$\begin{array}[]{lcl}|x_{k}-y_{k}|&\leq&\displaystyle|\delta_{k}|\frac{|\wp(t)-\wp(s)|}{\wp(s)\wp(t)}+\big{|}\sigma_{k}\mathcal{I}_{k}(s)-\sigma_{k}\mathcal{I}_{k}(t)\big{|}\\\\[11.38092pt]
&\leq&\displaystyle\left(\frac{\wp_{1}\mu_{1}}{\wp_{0}^{2}}\max_{\tau\in[0,T]}|\wp^{\prime}(\tau)|+\kappa\right)|t-s|=C|t-s|<\varrho.\end{array}$

In conclusion,

$\big{|}\mathcal{H}(w_{k},\sigma_{k})^{\prime}(t)-\mathcal{H}(w_{k},\sigma_{k})^{\prime}(s)\big{|}=\big{|}\varphi^{-1}(x_{k})-\varphi^{-1}(y_{k})\big{|}<\varepsilon$

provided that $|t-s|<\varrho/C$, independently of $k$. As an immediate
consequence of the Ascoli–Arzelà theorem $\mathcal{H}$ maps bounded sequences
of $X\times[0,1]$ into relatively compact sequences of $X$. Finally, if
$(w_{k},\sigma_{k})$ converges to some $(w,\sigma)$ in $X\times[0,1]$, then
$\delta_{k}\rightarrow\delta=\delta(w,\sigma)$ as $k\to\infty$. Hence,
$\mathcal{H}$ is continuous in $X\times[0,1]$. The proof of this fact is
similar to that given in Chapter 4 of [123]. The Leray-Schauder fixed point
theorem can therefore be applied and the mapping $\mathcal{H}(w,1)$ has a
fixed point $w$, which by (3.21) satisfies inequality (3.9).

To conclude, we prove that $w$ solves the ODE in (3.5) with the original
$f,l$. Differentiating (3.20) with $\sigma=1$, we see that $w$ is a weak
solution of

$\big{(}\wp\varphi(w^{\prime})\big{)}^{\prime}=\wp
a\bar{f}(w)\bar{l}(|w^{\prime}|)\qquad\text{on }\,(0,T),$ (3.24)

where now we have denoted with $\bar{f},\bar{l}$ the modifications of $f,l$ in
(3.13). Clearly, $\bar{l}(w^{\prime})=l(w^{\prime})$ on $(0,T)$, since $0\leq
w^{\prime}<\xi$. From $0\leq w\leq\eta$ and Lemma 3.2, there exists
$t_{0}\in[0,T)$ such that $w=0$ on $[0,t_{0}]$ and $w>0$, $w^{\prime}>0$ on
$(t_{0},T]$, and since $\bar{f}=f$ on $(0,\eta)$ we deduce

$\big{(}\wp\varphi(w^{\prime})\big{)}^{\prime}=\wp af(w)l(|w^{\prime}|)$
(3.25)

on $(t_{0},T)$. On the other hand, because of our assumption $f(0)l(0)=0$ the
function $w=0$ solves (3.25) on $(0,t_{0})$. Since $w\in C^{1}([0,T])$, (3.25)
holds weakly on all of $(0,T)$, as claimed. ∎

######  Remark 3.4.

_The requirement $f(0)l(0)=0$ is crucial for the validity of the above
theorem, because otherwise $w$ might be negative somewhere. _

Given $f,\varphi,l$ satisfying (3.2), define $K,F$ as in (1.19), (1.20). We
next present two auxiliary calculus results. The first is similar to Lemma
4.4.1–$(i)$ in [123]. We recall that the notion of a $C$-increasing function
is given in Definition 1.9.

######  Lemma 3.5.

Assume that $f$ is $C$-increasing in $(0,\eta_{0})$, for some $\eta_{0}>0$.
Then, for each $\sigma\in[0,1]$ we have $F(\sigma t)\leq C\,\sigma F(t)$ for
all $t\in[0,\eta_{0})$.

###### Proof.

Fix $\sigma\in[0,1]$. Since $f$ is $C$-increasing in $(0,\eta_{0})$, we have
$\sigma f(\sigma t)\leq C\,\sigma f(t)$ for each $t\in[0,\eta_{0})$, and thus

$F(\sigma t)=\int_{0}^{\sigma t}f(s)\mathrm{d}s=\int_{0}^{t}\sigma
f(\sigma\tau)\mathrm{d}\tau\leq
C\,\sigma\int_{0}^{t}f(\tau)\mathrm{d}\tau=C\sigma F(t),$

as claimed. ∎

The second lemma concerns the preservation of the validity of (KO0) and
($\mathrm{KO}_{\infty}$) when we replace $f$ with $\sigma f$,
$\sigma\in\mathbb{R}^{+}$. Similar results have been proved in [123, Lem.
4.1.2], [57] (remark on page 523) and [95].

######  Lemma 3.6.

Let $f,l$ satisfy (3.2), (3.3) and suppose $l>0$ on $\mathbb{R}^{+}$.

  * (i)

Assume that $f$ is positive and $C$-increasing in $(0,\eta_{0})$, for
$\eta_{0}>0$ in (3.2). Then

$\frac{1}{K^{-1}(F(s))}\in
L^{1}(0^{+})\qquad\Longleftrightarrow\qquad\frac{1}{K^{-1}(\sigma F(s))}\in
L^{1}(0^{+})$

for some (equivalently, any) $\sigma\in\mathbb{R}^{+}$.

  * (ii)

Assume that $f$ is positive and $C$-increasing on $(\bar{\eta}_{0},\infty)$,
for some $\bar{\eta}_{0}>0$. Having defined $F(t)=\int_{\bar{\eta}_{0}}^{t}f$,
it holds

$\frac{1}{K^{-1}(F(s))}\in
L^{1}(\infty)\qquad\Longleftrightarrow\qquad\frac{1}{K^{-1}(\sigma F(s))}\in
L^{1}(\infty)$

for some (equivalently, any) $\sigma\in\mathbb{R}^{+}$.

###### Proof.

For the ease of notation we denote with $\text{(KO${}_{0}$)}(\sigma)$ and
$\text{(KO${}_{\infty}$)}(\sigma)$, respectively, the integrability conditions

$\int_{0^{+}}\frac{\mathrm{d}s}{K^{-1}(\sigma
F(s))}<\infty,\qquad\int^{\infty}\frac{\mathrm{d}s}{K^{-1}(\sigma
F(s))}<\infty.$

We prove $(i)$, beginning with implication

$\eqref{KO_zero}\Rightarrow\text{(KO${}_{0}$)}(\sigma).$ (3.26)

If $\sigma\geq 1$, (3.26) is immediate from the monotonicity of $F$ and $K$.
If $\sigma\in(0,1)$, we apply Lemma 3.5 with $\sigma/C$ replacing $\sigma$
(note that $C\geq 1$) to deduce $F(\sigma t/C)\leq\sigma F(t)$ for each
$t\in(0,\eta_{0})$. Integrating and changing variables,

$\int_{0^{+}}\frac{\mathrm{d}s}{K^{-1}(\sigma
F(s))}\leq\int_{0^{+}}\frac{\mathrm{d}s}{K^{-1}(F(\sigma
s/C))}=\frac{C}{\sigma}\int_{0^{+}}\frac{\mathrm{d}\tau}{K^{-1}(F(\tau))},$

which proves (3.26). To show the reverse implication in (3.26), it is enough
to observe that $\text{(KO${}_{0}$)}(\sigma)$ is condition (KO0) for the
function $\bar{f}=\sigma f$, and to apply the previous estimates with
$\bar{f}$ replacing $f$ and $\sigma^{-1}$ replacing $\sigma$.
The proof of $(ii)$ is analogous. ∎

We are now ready to obtain further information on the solution of problem
(3.5) given in Theorem 3.1. First, we investigate sufficient conditions to
ensure $w^{\prime}(0)>0$.

######  Proposition 3.7.

Assume (3.2), (3.3) and (3.4). Suppose that

  * $f(0)l(0)=0$;

  * $f$ is $C$-increasing on $(0,\eta_{0})$, for $\eta_{0}>0$ as in (3.2),

and that one of the following two sets of conditions is met:

  * (i)

$f\equiv 0$ on $(0,\eta_{0})$;

  * (ii)

$f>0$ on $(0,\eta_{0})$, and also

    * -

$l$ is $C$-increasing on $(0,\xi_{0})$, for some $\xi_{0}>0$.

    * -

$\wp$ is monotone on $[0,T_{0}]$, either increasing or decreasing ($T_{0}$ as
in (3.4)),

and

$\frac{1}{K^{-1}\circ F}\not\in L^{1}(0^{+}).$ ($\neg$KO0)

Then, the solution $w$ of problem (3.5), with $\eta\in(0,\eta_{0})$,
constructed in Theorem 3.1, has the further properties

$w>0\quad\text{on }(0,T],\qquad w^{\prime}>0\quad\text{on }[0,T].$ (3.27)

###### Proof.

First, observe that if we prove that $w^{\prime}(0)>0$, then (3.27) follows by
a direct application of Lemma 3.2. We prove $w^{\prime}(0)>0$ for cases $(i)$
and $(ii)$ separately.

Case $(i)$.
Since $f\equiv 0$ on $(0,\eta_{0})$, then
$(\wp\varphi(w^{\prime}))^{\prime}=0$ in $(0,T)$ by (3.5) and the choice
$\eta<\eta_{0}$. Hence, integrating

$\wp(t)\varphi(w^{\prime}(t))=\wp(0)\varphi(w^{\prime}(0))\quad\text{for all
}t\in(0,T].$ (3.28)

Suppose by contradiction that $w^{\prime}(0)=0$. From (3.2) we have
$\varphi>0$ on $\mathbb{R}^{+}$, and also $\wp>0$ on $[0,T]$. Thus, (3.28)
would imply that $w^{\prime}$, hence $w$, is identically zero in $[0,T]$,
contradicting $w(T)=\eta$.

Case $(ii)$.
Theorem 3.1 guarantees that $\varphi(w^{\prime})\in C^{1}([0,T])$,
$w^{\prime}(0)\geq 0$ and $0\leq w\leq\eta$. Let us reason by contradiction
and suppose that $w^{\prime}(0)=0$. We shall then prove that

$\frac{1}{K^{-1}\circ F}\in L^{1}(0^{+}),$

which contradicts ($\neg$KO0), completing the proof.

First, by Lemma 3.2 there exists $t_{0}\in[0,T)$ such that $w(t)\equiv 0$ on
$[0,t_{0}]$ while $w>0$ on $(t_{0},T]$. If $t_{0}=0$ then
$w^{\prime}(t_{0})=0$ by our assumption, while if $t_{0}>0$ then
$w(t_{0})=w^{\prime}(t_{0})=0$ since $w$ is $C^{1}([0,T])$. From (3.20) and
since $\varphi^{\prime}>0$ on $\mathbb{R}^{+}$ we infer the existence of
$w^{\prime\prime}$ in $(t_{0},T)$. Thus, $w$ satisfies

$\wp\varphi^{\prime}(w^{\prime})w^{\prime\prime}+\wp^{\prime}\varphi(w^{\prime})=\wp
af(w)l(w^{\prime})\qquad\mbox{in }(t_{0},T).$ (3.29)

We first suppose that $\wp^{\prime}\geq 0$ on $[0,T_{0}]$. By (3.12) and
(3.4), $w$ is a solution of the inequality

$w^{\prime}\varphi^{\prime}(w^{\prime})w^{\prime\prime}\leq
af(w)w^{\prime}l(w^{\prime})\qquad\mbox{in }(t_{0},T).$

Integrating on $[t_{0},t)$, with $t\in(t_{0},T]$ we have

$\int_{t_{0}}^{t}\frac{w^{\prime}\varphi^{\prime}(w^{\prime})w^{\prime\prime}}{l(w^{\prime})}\mathrm{d}\tau\leq\int_{t_{0}}^{t}af(w)w^{\prime}\mathrm{d}\tau\leq
a_{1}\int_{t_{0}}^{t}f(w)w^{\prime}\mathrm{d}\tau.$

Changing variables and using $w^{\prime}(t_{0})=w(t_{0})=0$ we deduce

$K(w^{\prime}(t))=\int_{0}^{w^{\prime}(t)}\frac{s\varphi^{\prime}(s)}{l(s)}\mathrm{d}s\leq
a_{1}\int_{0}^{w(t)}f(s)\mathrm{d}s=a_{1}F(w(t)).$ (3.30)

Assumption $f>0$ on $(0,\eta_{0})$ implies that $F>0$ on $(0,\eta_{0})$.
Having chosen $T_{1}\in(t_{0},T]$ in such a way that
$a_{1}F(w(T_{1}))<K_{\infty}$, we apply $K^{-1}$, rearrange and integrate to
obtain

$\int_{t_{0}}^{t}\frac{w^{\prime}(s)\mathrm{d}s}{K^{-1}\big{(}a_{1}F(w(s))\big{)}}\leq(t-t_{0})\qquad\forall\,t\in(t_{0},T_{1}].$
(3.31)

Changing variables,

$\int_{0}^{w(t)}\frac{\mathrm{d}\tau}{K^{-1}(a_{1}F(\tau))}\leq(t-t_{0})\qquad\forall\,t\in(t_{0},T_{1}].$
(3.32)

By Lemma 3.6 property (KO0) holds, as claimed.

We are left to consider the case $\wp^{\prime}\leq 0$. Then, by (3.29) we
deduce that $w^{\prime\prime}\geq 0$ on $(t_{0},T)$, hence $w^{\prime}$ is
increasing there. Integrating (3.29) on $(t_{0},t)$ and using the
$C$-monotonicity of $f,l$ together with $w^{\prime}(t_{0})=0$, we get

$\begin{array}[]{lcl}\displaystyle\varphi\big{(}w^{\prime}(t)\big{)}&=&\displaystyle\frac{1}{\wp(t)}\int_{t_{0}}^{t}\wp(s)a(s)f(w(s))l(w^{\prime}(s))\mathrm{d}s\\\\[11.38092pt]
&\leq&\displaystyle
C^{2}a_{1}f(w(t))l(w^{\prime}(t))\left[\frac{1}{\wp(t)}\int_{0}^{t}\wp(s)\mathrm{d}s\right].\end{array}$
(3.33)

Now, consider the energy
$E(t)=K\big{(}w^{\prime}(t)\big{)}-a_{1}F\big{(}w(t)\big{)}$. Differentiating
and using (3.33) and the definition of $a_{1}$, we obtain

$\begin{array}[]{lcl}E^{\prime}(t)&=&\displaystyle\frac{w^{\prime}\varphi^{\prime}(w^{\prime})w^{\prime\prime}}{l(w^{\prime})}-a_{1}f(w)w^{\prime}=\frac{w^{\prime}}{l(w^{\prime})}\left[-\frac{\wp^{\prime}}{\wp}\varphi(w^{\prime})+af(w)l(w^{\prime})-a_{1}f(w)l(w^{\prime})\right]\\\\[11.38092pt]
&\leq&\displaystyle-\frac{\wp^{\prime}}{\wp}\frac{w^{\prime}\varphi(w^{\prime})}{l(w^{\prime})}\leq\left|\frac{\wp^{\prime}(t)}{\wp(t)^{2}}\int_{0}^{t}\wp(s)\mathrm{d}s\right|C^{2}a_{1}f(w)w^{\prime}\\\\[11.38092pt]
&\leq&\displaystyle\frac{\wp_{1}T_{0}\|\wp^{\prime}\|_{L^{\infty}([0,T_{0}])}}{\wp_{0}^{2}}C^{2}a_{1}f(w)w^{\prime}=\bar{c}f(w)w^{\prime}.\end{array}$

Integrating and using $w(t_{0})=w^{\prime}(t_{0})=0$,

$K\big{(}w^{\prime}(t)\big{)}\leq
a_{1}F\big{(}w(t)\big{)}+\bar{c}F\big{(}w(t)\big{)}=(a_{1}+\bar{c})F\big{(}w(t)\big{)}.$

Having obtained again an inequality like (3.30), to achieve the desired
contradiction it is sufficient to repeat verbatim the last steps of the proof
for $\wp^{\prime}\geq 0$. ∎

######  Remark 3.8.

_When $\wp^{\prime}\geq 0$, to reach the desired conclusion in $(ii)$ we do
not use the assumption that $l$ is $C$-increasing. _

Proposition 3.7 has a converse, at least if the threshold $\eta$ in (3.5) is
sufficiently small, namely (KO0) implies that $w^{\prime}(0)=0$. To reach the
goal, the idea is to compare $w$ with an explicit supersolution of (3.5),
whose construction generalizes the End Point Lemma in [123, Lem. 4.4.1].

######  Proposition 3.9.

Assume (3.2), (3.3) and (3.4). Suppose that

  * $f(0)l(0)=0$;

  * $f$ is $C$-increasing on $(0,\eta_{0})$, for $\eta_{0}>0$ as in (3.2);

  * $l$ is $C$-increasing on $(0,\xi_{0})$, for some $\xi_{0}>0$.

If $f>0$ on $(0,\eta_{0})$ and

$\frac{1}{K^{-1}\circ F}\in L^{1}(0^{+}),$ (KO0)

then there exists $\eta_{1}$ sufficiently small that, for each
$\eta\in(0,\eta_{1})$, the solution $w$ of problem (3.5) constructed in
Theorem 3.1 satisfies

$w^{\prime}(0)=0.$ (3.34)

###### Proof.

For $\sigma\in(0,1)$ to be determined, using (KO0) and Lemma 3.6 we implicitly
define $z(t)$ by setting

$t=\int_{0}^{z(t)}\frac{\mathrm{d}s}{K^{-1}(\sigma F(s))}\qquad\text{for
}\,t\in[0,T).$

Note that $z$ is positive on $(0,T)$. Differentiating,

$z^{\prime}=K^{-1}(\sigma F(z)\big{)},$ (3.35)

whence $z^{\prime}>0$ on $(0,T)$ and $z^{\prime}(0)=0$. Evaluating $K$ on both
sides of (3.35) and differentiating once more we get

$\frac{z^{\prime}\varphi^{\prime}(z^{\prime})z^{\prime\prime}}{l(z^{\prime})}=\sigma
f(z)z^{\prime}\qquad\text{on }\,(0,T).$

Since $\varphi^{\prime},l,z^{\prime}$ are positive, we deduce
$z^{\prime\prime}\geq 0$, and simplifying by $z^{\prime}$ we infer

$\big{(}\varphi(z^{\prime})\big{)}^{\prime}=\sigma f(z)l(z^{\prime}).$

Integrating, using $z^{\prime}(0)=0$ together with the $C$-increasing property
of $f$ and $l$ (note that $z^{\prime}$ is increasing) we obtain

$\varphi\big{(}z^{\prime}(t)\big{)}=\displaystyle\sigma\int_{0}^{t}f\big{(}z(s)\big{)}l\big{(}z^{\prime}(s)\big{)}\mathrm{d}s\leq\sigma
C^{2}Tf\big{(}z(t)\big{)}l\big{(}z^{\prime}(t)\big{)}.$

Summarizing, so far we have obtained,

$\begin{array}[]{lcl}\displaystyle\big{[}\wp\varphi(z^{\prime})\big{]}^{\prime}=\wp\big{(}\varphi(z^{\prime})\big{)}^{\prime}+\wp^{\prime}\varphi(z^{\prime})&\leq&\displaystyle\frac{\sigma}{a_{0}}\left[1+C^{2}T\frac{\|\wp^{\prime}\|_{L^{\infty}([0,T_{0}])}}{\wp_{0}}\right]\wp
af(z)l(z^{\prime})\\\\[11.38092pt]
\displaystyle&=&\displaystyle\frac{1}{2C}\wp af(z)l(z^{\prime}),\end{array}$

where we have defined $\sigma$ in order to satisfy the last equality. Next, we
fix $\eta_{1}\leq z(T)$ small enough in such a way that each
$\eta\in(0,\eta_{1})$ meets the requirements in Theorem 3.1, to guarantee the
existence of $w$. We claim that, for $\eta\in(0,\eta_{1})$, the solution $w$
in Theorem 3.1 satisfies $w\leq z$ on $(0,T)$. This, together with the already
established $z^{\prime}(0)=0$, forces $w^{\prime}(0)=0$ and concludes our
proof. By contradiction, suppose that $c=\max_{[0,T]}(w-z)>0$ and let
$\Gamma=\\{w-z=c\\}$. By construction, $\Gamma\Subset(0,T)$, and since $w,z$
are $C^{1}$ we deduce $w^{\prime}=z^{\prime}$ on $\Gamma$. By continuity and
since $w^{\prime}>0$ on $(0,T)$, we can choose $\delta\in(0,c)$ close enough
to $c$ in such a way that $l(z^{\prime})\leq 2l(w^{\prime})$ on the set
$I_{\delta}=\\{w-z>\delta\\}$. On $I_{\delta}$, using the $C$-increasing
property we therefore have

$\big{[}\wp\varphi(w^{\prime})\big{]}^{\prime}\geq a\wp
f(w)l(w^{\prime})\geq\frac{1}{2C}a\wp
f(z)l(z^{\prime})\geq\big{[}\wp\varphi(z^{\prime})\big{]}^{\prime}=\big{[}\wp\varphi((z+c)^{\prime})\big{]}^{\prime}$

and $w=z+c$ on $\partial I_{\delta}$. By standard comparison (one can apply,
for instance Proposition 4.1 below to an appropriate radial model), $w\leq
z+c$ on $U_{\delta}$, contradiction. ∎

###  3.2 The mixed Dirichlet-Neumann problem

We next move to investigate the problem

$\begin{cases}\big{[}\wp\,\varphi(w^{\prime})\big{]}^{\prime}=\wp
af(w)l(|w^{\prime}|)\qquad\text{on }\,(0,T),\\\\[5.69046pt]
w^{\prime}(0)=0,\qquad w(T)=\eta,\\\\[5.69046pt] 0\leq w\leq\eta,\qquad
w^{\prime}\geq 0\ \text{ on }\,(0,T),\end{cases}$ (3.36)

for given $\eta>0$, $T\in(0,T_{0})$. We here extend and generalize in several
directions the core of Corollary 1.4 of [57], _without requiring any
monotonicity on $l$_, as well as the results of Section 4 of [21]. We assume
(3.2) and (3.4), and we define $a_{0},a_{1},\wp_{0},\wp_{1},f_{\eta},l_{\xi}$
and $\Theta(T)$ as in (3.6).

