim 1 holds.

Now let us show that

(3.36)\begin{equation} \rho_\beta(x) = u_\beta^{q-1}(x), \quad \mbox{if}\ u_\beta(x) > \beta. \end{equation}

For this, let us define $E = [u_\beta > \beta ]$. Note that

(3.37)\begin{equation} \int_E \rho_{p,\beta} \,{\rm d}x = \int_{E \cap [u_{p,\beta} = \beta]}\rho_{p,\beta} \,{\rm d}x + \int_{E \cap [u_{p,\beta} > \beta]}u_{p,\beta}^{q-1} \,{\rm d}x. \end{equation}

As in (3.34), we can prove that

(3.38)\begin{equation} \chi_{[u_{p,\beta} = \beta]} \to 0, \quad \mbox{a.e. in}\ E,\ \mbox{as}\ p \to 1^+. \end{equation}

Then, from (3.38) and Lebesgue Convergence Theorem,

(3.39)\begin{equation} \int_{E \cap [u_{p,\beta} = \beta]}\rho_{p,\beta} \,{\rm d}x \leq \beta^{q-1} \int_E \chi_{[u_{p,\beta} = \beta]} \to 0, \quad \mbox{as}\ p \to 1^+. \end{equation}

On the other hand, since

\[ \int_{E \cap [u_{p,\beta} > \beta]}u_{p,\beta}^{q-1} \,{\rm d}x = \int_{E} u_{p,\beta}^{q-1}\chi_{[u_{p,\beta} > \beta]} \,{\rm d}x, \]

we have from (3.26) and the fact that $\chi _{[u_{p,\beta } > \beta ]} \to 1$, a.e. in $E$, as $p \to 1^+$, that

(3.40)\begin{equation} \int_{E \cap [u_{p,\beta} > \beta]}u_{p,\beta}^{q-1} \,{\rm d}x \to \int_{E} u_\beta^{q-1} \,{\rm d}x. \end{equation}

Then, from (3.37), (3.39) and (3.40), it follows that

(3.41)\begin{equation} \int_E \rho_{p,\beta} \,{\rm d}x \to \int_{E} u_\beta^{q-1} \,{\rm d}x, \quad \mbox{as} p \to 1^+. \end{equation}

Hence, from (3.30) and (3.41), we have that

(3.42)\begin{equation} \int_E \rho_\beta \,{\rm d}x = \int_E u_\beta^{q-1} \,{\rm d}x \end{equation}

Claim 2: $\rho _\beta (x) \leq u_\beta ^{q-1}(x)$ in $[u_\beta \geq \beta ]$.

Assuming that claim 2 holds, it follows from (3.42) that

\[ \rho_\beta(x) = u_\beta^{q-1}(x), \quad \mbox{in}\ E \]

and

\[ \rho_\beta(x) \in [0,\beta^{q-1}], \quad \mbox{in}\ [u_\beta = \beta] \]

and we are done.

In order to prove claim 2, let us assume by contradiction that there exists $E_* \subset [u_\beta \geq \beta ]$, with positive measure and such that

\[ \rho_\beta > u_\beta ^{q-1}, \quad \mbox{in}\ E_*. \]

Then,

(3.43)\begin{equation} \int_{E_*} \rho_\beta \,{\rm d}x > \int_{E_*} u_\beta^{q-1} \,{\rm d}x . \end{equation}

Then, from (3.26) and (3.30), there exists $p_n \to 1^+$, as $n \to +\infty$, such that

\[ \int_{E_*} \rho_{p_n,\beta} \,{\rm d}x > \int_{E_*} u_{p_n,\beta}^{q-1} \,{\rm d}x, \]

which contradicts the fact that $\rho _{p,\beta }(x) \leq u_{p,\beta }^{q-1}(x)$, a.e. in $\Omega$.

Then claim 2 holds and this finishes the proof.

