#### Proof.

For any $0<\rho <\mu $, let us consider the sets

${\mathrm{\Omega}}_{\rho}^{-}:=\{x\in \mathrm{\Omega}:\rho <d(x)<\mu \}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}{\mathrm{\Omega}}_{\rho}^{+}:=\{x\in \mathrm{\Omega}:0<d(x)<\mu -\rho \}.$

*Proof of (4.1).*
For an appropriate choice of a positive constant ${A}^{*}$, we will show that

${w}^{*}(x):={A}^{*}\varphi (d(x)-\rho ),x\in {\mathrm{\Omega}}_{\rho}^{-},$

is a super-solution of (1.3) on ${\mathrm{\Omega}}_{\rho}^{-}$ for all $0<\rho <\mu $ and sufficiently small μ.
To this end, we estimate (2.2) with $\varrho (x):=d(x)-\rho $ as follows.
Recalling that $c\le 0$ in Ω, we estimate

$L{w}^{*}\le f(\varphi (d(x)-\rho ))\left[{A}^{*}\mathrm{\Lambda}+\frac{{A}^{*}\sqrt{2F(\varphi (d-\rho ))}}{f(\varphi (d-\rho ))}(|{L}_{0}d|+{\parallel \mathbf{b}\parallel}_{{L}^{\mathrm{\infty}}})\right].$(4.3)

Let $\epsilon >0$, to be specified later.
By (2.1), we can take $\mu >0$ sufficiently small so that for all $x\in {\mathrm{\Omega}}_{\rho}^{-}$,

$\frac{{A}^{*}\sqrt{2F(\varphi (d-\rho ))}}{f(\varphi (d-\rho ))}(|{L}_{0}d|+{\parallel \mathbf{b}\parallel}_{{L}^{\mathrm{\infty}}})<\epsilon .$(4.4)

Therefore, from (4.3) and (4.4), we conclude that for $x\in {\mathrm{\Omega}}_{\mu}^{-}$,

$L{w}^{*}\le f(\varphi (d-\rho )))({A}^{*}\mathrm{\Lambda}+\epsilon ).$

Now, let ε be chosen so that ${\mathrm{\Theta}}_{*}-2\epsilon >{\mathrm{\Xi}}_{*}$.
We pick a number ${A}^{*}:={A}^{*}(\mathrm{\Lambda},f,{\mathrm{\Theta}}_{*}(h))>0$ such that

${\mathrm{\Theta}}_{*}-2\epsilon >{g}^{*}({A}^{*})={A}^{*}\mathrm{\Lambda}-\underset{0<d(x)-\rho \to 0}{lim\; inf}\frac{f({A}^{*}\varphi (d(x)-\rho ))}{f(\varphi (d(x)-\rho ))}.$

That is,

${A}^{*}\mathrm{\Lambda}+\epsilon <{\mathrm{\Theta}}_{*}-\epsilon +\underset{0<d(x)-\rho \to 0}{lim\; inf}\frac{f({A}^{*}\varphi (d(x)-\rho ))}{f(\varphi (d(x)-\rho ))}.$

We also suppose that μ is sufficiently small, so that for $(x,\rho )\in {\mathrm{\Omega}}_{\rho}^{-}\times (0,\mu )$,

${A}^{*}\mathrm{\Lambda}+\epsilon <{\mathrm{\Theta}}_{*}-\frac{\epsilon}{2}+\frac{f({A}^{*}\varphi (d(x)-\rho ))}{f(\varphi (d(x)-\rho ))}.$

Let us now suppose that ${\mathrm{\Theta}}_{*}(h)\le 0$.
In this case, for $(x,\rho )\in {\mathrm{\Omega}}_{\rho}^{-}\times (0,\mu )$, we have

$L{w}^{*}\le f(\varphi (d(x)-\rho ))({A}^{*}\mathrm{\Lambda}+\epsilon )$$\le f(\varphi (d(x)-\rho ))\left[{\mathrm{\Theta}}_{*}-{\displaystyle \frac{\epsilon}{2}}+{\displaystyle \frac{f({A}^{*}\varphi (d(x)-\rho ))}{f(\varphi (d(x)-\rho ))}}\right]$$=\left({\mathrm{\Theta}}_{*}-{\displaystyle \frac{\epsilon}{2}}\right)f(\varphi (d(x)-\rho ))+f({A}^{*}\varphi (d(x)-\rho ))$$\le \left({\mathrm{\Theta}}_{*}-{\displaystyle \frac{\epsilon}{2}}\right)f(\varphi (d(x)))+f({A}^{*}\varphi (d(x)-\rho ))$$\le h(x)+f({w}^{*}),$

from the definition of ${\mathrm{\Theta}}_{*}$.
Let ${B}^{*}:=\mathrm{max}\{u(x):d(x)\ge \mu \}$.
Then $u\le {w}^{*}+{B}^{*}$ on $\partial {\mathrm{\Omega}}_{\rho}^{-}$ and