######  Theorem 3.10.

Assume (3.2) and (3.4), and that

$\begin{array}[]{l}f>0\quad\text{on }\,\mathbb{R}^{+},\qquad
f(0)=0;\\\\[5.69046pt] l>0\quad\text{on }\,\mathbb{R}^{+}_{0}.\\\\[5.69046pt]
\end{array}$ (3.37)

Then, for each $\eta,\xi>0$ and $T\in(0,T_{0})$ satisfying

$\Theta(T)f_{\eta}l_{\xi}<\varphi(\xi),$ (3.38)

the problem

$\begin{cases}[\wp\,\varphi(w^{\prime})]^{\prime}=\wp
af(w)l(|w^{\prime}|)\qquad\text{on }\,(0,T)\\\\[5.69046pt]
w^{\prime}(0)=0,\qquad w(T)=\eta,\\\\[5.69046pt] 0\leq w\leq\eta,\qquad 0\leq
w^{\prime}<\xi\ \text{ on }\,(0,T),\end{cases}$ (3.39)

admits solution $w\in C^{1}([0,T])$, and there exists $t_{0}\in[0,T)$ such
that

$w(t)\equiv w(t_{0})\geq 0\ \ \text{ on }\,[0,t_{0}],\qquad w^{\prime}>0\ \
\text{ on }\,(t_{0},T].$ (3.40)

Moreover, if

$\varphi\in C^{1}(\mathbb{R}^{+}),\qquad\varphi^{\prime}>0\quad\text{ on
}\,\mathbb{R}^{+},$

then $w\in C^{2}\big{(}(t_{0},T]\big{)}$ and satisfies

$\frac{\varphi^{\prime}(w^{\prime})}{l(w^{\prime})}\,w^{\prime\prime}=a\,f(w)-\frac{\wp^{\prime}}{\wp}\cdot\frac{\varphi(w^{\prime})}{l(w^{\prime})}\qquad\text{on
}\,(t_{0},T).$ (3.41)

All of the above conclusions still hold if condition $\wp>0$ on $[0,T]$, in
(3.4), is replaced by

$\begin{array}[]{l}\displaystyle\wp>0\quad\text{on
}\,(0,T],\qquad\wp(0)=0\\\\[5.69046pt] \wp^{\prime}\geq 0\quad\text{on
}\,[0,\delta),\ \,\text{ for some }\,\delta>0.\end{array}$ (3.42)

######  Remark 3.11.

_Differently from the Dirichlet problem, if we allow $l$ to vanish at $t=0$ in
the Neumann case we cannot guarantee that the solution of (3.39) be non-
constant. This motivates the necessity to require $l>0$ on
$\mathbb{R}^{+}_{0}$. _

###### Proof.

The strategy goes along the same lines as that for the Dirichlet problem.
First, we redefine $f$ outside of $[0,\eta]$ and $l$ outside of $[0,\xi]$ in
such a way that

$\begin{array}[]{l}\displaystyle f\in C(\mathbb{R}),\qquad\displaystyle 0\leq
f(s)\leq f_{\eta}\quad\text{ for }\,s\geq\eta,\qquad f(s)=0\quad\text{ for
}\,s<0,\\\\[5.69046pt] l\in C(\mathbb{R}^{+}_{0}),\qquad 0<l(s)\leq
l_{\xi}\quad\text{ for }\,s\geq\xi\end{array}$ (3.43)

This will not affect the conclusion of the proposition, since any ultimate
solution $w$ of (3.39), with $w\geq 0$, $w^{\prime}\geq 0$ in $[0,T]$,
satisfies $0\leq w\leq\eta$ and $|w^{\prime}|<\xi$.

Denote with $X$ the Banach space $X=C^{1}\big{(}[0,T]\big{)}$, endowed with
the usual norm $\|w\|=\|w\|_{\infty}+\|w^{\prime}\|_{\infty}$. Define the
homotopy $\mathcal{H}:X\times[0,1]\to X$ by

${\mathcal{H}}[w,\sigma](t)=\sigma\eta-\int^{T}_{t}\varphi^{-1}\left(\frac{\sigma}{\wp(s)}\int^{s}_{0}\wp(\tau)a(\tau)f(w(\tau))l(|w^{\prime}(\tau)|)\mathrm{d}\tau\right)\mathrm{d}s.$
(3.44)

We claim that $\mathcal{H}$ is well defined and valued in $X$. Indeed, in our
assumptions

$0\leq\frac{\sigma}{\wp(t)}\int^{t}_{0}\wp(\tau)a(\tau)f(w(\tau))l(|w^{\prime}(\tau)|)\mathrm{d}\tau\leq\Theta(T)f_{\eta}l_{\xi}<\varphi(\xi),$
(3.45)

hence the term in round brackets in (3.44) lies in the domain of
$\varphi^{-1}$ and

$\mathcal{H}[w,\sigma]^{\prime}(t)=\varphi^{-1}\left(\frac{\sigma}{\wp(t)}\int^{t}_{0}\wp(\tau)a(\tau)f(w(\tau))l(|w^{\prime}(\tau)|)\mathrm{d}\tau\right)\in[0,\xi).$
(3.46)

Furthermore, $\mathcal{H}[w,\sigma]^{\prime}$ is continuous on $[0,T]$, hence
$\mathcal{H}$ is valued in $X$. By construction,
$\mathcal{H}[w,\sigma](T)=\sigma\eta$ and $\mathcal{H}[w,0]=0$. From

$0\leq\frac{1}{\wp(t)}\int^{t}_{0}\wp(\tau)a(\tau)f(w(\tau))l(|w^{\prime}(\tau)|)\mathrm{d}\tau\leq
f_{\eta}l_{\xi}\frac{1}{\wp(t)}\int^{t}_{0}\wp(\tau)a(\tau)\mathrm{d}\tau,$

we deduce that $\mathcal{H}[w,\sigma]^{\prime}(0)=0$. Fix $\sigma\in(0,1]$,
and let $w$ be a solution of $w=\mathcal{H}[w,\sigma]$. We claim that
$w(0)\geq 0$: otherwise, since $w(T)=\sigma\eta>0$ there would exist a first
point $t_{1}\in(0,T)$ such that $w<0$ on $[0,t_{1})$ and $w(t_{1})=0$, and
therefore $f(w(t))=0$ on $[0,t_{1}]$. Thus, $w^{\prime}\equiv 0$ on
$[0,t_{1}]$ by (3.46), that is, $w$ would be constant on $[0,t_{1})$,
contradicting $w(0)<0=w(t_{1})$. From $w\geq 0$ we also deduce $w^{\prime}\geq
0$ on $[0,T]$ by (3.46). Also, (3.46) implies that $\varphi(w^{\prime})$ is of
class $C^{1}\big{(}[0,T]\big{)}$, and then from (3.44) that $w$ is a classical
weak solution of the problem

$\begin{cases}[\wp\varphi(w^{\prime})]^{\prime}=\sigma\wp
af(w)l(|w^{\prime}|)\qquad\text{on }\,(0,T),\\\\[5.69046pt]
w^{\prime}(0)=0,\quad w(T)=\sigma\eta\\\\[5.69046pt] 0\leq
w\leq\sigma\eta,\qquad 0\leq w^{\prime}<\xi\qquad\text{on
}\,(0,T).\end{cases}$ (3.47)

In particular, for $\sigma=1$, $w$ is the desired solution of (3.39). To prove
(3.40), let $\sigma=1$. From $w(T)=\eta$ and $w^{\prime}\geq 0$ we infer the
existence of a minimal $t_{0}\in(0,T)$ such that $w>0$ on $(t_{0},T]$. Since
$f>0$ on $\mathbb{R}^{+}$, $l>0$ on $\mathbb{R}^{+}_{0}$ and $a>0$ on $[0,T]$,
a solution of $w=\mathcal{H}[w,1]$ satisfies

$w^{\prime}(t)=\mathcal{H}[w,1]^{\prime}(t)=\varphi^{-1}\left(\frac{1}{\wp(t)}\int^{t}_{0}\wp(\tau)a(\tau)f(w(\tau))l(|w^{\prime}(\tau)|)\mathrm{d}\tau\right)>0\qquad\forall\,t\in(t_{0},T].$

If $t_{0}\neq 0$, by the monotonicity and non-negativity of $w$ we get $w=0$
on $[0,t_{0}]$. To show (3.41), using $\varphi(w^{\prime})\in C^{1}([0,T])$,
$\varphi^{\prime}>0$ on $\mathbb{R}^{+}$ and $w^{\prime}>0$ on $(t_{0},T]$ in
(3.46) we deduce $w^{\prime}\in C^{1}((t_{0},T])$. Identity (3.41) immediately
follows by expanding the derivative in (3.39).

We assert that a solution of $w=\mathcal{H}[w,1]$ exists, using again the
Browder version of the Leray-Schauder theorem (see [58, Thm 11.6]).

To begin with, as already observed $\mathcal{H}[w,0]\equiv 0$ for all $w\in
X$. We next show that $\mathcal{H}$ is continuous on $X\times[0,1]$. Indeed,
consider a sequence $\\{(w_{k},\sigma_{k}\\}\subset X\times[0,1]$, with
$w_{k}\to w$ in $X$ and $\sigma_{k}\to\sigma$ as $k\rightarrow\infty$. By
continuity, $\sigma_{k}f(w_{k})l(|w_{k}^{\prime}|)\to\sigma
f(w)l(|w^{\prime}|)$, and so
$\mathcal{H}[w_{k},\sigma_{k}]\to\mathcal{H}[w,\sigma]$ by (3.44) and Lebesgue
convergence theorem, as required. Next we show that $\mathcal{H}$ is compact.
To this aim, let $\\{(w_{k},\sigma_{k})\\}$ be a bounded sequence in
$X\times[0,1]$. From (3.46),

$\|\mathcal{H}[w_{k},\sigma_{k}]^{\prime}\|_{\infty}<\xi,$ (3.48)

and thus, since $\mathcal{H}[w_{k},\sigma_{k}](T)=\sigma\eta\in[0,\eta]$,
$\\{\mathcal{H}[w_{k},\sigma_{k}]\\}$ is equi-bounded in $X$. We shall prove
that $\\{\mathcal{H}[w_{k},\sigma_{k}]^{\prime}\\}$ is equicontinuous. Set

$\begin{array}[]{l}I_{k}(t)=\displaystyle\int_{0}^{t}\wp(\tau)a(\tau)f(w_{k}(\tau))l(|w^{\prime}(\tau)|)\mathrm{d}\tau,\\\\[11.38092pt]
x_{k}=\displaystyle\frac{\sigma_{k}I_{k}(t)}{\wp(t)},\qquad
y_{k}=\frac{\sigma_{k}I_{k}(s)}{\wp(s)},\end{array}$ (3.49)

and note that

$\big{|}\mathcal{H}[w_{k},\sigma_{k}]^{\prime}(t)-\mathcal{H}[w_{k},\sigma_{k}]^{\prime}(s)\big{|}=\big{|}\varphi^{-1}(x_{k})-\varphi^{-1}(y_{k})\big{|},$
(3.50)

and that, by (3.45),

$0\leq x_{k},y_{k}<\varphi(\xi),\qquad|I_{k}(s)-I_{k}(t)|\leq
c|s-t|\quad\text{with }\,c=\wp_{1}a_{1}f_{\eta}l_{\xi}.$ (3.51)

Since $\wp>0$ on $[0,T]$ and there it is $C^{1}$, we deduce

$\begin{array}[]{lcl}\displaystyle|x_{k}-y_{k}|&=&\displaystyle\sigma_{k}\left|\frac{\wp(t)I_{k}(s)-\wp(s)I_{k}(t)}{\wp(t)\wp(s)}\right|\leq\left|\frac{\wp(t)I_{k}(s)-\wp(s)I_{k}(t)}{\wp(t)\wp(s)}\right|\\\\[14.22636pt]
&\leq&\displaystyle\frac{|\wp(t)-\wp(s)|}{\wp(s)\wp(t)}I_{k}(s)+\frac{|I_{k}(s)-I_{k}(t)|}{\wp(t)}\\\\[14.22636pt]
&\leq&\displaystyle\left[\frac{cT}{\wp_{0}^{2}}\left(\max_{[0,T]}\wp^{\prime}\right)+\frac{c}{\wp_{0}}\right]|s-t|=\kappa|s-t|.\end{array}$
(3.52)

Given $\varepsilon>0$, let $\varrho=\varrho(\varphi^{-1},\varepsilon)>0$ be
given by the uniform continuity of $\varphi^{-1}$ on $[0,\varphi(\xi)]$. If
$|s-t|<\varrho/\kappa$, then $|x_{k}-y_{k}|<\varrho$ and thus, by (3.50),

$\big{|}\mathcal{H}[w_{k},\sigma_{k}]^{\prime}(t)-\mathcal{H}[w_{k},\sigma_{k}]^{\prime}(s)\big{|}=\big{|}\varphi^{-1}(x_{k})-\varphi^{-1}(y_{k})\big{|}<\varepsilon,$
(3.53)

proving the (uniform) equicontinuity of
$\\{\mathcal{H}[w_{k},\sigma_{k}]^{\prime}\\}$. The compactness of
$\mathcal{H}$ then follows from Ascoli-Arzelá theorem.

To apply the Leray-Schauder theorem it remains to check the existence of a
constant $L>0$ such that $\|w\|\leq L$ for each solution of
$\mathcal{H}[w,\sigma]=w$. But this immediately follows from properties
(3.47), and indeed $\|w\|\leq\eta+\xi$.

It remains to consider the case when $\wp>0$ on $[0,T]$ is replaced by (3.42).
From (3.46) and the monotonicity of $\wp$ on $[0,\delta)$, we deduce that for
$t\in(0,\delta)$

$\begin{array}[]{lcl}\mathcal{H}[w,\sigma]^{\prime}(t)&=&\displaystyle\varphi^{-1}\left(\frac{\sigma}{\wp(t)}\int^{t}_{0}\wp(\tau)a(\tau)f(w(\tau))l(|w^{\prime}(\tau)|)\mathrm{d}\tau\right)\\\\[14.22636pt]
&\leq&\displaystyle\varphi^{-1}\left(\frac{f_{\eta}l_{\xi}a_{1}}{\wp(t)}\int^{t}_{0}\wp(\tau)\mathrm{d}\tau\right)\leq\varphi^{-1}\left(f_{\eta}l_{\xi}a_{1}t\right).\end{array}$

Hence, $\mathcal{H}[w,\sigma]^{\prime}$ is continuous up to $t=0$ (i.e.
$\mathcal{H}$ is valued in $X$) and $w^{\prime}(0)=0$ for each solution of
$w=\mathcal{H}[w,\sigma]$. The rest of the proof follows verbatim, except for
the equicontinuity of $\\{\mathcal{H}[w_{k},\sigma_{k}]^{\prime}\\}$ that we
now consider. By the monotonicity of $\wp$,

$|x_{k}|\leq\frac{|I_{k}(t)|}{\wp(t)}\leq a_{1}f_{\eta}l_{\xi}t,$

independently of $k$. Thus, given $\varepsilon>0$ and the corresponding
$\varrho=\varrho(\varphi^{-1},\varepsilon)$ of the uniform continuity of
$\varphi^{-1}$ on $[0,\varphi(\xi)]$, we can choose $\vartheta\in(0,T/2)$
independent of $k$ such that $|x_{k}|<\varrho/2$ if $t<2\vartheta$ and
$|y_{k}|<\varrho/2$ if $s<2\vartheta$. Set

$\hat{\wp}_{0}=\inf_{[\vartheta,T]}\wp>0,\qquad\hat{\kappa}=\frac{cT}{\hat{\wp}_{0}^{2}}\left(\max_{[0,T]}\wp^{\prime}\right)+\frac{c}{\hat{\wp}_{0}},$

with $c$ as in (3.51). Define

$\bar{\vartheta}=\min\left\\{\frac{\varrho}{\hat{\kappa}},\vartheta\right\\}.$

Let $s,t\in[0,T]$ with $|s-t|<\bar{\vartheta}$. If $s,t\geq\vartheta$, then
the chain of inequalities (3.52) holds verbatim with
$\hat{\kappa},\hat{\wp}_{0}$ replacing $\kappa,\wp_{0}$, respectively, and we
deduce

$|x_{k}-y_{k}|\leq\hat{\kappa}|s-t|<\varrho.$

On the other hand, if one between $s,t$, say $t$, is less than
$\bar{\vartheta}$, then from $\bar{\vartheta}\leq\vartheta$ we deduce
$s<t+|s-t|<2\vartheta$. Hence, $|x_{k}|<\varrho/2$ and $|y_{k}|<\varrho/2$,
and thus

$|x_{k}-y_{k}|\leq|x_{k}|+|y_{k}|<\varrho.$

In both the cases, $|x_{k}-y_{k}|<\varrho$ and therefore (3.53) holds, proving
the (uniform) equicontinuity of
$\\{\mathcal{H}[w_{k},\sigma_{k}]^{\prime}\\}$. ∎

The next result relates condition $w(0)=0$ to (KO0). In order to do so, we
shall further require (3.3) in order to define $K$, $F$ as in (1.19), (1.20).

######  Proposition 3.12.

In the assumptions of Theorem 3.10, suppose further (3.3) and that

$\begin{array}[]{l}\wp^{\prime}\geq 0\qquad\text{on }\,(0,T),\\\\[5.69046pt]
\text{$f$ is $C$-increasing on $(0,\eta_{0})$, for some constant
$\eta_{0}>0$.}\end{array}$

If $w(0)=0$, then (KO0) holds.

###### Proof.

Because of the monotonicity of $\wp^{\prime}$ and $w$, from (3.39) we deduce

$\big{(}\varphi(w^{\prime})\big{)}^{\prime}\leq af(w)l(w^{\prime})\leq
a_{1}f(w)l(w^{\prime})\qquad\text{for }\,t\in(0,T).$

We now follow the steps in Proposition 3.7: differentiating on $(t_{0},T)$, we
deduce

$K^{\prime}(w^{\prime})w^{\prime\prime}\leq a_{1}f(w)w^{\prime},$

and integrating on $(t_{0},t)$ with the aid of $w(t_{0})=w^{\prime}(t_{0})=0$
we infer $K(w^{\prime})\leq a_{1}F(w)$. Let $t_{1}\in(t_{0},t)$ be such that
$a_{1}F(w)<K_{\infty}$ for $t\in(t_{0},t_{1})$. This is possible, by
continuity, since $F(w(t_{0}))=0$. Applying $K^{-1}$, integrating and changing
variables we get

$\int_{0}^{w(t)}\frac{\mathrm{d}s}{K^{-1}(a_{1}F(s))}\leq(t-t_{0})\qquad\text{on
}\,(t_{0},t_{1}),$

and (KO0) follows from Lemma 3.6. ∎

We next investigate the maximal interval of definition of $w$. Assume the
validity of (3.2) and (3.37). To tie $w$ with (KO∞), we assume (3.3) and

$\frac{t\varphi^{\prime}(t)}{l(t)}\not\in L^{1}(\infty),$ (3.54)

in order for $K$ to be a homeomorphism of $\mathbb{R}^{+}_{0}$ onto itself. We
further replace (3.4) and (3.42) with

$\left\\{\begin{array}[]{ll}\displaystyle\wp\in
C^{1}(\mathbb{R}^{+}_{0}),&\qquad\wp>0,\ \ \wp^{\prime}\geq 0\quad\text{on
}\mathbb{R}^{+},\\\\[2.84544pt] a\in C(\mathbb{R}^{+}_{0}),&\qquad
a>0\quad\text{on }\,\mathbb{R}^{+}_{0}.\end{array}\right.$ (3.55)

Fix $T>0$. Applying Theorem 3.10 we infer the existence of $w$ solving (3.39)
for each $\eta>0$ sufficiently small (inequality (3.38) is always satisfied
for small $\eta$ since, by (3.37), $f_{\eta}\rightarrow 0$ as $\eta\rightarrow
0$). From $w^{\prime}(T)>0$, we conclude that $w$ can be extended on a maximal
interval $[0,R)$. Our next task is to prove that, if the Keller-Osserman
condition ($\mathrm{KO}_{\infty}$) is violated, then $R=\infty$.