Proof. The inequality (3.15) implies that (see [Reference Andreu, Ballester, Caselles and Mazón8, proposition 3] or [Reference Mercaldo, Rossi, Segura de León and Trombetti29, theorem 3.3]) there exists $\textbf {z}_{\beta } \in L^\infty (\Omega,\,\mathbb {R}^N)$, such that $\|\textbf {z}_{\beta }\|_\infty \leq 1$ and

(3.45)\begin{equation} |\nabla u_{p,\beta}|^{p-2}\nabla u_{p,\beta} \rightharpoonup \textbf{z}_{\beta} \mbox{ weakly in } L^r(\Omega,\mathbb{R}^N), \mbox{ as } p \to 1^+,\end{equation}

for all $1 \leq r < \infty$. In particular, as $p \to 1^+$,

(3.46)\begin{equation} |\nabla u_{p,\beta}|^{p-2}\nabla u_{p,\beta} \to \mbox{div}\, \textbf{z}_{\beta} \quad \mbox{in}\ \mathcal{D}'(\Omega).\end{equation}

Therefore, by using (3.2), (3.28) and (3.46) and the Lebesgue dominated convergence theorem, we conclude that

\[ -\mbox{div}\, \textbf{z}_{\beta} = \rho_{\beta}, \quad \mbox{in}\ \mathcal{D}'(\Omega), \]

which proves the lemma.

Proof. First of all, since $\|\textbf {z}_{\beta }\|_\infty \leq 1$, it follows that, $(\textbf {z}_{\beta },\,Du_\beta ) \leq |Du_\beta |$ in $\mathcal {M}(\Omega )$. In fact, for any Borel set $B$, by (2.1),

\begin{align*} \int_B (\textbf{z}_{\beta},Du_\beta) & \leq \left| \int_B (\textbf{z}_{\beta},Du_\beta) \right|\\ & \leq \|\textbf{z}_{\beta}\|_\infty \int_B |Du_\beta|\\ & \leq \int_B |Du_\beta|. \end{align*}

Hence, it is enough to show the opposite inequality, i.e., that for all $\varphi \in C^1_0(\Omega )$, $\varphi \geq 0$,

(3.47)\begin{equation} \langle (\textbf{z}_{\beta}, Du_\beta), \varphi \rangle \geq \int_\Omega \varphi |Du_\beta|.\end{equation}

In order to do so, let us consider $u_{p,\beta } \varphi \in W^{1,p}(\Omega )$ as a test function in (3.1). Thus we obtain,

(3.48)\begin{equation} \int_\Omega \varphi |\nabla u_{p,\beta}|^p \,{\rm d}x + \int_\Omega u_{p,\beta}|\nabla u_{p,\beta}|^{p-2}\nabla u_{p,\beta} \cdot \nabla \varphi \,{\rm d}x = \int_\Omega \rho_{p,\beta} \varphi \,{\rm d}x.\end{equation}

Now we shall calculate the lower limit as $p \to 1^+$ in both sides of (3.48). Before it, note that, Young's inequality and the lower semicontinuity of the map $v \mapsto \displaystyle \int _\Omega \varphi |Dv|$ with respect to the $L^r(\Omega )$ convergence, imply that

\begin{align*} \int_\Omega \varphi |Du_\beta| & \leq \liminf_{p \to 1^+} \int_\Omega \varphi |\nabla u_{p,\beta}| \,{\rm d}x\\ & \leq \liminf_{p \to 1^+}\left( \frac{1}{p} \int_\Omega \varphi |\nabla u_{p,\beta}|^p \,{\rm d}x + \frac{p-1}{p}\int_\Omega \varphi \,{\rm d}x \right)\\ & = \liminf_{p \to 1^+} \int_\Omega \varphi |\nabla u_{p,\beta}|^p \,{\rm d}x. \end{align*}

Moreover, by (3.46), it follows that

(3.49)\begin{equation} \lim_{p \to1^+}\int_\Omega u_{p,\beta}|\nabla u_{p,\beta}|^{p-2}\nabla u_{p,\beta} \nabla \varphi \,{\rm d}x = \int_\Omega u_\beta \textbf{z}_{\beta}\cdot \nabla \varphi \,{\rm d}x.\end{equation}

Finally, Lebesgue's dominated convergence theorem and (3.26) imply that

(3.50)\begin{equation} \lim_{p \to 1^+}\int_\Omega \rho_{p,\beta} \varphi \,{\rm d}x = \int_\Omega \rho_{\beta} \varphi \,{\rm d}x.\end{equation}