$Lu=f(u)+h,L({w}^{*}+{B}^{*})\le L{w}^{*}\le f({w}^{*})+h\le f({w}^{*}+{B}^{*})+h\mathit{\hspace{1em}}\text{on}{\mathrm{\Omega}}_{\rho}^{-}.$

By the Comparison Principle, Lemma 2.1, we conclude that $u\le {w}^{*}+{B}^{*}$ in ${\mathrm{\Omega}}_{\rho}^{-}$.
Therefore,

$\frac{u(x)}{\varphi (d(x)-\rho )}-\frac{{B}^{*}}{\varphi (d(x)-\rho )}\le {A}^{*}\mathit{\hspace{1em}}\text{for}x\in {\mathrm{\Omega}}_{\rho}^{-}.$

On letting $\rho \to {0}^{+}$, we see that the following holds on ${\mathrm{\Omega}}_{\mu}$:

$\frac{u(x)}{\varphi (d(x))}-\frac{{B}^{*}}{\varphi (d(x))}\le {A}^{*}.$

Now, let $d(x)\to 0$ to obtain (4.1) when ${\mathrm{\Theta}}_{*}(h)\le 0$.

We now consider the case ${\mathrm{\Theta}}_{*}(h)>0$.
Then *h* is non-negative near $\partial \mathrm{\Omega}$.
Let $\{{\mathcal{\mathcal{O}}}_{j}\}$ be the sequence of open subsets of Ω defined as in the proof of Lemma 3.6 and $m:=\mathrm{min}\{h(x):x\in \overline{\mathrm{\Omega}}\}$.
We note that $m>-\mathrm{\infty}$.
Let ${v}_{j}$ be a solution of (3.4) with $h\equiv m$.
If *u* is any solution of (1.3), then $Lu=f(u)+h\ge f(u)+m$.
By the Comparison Principle, $u\le {v}_{j}$ in ${\mathcal{\mathcal{O}}}_{j}$ for all *j*. Proceeding as in the proof of Theorem 3.7, one can show that $v={lim}_{j\to \mathrm{\infty}}{v}_{j}$ is a solution of (1.3) on Ω with $h\equiv m$ on Ω.
Clearly, $u\le v$ in Ω, and since *v* is a large solution of $Lv=f(v)+m$ on Ω and ${\mathrm{\Theta}}_{*}(m)=0$, the case considered above applied to *v* shows that

$\underset{d(x)\to 0}{lim\; sup}\frac{u(x)}{\varphi (d(x))}\le \underset{d(x)\to 0}{lim\; sup}\frac{v(x)}{\varphi (d(x))}\le {A}^{*},$

where ${A}^{*}={A}^{*}(\mathrm{\Lambda},f,{\mathrm{\Theta}}_{*}(h))$, and therefore (4.1) holds.

*Proof of (4.2).*

We consider the function

${w}_{*}(x):={A}_{*}\varphi (d(x)+\rho ),x\in {\mathrm{\Omega}}_{\rho}^{+}.$

For an appropriate choice of ${A}_{*}$, we show that ${w}_{*}$ is a sub-solution in ${\mathrm{\Omega}}_{\rho}^{+}$ for all $\rho <\mu $, assuming that μ is sufficiently small.
We assume first that ${\mathrm{\Xi}}^{*}>0$.
Let us fix $\epsilon >0$ such that $3\epsilon <{\mathrm{\Xi}}^{*}$.
Then, by definition, there exists a positive real number ${A}_{*}$ such that ${\mathrm{\Xi}}^{*}/2+3\epsilon /2<{g}^{*}({A}_{*})$.
That is,

$\frac{1}{2}{\mathrm{\Xi}}^{*}+\underset{d(x)+\rho \to 0}{lim\; sup}\frac{f({A}_{*}\varphi (d(x)+\rho ))}{f(\varphi (d(x)+\rho ))}<{A}_{*}-\frac{3}{2}\epsilon .$

Therefore, assuming that μ is sufficiently small, the following holds in ${\mathrm{\Omega}}_{\rho}^{+}$ for all $0<\rho <\mu $:

$\frac{1}{2}{\mathrm{\Xi}}^{*}+\frac{f({A}_{*}\varphi (d(x)+\rho ))}{f(\varphi (d(x)+\rho ))}<{A}_{*}\lambda -\epsilon .$

By shrinking μ if necessary, we can invoke (2.1) to estimate

${A}_{*}\left[\frac{\sqrt{2F(\varphi (d(x)+\rho ))}}{f(\varphi (d(x)+\rho ))}(|{L}_{0}d|+{\parallel \mathbf{b}\parallel}_{{L}^{\mathrm{\infty}}})+\frac{\varphi (d(x)+\rho )}{f(\varphi (d(x)+\rho ))}{\parallel c\parallel}_{{L}^{\mathrm{\infty}}}\right]<\epsilon \mathit{\hspace{1em}}\text{for all}x\in {\mathrm{\Omega}}_{\rho}^{+}\text{and}0\rho \mu .$

Thus, for any $0<\rho <\mu $, the following chain of inequalities hold on ${\mathrm{\Omega}}_{\rho}^{+}$:

$L{w}_{*}\ge f(\varphi (d(x)+\rho ))\left[{A}_{*}-{A}_{*}{\displaystyle \frac{\sqrt{2F(\varphi (d(x)+\rho ))}}{f(\varphi (d(x)+\rho ))}}(|{L}_{0}d|+{\parallel \mathbf{b}\parallel}_{{L}^{\mathrm{\infty}}})-{\displaystyle \frac{{A}_{*}{\parallel c\parallel}_{{L}^{\mathrm{\infty}}}\varphi (d(x)+\rho )}{f(\varphi (d(x)+\rho ))}}\right]$$\ge f(\varphi (d(x)+\rho ))({A}_{*}-\epsilon )$$\ge f(\varphi (d(x)+\rho ))\left({\displaystyle \frac{1}{2}}{\mathrm{\Xi}}^{*}+{\displaystyle \frac{f({A}_{*}\varphi (d(x)+\rho ))}{f(\varphi (d(x)+\rho ))}}\right)$$\ge {\displaystyle \frac{1}{2}}{\mathrm{\Xi}}^{*}f(\varphi (d(x)+\rho ))+f({A}_{*}\varphi (d(x)+\rho )).$(4.5)

Recall that ${\mathrm{\Xi}}^{*}>0$.
Since *h* is bounded from above on Ω, we can assume μ is small enough so that

$\frac{1}{2}{\mathrm{\Xi}}^{*}f(\varphi (d(x)+\rho ))\ge h(x),x\in {\mathrm{\Omega}}_{\rho}^{+}.$

Thus, for ${\mathrm{\Xi}}^{*}>0$ and sufficiently small $\mu >0$, we have shown that

$L{w}_{*}\ge f({w}^{*})+h(x),x\in {\mathrm{\Omega}}_{\rho}^{+},\text{for all}0\rho \mu .$

Now let us suppose that ${\mathrm{\Xi}}^{*}=0$.
Then there exists ${A}_{*}>0$ such that for $\epsilon >0$ and $d+\rho $ small, we have

${A}_{*}-\frac{f({A}_{*}\varphi (d+\rho ))}{f(\varphi (d+\rho ))}>-\epsilon ,$

which can be rewritten as

$\frac{f({A}_{*}\varphi (d+\rho ))}{f(\varphi (d+\rho ))}-2\epsilon <{A}_{*}-\epsilon .$

Using this in estimate (4.5), we find

$L{w}_{*}\ge f({A}_{*}\varphi (d+\rho ))-2\epsilon f(\varphi (d+\rho )).$(4.6)

By (h1), we recall that $-\mathrm{\infty}\le {\mathrm{\Theta}}^{*}<{\mathrm{\Xi}}^{*}=0$.
There is no loss in generality if we assume that ${\mathrm{\Theta}}^{*}>-\mathrm{\infty}$.
At this point we use condition (h1), that is,

$\underset{d(x)\to 0}{lim\; sup}\frac{h(x)}{f(\varphi (d))}={\mathrm{\Theta}}^{*}<0.$(4.7)

On noting that $f(\varphi (d+\rho ))\le f(\varphi (d))$, and using (4.7), we find