######  Proposition 3.13.

Assume (3.2), (3.3), (3.37) and (3.54). Let $a,\wp$ satisfy (3.55). For a
fixed $T>0$, consider the solution $w$ of (3.39) for small positive $\eta$ and
let $[0,R)$ be the maximal interval where $w$ is defined. If $t_{0}$ is as in
Theorem 3.10, then

$w=w(t_{0})\quad\text{on }\,[0,t_{0}],\qquad w>0,\quad
w^{\prime}>0\quad\text{on }\,(t_{0},R).$ (3.56)

Furthermore, suppose that $f$ is $C$-increasing on $(\bar{\eta}_{0},\infty)$
for some $\bar{\eta}_{0}\geq 0$. If

$\frac{1}{K^{-1}\circ F}\not\in L^{1}(\infty),$ ($\neg$KO∞)

then $R=\infty$.

###### Proof.

Taking into account the sign of $f,l,a$, by (3.39) $\wp\varphi(w^{\prime})$ is
$C^{1}$ and strictly increasing where $w$ is positive, and from (3.40) it
readily follows that $\wp\varphi(w^{\prime})>0$ on $(t_{0},R)$. Properties
(3.56) are then immediate from (3.40), $w^{\prime}(T)>0$ and our assumptions
on $\varphi$. Next, suppose by contradiction that $R<\infty$. We first claim
that necessarily

$w^{*}=\lim_{t\to R^{-}}w(t)=\infty,$ (3.57)

where the existence of the limit is guaranteed by the monotonicity of $w$. To
prove (3.57), assume by contradiction that $w^{*}<\infty$. Because of (3.41)
and since $\wp^{\prime}\geq 0$,

$\frac{\varphi^{\prime}(w^{\prime})w^{\prime\prime}}{l(w^{\prime})}\leq
af(w)\leq a_{1}f(w)\qquad\text{on }\,(t_{0},R),$

where we set $a_{1}=\|a\|_{L^{\infty}([0,R])}$. Multiplying by $w^{\prime}$,
integrating on $(t_{0},t)$ and changing variables we deduce $K(w^{\prime})\leq
a_{1}F(w)$ (we recall that $w^{\prime}(t_{0})=0$ by (3.39)). Thus,
$w^{\prime}$ is bounded in $(R/2,R)$, namely $\|w^{\prime}\|_{\infty}\leq
K^{-1}(a_{1}F(w^{*}))=L$. For $t,s\in(R/2,R)$, define $I_{k},x_{k},y_{k}$ as
in (3.49), and note that

$|x_{k}|+|y_{k}|\leq 2a_{1}f_{w^{*}}l_{L}R=\bar{C}$

Given $\varepsilon>0$ let $\varrho=\varrho(\varphi^{-1},\varepsilon)$ be given
by the uniform continuity of $\varphi^{-1}$ on $[0,\bar{C}]$. Proceeding as in
(3.52) we deduce the existence of $\kappa>0$ such that

$|x_{k}-y_{k}|\leq\kappa|s-t|\qquad\text{for each
}\,s,t\in\left(\frac{R}{2},R\right).$

If $|s-t|<\varrho/\kappa$, then $|x_{k}-y_{k}|<\varrho$ and so

$|w^{\prime}(t)-w^{\prime}(s)|=\big{|}\varphi^{-1}(x_{k})-\varphi^{-1}(y_{k})\big{|}<\varepsilon.$

In conclusion, $w^{\prime}$ is uniformly continuous on $(R/2,R)$, and can be
therefore extended by continuity at $t=R$. By the existence theory for ODEs,
$w$ would be further extendible beyond $R$, contradiction. This proves that
$w(R^{-})=\infty$. Now, fix $T_{1}\in(0,R)$ large enough that
$w(T_{1})>\bar{\eta}_{0}$. Applying $K^{-1}$ to inequality $K(w^{\prime})\leq
a_{1}F(w)$ on $(T_{1},R)$, rearranging, integrating on $[T_{1},t)$ and
changing variables we get

$\int_{w(T_{1})}^{w(t)}\frac{\mathrm{d}s}{K^{-1}(a_{1}F(s))}\leq t-T_{1}.$
(3.58)

Since ($\neg$KO∞) holds and, by assumption, $f$ is $C$-increasing on
$(\bar{\eta}_{0},\infty)$, applying Lemma 3.6 we deduce that the left-hand
side of (3.58) is unbounded as $t\rightarrow R^{-}$ while the right-hand side
is not, contradiction. ∎

##  4 Comparison results and the finite maximum principle

###  4.1 Basic comparisons and a pasting lemma

In this subsection, we collect two comparison theorems and a “pasting lemma”
for $\mathrm{Lip}_{\mathrm{loc}}$ solutions that will be repeatedly used in
the sequel. Throughout the section we assume

$\varphi\in C(\mathbb{R}^{+}_{0}),\qquad\varphi(0)=0,\qquad\varphi>0\ \text{
on }\,\mathbb{R}^{+}.\\\ $ (4.1)

The first comparison is Proposition 6.1 of [110], see also Theorem 2.4.1 of
[123].

######  Proposition 4.1.

Assume (4.1) and that $\varphi$ is strictly increasing on $\mathbb{R}^{+}$.
Let $\Omega\subset M$ be open, and suppose that $u$,
$v\in\mathrm{Lip}_{\mathrm{loc}}(\Omega)\cap C(\overline{\Omega})$ solve

$\left\\{\begin{array}[]{l}\Delta_{\varphi}u\geq\Delta_{\varphi}v\qquad\text{weakly
in }\Omega,\\\\[8.5359pt] \displaystyle u\leq v\quad\text{on
}\partial\Omega,\end{array}\right.$ (4.2)

and
$\displaystyle\limsup_{x\in\Omega,\,x\rightarrow\infty}\big{(}u(x)-v(x)\big{)}\leq
0$ if $\Omega$ has non-compact closure. Then $u\leq v$ in $\Omega$.

Our second comparison result is a special case of [7, Thm. 5.6], see also
[123, Thm. 3.6.5].

######  Proposition 4.2.

Let $\varphi,f,l$ satisfy

$\begin{array}[]{l}\varphi\in C(\mathbb{R}^{+}_{0})\cap
C^{1}(\mathbb{R}^{+}),\qquad\varphi(0)=0,\qquad\varphi^{\prime}>0\quad\text{on
}\,\mathbb{R}^{+}\\\\[5.69046pt] f\in C(\mathbb{R}),\qquad\text{$f$ is non-
decreasing on $\mathbb{R}$,}\\\\[5.69046pt] l\in
C(\mathbb{R}^{+}_{0})\cap\mathrm{Lip}_{\mathrm{loc}}(\mathbb{R}^{+}),\qquad
l>0\ \text{ on }\,\mathbb{R}^{+}.\end{array}$

Let $\Omega\subset M$ be an open subset of a Riemannian manifold $M$, and fix
$0<b(x)\in C(\overline{\Omega})$. If
$u,v\in\mathrm{Lip}_{\mathrm{loc}}(\Omega)\cap C(\overline{\Omega})$ solve

$\left\\{\begin{array}[]{l}\Delta_{\varphi}u\geq b(x)f(u)l(|\nabla
u|)\qquad\text{on }\,\Omega,\\\\[5.69046pt] \Delta_{\varphi}v\leq
b(x)f(v)l(|\nabla v|)\qquad\text{on }\,\Omega,\\\\[5.69046pt] u\leq
v\qquad\text{on }\,\partial\Omega\\\\[5.69046pt]
\mathrm{ess}\inf_{K}\Big{\\{}|\nabla v|+|\nabla u|\Big{\\}}>0\qquad\text{for
each }\,K\Subset\Omega.\end{array}\right.$ (4.3)

Then, $u\leq v$ on $\Omega$.

######  Remark 4.3.

_Condition $\mathrm{ess}\inf_{K}\big{\\{}|\nabla v|+|\nabla u|\big{\\}}>0$ for
each $K\Subset\Omega$ cannot be avoided, as the counterexample in Remark 1, p.
79 of [123] shows. However, the restriction can be removed if
$\Delta_{\varphi}$ is strictly elliptic, see Section 3.5 of [123] for
definitions and relevant results. _

######  Remark 4.4.

_The underlying metric is not required to be smooth, and indeed a metric whose
local matrix $g_{ij}$ is continuous is sufficient. _

######  Remark 4.5.

_It is worth to comment on[7, Thm. 5.6]. There, the authors consider solutions
of more general quasilinear inequalities of the form_

$\mathrm{div}\mathbf{A}(x,\nabla u)\geq\mathcal{B}(x,u,\nabla
u)\qquad\text{and}\qquad\mathrm{div}\mathbf{A}(x,\nabla
v)\leq\mathcal{B}(x,v,\nabla v),$

_for suitable Caratheódory maps $\mathbf{A},\mathcal{B}$. In our setting,_

$\mathbf{A}(x,\xi)=\frac{\varphi(|\xi|)}{|\xi|}\xi,$

_thus the positivity of $\varphi(s)/s$ and $\varphi^{\prime}(s)$ on
$\mathbb{R}^{+}$ imply that the tangent map $\mathbf{A}_{*}$ of $\mathbf{A}$
is uniformly positive definite on compacta of fibers of
$T\Omega\backslash\\{\mathbf{0}\\}\rightarrow\Omega$, a condition needed to
apply Lemma 5.7 in [7]. The regularity of $\mathcal{B}$, defined at the end of
p.592 therein, is equivalent to condition
$l\in\mathrm{Lip}_{\mathrm{loc}}(\mathbb{R}^{+})$. _

To conclude, we discuss the pasting lemma. It is well-known that the maximum
of two subharmonic functions is still subharmonic. For subsolutions of more
general operators the situation is more delicate and we refer to [85] for a
very general result, and to [17, Appendix] for an alternative approach via
obstacle problems, in the setting of homogeneous operators. For our purposes,
it is sufficient to consider the case in which one of the solutions is
constant. The technique goes back to T. Kato in [79], and has been generalized
to a large class of quasilinear operators in [40]. The next result is special
case of [40, Thm. 2.1].

######  Lemma 4.6.

Assume (4.1), and let $f\in C(\mathbb{R})$, $l\in C(\mathbb{R}^{+}_{0})$ with
$f(0)l(0)=0$. Suppose furthermore than $b\in C(M)$. If
$u\in\mathrm{Lip}_{\mathrm{loc}}(M)$ is a nontrivial solution of

$\Delta_{\varphi}u\geq b(x)f(u)l(|\nabla u|)\qquad\text{on }\,M,$ (4.4)

then $u_{+}=\max\\{u,0\\}$ is a $\mathrm{Lip}_{\mathrm{loc}}(M)$, non-negative
solution of

$\Delta_{\varphi}u_{+}\geq b(x)f(u_{+})l(|\nabla u_{+}|)\qquad\text{on }\,M.$

######  Remark 4.7.

_Note that $u=0$ is a solution of (4.4) since $f(0)l(0)=0$. _

###  4.2 The finite maximum principle

We now prove Theorem 1.10 in the Introduction. For the convenience of the
reader, we rewrite the assumptions and restate the result. We require

$\left\\{\begin{array}[]{l}\displaystyle\varphi\in C(\mathbb{R}^{+}_{0})\cap
C^{1}(\mathbb{R}^{+}),\qquad\varphi(0)=0,\qquad\varphi^{\prime}>0\ \text{ on
}\,\mathbb{R}^{+};\\\\[5.69046pt]
\displaystyle\frac{t\varphi^{\prime}(t)}{l(t)}\in L^{1}(0^{+})\\\\[11.38092pt]
f\in C(\mathbb{R}),\qquad f\geq 0\quad\text{on }\,(0,\eta_{0});\\\\[8.5359pt]
l\in C(\mathbb{R}_{0}^{+}),\qquad l>0\ \text{on
}\,\mathbb{R}^{+}.\end{array}\right.$ (4.5)

######  Theorem 4.8.

Let $M$ be a Riemannian manifold, and assume that $\varphi,f$ and $l$ satisfy
(4.5) and moreover

  * -

$f(0)l(0)=0$;

  * -

$f$ is $C$-increasing on $(0,\eta_{0})$, with $\eta_{0}$ as in (4.5);

  * -

$l$ is $C$-increasing on $(0,\xi_{0})$, for some $\xi_{0}>0$;

Fix a domain $\Omega\subset M$ and let $0<b\in C(\Omega)$. Then,
$(\mathrm{FMP})$ holds on $\Omega$ if and only if either

$f\equiv 0\qquad\text{on }\,(0,\eta_{0}),$ (4.6)

for some $\eta_{0}>0$, or

$f>0\quad\text{on }\,(0,\eta_{0}),\quad\text{and}\qquad\frac{1}{K^{-1}\circ
F}\not\in L^{1}(0^{+}).$ (4.7)

###### Proof.

We recall that the validity of $(\mathrm{FMP})$ means that for any solution
$u\in C^{1}(\Omega)$ of

$\left\\{\begin{array}[]{l}\Delta_{\varphi}u\leq b(x)f(u)l(|\nabla
u|)\qquad\text{on }\,\Omega\\\\[5.69046pt] u\geq 0\qquad\text{on
}\,\Omega,\end{array}\right.$ (4.8)

if $u(x_{0})=0$ for some $x_{0}\in\Omega$ then $u\equiv 0$. The argument
follows the lines of the proof in [123], with the help of a trick from [95].
We prove separately the sufficiency and necessity of (4.6),(4.7). First,
having fixed a point $o\in\Omega$ to be specified later, we choose
$R,\kappa>0$ such that $\overline{B_{2R}(o)}$ does not intersect
$\mathrm{cut}(o)$ and

$\mathrm{Ric}(\nabla r,\nabla r)\geq-(m-1)\kappa^{2}\qquad\text{on
}\,\overline{B_{2R}(o)}\backslash\\{o\\},$

where $r(x)=\mathrm{dist}(x,o)$. By the Laplacian comparison theorem (see Thm.
9.3 in Appendix A), denoting with $v_{g_{\kappa}}(r)$ the volume of a geodesic
sphere of radius $r$ in a model manifold of sectional curvature $-\kappa^{2}$

$\Delta
r\leq\frac{v_{g_{\kappa}}^{\prime}(r)}{v_{g_{\kappa}}(r)}=(m-1)\kappa\coth(\kappa
r)\qquad\text{on }\,\overline{B_{2R}(o)}\backslash\\{o\\}.\\\ $ (4.9)

_Sufficiency of conditions ( 4.6) and (4.7)_.
For $o\in\Omega$ and having set $R,\kappa$ as above, fix
$a_{1}\in\mathbb{R}^{+}$ such that $b(x)\leq a_{1}$ on $B_{2R}(o)$. Let $C\geq
1$ be the constant defining the $C$-increasing property of $f$. Define
$\wp(t)=v_{g_{\kappa}}(2R-t)$, $a(t)=a_{1}$, $T_{0}=3R/2$ and $T=R$. We claim
that we can suitably reduce $R$, and choose $\eta\in(0,\eta_{0})$ small
enough, in such a way that

$\frac{\wp_{1}}{\wp_{0}}\varphi\left(\frac{\eta}{R}\right)+4C\Theta(R)f_{\eta}l_{\xi}<\varphi(\xi),$
(4.10)

where $f_{\eta}$, $l_{\xi}$, $\wp_{0}$, $\wp_{1}$ and $\Theta$ are defined as
in (3.6). Indeed, since $v_{g_{\kappa}}^{\prime}\geq 0$, by definition
$\wp_{1}=v_{g_{\kappa}}(2R)$, $\wp_{0}=v_{g_{\kappa}}(R/2)$, and thus
$\wp_{1}/\wp_{0}\rightarrow 4^{m}$ and $\Theta(R)\rightarrow 0$ as
$R\rightarrow 0$. Hence, we can first reduce $R$ to guarantee

$4C\Theta(R)f_{\eta_{0}}l_{\xi}<\frac{\varphi(\xi)}{2},$

and then choose $\eta\in(0,\eta_{0})$ small enough to satisfy (4.10). Applying
Theorem 3.1, there exists a solution $z(t)$ of

$\left\\{\begin{array}[]{l}\big{(}\wp(t)\varphi(z_{t})\big{)}_{t}=2C\wp
a_{1}f(z)l(z_{t})\qquad\text{on }\,(0,R),\\\\[5.69046pt] z(0)=0,\qquad
z(R)=\eta,\end{array}\right.$ (4.11)

where the subscript $t$ indicates differentiation with respect to $t$.
Furthermore, combining Theorem 3.1 and Proposition 3.7 (note that here
$\wp^{\prime}<0$), the solution $z$ satisfies the following properties:

$\begin{gathered}z>0\ \text{ on }\,(0,R],\qquad z_{t}>0\ \text{ on
}\,[0,R],\qquad\|z_{t}\|_{\infty}<\xi.\end{gathered}$ (4.12)

Taking into account that we have extended $\varphi$ on $\mathbb{R}$ in such a
way that $\varphi(-s)=-\varphi(s)$, the function $w(r)=z(2R-r)$ satisfies

$\left\\{\begin{array}[]{l}\big{[}v_{g}\varphi(w^{\prime})\big{]}^{\prime}=2Cv_{g}a_{1}f(w)l(|w^{\prime}|)\qquad\text{on
}\,\left(R,2R\right),\\\\[5.69046pt] w(2R)=0,\quad w(R)=\eta,\quad
w^{\prime}<0\ \text{ on }\,\left[R,2R\right].\end{array}\right.$ (4.13)

Define $v(x)=w(r(x))$. Using (4.9), together with $\varphi(w^{\prime})<0$ and
(4.13), we deduce that $v$ solves

$\displaystyle\Delta_{\varphi}v$
$\displaystyle=\big{(}\varphi(w^{\prime})\big{)}^{\prime}+\varphi(w^{\prime})\Delta
r\geq v_{g}^{-1}(r)\big{[}v_{g}(r)\varphi(w^{\prime})\big{]}^{\prime}$ (4.14)
$\displaystyle=v_{g}(r)^{-1}\big{(}2Cv_{g}(r)a_{1}f(w)l(|w^{\prime}|)\geq
2Cb(x)f(v)l(|\nabla v|)$

on the annulus $E_{R}(o)=B_{2R}(o)\backslash\overline{B_{R}(o)}$. Moreover,
denoting with $\nu$ the outward pointing unit normal from $\partial
B_{2R}(o)$,

$\langle\nabla v,\nu\rangle=w^{\prime}(2R)<0\qquad\text{on }\,\partial
B_{2R}(o).$ (4.15)

Following E. Hopf’s argument, we now prove that $u$ solving (4.8) is
identically zero provided that $u(x_{0})=0$ at some $x_{0}$. Suppose by
contradiction that this is not the case, and let
$\Omega_{+}=\\{x\in\Omega:u(x)>0\\}$. Choose $x_{1}\in\Omega_{+}$ in such a
way that
$\mathrm{dist}(x_{1},\partial\Omega_{+})>\mathrm{dist}(x_{1},\partial\Omega)$,
and let $B(x_{1})\subset\Omega_{+}$ be the largest ball contained in
$\Omega_{+}$. Then, $u>0$ in $B(x_{1})$, while $u(\bar{x})=0$ for some
$\bar{x}\in\partial B(x_{1})\cap\Omega$. Clearly $\nabla u(\bar{x})=0$, since
$\bar{x}$ is an absolute minimum for $u$. Take a unit speed minimizing
geodesic $\gamma:[0,\mathrm{dist}(x_{1},\bar{x})]\rightarrow\Omega$ from
$x_{1}$ to $\bar{x}$. Up to choosing the arclength parameter $s$ sufficiently
close to $\mathrm{dist}(x_{1},\bar{x})$ and setting
$2R=\mathrm{dist}(x_{1},\bar{x})-s$, the closure of the ball $B_{2R}(o)$
centered at $o=\gamma(s)$ does not intersect $\mathrm{cut}(o)$, and
$B_{2R}(o)\subset B(x_{1})$, with $\bar{x}\in\partial B_{2R}(o)$. We consider
the function $v$ constructed above on $E_{R}(o)\subset B_{2R}(o)$, with $\eta$
small enough to satisfy (4.10) and also