Then, from (3.44), (3.48), (3.49) and (3.50), it follows that

\begin{align*} \langle (\textbf{z}_{\beta}, Du_\beta), \varphi\rangle & ={-}\int_\Omega \varphi u_\beta \mbox{div}\textbf{z}_{\beta} - \int_\Omega u_\beta \textbf{z}_{\beta}\cdot\nabla \varphi \,{\rm d}x\\ & = \int_\Omega \rho_{\beta} u_\beta \varphi \,{\rm d}x - \int_\Omega u_\beta \textbf{z}_{\beta}\cdot\nabla \varphi \,{\rm d}x\\ & = \lim_{p \to 1^+}\left(\int_\Omega \rho_{p,\beta} u_{p,\beta} \varphi \,{\rm d}x - \int_\Omega u_{p,\beta}|\nabla u_{p,\beta}|^{p-2}\nabla u_{p,\beta} \cdot\nabla \varphi \,{\rm d}x\right)\\ & = \liminf_{p \to 1^+}\int_\Omega \varphi|\nabla u_{p,\beta}|^p \,{\rm d}x\\ & \geq \int_\Omega \varphi |Du_\beta|. \end{align*}

Then, (3.47) holds and this finishes the proof.

Proof. To check that $[\textbf {z}_{\beta },\,\nu ]\in sign(-u_\beta )$ it is enough to show that

(3.51)\begin{equation} \int_\Omega (| u_\beta | + u_\beta[\textbf{z}_{\beta},\nu])d\mathcal{H}^{N-1} = 0.\end{equation}

Indeed, since

\begin{align*} -u_\beta[\textbf{z}_{\beta},\nu] & \leq \| \textbf{z}_{\beta} \|_{L^{\infty}(\Omega)}| u_\beta | \\ & \leq | u_\beta |, \end{align*}

the integrand in (4.18) is nonnegative. Then, (4.18) holds if and only if $[\textbf {z}_{\beta },\,\nu ](-u_\beta ) = |u_\beta |$ $\mathcal {H}^{N-1}$a.e. on $\partial \Omega$.

In order to verify (4.18), let us consider $(u_{p,\beta } - \varphi ) \in W^{1,p}_{0}(\Omega )$ as test function in (3.1) with $\varphi \in C^1_{0}(\Omega )$. Then we get

(3.52)\begin{equation} \int_\Omega |\nabla u_{p,\beta}|^p \,{\rm d}x = \int_\Omega |\nabla u_{p,\beta}|^{p-2}\nabla u_{p,\beta} \nabla \varphi \,{\rm d}x + \int_\Omega \rho_{p,\beta} (u_{p,\beta} - \varphi)\,{\rm d}x.\end{equation}

From Young's inequality, Green's Formula, (3.44), (3.46), lemma 3.7 and (3.52), we have that, as $p \to 1^+$,

(3.53)\begin{align} p \int_\Omega |\nabla u_{p,\beta}| \,{\rm d}x & \leq \int_\Omega |\nabla u_{p,\beta}|^p \,{\rm d}x + (p-1)|\Omega|\nonumber\\ & =\int_\Omega |\nabla u_{p,\beta}|^{p-2}\nabla u_{p,\beta} \nabla \varphi \,{\rm d}x + \int_\Omega \rho_{p,\beta}(u_{p,\beta} - \varphi)\,{\rm d}x + (p-1)|\Omega|\nonumber\\ & =\int_\Omega \textbf{z}_{\beta}\cdot \nabla \varphi \,{\rm d}x + \int_\Omega \rho_{\beta}(u_\beta - \varphi)\,{\rm d}x + o_p(1)\nonumber\\ & ={-}\int_\Omega \varphi \mbox{div}\textbf{z}_{\beta} -\int_\Omega \rho_{\beta} \varphi \,{\rm d}x + \int_\Omega \rho_{\beta} u_\beta \,{\rm d}x + o_p(1)\nonumber\\ & = \int_\Omega \rho_{\beta} u_\beta \,{\rm d}x + o_p(1)\nonumber\\ & ={-}\int_\Omega u_\beta \mbox{div}\textbf{z}_{\beta} + o_p(1)\nonumber\\ & =\int_\Omega (\textbf{z}_{\beta}, Du_\beta) - \int_{\partial \Omega} \left[\textbf{z}_{\beta},\nu \right] u_\beta d\mathcal{H}^{N-1} + o_p(1)\nonumber\\ & =\int_\Omega |Du_\beta| - \int_{\partial \Omega} \left[\textbf{z}_{\beta},\nu \right] u_\beta d\mathcal{H}^{N-1} + o_p(1). \end{align}

Hence, from (3.53) and the lower semicontinuity of the norm in $BV(\Omega )$, it follows that

(3.54)\begin{equation} \int_{\partial B}\left(|u_\beta| + \left[\textbf{z}_{\beta},\nu \right] u_\beta \right) d\mathcal{H}^{N-1} \leq 0.\end{equation}

But the last inequality implies in (4.18) and we are done.