$\frac{h(x)}{f(\varphi (d+\rho ))}\le \frac{1}{2}{\mathrm{\Theta}}^{*}.$

Choosing $\epsilon =-{\mathrm{\Theta}}^{*}/4$ in the above inequality, we find

$h(x)\le -2\epsilon f(\varphi (d+\rho )).$

Inserting the latter estimate into (4.6), for sufficiently small $\rho >0$, yields

$L{w}_{*}\ge f({w}_{*})+h(x),x\in {\mathrm{\Omega}}_{\rho}^{+}.$

In conclusion, we have shown that in either of the cases, the following holds:

$L{w}_{*}\ge f({w}_{*})+h(x),x\in {\mathrm{\Omega}}_{\rho}^{+}.$

Let ${B}_{*}:={A}_{*}\varphi (\mu )$.
Note that ${w}_{*}-{B}_{*}\le u$ on $\partial {\mathrm{\Omega}}_{\rho}^{+}$ and

$L({w}_{*}-{B}_{*})\ge L{w}_{*}\ge f({w}_{*})+h\ge f({w}_{*}-{B}_{*})+h\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}L(u)=f(u)+h\mathit{\hspace{1em}}\text{in}{\mathrm{\Omega}}_{\rho}^{+}.$

Therefore, ${w}_{*}-{B}_{*}\le u$ in ${\mathrm{\Omega}}_{\rho}^{+}$ and

${A}_{*}\le \frac{u(x)}{\varphi (d(x)+\rho )}+\frac{{B}_{*}}{\varphi (d(x)+\rho )}\mathit{\hspace{1em}}\text{for}x\in {\mathrm{\Omega}}_{\rho}^{+}.$

On letting $\rho \to {0}^{+}$, we see that

${A}_{*}\le \frac{u(x)}{\varphi (d(x))}+\frac{{B}_{*}}{\varphi (d(x))}\mathit{\hspace{1em}}\text{on}{\mathrm{\Omega}}_{\mu}.$

On recalling that $\varphi (d(x))\to \mathrm{\infty}$ as $d(x)\to 0$, we get

${A}_{*}\le \underset{d(x)\to 0}{lim\; inf}\frac{u(x)}{\varphi (d(x))},$

and the proof is complete.
∎

#### Proof.

Estimate (4.1) is proved in Lemma 4.2.
Thus, it only remains to show (4.2).
For $j=1,2,\mathrm{\dots}$, let ${v}_{j}$ be the solution of

$L{v}_{j}=f({v}_{j})-\mathrm{min}\{j,{h}^{-}\}\mathit{\hspace{1em}}\text{in}\mathrm{\Omega},{v}_{j}=j\mathit{\hspace{1em}}\text{on}\partial \mathrm{\Omega}.$

The Comparison Principle shows that $\{{v}_{j}\}$ is an increasing sequence.
Let *w* be the unique solution of (4.8).
Since $w<0$ on Ω, we find that

$L({v}_{j}+w)=f({v}_{j})-\mathrm{min}\{j,{h}^{-}\}+{h}^{+}\ge f({v}_{j}+w)+h\mathit{\hspace{1em}}\text{in}\mathrm{\Omega},{v}_{j}+w=j\mathit{\hspace{1em}}\text{on}\partial \mathrm{\Omega}.$

Now, if *u* is any solution of (1.3) in Ω, then again by the Comparison Principle, we find

${v}_{j}+w(x)\le u(x)\mathit{\hspace{1em}}\text{for all}j\text{.}$

Letting $j\to \mathrm{\infty}$, we obtain

$v+w\le u\mathit{\hspace{1em}}\text{in}\mathrm{\Omega},$

where

$v(x):=\underset{j\to \mathrm{\infty}}{lim}{v}_{j}(x),x\in \mathrm{\Omega}.$

Then *v* is a large solution of $Lv=f(v)-{h}^{-}$ on Ω.

Since *h* satisfies (h1) and (h3), we see that ${\mathrm{\Theta}}^{*}(-{h}^{-})={\mathrm{\Theta}}^{*}(h)<{\mathrm{\Xi}}^{*}$.
Since, in addition, $-{h}^{-}$ is bounded from above we invoke Lemma 4.2 (ii) to infer that there exists a constant ${A}_{*}>0$ such that *v* satisfies the asymptotic estimate (4.2).
Since *w* is bounded on Ω, we have

${A}_{*}\le \underset{d(x)\to 0}{lim\; inf}\frac{v(x)}{\varphi (d(x))}\le \underset{d(x)\to 0}{lim\; sup}\frac{u(x)}{\varphi (d(x))},$

and this concludes the proof of (4.2).
∎

The following corollary is noteworthy.

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