$\eta<\inf_{\partial B_{R}(o)}u.$ (4.16)

We claim that $u\geq v$ on $E_{R}(o)$. Otherwise, suppose that

$\displaystyle\max_{E_{R}(o)}(v-u)=\bar{\delta}>0,$

and let $\Gamma$ be the set of maximum points of $v-u$. Note that
$\Gamma\Subset E_{R}(o)$ and that $\nabla u=\nabla v$ for each $x\in\Gamma$.
For $\delta\in(0,\bar{\delta})$, set $U_{\delta}=\\{v-u>\delta\\}$. By
construction, there exists $\varepsilon>0$ such that $\varepsilon\leq|\nabla
v|\leq 1$ on $E_{R}(o)$, and since $l>0$ on $\mathbb{R}^{+}$ we deduce that
the quotient $l(|\nabla u|)/l(|\nabla v|)$ is continuous on $E_{R}(o)$ and
equal to $1$ on $\Gamma$. A compactness argument shows that, for $\delta$
sufficiently close to $\bar{\delta}$, $l(|\nabla u|)/l(|\nabla v|)\leq 2$ on
$U_{\delta}$. Taking into account that the $C$-increasing property of $f$
implies $f(v)\geq C^{-1}f(u)$ on $U_{\delta}$, we deduce

$\Delta_{\varphi}u\leq b(x)f(u)l(|\nabla u|)\leq 2Cb(x)f(v)l(|\nabla
v|)\leq\Delta_{\varphi}v\qquad\text{on }\,U_{\delta}.$

From $v=u+\delta$ on $\partial U_{\delta}$, by the comparison Proposition 4.1
we get $v\leq u+\delta$ on $U_{\delta}$, contradicting the very definition of
$U_{\delta}$. Hence, $v\leq u$ on $E_{R}(o)$, and in particular
$\langle\nabla(u-v),\nu\rangle\leq 0$ at $\bar{x}$, which is impossible by
(4.15) and by the fact that $\nabla u(\bar{x})=0$. This contradiction
concludes the proof of the sufficiency part.
_Necessity of conditions ( 4.6) and (4.7)_.
Suppose the failure of both (4.6) and (4.7), or equivalently (recall that $f$
is $C$-increasing) the validity of

$f>0\quad\text{on }\,(0,\eta_{0}),\qquad\frac{1}{K^{-1}\circ F}\in
L^{1}(0^{+}).$ (KO0)

For each $o\in M$, we shall now construct on $B_{2R}(o)$ (with $R$ as in the
beginning of the proof) a $C^{1}$ non-negative, nonzero solution $u$ of
$(P_{\leq})$ with $u=0$ on $\overline{B_{R}(o)}$, contradicting
$(\mathrm{FMP})$. Set $T_{0}=2R$, $T=R$,

$\wp(t)=v_{g_{\kappa}}(t+R),\quad a(t)=\inf_{B_{2R}(o)\backslash B_{R}(o)}b,$

where $v_{g_{\kappa}}(r)$ is the volume of a geodesic sphere of radius $r$ in
a model of curvature $-\kappa^{2}$, as defined at the beginning of the proof.
Since $f(0)l(0)=0$, we can choose $\eta$ small enough to satisfy (3.8) in
Theorem 3.1, whence there exists $w\in C^{1}([0,T])$ non-decreasing and
solving

$\left\\{\begin{array}[]{l}\displaystyle\big{[}\wp\varphi(w^{\prime})\big{]}^{\prime}=\wp
af(w)l(w^{\prime})\qquad\text{on }\,(0,T),\\\\[5.69046pt] 0\leq
w\leq\eta,\quad w^{\prime}\geq 0\qquad\text{on }\,[0,T],\\\\[5.69046pt]
w(0)=0,\qquad w(T)=\eta.\end{array}\right.$ (4.17)

Up to reducing $\eta$ further, we can apply Proposition 3.9 to deduce that
$w^{\prime}(0)=0$. Set $u(x)=w\big{(}r(x)-R\big{)}$. By the Laplacian
comparison theorem and since $w^{\prime}\geq 0$, on $B_{2R}(o)\backslash
B_{R}(o)$ it holds

$\begin{array}[]{lcl}\Delta_{\varphi}u&=&\displaystyle\big{(}\varphi(w^{\prime})\big{)}^{\prime}+\varphi(w^{\prime})\Delta
r\leq\displaystyle\big{(}\varphi(w^{\prime})\big{)}^{\prime}+\varphi(w^{\prime})\frac{v_{g_{\kappa}}^{\prime}(r)}{v_{g_{\kappa}}(r)}\\\\[8.5359pt]
&\leq&\displaystyle\Big{(}\wp^{-1}\big{[}\wp\varphi(w^{\prime})\big{]}^{\prime}\Big{)}_{r(x)-R}\leq
af(w)l(w^{\prime})\\\\[8.5359pt] &\leq&b(x)f(u)l(|\nabla u|).\end{array}$

Extending $u$ to be zero on $B_{R}(o)$ defines a nonzero, $C^{1}$-solution of
$(P_{\leq})$ on all of $B_{2R}(o)$, which clearly violates $(\mathrm{FMP})$. ∎

######  Remark 4.9.

_The function $u$ in the proof of the necessity part, defined on $B_{2R}(o)$
and vanishing identically on a smaller ball $B_{R}(o)$, is an example of a
_dead core_ (super)solution. For a thorough investigation of dead core
problems, we refer the reader to [123] and the references therein. We mention
that $u$ can even be constructed to be positive on
$B_{2R}(o)\backslash\overline{B_{R}(o)}$. Indeed, it is enough to replace the
solution $w$ of (4.17) with the supersolution $z$ defined in the proof of
Proposition 3.9, which is known to be positive on $(0,T]$. _

######  Remark 4.10.

_With a similar technique, one could consider the $(\mathrm{FMP})$ for more
general equations of the type_

$\Delta_{\varphi}u\leq b(x)f(u)l(|\nabla
u|)+\bar{b}(x)\bar{f}(u)\bar{l}(|\nabla u|),$

_for suitable $\bar{b},\bar{f},\bar{l}$. The prototype case_

$\Delta_{p}u\leq f(u)+|\nabla u|^{q},\qquad\text{with }\,p>1$

_has been considered in[123, Thm. 5.4.1] and [125] when $q\geq p-1$, and in
[51] for $q<p-1$. _

##  5 Weak maximum principle, a-priori estimates and Liouville’s property

###  5.1 The equivalence between $(\mathrm{WMP}_{\infty})$ and $(\mathrm{L})$

In this section, we prove a more general version of Proposition 1.16, that
describes the relationship between $(\mathrm{WMP}_{\infty})$ and the Liouville
property $(\mathrm{L})$. We begin by introducing another form of the weak
maximum principle.

######  Definition 5.1.

_Assume ( 1.3) and fix $b,l$ satisfying (1.5). We say that_

  * •

the _open weak maximum principle_ $(\mathrm{OWMP}_{\infty})$ holds for
$(bl)^{-1}\Delta_{\varphi}$ if, for each $f\in C(\mathbb{R})$, for each open
set $\Omega\subset M$ with $\partial\Omega\neq\emptyset$ and for each $u\in
C(\overline{\Omega})\cap\mathrm{Lip}_{\mathrm{loc}}(\Omega)$ solving

$\left\\{\begin{array}[]{l}\Delta_{\varphi}u\geq b(x)f(u)l(|\nabla
u|)\qquad\text{on }\,\Omega,\\\\[5.69046pt]
\sup_{\Omega}u<\infty,\end{array}\right.$ (5.1)

we have that

$\text{either}\quad\sup_{\Omega}u=\sup_{\partial\Omega}u,\qquad\text{or}\quad
f\big{(}\sup_{\Omega}u\big{)}\leq 0.$ (5.2)

The open weak maximum principle at infinity has been introduced in [4] in the
study of immersed submanifolds of warped product ambient spaces, and parallels
Ahlfors’ definition of parabolicity. For a detailed investigation and an
extensive bibliography, we refer to Chapters 3 and 4 of [2].

######  Remark 5.2.

_The recent[93] contains a different approach to maximum principles at
infinity in the spirit of $(\mathrm{OWMP}_{\infty})$ (called there _the
Ahlfors property_), which is based on viscosity theory and enables to consider
classes of fully nonlinear operators of geometric interest. A systematic
approach via viscosity theory is very well suited to treat weak and strong
maximum principles in a unified way, and especially to investigate principles
like the classical Ekeland or $(\mathrm{SMP}_{\infty})$, where a gradient
condition on $u$ appears. _

######  Proposition 5.3.

Let $\varphi$ and $b,f,l$ satisfy respectively the assumptions in (1.5),
(1.3). Then, the following properties are equivalent:

  * (i)

$(bl)^{-1}\Delta_{\varphi}$ satisfies $(\mathrm{WMP}_{\infty})$;

  * (ii)

$(\mathrm{L})$ holds for some (equivalently, every) $f$ satisfying

$f(0)=0,\qquad f>0\quad\text{on }\,\mathbb{R}^{+};$

  * (iii)

each non-constant $u\in\mathrm{Lip}_{\mathrm{loc}}(M)$ solving $(P_{\geq})$ on
$M$ and bounded above satisfies $f(u^{*})\leq 0$, with $u^{*}=\sup_{M}u$;

  * (iv)

$(bl)^{-1}\Delta_{\varphi}$ satisfies $(\mathrm{OWMP}_{\infty})$.

###### Proof.

We prove the chain of implications
$(i)\Rightarrow(iii)\Rightarrow(ii)\Rightarrow(i)$ and then
$(iv)\Leftrightarrow(i)$. When the “some-every” alternative occurs, we always
assume the weaker property and prove the stronger.
$(i)\Rightarrow(iii)$.
Let $u\in\mathrm{Lip}_{\mathrm{loc}}(M)$ be a non-constant solution of
$(P_{\geq})$ that is bounded from above, and assume by contradiction that
$f(u^{*})=2K>0$. By continuity, there exists $\eta<u^{*}$ sufficiently close
to $u^{*}$ such that $f(u)\geq K$ on $\Omega_{\eta}=\\{u>\eta\\}$, thus

$\Delta_{\varphi}u\geq Kb(x)l(|\nabla u|)\qquad\text{on }\,\Omega_{\eta}.$

The definition of $(\mathrm{WMP}_{\infty})$ then implies $K\leq 0$,
contradiction.
$(iii)\Rightarrow(ii)$.
Let $u\in\mathrm{Lip}_{\mathrm{loc}}(M)$ be a non-constant, bounded, non-
negative solution of $(P_{\geq})$. Then $f(u^{*})\leq 0$ by $(iii)$. However,
from $f>0$ on $\mathbb{R}^{+}$ we get $u^{*}\equiv 0$, that is, $u\equiv 0$ is
constant, contradiction.
$(ii)\Rightarrow(i)$.
Let us consider a problem $(P_{\geq})$ with $f>0$ on $\mathbb{R}^{+}$,
$f(0)=0$ for which $(\mathrm{L})$ holds. Suppose by contradiction that $(i)$
is not satisfied, that is, there exists a non-constant
$u\in\mathrm{Lip}_{\mathrm{loc}}(M)$ with

$\Delta_{\varphi}u\geq Kb(x)l(|\nabla u|)\qquad\text{on some
}\,\Omega_{\bar{\eta}},$

for some $K>0$ and $\bar{\eta}<u^{*}$. Since $f(0)=0$, we can choose
$\eta\in(\bar{\eta},u^{*})$ in such a way that $f<K$ on $(0,u^{*}-\eta)$.
Hence, $u_{\eta}=u-\eta$ solves

$\Delta_{\varphi}u_{\eta}\geq Kb(x)l(|\nabla u_{\eta}|)\geq
b(x)f(u_{\eta})l(|\nabla u_{\eta}|)\qquad\text{on }\Omega_{\eta}.$

Thanks to Lemma 4.6, $w=\max\\{u_{\eta},0\\}$ is a non-constant, non-negative,
bounded solution of $\Delta_{\varphi}w\geq b(x)f(w)l(|\nabla w|)$ on $M$,
contradicting property $(\mathrm{L})$ for such an $f$.
$(i)\Rightarrow(iv)$.
If $u$ solves (5.1) but none of the properties in (5.2) holds, then by
continuity we can choose $\eta\in(\sup_{\partial\Omega}u,\sup_{\Omega}u)$ such
that $f(u)\geq K>0$ on $\\{u>\eta\\}$. Thus, $u$ solves $\Delta_{\varphi}u\geq
Kb(x)l(|\nabla u|)$ on $\\{u>\eta\\}\subset\Omega$, contradicting
$(\mathrm{WMP}_{\infty})$.
$(iv)\Rightarrow(i)$.
Assuming that $(\mathrm{WMP}_{\infty})$ does not hold, take a non-constant $u$
which is bounded above and solves $\Delta_{\varphi}u\geq Kb(x)l(|\nabla u|)$
on some set $\Omega_{\eta}=\\{u>\eta\\}$, for some $K>0$. If $\eta$ is close
enough to $u^{*}$, $\partial\Omega_{\eta}\neq\emptyset$, thus clearly $u$
contradicts $(\mathrm{OWMP}_{\infty})$ on $\Omega_{\eta}$ with the choice
$f=K$.

∎

###  5.2 Volume growth and $(\mathrm{WMP}_{\infty})$

In this section, we explore geometric conditions that ensure the validity of
$(\mathrm{WMP}_{\infty})$, in the form given by $(iii)$ of Proposition 1.16,
that is, each non-constant $u\in\mathrm{Lip}_{\mathrm{loc}}(M)$ solving
$(P_{\geq})$ and bounded above satisfies $f(u^{*})\leq 0$. Here, as usual,
$u^{*}=\sup_{M}u$, and similarly we will use the notation $u_{*}=\inf_{M}u$.
When $l\equiv 1$, the problem has been tackled in a series of papers [77, 134,
109, 110] by means of integral methods, and in particular we refer to [110]
for a thorough discussion. Since then, in a manifolds setting, the first
results that we are aware of allowing a nontrivial $l$ appeared in [95], where
$l$ is assumed to be $C$-increasing (in fact, polynomial in Thm. 5.1 therein).
In particular, the relevant case when $l$ can vanish both as $t\rightarrow
0^{+}$ and as $t\rightarrow\infty$ seems to be still open even in the
Euclidean space, although it has been recently considered for Carnot groups in
[2]. It is a remarkable feature that the results quoted above ensure
$f(u^{*})\leq 0$ not only when $u$ is a-priori bounded above, but also when
$u$ does not grow too fast at infinity, in the sense that in this case $u^{*}$
is also shown to be finite. This is a natural condition, and its origin is
related to the growth of an explicit Khas’minskii-type potential for the
operator considered, see Section 4 of [110] for more information. Related
interesting results on $\mathbb{R}^{m}$ can be found in [49, 50, 124, 38] and
will be described in more detail later.

The next theorem improves on [95, Thm 5.1] and [2, Thm. 2.1]. Throughout this
section, we assume the following growth conditions:

$\begin{array}[]{ll}\text{there exist constants $p,\bar{p}>1$, $C,\bar{C}>0$
such that }\\\\[5.69046pt] \varphi(t)\leq Ct^{p-1}\ \text{ on
}\,[0,1],\qquad\varphi(t)\leq\bar{C}t^{\bar{p}-1}\quad\text{on
}\,[1,\infty).\end{array}$ (5.3)

If $p=\bar{p}$, (5.3) is called the _weak $p$-coercivity_ of
$\Delta_{\varphi}$ in [38]. We will explain in depth the different role played
by $p$ and $\bar{p}$ for the validity of $(\mathrm{WMP}_{\infty})$.

######  Theorem 5.4.

Let $M$ be a complete Riemannian manifold, and consider $\varphi,b,f$ and $l$
meeting the requirements in (1.5), (1.3) and the bounds (5.3) for some
$p,\bar{p}>1$. Assume that, for some $\mu,\chi\in\mathbb{R}$ satisfying

$\chi\geq 0,\qquad\mu\leq\chi+1,$ (5.4)

the following inequalities hold:

$\begin{array}[]{ll}b(x)\geq C\big{(}1+r(x)\big{)}^{-\mu}&\quad\text{on
}\,M,\\\\[5.69046pt] f(t)\geq C&\quad\text{for }\,t\,\gg 1\\\\[5.69046pt]
\displaystyle l(t)\geq C\frac{\varphi(t)}{t^{\chi}}&\quad\text{on
}\,\mathbb{R}^{+},\end{array}$ (5.5)

for some constant $C>0$. Let $u\in\mathrm{Lip}_{\mathrm{loc}}(M)$ be a non-
constant, weak solution of $(P_{\geq})$ such that either

  * (i)

$u$ is bounded above and one of the following properties hold:

$\begin{array}[]{lll}\mu<\chi+1&\text{and}&\displaystyle\qquad\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{\chi+1-\mu}}<\infty\quad(=0\,\text{
if }\,\chi=0);\\\\[11.38092pt]
\mu=\chi+1&\text{and}&\displaystyle\qquad\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{\log
r}<\infty\quad(\leq p\,\text{ if }\,\chi=0).\end{array}$ (5.6)

  * (ii)

$u$ satisfies

$u_{+}(x)=o\big{(}r(x)^{\sigma}\big{)}\qquad\text{as
}\,r(x)\rightarrow\infty,$ (5.7)

for some $\sigma>0$ such that

$\chi\sigma\leq\chi+1-\mu,$ (5.8)

and one of the following properties hold:

$\begin{array}[]{ll}\chi\sigma<\chi+1-\mu,&\quad\displaystyle\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{\chi+1-\mu-\chi\sigma}}<\infty\quad(=0\,\text{
if }\,\chi=0);\\\\[11.38092pt] \chi\sigma=\chi+1-\mu,\ \ \
\chi>0,&\quad\displaystyle\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{\log
r}<\infty;\\\\[11.38092pt]
\chi\sigma=\chi+1-\mu,&\quad\displaystyle\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{\log
r}\leq\left\\{\begin{array}[]{ll}p-\sigma(p-1)&\text{if }\,\sigma\leq
1,\\\\[2.84544pt] \bar{p}-\sigma(\bar{p}-1)&\text{if
}\,\sigma>1.\end{array}\right.\end{array}$ (5.9)

Then, $u$ is bounded above on $M$ and $f(u^{*})\leq 0$. If moreover $u$
satisfies $(P_{=})$,

$tf(t)\geq C|t|\qquad\text{for }\,|t|>>1$ (5.10)

and either $(i)$ or $(ii)$ holds both for $u_{+}$ and for $u_{-}$, then

$u\in L^{\infty}(M)\qquad\text{and}\qquad f(u^{*})\leq 0\leq f(u_{*}).$ (5.11)

######  Remark 5.5.

_In the third case of ( 5.9), that is, when $\chi\sigma=\chi+1-\mu$ and the
volume growth of $B_{r}$ is suitably small with respect to $p,\bar{p},\sigma$,
the result still holds under the weaker assumption_

$u_{+}(x)=O\big{(}r(x)^{\sigma}\big{)}\qquad\text{as
}\,r(x)\rightarrow\infty.$ (5.12)

As a consequence of Theorem 5.4, we deduce Theorem 1.19 of the Introduction.