Proof. First of all, let us prove that, if $0 < \beta _1 < \beta _2 < \beta _0$, then

(4.5)\begin{equation} I_{\beta_1}(u_{\beta_1}) \leq I_{\beta_2}(u_{\beta_2}). \end{equation}

In order to do so, let us prove that, for $p > 1$ fixed,

(4.6)\begin{equation} I_{p,\beta_1}(u_{p,\beta_1}) < I_{p,\beta_2}(u_{p,\beta_2}). \end{equation}

Note that, for $u \in W^{1,p}_0(\Omega )$, since $F_{\beta _1}(u) \geq F_{\beta _2}(u)$ a.e. in $\Omega$, it follows that

(4.7)\begin{equation} I_{p,\beta_1}(u) \leq I_{p,\beta_2}(u). \end{equation}

Moreover, let us assume that the function $e$ in lemma 3.1, is $e(\beta _0)$, i.e., that satisfies $I_{p,\beta _0}(e) < p/(p-1)$. Hence, from (4.7), we have that

\[ I_{p,\beta}(e) \leq I_{p,\beta_0}(e) < \frac{p}{p-1}, \]

for all $0 < \beta < \beta _0$. Hence, in the definition of $c_{p,\beta }$, for $0 < \beta < \beta _0$, we can assume without loss of generality that $e = e(\beta _0)$ and then the class of paths $\Gamma$ does not depend on $\beta$. Then, from (4.7), it follows that

\begin{align*} I_{p,\beta_1}(u_{p,\beta_1}) & = c_{p,\beta_1}\\ & =\inf_{\gamma \in \Gamma}\sup_{t \in [0,1]} I_{p,\beta_1}(\gamma(t))\\ & \leq \inf_{\gamma \in \Gamma}\sup_{t \in [0,1]} I_{p,\beta_2}(\gamma(t))\\ & = c_{p,\beta_2}\\ & = I_{p,\beta_2}(u_{p,\beta_2}). \end{align*}

This, in turn, proves (4.6).

Hence, from (4.4), passing the limit as $p \to 1^+$ in (4.6), we have that (4.5) holds.

Then, for all $0 < \beta < \beta _0$,

\[ I_\beta(u_\beta) \leq I_{\beta_0}(u_{\beta_0}) =: C. \]

Note that, by using $u_\beta$ as test function in (4.1), from Green's Formula, we have that

(4.8)\begin{align} \|u_\beta\| & = \int_\Omega |Du_\beta| + \int_{\partial \Omega}|u_\beta| d\mathcal{H}^{N-1}\nonumber\\ & =\int_\Omega ({\bf z}_\beta,Du_\beta) + \int_{\partial \Omega}[{\bf z}_\beta,\nu]u_\beta d\mathcal{H}^{N-1}\nonumber\\ & ={-}\int_\Omega u_\beta \mbox{div}\, {\bf z}_\beta \,{\rm d}x\\ & = \int_\Omega u_\beta \rho_\beta \,{\rm d}x. \nonumber \end{align}

Then, from (3.29), (4.8) and the definition of $F_\beta$, we have that

\begin{align*} I_\beta(u_\beta) & = I_\beta(u_\beta) - \frac{1}{q}\left(\|u_\beta\| - \int_\Omega u_\beta \rho_\beta \,{\rm d}x\right)\\ & =\left(1 - \frac{1}{q}\right)\|u_\beta\| + \int_\Omega \left(\frac{1}{q} u_\beta \rho_\beta - F_\beta(u_\beta)\right) \,{\rm d}x\\ & \geq\left(1 - \frac{1}{q}\right)\|u_\beta\| + \frac{\beta^q}{q}|\{u_\beta > \beta\}|\\ & \geq \left(1 - \frac{1}{q}\right)\|u_\beta\|. \end{align*}

Hence, since $(I_\beta (u_{\beta }))_{0 < \beta < \beta _0}$ is bounded, it follows from the last inequality that $(u_\beta )_{0 < \beta < \beta _0}$ is also bounded.