###### Proof of Theorem 1.19.

Consider a non-constant solution $u\in\mathrm{Lip}_{\mathrm{loc}}(M)$ with
$u^{*}<\infty$ of $\Delta_{\varphi}u\geq Kb(x)l(|\nabla u|)$ on some upper
level set $\Omega_{\eta}=\\{x\in M\ :\ u(x)>\eta\\}$. We shall prove that
$K\leq 0$. Suppose that this is not the case. By adding a constant to $u$, we
can suppose that $\eta<0$ but sufficiently near to $0$ so that
$\Omega_{0}=\\{x\in M\ :\ u(x)>0\\}\not=\emptyset$. Choose $f\in
C(\mathbb{R})$ such that $0\leq f\leq K$, $f(0)=0$ and $f(t)=K$ for
$t>u^{*}/2$. Then, $u$ solves

$\Delta_{\varphi}u\geq b(x)f(u)l(|\nabla u|)$ (5.13)

on $\Omega_{\eta}\supset\Omega_{0}$, and with the aid of Lemma 4.6 we can
assume that $u\geq 0$ solves (5.13) on the whole of $M$. Moreover,
$f(u^{*})=K$. To reach a contradiction, we just need to check the requirements
to apply Theorem 5.4, case $(i)$ and conclude $f(u^{*})\leq 0$. This, by
Proposition 1.16, implies the $(\mathrm{WMP}_{\infty})$.
First, observe that $\mu\leq\chi-\alpha/2$ in (1.47) can be rewritten as

$1+\frac{\alpha}{2}\leq\chi+1-\mu,$ (5.14)

and from $\alpha\geq-2$ it implies $\mu\leq\chi+1$. We exhamine the validity
of (5.6). If $\alpha>-2$ and $\chi>0$, then $\mu<\chi+1$ by (5.14) and the
volume assumption (1.46) imply the first in (5.6). If $\alpha=-2$ and
$\chi>0$, then again by (1.46) both of (5.6) are met, respectively, when
$\mu<\chi+1$ and $\mu=\chi+1$. Suppose now that $\chi=0$ and
$\mu<\chi-\alpha/2$. Then, $\mu<\chi+1$ and, for each $\alpha\geq-2$, the
strict inequality in (5.14) coupled with (1.46) guarantees the first in (5.6).
If $\chi=0$ and $\mu=\chi-\alpha/2$, according to whether $\alpha>-2$ or
$\alpha=-2$ the requirement $V_{\infty}=0$, respectively $V_{\infty}\leq p$,
in (1.47) is precisely what is needed to deduce the validity of (5.6),
concluding the proof. ∎

It is worth to postpone the proof of Theorem 5.4 and comment on various
aspects of its statement.

  * -

$p,\bar{p}$ play no explicit role in (5.4). However, a bound on $\chi$ in
terms of $p$ alone is hidden, in some cases, in the requirement
$l(t)\geq\varphi(t)/t^{\chi}$ on $\mathbb{R}^{+}$: indeed, if
$\varphi(t)\asymp t^{p-1}$ near $t=0$, the continuity of $l$ at zero forces

$\chi\leq p-1.$

  * -

For $l\equiv 1$, that is, when no gradient appears, the third in (5.5) forces
$\varphi(t)\leq C^{-1}t^{\chi}$ on $\mathbb{R}^{+}$. If we suppose that
$\varphi(t)\asymp t^{p-1}$ on $[0,1]$ and $\varphi(t)\asymp t^{\bar{p}-1}$ on
$[1,\infty)$, the above theorem can be applied provided

$\bar{p}-1\leq\chi\leq p-1.$

Therefore, when the operator is the $p$-Laplacian operator, the gradientless
case $l\equiv 1$ is recovered with the choice $\chi=p-1$. On the other hand,
for the mean curvature operator, $p\leq 2$ and $\bar{p}>1$ can be chosen
arbitrarily close to $1$, and the gradientless case $l\equiv 1$ is recovered
for any choice of $\chi\in[0,1]$. For a fixed $\mu$, in case $(i)$ or in
$(ii)$ with $\sigma\leq 1$, it is evident that the choice $\chi=1$, $p=2$
gives the best result, while in case $(ii)$ for $\sigma>1$ the best choice is
$\chi=0$ and $\bar{p}$ approaching $1$.

  * -

The second in (5.9) is the only place where $p$ and $\bar{p}$ appear. Its
validity forces an upper bound for $\sigma$, since the right-hand side in the
second of (5.9) is non-negative if and only if

$\sigma\leq\frac{\bar{p}}{\bar{p}-1}.$

  * -

The third in (5.9) supports and makes rigorous the next idea: in the sublinear
range $\sigma<1$, the region where $|\nabla u|$ is close to zero should be,
somehow, larger than the one where $|\nabla u|$ is big, and consequently the
growth of $\varphi$ on $[1,\infty)$ (if still polynomial) might be neglectable
with respect to the behaviour of $\varphi$ on $[0,1]$. If $\sigma>1$ the
situation reverses, and now the main contribution should be given by $\varphi$
on $[1,\infty)$.

We feel interesting and a bit surprising that the method to prove Theorem 5.4
is able to detect, in some sense, the size of the regions where $|\nabla u|$
is small or large. In particular, if $u$ is bounded above this might suggest
that the above proof could be refined to show that, under the same
assumptions, the _strong_ maximum principle $(\mathrm{SMP}_{\infty})$ is true,
see Problem 2 in the Introduction.

As we will show at the end of the section, Theorem 5.4 is sharp in the
following sense: under the validity of the range of the parameters $\chi,\mu$
in (5.4), for almost each condition on $\sigma$ and $\mathrm{vol}(B_{r})$ we
are able to find a non-constant solution of $(P_{\geq})$ with $f\equiv 1$
satisfying all the remaining assumptions but the chosen one.

If $f$ has a unique zero, from (5.11) we deduce a Liouville type theorem for
slowly growing solutions $u$ of $(P_{=})$, that fits very well with some
results obtained, in the Euclidean setting, by Farina and Serrin in [49, 50].
For the sake of comparison, we state their theorems renaming their parameters
to agree with our notation:

######  Theorem 5.6.

[50, Thm. 11 and 12] On Euclidean space, consider $\varphi,b,f$ and $l$
meeting the requirements in (1.5), (1.3) and the bounds (5.3) for some
$p=\bar{p}>1$. Assume that, for some $\mu,\chi,\omega\in\mathbb{R}$ satisfying

$0<\omega\leq\chi\leq p-1,$

the following inequalities hold:

$\begin{array}[]{ll}b(x)\geq C\big{(}1+r(x)\big{)}^{-\mu}&\quad\text{on
}\,\mathbb{R}^{m},\\\\[5.69046pt] tf(t)\geq C|t|^{\omega+1}&\quad\text{for
}\,t\in\mathbb{R},\\\\[5.69046pt] \displaystyle l(t)\geq
Ct^{p-1-\chi}&\quad\text{on }\,\mathbb{R}^{+},\end{array}$ (5.15)

for some constant $C>0$. Then, a solution
$u\in\mathrm{Lip}_{\mathrm{loc}}(\mathbb{R}^{m})$ of $(P_{=})$ satisfying

$|u(x)|=O\big{(}r(x)^{\sigma}\big{)}\qquad\text{as }\,r(x)\rightarrow\infty,$

for some $\sigma>0$, is constant provided one of the following cases occur:

  * (i)

$\omega<\chi$, $\mu<\chi+1$,

$(p-1)(1+\chi-\mu)\geq(p-m)(\chi-\omega)\qquad\text{and}\qquad\sigma\in\left(0,\frac{\chi+1-\mu}{\chi-\omega}\right);$
(5.16)

  * (ii)

$\omega<\chi$, $m<p$,

$(p-1)(1+\chi-\mu)\leq(p-m)(\chi-\omega)\qquad\text{and}\qquad\sigma\in\left(0,\frac{p-m}{p-1}\right);$
(5.17)

  * (iii)

$\omega=\chi$, $\mu<\chi+1$, independently of $\sigma>0$.

  * (iv)

$\omega=\chi$, $m<p$,

$\mu\geq\chi+1\qquad\text{and}\qquad\sigma\in\left(0,\frac{p-m}{p-1}\right).$

######  Remark 5.7.

_Note that the bounds in $(i)$ and $(ii)$ well match when equality holds in
the first of (5.16) and (5.17). If $l\equiv 1$, that is, $\chi=p-1$, $(i)$ and
$(iii)$ have been proved, respectively, in Theorems B and A in [49], and their
sharpness is discussed in Examples 5 and 4, Section 11 therein. It is
interesting to observe that the first in (5.16) is not required in [49, Thm.
B], but appears in discussing the sharpness of $(i)$. More precisely, Example
5 in [49] stresses that the conclusion of Theorem 5.6 in case $(i)$ does not
hold when $\sigma=\frac{\chi+1-\mu}{\chi-\omega}$ in (5.16), provided that the
first in (5.16) is strengthened to_

$(p-1)(1+\chi-\mu)>(p-m)(\chi-\omega)\qquad\text{with }\,\chi=p-1.$ (5.18)

_On the contrary, perhaps surprisingly, the conclusion of $(ii)$ still holds
for $\sigma=\frac{p-m}{p-1}$, see the next Corollary 5.8. This is, clearly,
not in contradiction with [49] in view of the incompatibility of (5.18) with
(5.17), see also the discussion [124, p. 677]. Further results for solutions
of $(P_{\geq})$ which are a-priori bounded or vanishing at infinity are given
in Theorems D,E,F in [49] and Theorems 1 and 2 in [141]. Inspection shows that
they fit very well with the case when $u$ is bounded above in Theorem 5.4. _

First, we compare Theorem 5.4 with case $(i)$ in Theorem 5.6, and we therefore
assume $M=\mathbb{R}^{m}$, $p=\bar{p}$ in (5.3), $0<\chi\leq p-1$ and
$\mu<\chi+1$. It is apparent that, for each $\omega>0$, condition $tf(t)\geq
C|t|^{\omega+1}$ implies both (5.10) and $tf(t)>0$ on $\mathbb{R}^{+}$.
Theorem 5.4 then gives the constancy of solutions of $(P_{=})$ on
$\mathbb{R}^{m}$ under the assumption

$u(x)=o\big{(}r(x)^{\sigma}\big{)}\qquad\text{as
}\,r(x)\rightarrow\infty\qquad\text{and}\qquad\sigma\in\left[0,\frac{\chi+1-\mu}{\chi}\right),$
(5.19)

for any dimension $m$ and any $p>1$. The upper bound for $\sigma$ is smaller
than the one in (5.16), as a counterpart of the stronger requirement
$tf(t)\geq C|t|^{\omega+1}$, but (5.16) converges to (5.19) as
$\omega\rightarrow 0^{+}$. Moreover, the first in (5.16) is not needed, in
accordance with [49, Thm. B] and [141, Thm 2 $(i)$]. Next, we investigate the
relationship with $(iii),(iv)$ in Theorem 5.6, where $\omega=\chi$ is assumed.
Since in our case $\omega=0$, $(iii)$ and $(iv)$ should be compared with case
$\chi=0$ of Theorem 5.4. Observe that $\chi=0$ includes an interesting class
of borderline inequalities such as

$\Delta_{p}u\geq b(x)f(u)|\nabla u|^{p-1},$ (5.20)

for which we have the next corollary; since, for $\chi=0$, any $\sigma>0$
satisfies (5.8), when $\mu<1$ we obtain a Liouville theorem for solutions $u$
with polynomial growth (i.e. satisfying $|u|=O(r^{\sigma})$ as
$r\rightarrow\infty$ for some $\sigma>0$).

######  Corollary 5.8.

Let $M$ be a complete Riemannian manifold, and consider $\varphi,b,f$ meeting
the requirements in (1.5), (1.3) and the bounds (5.3) for some $p,\bar{p}>1$.
Assume that, for some $\mu\leq 1$, the following inequalities hold:

$\begin{array}[]{ll}b(x)\geq C\big{(}1+r(x)\big{)}^{-\mu}&\quad\text{on
}\,M,\\\\[5.69046pt] f(t)\geq C&\quad\text{for }\,t\gg 1\end{array}$ (5.21)

and for some constant $C>0$. Let $u\in\mathrm{Lip}_{\mathrm{loc}}(M)$ be a
non-constant, weak solution of

$\Delta_{\varphi}u\geq b(x)f(u)\varphi(|\nabla u|)\qquad\text{on }\,M.$ (5.22)

Suppose that $u_{+}(x)=O\big{(}r(x)^{\sigma}\big{)}$, for some $\sigma>0$, and
that

$\begin{array}[]{rlll}(1)&\quad\mu<1,&\displaystyle\qquad\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{1-\mu}}=0,&\quad\text{
and }\,\sigma>0,\ \text{ or}\\\\[11.38092pt]
(2)&\quad\mu=1,&\displaystyle\qquad\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{\log
r}=d_{0},&\quad\text{ and
}\,0<\sigma\leq\left\\{\begin{array}[]{ll}\frac{p-d_{0}}{p-1},&\text{if
}\,d_{0}\geq 1,\\\\[8.5359pt] \frac{\bar{p}-d_{0}}{\bar{p}-1}&\text{if
}\,d_{0}<1.\end{array}\right.\end{array}$ (5.23)

Then, $u$ is bounded above on $M$ and $f(u^{*})\leq 0$. If moreover $u$
satisfies $(P_{=})$,

$tf(t)\geq C|t|\qquad\text{for }\,|t|>>1$ (5.24)

and either $(i)$ or $(ii)$ holds for $u_{+}$ and for $u_{-}$, then

$u\in L^{\infty}(M)\qquad\text{and}\qquad f(u^{*})\leq 0\leq f(u_{*}).$ (5.25)

###### Proof of Corollary 5.8, assuming Theorem 5.4.

It is enough to choose $\chi=0$ and $\sigma>0$ in Theorem 5.4. An algebraic
manipulation shows that the condition on $\sigma$ appearing in $(2)$ of (5.23)
is equivalent to

$d_{0}\leq\left\\{\begin{array}[]{ll}p-\sigma(p-1)&\text{if }\,\sigma\leq
1;\\\\[5.69046pt] \bar{p}-\sigma(\bar{p}-1)&\text{if
}\,\sigma>1.\end{array}\right.$

While Theorem 5.4 requires $u_{+}=o(r^{\sigma})$, when $\mu>1$ no problem
arise as we can enlarge $\sigma$ a bit to match this last requirement. On the
other hand, if $\mu=1$, thanks to Remark 5.5 we can still reach the conclusion
in Theorem 5.4 when $u_{+}=O(r^{\sigma})$. In particular, the upper bound for
$\sigma$ in (5.23) can be achieved. ∎

######  Remark 5.9.

_Comparing with $(iii),(iv)$ in Theorem 5.6, we readily see that (5.23) fits
very well with their assumptions, and we can also capture the case
$\sigma=\frac{p-m}{p-1}$. On the other hand, it should be remarked that our
result is restricted to $\mu\leq 1$. _

######  Remark 5.10.

_In the Euclidean setting $M=\mathbb{R}^{m}$ and for $p=\bar{p}\geq m$, other
interesting Liouville theorems for slowly growing solutions of $(P_{=})$ can
be found in [124, Thm. 1.1] and [50, Thm. 10], where the case $tf(t)\geq 0$
($\equiv 0$ in the second) is considered. There, the authors obtain the
constancy of solutions of $(P_{=})$ on $\mathbb{R}^{m}$ whenever $p>m$
and666It should be observed that assumption (1.3) in [124], when rephrased for
$(P_{\geq})$, gives necessarily $\varphi(t)=Ct^{p-1}$. However, the above
restriction does not appear in Theorem 10 of [50], which considers the case
$f(t)\equiv 0$._

$u(x)=o\left(r(x)^{\frac{p-m}{p-1}}\right)\qquad\text{as
}\,r(x)\rightarrow\infty.$ (5.26)

_Condition ( 5.26) is sharp and related to the growth of the fundamental
solution for the $p$-Laplacian (see [124] and Remark 10.3 in [40]). Further
interesting results covering $p\geq m$ can be found in [40] (Theorems 10.1 and
10.4 therein). _

Next, let us show how Corollary 5.8 applies in the setting of the mean
curvature operator to give the next

######  Theorem 5.11.

Let $M$ be a complete manifold satisfying

$\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{1-\mu}}=0,$
(5.27)

for some $\mu<1$. Let $b,f,l$ satisfy (1.5) and

$\begin{array}[]{l}\displaystyle b(x)\geq
C\big{(}1+r(x)\big{)}^{-\mu}\qquad\text{on }\,M,\\\\[5.69046pt] \displaystyle
l(t)\geq C\frac{t}{\sqrt{1+t^{2}}}\qquad\text{on
}\,\mathbb{R}^{+},\\\\[11.38092pt] \text{$f$ is non-decreasing on $\mathbb{R}$
and $f\not\equiv 0$}\\\\[5.69046pt] \end{array}$ (5.28)

for some $C>0$. If $u\in\mathrm{Lip}_{\mathrm{loc}}(M)$ is a non-constant
solution of

$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}}\right)=b(x)f(u)l(|\nabla u|)\qquad\text{on }\,M$ (5.29)

with a polynomial growth, then $u$ solves the minimal surface equation

$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}}\right)=0\qquad\text{on }\,M$

and it is bounded on one side. In particular, if (5.27) is strengthened to

$\mathrm{Ric}\geq 0,\quad\qquad K\geq-\kappa^{2}\qquad\text{on M,}$ (5.30)

for some constant $\kappa>0$ ($K$ the sectional curvature of $M$), then a
solution of (5.29) with a polynomial growth, if any, is constant for each
$\mu<1$.

######  Remark 5.12.

_The assumption $f\not\equiv 0$ is necessary, as the example of affine
functions solving the minimal graph equation on $\mathbb{R}^{m}$ shows. _

######  Remark 5.13.

_When $l\equiv 1$, $b\equiv 1$ and $M=\mathbb{R}^{m}$, V. Tkachev [144] proved
Theorem 5.11 for $C^{2}$ solutions of (5.29) without any growth restriction,
see also Thm. 10.4 of A. Farina’s survey [48]. The result has first been
extended for mean curvature type operators by Y. Naito and Y. Usami [100, Thm.
1], with a different argument using radialization in a way related to the one
in Section 8. An improvement of [144, 100] for (5.29) with a nontrivial
gradient dependence is shown in Theorem 8.31 below, while generalizations to a
larger class of quasilinear operators can be found in [141, Thm. 3], with the
same method as in [144], and in [123, Thm. 8.1.3] with an approach, by means
of Khas’minskii potentials, close to the one in [100]. _

###### Proof of Theorem 5.11.

Since one of the sets $\\{t\in\mathbb{R}\ :\ f(t)>0\\}$ and
$\\{t\in\mathbb{R}\ :\ f(t)<0\\}$ is non-empty, the monotonicity of $f$
implies that either $f(t)\geq C_{1}$ for $t>>1$, or $f(t)\leq-C_{1}$ for
$t<<-1$, for some constant $C_{1}>0$. Without loss of generality, we can
suppose $f(t)\geq C_{1}$ for $t>>1$. Using Corollary 5.8, since $u$ is non-
constant we deduce $u^{*}<\infty$ and $f(u^{*})\leq 0$. Again by the
monotonicity of $f$, either $f=0$ on $(-\infty,u^{*}]$ or $f<0$ somewhere. In
the second case, $f(t)\leq-C$ for $t<<-1$ and thus, applying again Corollary
5.8 with $-u$ replacing $u$ and $-f(-t)$ replacing $f(t)$ we obtain
$f(u_{*})\geq 0$. Therefore, $f=0$ on $[u_{*},u^{*}]$. Summarizing, in both
cases $u$ solves the minimal surface equation

$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}}\right)=0\qquad\text{on }\,M$ (5.31)

(hence, $u$ is smooth), and $u$ is bounded on one side. If (5.27) is replaced
by (5.30), the volume comparison implies $\mathrm{vol}(B_{r})\leq Cr^{m}$ and
therefore (5.27) holds for each $\mu<1$. The constancy of $u$ is then a
consequence of Theorem 1.3 in [137], contradicting our assumption. ∎

######  Remark 5.14.

_It is an open (and, we believe, very interesting) problem to prove the
constancy of positive solutions of_

$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+|\nabla u|^{2}}}\right)=0$

_under the only condition that $\mathrm{Ric}\geq 0$, that is, removing the
requirement on the sectional curvature in (5.30). Indeed, the technique in
[137] relies on uniform gradient estimates for $u$ achieved via the Korevaar’s
method, for which an assumption on the sectional curvature seems challenging
to remove. _

We now come to the proof of Theorem 5.4. The result follows in a more or less
direct way from the following refined maximum principle for slowly growing
solutions of

$\Delta_{\varphi}u\geq K(1+r)^{-\mu}\frac{\varphi(|\nabla u|)}{|\nabla
u|^{\chi}}\qquad\text{on }\,\Omega_{\gamma},$ (5.32)

where $\Omega_{\gamma}=\\{x\in M\,:\,u(x)>\gamma\\}$ is the superlevel set of
$u$ at height $\gamma\in\mathbb{R}$.

######  Theorem 5.15.

Let $M$ be a complete Riemannian manifold and let the growth (5.3) be met for
some $p,\bar{p}>1$. Let $\mu,\chi\in\mathbb{R}$ verify (5.4) and let
$u\in\mathrm{Lip}_{\mathrm{loc}}(M)$ be a function for which

$\hat{u}=\limsup_{r(x)\rightarrow\infty}\frac{u_{+}(x)}{r(x)^{\sigma}}<\infty,$
(5.33)

for some $\sigma\in\mathbb{R}^{+}_{0}$ satisfying

$0\leq\chi\sigma\leq\chi+1-\mu.$ (5.34)

Suppose that either one of the following assumptions is met:

$\begin{array}[]{lll}\chi\sigma<\chi+1-\mu&\quad\text{and}&\displaystyle\qquad\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{\chi+1-\mu-\chi\sigma}}=d_{0}<\infty;\\\\[5.69046pt]
\text{or}\\\\[5.69046pt]
\chi\sigma=\chi+1-\mu&\quad\text{and}&\displaystyle\qquad\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{\log
r}=d_{0}<\infty;\end{array}$ (5.35)

If, for some $\gamma\in\mathbb{R}$, the open set $\Omega_{\gamma}$ is non-
empty and $u$ is a non-constant, weak solution of (5.32) in $\Omega_{\gamma}$,
then

$K\leq H\cdot\hat{u}^{\chi},$ (5.36)

where, setting

$\zeta=\chi+1-\mu,\qquad{\rm and}\quad d^{*}=\min\Big{\\{}p-\sigma(p-1),\
\bar{p}-\sigma(\bar{p}-1)\Big{\\}},$ (5.37)

the constant $H=H(\sigma,\chi,p,\mu,d_{0})$ is given by

$\begin{array}[]{l}H=\left\\{\begin{array}[]{rll}(i)&\ 0&\quad\text{if
}\quad\chi>0,\quad\sigma=0;\\\\[5.69046pt] (ii)&\
d_{0}\big{[}\zeta-\chi\sigma\big{]}^{\chi+1}&\quad\displaystyle\text{if
}\quad\chi>0,\quad 0<\chi\sigma<\frac{\chi}{\chi+1}\zeta\\\\[8.5359pt] (iii)&\
d_{0}\sigma^{\chi}(\zeta-\chi\sigma)&\quad\displaystyle\text{if
}\quad\chi>0,\quad\frac{\chi}{\chi+1}\zeta\leq\chi\sigma<\zeta\\\\[11.38092pt]
(iv)&\ d_{0}(1-\mu)&\quad\text{if }\quad\chi=0,\quad\mu<1,\quad\sigma\geq
0\\\\[5.69046pt] (v)&\
\sigma^{\chi}(d_{0}-d^{*})_{+}&\quad\displaystyle\text{if
}\quad\chi\sigma=\zeta>0\ \ \text{ or }\ \ \chi\sigma=\zeta=0,\
\chi=0.\end{array}\right.\end{array}$ (5.38)

###### Proof.

We can suppose $K>0$, otherwise the estimate is trivial. Note that (5.32) is
invariant with respect to translations $u\mapsto u_{s}=u+s$. Fix
$\beta>\hat{u}$. We claim that a suitable translated $u_{s}$ satisfies

$u_{s}\leq\beta(1+r)^{\sigma}\qquad\text{on }\,M,\qquad u_{s}>0\ \text{
somewhere.}$ (5.39)

Indeed, if $\sigma>0$, (5.33) implies that $u<\beta(1+r)^{\sigma}$ outside a
large compact set $\Omega$, and translating $u$ downwards we can achieve the
same inequality also in $\Omega$, still keeping $u_{s}>0$ somewhere. On the
other hand, the claim is obvious if $\sigma=0$. In this last case, note that
here we do _not_ claim that $\hat{u}$ is not attained: this would follow from
a strong maximum principle, that to the best of our knowledge is unknown under
the sole assumption (5.3). Using that the resulting $u_{s}$ is positive
somewhere, we can also assume $\gamma>0$. Hereafter, computations will be
performed with $u=u_{s}$. Choose $\alpha>\beta$ and define

$v(x)=\alpha\big{(}1+r(x)\big{)}^{\sigma}-u(x),$

so that

$(\alpha-\beta)(1+r)^{\sigma}\leq v\leq\alpha(1+r)^{\sigma}\qquad\text{on
}\,\Omega_{\gamma}.$ (5.40)

Hereafter, with $C_{1},C_{2},C_{3},\ldots$ we will denote positive absolute
constants, that is, independent of $\sigma,\mu,\chi$. Fix a function
$\lambda\in C^{1}(\mathbb{R})$ such that

$0\leq\lambda\leq 1,\quad\lambda\equiv 0\ \ \text{ on
}\,(-\infty,\gamma],\qquad\lambda>0\ \ \text{on
}\,(\gamma,\infty),\quad\lambda^{\prime}\geq 0,$

and a cut-off function $\psi\in\mathrm{Lip}_{c}(M)$, $0\leq\psi\leq 1$, whose
support has nontrivial intersection with $\Omega_{\gamma}$. Next, consider
$F\in C^{1}(\mathbb{R}^{2})$, $F=F(r,v)$, satisfying

$F(r,v)>0,\qquad F_{v}=\frac{\partial F}{\partial v}(r,v)<0.$ (5.41)

Suppose first that $\bar{p}\geq p$. We insert the test function

$\psi^{\bar{p}}\lambda(u)F(v,r)\in\mathrm{Lip}_{c}(\overline{\Omega}_{\gamma})$
(5.42)

in the weak definition of (5.32). Using $\lambda^{\prime}\geq 0$ together with
the Cauchy-Schwarz inequality we obtain

$\begin{array}[]{l}\displaystyle K\int\psi^{\bar{p}}\lambda
F(1+r)^{-\mu}\frac{\varphi(|\nabla u|)}{|\nabla
u|^{\chi}}\leq\displaystyle\bar{p}\int\psi^{\bar{p}-1}\lambda F\varphi(|\nabla
u|)|\nabla\psi|+\int\psi^{\bar{p}}\lambda F_{v}\varphi(|\nabla u|)|\nabla
u|\\\\[14.22636pt]
\displaystyle\qquad\qquad+\int\psi^{\bar{p}}\lambda\varphi(|\nabla
u|)\left|\alpha\sigma(1+r)^{\sigma-1}F_{v}+F_{r}\right|.\end{array}$ (5.43)

Rearranging,

$\begin{array}[]{l}\displaystyle\int\psi^{\bar{p}}\lambda\left|F_{v}\right|\frac{\varphi(|\nabla
u|)}{|\nabla u|^{\chi}}B(x,u)\leq\bar{p}\int\psi^{\bar{p}-1}\lambda
F\varphi(|\nabla u|)|\nabla\psi|,\end{array}$ (5.44)

with

$\displaystyle B(x,u)=\displaystyle K(1+r)^{-\mu}\frac{F}{|F_{v}|}+|\nabla
u|^{\chi+1}-|\nabla
u|^{\chi}\left|-\alpha\sigma(1+r)^{\sigma-1}+\frac{F_{r}}{|F_{v}|}\right|.$
(5.45)

Let us assume the validity of the following

$\emph{claim: }\quad B(x,u)\geq\Lambda|\nabla u|^{\chi+1}\qquad\text{for some
$\Lambda\in(0,1]$ independent of $r$.}$ (5.46)

Plugging into (5.44) gives

$\begin{array}[]{l}\displaystyle\frac{\Lambda}{\bar{p}}\int\psi^{\bar{p}}\lambda\left|F_{v}\right|\varphi(|\nabla
u|)|\nabla u|\leq\int\psi^{\bar{p}-1}\lambda F\varphi(|\nabla
u|)|\nabla\psi|,\end{array}$ (5.47)

We now split the integrals on the subsets $\\{|\nabla u|<1\\}$ and $\\{|\nabla
u|\geq 1\\}$, where we apply different Young inequalities. Letting
$p^{\prime},\bar{p}^{\prime}$ be the conjugate exponents to $p$ and $\bar{p}$,
we can rewrite (5.49) as follows:

$\begin{array}[]{l}\displaystyle\frac{\Lambda}{\bar{p}}\int_{\\{|\nabla
u|<1\\}}\psi^{\bar{p}}\lambda\left|F_{v}\right|\left[\frac{\varphi(|\nabla
u|)}{|\nabla u|^{p-1}}\right]|\nabla
u|^{p}+\displaystyle\frac{\Lambda}{\bar{p}}\int_{\\{|\nabla u|\geq
1\\}}\psi^{\bar{p}}\lambda\left|F_{v}\right|\left[\frac{\varphi(|\nabla
u|)}{|\nabla u|^{\bar{p}-1}}\right]|\nabla u|^{\bar{p}}\\\\[11.38092pt]
\displaystyle\qquad\leq\int_{\\{|\nabla
u|<1\\}}\left[\psi^{\bar{p}}\lambda\left|F_{v}\right|\left(\frac{\varphi(|\nabla
u|)}{|\nabla u|^{p-1}}\right)|\nabla
u|^{p}\right]^{\frac{1}{p^{\prime}}}\left[\psi^{\bar{p}-1-\frac{\bar{p}}{p^{\prime}}}\lambda^{\frac{1}{p}}F|F_{v}|^{-\frac{1}{p^{\prime}}}\left(\frac{\varphi(|\nabla
u|)}{|\nabla u|^{p-1}}\right)^{\frac{1}{p}}|\nabla\psi|\right]\\\\[11.38092pt]
\displaystyle\qquad\ \ +\int_{\\{|\nabla u|\geq
1\\}}\left[\psi^{\bar{p}}\lambda\left|F_{v}\right|\left(\frac{\varphi(|\nabla
u|)}{|\nabla u|^{\bar{p}-1}}\right)|\nabla
u|^{\bar{p}}\right]^{\frac{1}{\bar{p}^{\prime}}}\left[\lambda^{\frac{1}{\bar{p}}}F|F_{v}|^{-\frac{1}{\bar{p}^{\prime}}}\left(\frac{\varphi(|\nabla
u|)}{|\nabla
u|^{\bar{p}-1}}\right)^{\frac{1}{\bar{p}}}|\nabla\psi|\right]\end{array}$
(5.48)

Observe that $\bar{p}\geq p$ implies the non-negativity of the exponent
$\bar{p}-1-\frac{\bar{p}}{p^{\prime}}\geq 0$ for $\psi$ above. We apply Young
inequality
$ab\leq(a\varepsilon)^{p^{\prime}}/p^{\prime}+(b/\varepsilon)^{p}/p$ to the
first term in the right-hand side of (5.49), and an analogous one with
$\bar{\varepsilon},\bar{p},\bar{p}^{\prime}$ to the second one. Rearranging,
we obtain

$\begin{array}[]{l}\displaystyle\left(\frac{\Lambda}{\bar{p}}-\frac{\varepsilon^{p^{\prime}}}{p^{\prime}}\right)\int_{\\{|\nabla
u|<1\\}}\psi^{\bar{p}}\lambda\left|F_{v}\right|\left[\frac{\varphi(|\nabla
u|)}{|\nabla u|^{p-1}}\right]|\nabla u|^{p}\\\\[11.38092pt]
+\displaystyle\left(\frac{\Lambda}{\bar{p}}-\frac{\bar{\varepsilon}^{\bar{p}^{\prime}}}{\bar{p}^{\prime}}\right)\int_{\\{|\nabla
u|\geq 1\\}}\psi^{\bar{p}}\lambda\left|F_{v}\right|\left[\frac{\varphi(|\nabla
u|)}{|\nabla u|^{\bar{p}-1}}\right]|\nabla u|^{\bar{p}}\\\\[11.38092pt]
\displaystyle\qquad\displaystyle\leq\frac{\varepsilon^{-p}}{p}\int_{\\{|\nabla
u|<1\\}}\psi^{\bar{p}-p}\lambda
F\left[\frac{F}{|F_{v}|}\right]^{p-1}\left(\frac{\varphi(|\nabla u|)}{|\nabla
u|^{p-1}}\right)|\nabla\psi|^{p}\\\\[11.38092pt] \displaystyle\qquad\ \
+\frac{\bar{\varepsilon}^{-\bar{p}}}{\bar{p}}\int_{\\{|\nabla u|\geq
1\\}}\lambda
F\left[\frac{F}{|F_{v}|}\right]^{\bar{p}-1}\left(\frac{\varphi(|\nabla
u|)}{|\nabla u|^{\bar{p}-1}}\right)|\nabla\psi|^{\bar{p}}\end{array}$ (5.49)

Choose $\varepsilon,\bar{\varepsilon}$ in such a way that both the
coefficients in round brackets in the left-hand side are $\Lambda/(2\bar{p})$.
From $\bar{p}\geq p$, $\psi\leq 1$ and $\Lambda\leq 1$, using (5.3) we infer
the inequality

$\int\psi^{\bar{p}}\lambda\left|F_{v}\right|\varphi(|\nabla u|)|\nabla u|\leq
C_{1}\left[\int\lambda
F\left(\frac{F}{|F_{v}|}\right)^{p-1}|\nabla\psi|^{p}+\int\lambda
F\left(\frac{F}{|F_{v}|}\right)^{\bar{p}-1}|\nabla\psi|^{\bar{p}}\right],$
(5.50)

where $C_{1}$ is some constant depending on $\Lambda$ and on
$p,\bar{p},C,\bar{C}$ in (5.3). Fix $R_{0}\geq 1$ large enough that $u$ is not
constant on $\Omega_{\gamma}\cap B_{R_{0}}\neq\emptyset$. Then, clearly
$\nabla u$ is not identically zero on $\Omega_{\gamma}\cap B_{R_{0}}$, because
otherwise $u$ would be constant on connected components of $\Omega_{\gamma}$,
which would imply that $\Omega_{\gamma}\equiv M$ and $u$ be constant,
contradiction. Fix $\delta\in(1/2,1)$. For $R>2R_{0}\geq 2$, we choose $\psi$
in such a way that

$0\leq\psi\leq 1,\quad\psi\equiv 1\ \text{ on }B_{\delta
R},\quad\operatorname{supp}(\psi)\subset
B_{R},\quad|\nabla\psi|\leq\frac{C_{2}}{(1-\delta)R},$ (5.51)

for some absolute constant $C_{2}$. Inserting into (5.50) and recalling that
$\lambda\leq 1$ we obtain

$\begin{array}[]{l}\displaystyle\int_{\Omega_{\gamma}\cap
B_{R_{0}}}\lambda\left|F_{v}\right|\varphi(|\nabla u|)|\nabla
u|\\\\[14.22636pt] \qquad\leq\quad\displaystyle
C_{3}\left[\frac{1}{R^{p}}\int_{(B_{R}\backslash B_{\delta
R})\cap\Omega_{\gamma}}F\left(\frac{F}{|F_{v}|}\right)^{p-1}+\frac{1}{R^{\bar{p}}}\int_{(B_{R}\backslash
B_{\delta
R})\cap\Omega_{\gamma}}F\left(\frac{F}{|F_{v}|}\right)^{\bar{p}-1}\right],\end{array}$
(5.52)

for some $C_{3}=C_{3}(p,\bar{p},C,\bar{C},\Lambda,\delta)$. In the
complementary case $\bar{p}\leq p$, we achieve (5.52) by simply exchanging the
role of $p,\bar{p}$ and the related inequalities on $\\{|\nabla u|<1\\}$,
$\\{|\nabla u|\geq 1\\}$.

We now need to check the validity of the claim in (5.46), for a suitable
choice of $F$. Observe that the expression of $B$ in (5.45) is a function of
the type

$g(s)=P+s^{\chi+1}-Qs^{\chi},$

for $s=|\nabla u|$ and positive parameters

$P=K(1+r)^{-\mu}\frac{F}{|F_{v}|},\qquad
Q=\left|-\alpha\sigma(1+r)^{\sigma-1}+\frac{F_{r}}{|F_{v}|}\right|$

depending on $r$. It is a calculus exercise to check that $g(s)\geq\Lambda
s^{\chi+1}$ on $\mathbb{R}^{+}_{0}$ when either

$\left\\{\begin{array}[]{lll}\chi=0,&\quad Q\leq
P&\quad\text{and}\qquad\Lambda\leq 1,\quad\text{or}\\\\[5.69046pt]
\chi>0,&\displaystyle\quad\frac{Q^{\chi+1}}{P}\leq\frac{(\chi+1)^{\chi+1}}{\chi^{\chi}}&\quad\text{and}\displaystyle\qquad\Lambda\leq
1-\frac{\chi}{(\chi+1)^{\frac{\chi+1}{\chi}}}\left(\frac{Q^{\chi+1}}{P}\right)^{1/\chi}.\end{array}\right.,$
(5.53)

Having fixed a parameter $\theta\in(0,1)$, we will choose $F$ in order to
satisfy the next relations between $Q$ and $P$:

$\left\\{\begin{array}[]{ll}\text{if }\,\chi=0\,\text{ we want }\ \
Q=P,&\quad\text{and in this case set}\quad\Lambda=1;\\\\[5.69046pt] \text{if
}\,\chi>0\,\text{ we want }\ \
\frac{Q^{\chi+1}}{P}=\theta^{\chi}\frac{(\chi+1)^{\chi+1}}{\chi^{\chi}}&\quad\text{and
in this case set}\displaystyle\quad\Lambda=1-\theta.\end{array}\right.$ (5.54)

In this way, (5.53) and thus (5.46) is met. Observe that the first case in
(5.54) can be obtained by letting $\chi\rightarrow 0$ and then
$\theta\rightarrow 0$ in the second one. To meet the identities in (5.54), we
necessarily need an upper bound for $Q/P$ or $Q^{\chi+1}/P$ that does not
depend on $r$, and this suggests our choice of $F$, that will be different
from case to case. Set for convenience

$\eta=\mu+(\sigma-1)(\chi+1)=(\chi+1)\sigma-\zeta,$ (5.55)

where $\zeta$ is as in (5.37), and note that

$\displaystyle\sigma-\eta=\zeta-\chi\sigma\geq 0.$ (5.56)

Analysis of case $(ii)$, case $(i)$ for $\zeta>0$, and case $(iv)$ for
$\sigma<1-\mu$:

$\chi>0,\quad
0\leq\chi\sigma<\frac{\chi}{\chi+1}\zeta,\qquad\text{or}\qquad\chi=0,\quad\mu<1,\quad
0\leq\sigma<1-\mu.$

Using the definition of $\eta$, these cases correspond to

$\chi>0,\qquad\sigma\geq 0,\quad\eta<0.$

Note that $\sigma>\eta$. We choose

$F(v,r)=\exp\left\\{-\tau v(1+r)^{-\eta}\right\\},$ (5.57)

for a real number $\tau>0$ that will be specified later in order to satisfy
the identity for $P,Q$ in (5.54). Then, on $\Omega_{\gamma}$

$\frac{F}{|F_{v}|}=\frac{(1+r)^{\eta}}{\tau},\qquad\frac{F_{r}}{|F_{v}|}=\frac{v\eta}{(1+r)},$

and hence, by (5.40) and using $\sigma>\eta$, $\eta<0$

$\displaystyle-\alpha(\sigma-\eta)(1+r)^{\sigma-1}\leq-\alpha\sigma(1+r)^{\sigma-1}+\frac{F_{r}}{|F_{v}|}\leq
0$

Plugging into (5.45) we get

$B(x,u)\geq\displaystyle\frac{K}{\tau}(1+r)^{\eta-\mu}+|\nabla
u|^{\chi+1}-|\nabla u|^{\chi}\alpha(\sigma-\eta)(1+r)^{\sigma-1},$ (5.58)

In view of (5.55), to satisfy (5.54) with

$P=\frac{K}{\tau}(1+r)^{\eta-\mu},\qquad
Q=\alpha(\sigma-\eta)(1+r)^{\sigma-1}$ (5.59)

we need the identities

$\begin{array}[]{ll}\displaystyle\frac{[\alpha(\sigma-\eta)]^{\chi+1}\tau}{K}=\theta^{\chi}\frac{(\chi+1)^{\chi+1}}{\chi^{\chi}}&\quad\text{if
}\,\chi>0;\\\\[11.38092pt]
\displaystyle\frac{\alpha(\sigma-\eta)\tau}{K}=1&\quad\text{if
}\,\chi=0.\end{array}$ (5.60)

According to whether $\chi=0$ or $>0$, we then define $\tau$ as the value such
that (5.60) holds. With this choice, (5.46) is satisfied with
$\Lambda=1-\theta$ (if $\chi>0$) or $\Lambda=1$ (if $\chi=0$), and in view of
our choice of $F$, (5.52) becomes

$\begin{array}[]{lcl}\displaystyle\int_{B_{R_{0}}\cap\Omega_{\gamma}}\lambda
F(1+r)^{-\eta}\varphi(|\nabla u|)|\nabla u|&\leq&\displaystyle
C_{4}\left[\frac{1}{R^{p}}\int_{(B_{R}\backslash B_{\delta
R})\cap\Omega_{\gamma}}F(1+r)^{\eta(p-1)}\right.\\\\[14.22636pt]
&&\displaystyle\left.+\frac{1}{R^{\bar{p}}}\int_{(B_{R}\backslash B_{\delta
R})\cap\Omega_{\gamma}}F(1+r)^{\eta(\bar{p}-1)}\right]\end{array}$ (5.61)

Where $C_{4}$ also depends on $\tau$. Up to increasing $\bar{p}$, a change
that does not alter the validity of (5.3), we can suppose that $\bar{p}\geq
p$. On $(B_{R}\backslash B_{\theta R})\cap\Omega_{\gamma}$, (5.40) and
$\sigma-\eta>0$, $\eta<0$ give