Proof. First of all, note that, from (2.1),

\[ ({\bf z}_0,Du_0) \leq |Du_0| \quad \mbox{in}\ \mathcal{M}(\Omega). \]

For the inverse inequality, let $\varphi \in \mathcal {D}(\Omega )$, $\varphi \geq 0$. In (4.1), let us take $\varphi u_\beta$ as test function in (4.1). Then,

\[ \int_\Omega \varphi ({\bf z}_\beta,Du_\beta) = \int_\Omega \varphi u_\beta \rho_\beta \,{\rm d}x - \int_\Omega u_\beta {\bf z}_\beta\cdot \nabla \varphi \,{\rm d}x. \]

Taking into account that $({\bf z}_\beta,\,Du_\beta ) = |Du_\beta |$ in $\mathcal {M}(\Omega )$,

\[ \int_\Omega \varphi |Du_\beta| = \int_\Omega \varphi u_\beta \rho_\beta \,{\rm d}x - \int_\Omega u_\beta {\bf z}_\beta\cdot \nabla \varphi \,{\rm d}x. \]

Taking the $\liminf$ as $\beta \to 0^+$, from the lower semicontinuity of the norm in $BV(\Omega )$ with respect to the $L^r$ convergence, (4.9), (4.11) and (4.16), it follows that

\begin{align*} \int_\Omega \varphi |Du_0| & \leq \liminf_{\beta \to 0^+} \left(-\int_\Omega \varphi u_\beta \mbox{div}\, {\bf z}_\beta \,{\rm d}x - \int_\Omega u_\beta {\bf z}_\beta\cdot \nabla \varphi \,{\rm d}x\right)\\ & ={-}\int_\Omega \varphi u_0 \mbox{div}\, {\bf z}_0 \,{\rm d}x - \int_\Omega u_0 {\bf z}_0\cdot \nabla \varphi \,{\rm d}x\\ & = \int_\Omega \varphi ({\bf z}_0, Du_0). \end{align*}

This, in turn, proves that $|Du_0| \leq ({\bf z}_0,\,Du_0)$ and this finishes the proof.

Proof. As in lemma 3.8, it is enough to show that

(4.18)\begin{equation} \int_\Omega (| u_0 | + u_0[\textbf{z}_0,\nu])d\mathcal{H}^{N-1} = 0.\end{equation}

In order to verify (4.18), let us consider $(u_\beta - \varphi ) \in BV(\Omega )\cap L^\infty (\Omega )$ as test function in (4.1), where $\varphi \in \mathcal {D}(\Omega )$. Then, from (2.3) and (4.1), we get

(4.19)\begin{align} \int_\Omega |Du_\beta| + \int_{\partial \Omega} |u_\beta| d\mathcal{H}^{N-1} & = \int_\Omega ({\bf z}_\beta,Du_\beta) - \int_{\partial \Omega} u_\beta [{\bf z}_\beta,\nu] d\mathcal{H}^{N-1}\nonumber\\ & ={-}\int_\Omega u_\beta \mbox{div}\, {\bf z}_\beta\\ & =\int_\Omega \varphi \mbox{div}\, {\bf z}_\beta + \int_\Omega u_\beta \rho_\beta \,{\rm d}x - \int_\Omega \varphi \rho_\beta\nonumber\\ & =\int_\Omega u_\beta \rho_\beta \,{\rm d}x. \nonumber \end{align}

Then, calculating the $\liminf$ in (4.19), from the lower semicontinuity of the norm in $BV(\Omega )$, (4.17) and lemma 4.2, we have that

\begin{align*} \int_\Omega |Du_0| \int_{\partial \Omega}|u_0| d\mathcal{H}^{N-1} & \leq \int_\Omega u_0 \rho_0 \,{\rm d}x\\ & ={-}\int_\Omega u_0 \mbox{div}\, {\bf z}_0 \\ & =\int_\Omega ({\bf z}_0,Du_0) - \int_{\partial \Omega} u_0 [{\bf z}_0,\nu] \mathcal{H}^{N-1} \\ & =\int_\Omega |Du_0| - \int_{\partial \Omega} u_0 [{\bf z}_0,\nu] \mathcal{H}^{N-1}. \end{align*}

From the last inequality, it follows that

\[ | u_0 | + u_0[\textbf{z}_0,\nu] \leq 0 \quad \mathcal{H}^{N-1}-\mbox{a.e. on }\partial \Omega. \]

Since the inverse inequality is trivial, it follows that (4.18) holds.