$F(v,r)\leq\exp\left(-\tau(\alpha-\beta)(\delta
R)^{\sigma-\eta}\right),\qquad(1+r)^{\eta(\bar{p}-1)}\leq(1+r)^{\eta(p-1)}\leq
1,$

Inserting into (5.61) and using $R^{\bar{p}}\geq R^{p}$, we eventually get

$0<\displaystyle\int_{\Omega_{\gamma}\cap B_{R_{0}}}\lambda
F(1+r)^{-\eta}\varphi(|\nabla u|)|\nabla
u|\leq\frac{C_{5}}{R^{p}}\exp\left(-\tau(\alpha-\beta)(\delta
R)^{\sigma-\eta}\right)\mathrm{vol}(B_{R}).$ (5.62)

Because of (5.35) and (5.56), for each $d>d_{0}$ there exists a sequence
$\\{R_{k}\\}\uparrow\infty$ such that

$\mathrm{vol}(B_{R_{k}})\leq\exp\left\\{dR_{k}^{\sigma-\eta}\right\\}.$

Substituting into (5.62) and letting $k\rightarrow\infty$,

$\begin{array}[]{lcl}0&<&\displaystyle\int_{\Omega_{\gamma}\cap
B_{R_{0}}}\lambda F(1+r)^{-\eta}\varphi(|\nabla u|)|\nabla u|\\\\[14.22636pt]
&\leq&\displaystyle
C_{5}\limsup_{k\rightarrow\infty}\left(\frac{\exp\left\\{-\tau(\alpha-\beta)\delta^{\sigma-\eta}R_{k}^{\sigma-\eta}+dR_{k}^{\sigma-\eta}\right\\}}{R_{k}^{p}}\right).\end{array}$
(5.63)

Being the left-hand side of the above inequality strictly bigger than zero, we
deduce that necessarily $d\geq\tau(\alpha-\beta)\delta^{\sigma-\eta}$, and
letting $\delta\rightarrow 1$ and $d\downarrow d_{0}$. we get

$d_{0}\geq\tau(\alpha-\beta)$ (5.64)

Substituting the expression of $\tau$ in (5.60), setting $\alpha=t\beta$ with
$t>1$ and letting $\theta\rightarrow 1$ we deduce

$\begin{array}[]{ll}\displaystyle K\leq
d_{0}\frac{\chi^{\chi}}{(\chi+1)^{\chi+1}}(\sigma-\eta)^{\chi+1}\frac{t^{\chi+1}}{t-1}\beta^{\chi}&\quad\text{if
}\,\chi>0;\\\\[11.38092pt] \displaystyle K\leq
d_{0}(\sigma-\eta)\frac{t}{t-1}&\quad\text{if }\,\chi=0.\\\\[11.38092pt]
\end{array}$

Minimizing with respect to $t\in(1,\infty)$ and letting
$\beta\downarrow\hat{u}$, we eventually get

$\begin{array}[]{ll}K\leq
d_{0}(\sigma-\eta)^{\chi+1}\hat{u}^{\chi}=d_{0}\big{[}\zeta-\chi\sigma\big{]}^{\chi+1}\hat{u}^{\chi}&\quad\text{if
}\,\chi>0;\\\\[5.69046pt] K\leq d_{0}(\sigma-\eta)=d_{0}(1-\mu)&\quad\text{if
}\,\chi=0,\\\\[5.69046pt] \end{array}$ (5.65)

as claimed. We conclude by investigating part of case $(i)$, that is, when

$\chi>0,\quad\sigma=0<\zeta.$

Observe that a downward translation $u_{s}$ of $u$ still satisfies (5.32) with
the same constant $K$ (without loss of generality, we can suppose $\gamma=0$).
Hence, by (5.65),

$K\leq H\hat{u}_{s}^{\chi}\qquad\text{with }\quad H\leq
d_{0}\big{[}\zeta-\chi\sigma\big{]}^{\chi+1}.$ (5.66)

If $K>0$, since $u$ is bounded above and $\chi>0$ we can choose $u_{s}^{\chi}$
positive and small enough to contradict (5.66). Hence, necessarily $K=0$, and
a-posteriori we can choose $H=0$ in (5.38) as required. At the end of the
present proof, with the same trick we investigate the remaining case of $(i)$,
that is, when $\sigma=\zeta=0$. Note that the trick is not possible if
$\chi=0$, being $u_{s}^{\chi}\equiv 1$.

Analysis of case $(iii)$, and case $(iv)$ for $\sigma\geq 1-\mu$:

$\chi>0,\quad\frac{\chi}{\chi+1}\zeta\leq\chi\sigma<\zeta,\qquad\text{or}\qquad\chi=0,\quad\sigma\geq
1-\mu>0.$

Again from the definition of $\eta$, these cases correspond to the range

$0\leq\eta<\sigma,$

for which we choose

$F(v,r)=\exp\left\\{-\tau v^{\frac{\sigma-\eta}{\sigma}}\right\\}.$ (5.67)

Also in these cases, we increase $\bar{p}$ in order for $\bar{p}\geq p$ to
hold, since ultimately the size of $\bar{p}$ will not affect the conclusion of
the theorem. Performing computations analogous to the ones giving $(ii)$, we
obtain the desired estimate

$\begin{array}[]{ll}\displaystyle K\leq
d_{0}\sigma^{\chi}(\sigma-\eta)\hat{u}^{\chi}=d_{0}\sigma^{\chi}\big{[}\zeta-\chi\sigma\big{]}\hat{u}^{\chi}&\quad\text{if
}\,\chi>0;\\\\[8.5359pt] \displaystyle K\leq
d_{0}(\sigma-\eta)=d_{0}(1-\mu)&\quad\text{if }\,\chi=0.\\\\[8.5359pt]
\end{array}$

Analysis of case $(v)$, and case $(i)$ for $\zeta=0$:

$\chi\sigma=\zeta>0,\quad\text{or}\quad\chi\sigma=\zeta=0.$

In this case, by (5.55) it holds $\sigma-\mu=(\sigma-1)(\chi+1)$. We choose

$F(v,r)=v^{-\tau},$ (5.68)

$\tau>0$ to be determined. Then, using (5.40),

$\begin{array}[]{lcl}B(x,u)&\geq&\displaystyle\frac{K}{\tau}(1+r)^{-\mu}v+|\nabla
u|^{\chi+1}-|\nabla u|^{\chi}\alpha\sigma(1+r)^{\sigma-1}\\\\[14.22636pt]
&\geq&\displaystyle\frac{K(\alpha-\beta)}{\tau}(1+r)^{\sigma-\mu}+|\nabla
u|^{\chi+1}-|\nabla u|^{\chi}\alpha\sigma(1+r)^{\sigma-1},\end{array}$ (5.69)

and (5.54) applied with

$P=\frac{K(\alpha-\beta)}{\tau}(1+r)^{\sigma-\mu},\qquad
Q=\alpha\sigma(1+r)^{\sigma-1}$

implies the identities

$\begin{array}[]{ll}\displaystyle\frac{(\alpha\sigma)^{\chi+1}\tau}{K(\alpha-\beta)}=\theta^{\chi}\frac{(\chi+1)^{\chi+1}}{\chi^{\chi}}&\quad\text{if
}\,\chi>0;\\\\[11.38092pt]
\displaystyle\frac{\alpha\sigma\tau}{K(\alpha-\beta)}=1&\quad\text{if
}\,\chi=0.\end{array}$ (5.70)

Having specified $\tau$ to satisfy (5.70), (5.46) is met and (5.52) with
$\delta=3/4$ now reads

$0<\displaystyle\int_{\Omega_{\gamma}\cap
B_{R_{0}}}\lambda\left|F_{v}\right|\varphi(|\nabla u|)|\nabla
u|\leq\displaystyle C_{4}\int_{(B_{R}\backslash
B_{3R/4})\cap\Omega_{\gamma}}\frac{v^{p-1-\tau}}{R^{p}}+\frac{v^{\bar{p}-1-\tau}}{R^{\bar{p}}}.$
(5.71)

Estimating $v$ with the aid of (5.40), choosing the upper or lower bound
according to the sign of $p-1-\tau$ and $\bar{p}-1-\tau$, the right-hand side
is bounded from above by

$C_{5}\mathrm{vol}(B_{R})\Big{(}R^{-p+\sigma(p-1-\tau)}+R^{-\bar{p}+\sigma(\bar{p}-1-\tau)}\Big{)},$

for a suitable constant $C_{5}>0$. Because of $\eqref{volgrowth}$, for
$d>d_{0}$ we consider a sequence $\\{R_{k}\\}$ for which
$\mathrm{vol}(B_{R_{k}})\leq R_{k}^{d}$. Evaluating (5.52) on $R_{k}$, and
letting $k\rightarrow\infty$ in (5.71) and then $d\downarrow d_{0}$, we deduce
that necessarily

$d_{0}\geq\min\big{\\{}p-\sigma(p-1-\tau),\bar{p}-\sigma(\bar{p}-1-\tau)\big{\\}},$

that is, by (5.37),

$\tau\sigma\leq
d_{0}-\min\Big{\\{}p-\sigma(p-1),\bar{p}-\sigma(\bar{p}-1)\Big{\\}}=d_{0}-d^{*}.$
(5.72)

  * (i)

If $d_{0}<d^{*}$ or $d_{0}=d^{*}$ and $\sigma>0$, then there exists no
$\tau>0$ satisfying (5.72). Thus, $K>0$ leads to a contradiction, and we can
therefore choose $H=0$. This proves $(v)$ for $d_{0}<d_{*}$ and for
$d_{0}=d^{*}$, $\sigma>0$, as well as $(i)$ for $\chi>0$, $\sigma=\zeta=0$,
$d_{0}<d^{*}$.

  * (ii)

If $d_{0}\geq d^{*}$ then, inserting the expression of $\tau$ obtained from
(5.70) in (5.72), setting $\alpha=t\beta$ for $t>1$, solving (5.72) with
respect to $K$, letting $\theta\uparrow 1$ and $\beta\downarrow\hat{u}$ we
deduce

$\begin{array}[]{ll}\displaystyle
K\leq\left[d_{0}-d^{*}\right]\frac{\chi^{\chi}}{(\chi+1)^{\chi+1}}\hat{u}^{\chi}\sigma^{\chi}\frac{t^{\chi+1}}{t-1}&\quad\text{if
}\,\chi>0;\\\\[11.38092pt] \displaystyle
K\leq\left[d_{0}-d^{*}\right]\frac{t}{t-1}&\quad\text{if
}\,\chi=0,\end{array}$

and minimizing over $t\in(1,\infty)$ we get for both $\chi>0$ and $\chi=0$

$K\leq\left[d_{0}-d^{*}\right]\sigma^{\chi}\hat{u}^{\chi}.$ (5.73)

This concludes the cases $d_{0}>d^{*}$ and $\sigma>0$, and $d_{0}\geq d^{*}$
and $\chi=0$. To deal with the remaining part of $(i)$, that is, $\chi>0$,
$\sigma=\zeta=0$ and $d_{0}\geq d^{*}$, we can consider a downward translation
$u_{s}$ of $u$ in place of $u$, and $\gamma=0$. Then, $u_{s}$ satisfies (5.32)
with the same constant $K$, hence (5.73) holds for each $\hat{u}_{s}$.
However, from $\chi>0$, $\hat{u}_{s}^{\chi}$ can be made as small as we wish,
and since we have assumed $K>0$ this would contradict (5.73). Concluding,
necessarily $K\leq 0$, and $H$ can be chosen to be zero, as required.

∎

We now prove Theorem 5.4

###### Proof of Theorem 5.4.

Suppose, by contradiction, that either $u^{*}=\infty$ or $f(u^{*})>0$. Because
of the second in (5.21) and the continuity of $f$, in both of the cases there
exists $\gamma<u^{*}$ sufficiently close to $u^{*}$ such that $f(t)\geq C>0$
on $(\gamma,\infty)$, for some constant $C>0$. By (5.21), $u$ would solve

$\Delta_{\varphi}u\geq K(1+r)^{-\mu}\frac{\varphi(|\nabla u|)}{|\nabla
u|^{\chi}}\qquad\text{on }\,\Omega_{\gamma},$

For some $K>0$. To apply Theorem 5.15, we shall consider

$d^{*}=\min\Big{\\{}p-\sigma(p-1),\bar{p}-\sigma(\bar{p}-1)\Big{\\}}.$

As said before, $p$ (respectively, $\bar{p}$) in (5.3) can be reduced (resp.
increased) as much as we wish, still keeping the validity of (5.3). Therefore,

  * -

if $\sigma\leq 1$, we can increase $\bar{p}$ up to satisfy $\bar{p}\geq p$,
that gives $d^{*}=p-\sigma(p-1)$. In particular, $d^{*}=p$ when $\sigma=0$;

  * -

if $\sigma>1$, we can reduce $p$ to satisfy $\bar{p}\geq p$, that now implies
$d^{*}=\bar{p}-\sigma(\bar{p}-1)$.

The volume growth conditions in the second lines of (5.6) and (5.9) can
therefore be rewritten as

$\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{\log r}\leq
d^{*}.$

We are in the position to apply Theorem 5.15 and deduce $0<K\leq
H\hat{u}^{\chi}$. We reach a contradiction by proving that either $H=0$ or
$\hat{u}^{\chi}=0$. We split into cases:

  * -

If $\sigma=0$ and $\chi>0$, then we are in case $(i)$ of Theorem 5.15 and
(5.35) is satisfied because of (5.6), thus $H=0$, contradiction.

  * -

If $\sigma\geq 0$ and $\chi=0$, then we are either in case $(iv)$ or in case
$(v)$ of Theorem 5.15, according to whether $\mu<1$ or $\mu=1$. In case
$(iv)$, our growth requirements (5.6) and (5.9) for $\chi=0$, $\mu<1$ imply
that $d_{0}=0$ in (5.35). Applying Theorem 5.15 we get $H\leq d_{0}(1-\mu)=0$,
contradiction. In case $(v)$, as said conditions (5.6) and (5.9) for $\chi=0$,
$\mu=1$ are equivalent to $d_{0}\leq d^{*}$. By Theorem 5.15,
$H\leq\sigma^{\chi}(d_{0}-d^{*})_{+}=0$, contradiction.

  * -

If $\sigma>0$, $\chi>0$, then by (5.7) we get $\hat{u}^{\chi}=0$,
contradiction.

We have thus proved $u^{*}<\infty$ and $f(u^{*})\leq 0$. If now $u$ satisfies
$(P_{=})$ and (5.10) is in force, we can apply the result both to $u$ and to
$\bar{u}=-u$, noting that

$\Delta_{\varphi}\bar{u}=b(x)\bar{f}(\bar{u})l(|\nabla\bar{u}|),\qquad\bar{f}(t)=-f(-t),$

and that $\bar{f}(t)\geq C$ for $t$ large enough. From
$\bar{f}(\bar{u}^{*})\leq 0$ we get $f(u_{*})\geq 0$, and (5.11) follows. ∎

######  Remark 5.16.

_We now check Remark 5.5. Since the third in (5.9) corresponds to case $(v)$
of Theorem 5.15, if $d_{0}\leq d^{*}$ we achieve the contradiction $0<K\leq
H\hat{u}^{\chi}$ irrespectively to the vanishing of $\hat{u}$, as claimed. _

####  5.2.1 Bernstein theorems for minimal and MCF soliton graphs

We now apply Theorem 5.4 to deduce some Bernstein type results for prescribed
mean curvature graphs. We consider an ambient space $(\bar{M}^{m+1},(\,,\,))$
with the warped product structure

$\bar{M}=\mathbb{R}\times_{h}M,\qquad(\,,\,)=\mathrm{d}s^{2}+h(s)^{2}\langle\,,\,\rangle$
(5.74)

for some complete manifold $(M^{m},\langle\,,\,\rangle)$ and some $0<h\in
C^{\infty}(\mathbb{R})$. Given $v:M\rightarrow\mathbb{R}$, we consider the
graph

$\Sigma=\Big{\\{}(v(x),x)\in\bar{M}\ \ :\ \ x\in M\Big{\\}}.$

In the next theorems we always consider _entire_ graphs, that is, graphs
defined on all of $M$. As in Subsection 1.1 in the Introduction, we let
$\Phi_{t}$ be the flow of the conformal field $X=h(s)\partial_{s}$, and we
note that its flow parameter $t$ starting from the slice $\\{s=0\\}$ satisfies
(1.9):

$t=\int_{0}^{s}\frac{\mathrm{d}\sigma}{h(\sigma)},\qquad
t:\mathbb{R}\rightarrow t(\mathbb{R})=I.$ (5.75)

We define $\lambda(t)=h(s(t))$, $u(x)=t(v(x))$, and note that $u:M\rightarrow
I$.

######  Theorem 5.17.

Let $(M^{m},\langle\,,\,\rangle)$ be a complete manifold, and consider the
warped product $\bar{M}=\mathbb{R}\times_{h}M$. Suppose that $h$ satisfies:

$\left\\{\begin{array}[]{l}h^{\prime}(s)s\geq Cs\qquad\text{for }\,|s|\geq
r_{0};\\\\[8.5359pt] h^{-1}\in L^{1}(+\infty)\cap
L^{1}(-\infty),\end{array}\right.$ (5.76)

for some constants $C,s_{0}>0$. If the volume growth of geodesic balls in $M$
satisfies

$\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{2}}<\infty,$
(5.77)

Then every entire, geodesic minimal graph $\Sigma$ over $M$ is bounded and,
letting $v:M\rightarrow\mathbb{R}$ be the graph function,

$h^{\prime}(v^{*})\leq 0\leq h^{\prime}(v_{*}).$ (5.78)

In particular, if $h$ is strictly convex, $\Sigma$ is the totally geodesic
slice $\\{s=s_{1}\\}$, where $s_{1}$ is the unique minimum of $h$.

###### Proof.

By (1.11) and the minimality of $\Sigma$, $u$ solves

$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}}\right)=m\frac{\lambda_{t}(u)}{\lambda(u)}\frac{1}{\sqrt{1+|\nabla
u|^{2}}}$

The second in (5.76) implies that $I$ is a bounded interval, and thus
$u:M\rightarrow I$ is bounded. Taking into account that

$f(t)=\frac{\lambda_{t}(t)}{\lambda(t)}=h^{\prime}(s(t))\geq C\qquad\text{for
}\,t\geq t(s_{0}),$ (5.79)

we get

$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}}\right)\geq\frac{mC}{\sqrt{1+|\nabla u|^{2}}}\qquad\text{on
}\,\\{u>t(s_{0})\\}.$

If the set $\\{u>t(s_{0})\\}$ were non-empty, we can apply Theorem 5.15 with
$\mu=0$, $\chi=1$, $p=\bar{p}=2$ and $K=mC$ to obtain a contradiction.
Therefore, $u$ is bounded from above and from Theorem 5.4 we deduce
$f(u^{*})\leq 0$. Analogously, considering $-u$ we infer $f(u_{*})\geq 0$, and
thus (5.78) follows by changing variables. Note that, because of the first in
(5.76), (5.78) implies the boundedness of $v$. If $h$ is strictly convex, then
by (5.76) it has a unique stationary point (a minimum) $s_{1}$, and (5.78)
gives $v^{*}\leq s_{1}\leq v_{*}$, that is, $v\equiv s_{1}$. We remark that
each slice $\\{s=s_{2}\\}$ is totally umbilical in $\bar{M}$, with second
fundamental form $h^{\prime}(s_{2})/h(s_{2})\langle\,,\,\rangle$ in the
direction $-\partial_{s}$. ∎

######  Theorem 5.18.

Let $\bar{M}=\mathbb{R}\times_{h}M$ be as above, and suppose that $h$
satisfies:

$\left\\{\begin{array}[]{l}h^{\prime}>0\quad\text{on
}\,\mathbb{R},\\\\[8.5359pt] h^{\prime}(s)\geq C\qquad\text{for }\,s\geq
s_{0};\\\\[8.5359pt] h^{-1}\in L^{1}(+\infty),\end{array}\right.$ (5.80)

for some constants $C,s_{0}>0$. If $M$ is complete and the volume growth of
geodesic balls in $M$ satisfies

$\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{2}}<\infty,$
(5.81)

Then there are no entire, geodesic minimal graphs over $M$.

###### Proof.

The proof is analogous. By the third in (5.80), $u:M\rightarrow I$ is bounded
from above, hence applying Theorems 5.15 and 5.4 we get $h^{\prime}(v^{*})\leq
0$, contradicting the first in (5.80). ∎

###### Proof of Theorem 1.22.

It is immediate from Theorems 5.17 and 5.18, respectively for cases $(i)$ and
$(ii)$. Note that, in $(i)$, the strict convexity of $h$ and $h^{-1}\in
L^{1}(-\infty)\cap L^{1}(+\infty)$ imply that $sh^{\prime}(s)\geq Cs$ for
$|s|$ large enough. ∎

######  Remark 5.19.

_As a direct corollary we deduce Do Carmo-Lawson’s result[43] in the case
$H=0$, Theorem 1.3 in the Introduction: there are no geodesic, entire minimal
graphs over horospheres, and the only geodesic, entire minimal graph over a
totally geodesic hypersphere $M$ is $M$ itself. To see the first claim, apply
Theorem 5.18 to the warped product
$\mathbb{H}^{m+1}=\mathbb{R}\times_{e^{s}}\mathbb{R}^{m}$, and note that
$h(s)=e^{s}$ satisfies all the requirements in (5.80). For the second claim,
apply Theorem 5.17 to $\mathbb{H}^{m+1}=\mathbb{R}\times_{\cosh
s}\mathbb{H}^{m}$. _

The case of non-constant mean curvature will be investigated in Section 6.
Now, we focus on MCF solitons, starting with product ambient manifolds. Our
first result is for self-translators, Theorem 1.23 in the Introduction:

###### Proof of Theorem 1.23.

In the product case, $h(s)=1$ and thus $t=s$, $\lambda(t)=1$,
$u(x)=t(v(x))=v(x)$. By (1.15), a soliton for the field $\partial_{s}$
satisfies the equation

$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}}\right)=\frac{1}{\sqrt{1+|\nabla u|^{2}}}\qquad\text{on }\,M.$

We apply the first part of Theorem 5.4 with the choices $f(t)\equiv 1$,
$b\equiv 1$, $\chi=1$, $\mu=0$, $p=\bar{p}=2$: since the first two in (5.9)
correspond to (1.48), we infer from Theorem 5.4 that $u$ is bounded from above
and $f(u^{*})\leq 0$, contradiction. ∎

We next examine more closely the case of self-translators in Euclidean space,
in particular the case when the translation vector field $Y$ differs from the
vertical field $\partial_{s}$.

######  Theorem 5.20.

Let $(\mathbb{R}^{m+1},(\,,\,))=\mathbb{R}\times\mathbb{R}^{m}$ with
coordinates $(s,x)$, and let $\Sigma=\\{(v(x),x):x\in\mathbb{R}^{m}\\}$ be an
entire graph. Assume that

$\limsup_{r(x)\rightarrow\infty}\frac{|v(x)|}{r(x)}=\hat{v}<\infty$ (5.82)

Then, $\Sigma$ cannot be a self-translator with respect to any vector $Y$
whose angle $\vartheta\in(0,\pi/2)$ with the horizontal hyperplane
$\mathbb{R}^{m}$ satisfies

$\tan\vartheta>\hat{v}.$ (5.83)

In particular, if $\hat{v}=0$, $\Sigma$ cannot be a self-translator with
respect to a vector $Y$ which is not tangent to the horizontal
$\mathbb{R}^{m}$.

###### Proof.

If we reflect $\Sigma$ with respect to the horizontal hyperplane, the
reflected graph is a self-translator with respect to the reflection of $Y$.
Therefore, without loss of generality we can assume that $(Y,\partial_{t})>0$,
and that $Y$ has unit norm, by time rescaling. Moreover, $Y\neq\partial_{t}$
since $\vartheta<\pi/2$. Therefore, up to a rotation of coordinates on
$\mathbb{R}^{m}$, $Y=\cos\vartheta e_{1}+\sin\vartheta\partial_{t}$, where
$e_{1}$ is the gradient of the first coordinate function $x_{1}$ and
$\sin\vartheta>0$. In view of (1.10) and (1.11), and since $h=1$, $t=s$, the
soliton equation $mH=(Y,\nu)$ satisfied by $u(x)=t(v(x))=v(x)$ reads

$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}}\right)=mH=\frac{1}{\sqrt{1+|\nabla
u|^{2}}}\left[\sin\vartheta-\cos\vartheta\langle\nabla u,e_{1}\rangle\right],$

Rearranging,

$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}}\right)+\left\langle\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}},\nabla(x_{1}\cos\vartheta)\right\rangle=\frac{\sin\vartheta}{\sqrt{1+|\nabla
u|^{2}}},$

that we rewrite as

$e^{-x_{1}\cos\vartheta}\mathrm{div}\left(e^{x_{1}\cos\vartheta}\frac{\nabla
u}{\sqrt{1+|\nabla u|^{2}}}\right)=\frac{\sin\vartheta}{\sqrt{1+|\nabla
u|^{2}}}.$ (5.84)

The operator in the left-hand side is in divergence form if we consider the
weighted volume measure $e^{x_{1}\cos\vartheta}\mathrm{d}x$, with
$\mathrm{d}x$ the Euclidean volume. For these weighted operators, the proof of
Theorems 5.15 and 5.4 follow verbatim by replacing the Euclidean volume with
the weighted volume

$\mathrm{vol}_{x_{1}\cos\vartheta}(B_{r})=\int_{B_{r}}e^{x_{1}\cos\vartheta}\mathrm{d}x.$

Explicit computation gives

$\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}_{x_{1}\cos\theta}(B_{r})}{r}=\cos\vartheta<\infty.$
(5.85)

Suppose by contradiction that (5.83) holds. In particular, $u$ is non-
constant. We apply Theorem 5.15 to (5.86) with the choices

$K=\sin\vartheta,\ \ \sigma=1,\ \ \chi=1,\ \ \mu=0,\ \ d_{0}=\cos\vartheta,\ \
p=\bar{p}=2,$

to conclude from case $(iii)$ in (5.38) that
$\sin\vartheta\leq(\cos\vartheta)\hat{u}$. Since $u=v$, we eventually
contradict (5.83). ∎

######  Remark 5.21.

_The requirement $\vartheta>0$, that is, $(Y,\partial_{t})\neq 0$, is
essential in the above theorem because otherwise the slices
$\\{s=\mathrm{const}\\}$ are trivial self-translators. Furthermore, condition
(5.83) is sharp. Indeed, a totally geodesic hyperplane $\Sigma$ corresponding
to an affine, non-constant solution $v$ is a self-translator with respect to
each vector $Y\in T\Sigma$, and in this case $\tan\vartheta=\hat{v}$. _

######  Remark 5.22.

_A result related to Theorem 5.20 appears in [10], where the authors proved
that there exist no nontrivial complete self-translator (i.e. not a
hyperplane) whose Gauss image lies in a geodesic ball of $\mathbb{S}^{m}$ of
radius $<\pi/2$. Their assumption implies that $\Sigma$ is a graph with
respect to the plane orthogonal to the center of the ball (not necessary the
direction of translation), and the graph function $v$ has bounded
gradient777In fact, they also require that $H$ be bounded, but this
automatically follows from the self-translator equation
$\overrightarrow{H}=Y^{\perp}$, being $Y$ constant.. Our requirements (5.82)
and (5.83) seem to be skew with their one. However, it might be possible that
a suitable gradient estimate guarantees that a self-translator satisfying
(5.82) has automatically bounded gradient. If this is the case, the main
result in [10] would imply that the graph $\Sigma$ of $\hat{v}$ be a
hyperplane, which would prove Theorem 5.20 in view of Remark 5.21. _

Our last application is for entire graphs
$\Sigma^{m}\rightarrow\mathbb{R}^{m+1}$ which are self-expanders for the mean
curvature flow, that is, they move under MCF along the integral curves of the
position vector field

$Y(\bar{x})=(\bar{x}^{j}-q^{j})\partial_{j}=\frac{1}{2}\bar{\nabla}|\bar{x}-q|^{2}\qquad\forall\,\bar{x}\in\mathbb{R}^{m+1},$

for some fixed origin $q\in\mathbb{R}^{m+1}$.

######  Theorem 5.23.

Let $\Sigma=\\{(v(x),x):x\in\mathbb{R}^{m}\\}$ be an entire graph in
$(\mathbb{R}^{m+1},(\,,\,))$. If $\Sigma$ is a self-expander for the MCF and
$v$ is bounded, then $\Sigma$ is a hyperplane.

###### Proof.

Without loss of generality, we can assume that the center $q$ of the
homothetic field $Y$ is placed at the origin. We let
$(s,x)\in\mathbb{R}\times\mathbb{R}^{m}=\mathbb{R}^{m+1}$ be coordinates on
$\mathbb{R}^{m+1}$. If $\rho(x):\mathbb{R}^{m}\rightarrow\mathbb{R}$ denotes
the distance to the origin in $\mathbb{R}^{m}$, then

$Y\big{(}v(x),x\big{)}=x^{j}\partial_{j}+v\partial_{s}=\frac{1}{2}\nabla(\rho^{2})+v\partial_{s}.$

In view of (1.10) and (1.11), and since $h=1$, $t=s$, the soliton equation
$mH=(Y,\nu)$ satisfied by $u(x)=t(v(x))=v(x)$ reads

$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}}\right)=mH=\frac{1}{\sqrt{1+|\nabla u|^{2}}}\left[u-\langle\nabla
u,\nabla\frac{\rho^{2}}{2}\rangle\right].$

Rearranging,

$\mathrm{div}\left(\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}}\right)+\left\langle\frac{\nabla u}{\sqrt{1+|\nabla
u|^{2}}},\nabla\frac{\rho^{2}}{2}\right\rangle=\frac{u}{\sqrt{1+|\nabla
u|^{2}}},$

that we rewrite as

$e^{-\rho^{2}/2}\mathrm{div}\left(e^{\rho^{2}/2}\frac{\nabla
u}{\sqrt{1+|\nabla u|^{2}}}\right)=\frac{u}{\sqrt{1+|\nabla u|^{2}}}.$ (5.86)

Suppose that $v$ is non-constant. An explicit computation shows that

$\liminf_{r\rightarrow\infty}\frac{\log\mathrm{vol}_{e^{\rho^{2}/2}}(B_{r})}{r^{2}}<\infty.$

Since, by assumption, $v$ is bounded, we can apply Theorem 5.4 (adapted to
weighted volumes) with the choices $\chi=1$, $\mu=0$, $p=\bar{p}=2$, $f(t)=t$
to deduce $f(u^{*})\leq 0$, that is, $u\leq 0$. Applying the same theorem to
$-u$ we infer $u\equiv 0$, that is, $\Sigma$ is a hyperplane containing the
origin of $Y$. ∎

######  Remark 5.24.

_It is worth to stress that, differently from[46, 152, 10], Theorems 1.23,
5.20 and 5.23 _do not use the conformality of the soliton vector field $Y$._
In fact, what is needed to apply our results, possibly with a weighted volume,
is just that the field $Y$ can be decomposed as_

$Y(s,x)=h^{2}(s)\nabla^{\Sigma_{s}}g+\beta(s,x)\partial_{s}$

_where $\Sigma_{s_{0}}=\\{s=s_{0}\\}$, $g\in C^{1}(\mathbb{R}\times M)$ just
depends on the second variable, and $\beta\in C(M\times\mathbb{R})$. _

###  5.3 Counterexamples

To show the sharpness of Theorem 5.4, we consider a model manifold $M_{g}^{m}$
with radial sectional curvature

$K_{\mathrm{rad}}=-\kappa^{2}\big{(}1+r^{2}\big{)}^{\alpha/2}\qquad\text{for
}\,r(x)\geq 1,$

for some $\kappa>0$ and $\alpha\geq-2$. By Propositions 2.1 and 2.11 in [113],
as $r(x)\rightarrow\infty$

$\Delta r\geq\left\\{\begin{array}[]{ll}\displaystyle(m-1)\kappa
r^{\alpha/2}(1+o(1))&\quad\text{if }\,\alpha>-2;\\\\[8.5359pt]
\displaystyle\frac{(m-1)\bar{\kappa}}{r}(1+o(1))&\quad\text{if
}\,\alpha=-2,\end{array}\right.$ (5.87)

where $\bar{\kappa}=(1+\sqrt{1+4\kappa^{2}})/2$, and thus

$\log\mathrm{vol}B_{r}\sim\left\\{\begin{array}[]{ll}r^{1+\frac{\alpha}{2}}&\quad\text{if
}\,\alpha>-2;\\\\[5.69046pt] \big{[}(m-1)\bar{\kappa}+1\big{]}\log
r&\quad\text{if }\,\alpha=-2,\end{array}\right.$ (5.88)

as $r\rightarrow\infty$. If $\alpha=-2$, letting $\kappa\rightarrow 0$ we
deduce the classical expressions for $\Delta r$ and $\mathrm{vol}(B_{r})$ for
the Euclidean space. We shortly write the inequalities in (5.87) as $\Delta
r\geq\zeta r^{\alpha/2}$, where $\zeta$ tends to $(m-1)\kappa$ (if
$\alpha>-2$) or to $(m-1)\bar{\kappa}$ (if $\alpha=-2$) as
$r(x)\rightarrow\infty$. We are going to find an operator $\Delta_{\varphi}$
meeting (1.3) and (5.3) with the following property: given $\mu\in\mathbb{R}$,
$0\leq\chi\leq p-1$ and $\sigma>0$, we will construct an unbounded, radial
solution $u\in C^{\infty}(M_{g}\backslash\\{o\\})$, increasing as a function
of $r$, solving

$\Delta_{\varphi}u\geq K(1+r)^{-\mu}\frac{\varphi(|\nabla u|)}{|\nabla
u|^{\chi}}$ (5.89)

if $r$ is large enough (equivalently, for a high enough upper level set). The
solution is $u(x)=r(x)^{\sigma}$ whenever one of the following conditions
hold:

$\begin{array}[]{ll}1)&\quad\chi\sigma>\chi+1-\mu;\\\\[5.69046pt]
2)&\quad\chi\sigma=\chi+1-\mu,\qquad\text{and}\quad\alpha>-2;\\\\[5.69046pt]
3)&\quad\chi\sigma=\chi+1-\mu,\qquad\alpha=-2,\qquad\sigma\in(0,1]\qquad\text{and}\\\\[5.69046pt]
&\displaystyle\lim_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{\log
r}>p-\sigma(p-1);\\\\[11.38092pt]
4)&\quad\chi\sigma<\chi+1-\mu,\qquad\alpha>-2,\qquad\text{and}\\\\[5.69046pt]
&\displaystyle\lim_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{\chi+1-\mu-\chi\sigma}}>0.\end{array}$
(5.90)

Moreover, the solution is $u(x)=r(x)^{\sigma}/\log r(x)$ when either

$\begin{array}[]{ll}5)&\quad\chi\sigma<\chi+1-\mu,\qquad\chi>0,\qquad\text{and}\\\\[5.69046pt]
&\displaystyle\lim_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{\chi+1-\mu-\chi\sigma}}=\infty,\qquad\text{or}\\\\[11.38092pt]
6)&\quad\chi\sigma<\chi+1-\mu,\qquad\chi=0,\qquad\text{and}\\\\[5.69046pt]
&\displaystyle\lim_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{\chi+1-\mu-\chi\sigma}}\in(0,\infty).\end{array}$
(5.91)

In any of $1),\ldots,6)$, observe that the bound $\mu\leq\chi+1$ s not needed.
Once we establish (5.89) on $\Omega_{\gamma}$ for large enough $\gamma>0$, we
can choose $f\in C(\mathbb{R})$ increasing and satisfying $f\equiv 0$ on
$(0,\gamma)$, $f=K$ on $(2\gamma,\infty)$, $0\leq f\leq K$ on $\mathbb{R}$. By
the pasting Lemma, $\bar{u}=\max\\{u,\gamma\\}$ is a
$\mathrm{Lip}_{\mathrm{loc}}$ solution of

$\Delta_{\varphi}\bar{u}\geq(1+r)^{-\mu}f(\bar{u})\frac{\varphi(|\nabla\bar{u}|)}{|\nabla\bar{u}|^{\chi}}\qquad\text{on
}\,M.$ (5.92)

(here we used $\chi\leq p-1$, to guarantee that $\varphi(t)/t^{\chi}$ does not
diverge as $t\rightarrow 0$ and hence that the constant $\gamma$ solves
(5.92)). The existence of $\bar{u}$ under any of $1),\ldots,6)$ above shows
the sharpness of the parameter ranges (5.4) and (5.8), and of the growth
conditions (5.7) for $u$ and (5.9) for $\mathrm{vol}(B_{r})$. In particular,

  * -

in $(2)$, all the assumptions are satisfied but the third in (5.9), where the
liminf is $\infty$, while in $(3)$, the liminf is finite but bigger than the
threshold $p-\sigma(p-1)$ for $\sigma\leq 1$;

  * -

in $(4)$ the requirements in the first of (5.9) are all met, but
$u_{+}=O(r^{\sigma})$ instead of $u_{+}=o(r^{\sigma})$.

  * -

in $(5)$ and $(6)$, $u_{+}=o(r^{\sigma})$ but the requirements in the first of
(5.9) barely fail.

To show (5.89) first note that, because of (5.88), the volume growth
conditions in $3)$ and $4)$ are equivalent, respectively, to

$(m-1)\bar{\kappa}+1>p-\sigma(p-1),\qquad\text{and}\qquad\frac{\alpha}{2}+1\geq\chi+1-\mu-\chi\sigma.$
(5.93)

Consider

$\varphi(t)=\frac{t^{p-1}}{(1+t)^{q-1}},\quad p>1,\ 1\leq q\leq p,$ (5.94)

and we search for solutions of the form $u(x)=h\big{(}r(x)\big{)}$, for some
increasing $0<h\in C^{2}(\mathbb{R}^{+})$. Then,

$\begin{array}[]{lcl}\Delta_{\varphi}u&=&\displaystyle\varphi^{\prime}(h^{\prime})h^{\prime\prime}+\varphi(h^{\prime})\Delta
r\geq\varphi^{\prime}(h^{\prime})h^{\prime\prime}+\zeta\varphi(h^{\prime})r^{\alpha/2}\\\\[11.38092pt]
&=&\displaystyle\frac{(h^{\prime})^{p-2}}{(1+h^{\prime})^{q}}\Big{\\{}\left[(p-1)+(p-q)h^{\prime}\right]h^{\prime\prime}+\zeta
r^{\alpha/2}h^{\prime}(1+h^{\prime})\Big{\\}}.\end{array}$ (5.95)

Set $h(t)=t^{\sigma}$, $\sigma>0$. Then,

$\begin{array}[]{lcl}\Delta_{\varphi}u&\geq&\displaystyle
C_{1}\frac{r^{(\sigma-1)(p-1)-1}}{(1+\sigma
r^{\sigma-1})^{q}}\cdot\\\\[11.38092pt]
&&\displaystyle\cdot\Big{\\{}\left[(p-1)+\sigma(p-q)r^{\sigma-1}\right](\sigma-1)+\zeta
r^{\alpha/2+1}(1+\sigma r^{\sigma-1})\Big{\\}}\end{array}$ (5.96)

for some $C_{1}(p,\sigma)>0$. If $\sigma\geq 1$, getting rid of the first term
in brackets and using the definition of $\zeta$ we obtain

$\Delta_{\varphi}u\geq
C_{2}\frac{r^{(\sigma-1)(p-1)-1+\left(\alpha/2+1\right)}}{(1+\sigma
r^{\sigma-1})^{q-1}}$ (5.97)

On the other hand, if $\sigma\in(0,1]$ and $\alpha>-2$, the term in between
brackets in (5.96) is

$(p-1)(\sigma-1)(1+o(1))+\zeta r^{\alpha/2+1}(1+o(1))\geq C_{3}\zeta
r^{\alpha/2+1},$ (5.98)

for large enough $r$, and the last inequality of (5.97) still holds. If
$\sigma\in(0,1)$ and $\alpha=-2$, the term is

$\big{[}(p-1)(\sigma-1)+\zeta\big{]}(1+o(1)),$

while if $\sigma=1$, $\alpha=-2$, the term is simply $2\zeta$. If $r$ is large
enough, the last expression is bounded from below by a positive constant
whenever $\lim_{r\rightarrow\infty}\zeta>-(\sigma-1)(p-1)$, that is, when the
first in (5.93) is met (i.e. the volume growth in $3)$ of (5.90)).
Summarizing, for each $\sigma>0$ (if $\sigma\leq 1$ and $\alpha=-2$, under the
growth condition in $3)$ of (5.90)), the function $u$ turns out to solve
(5.97) for suitable constant $C_{2}>0$. Since

$(1+r)^{-\mu}\frac{\varphi(|\nabla u|)}{|\nabla u|^{\chi}}\leq
C_{4}\frac{r^{-\mu+(\sigma-1)(p-1-\chi)}}{(1+\sigma r^{\sigma-1})^{q-1}},$

(5.89) holds for $r$ large enough provided that

$-\mu+(\sigma-1)(p-1-\chi)\leq(\sigma-1)(p-1)-1+\left(\alpha/2+1\right),$

that is, simplifying, if $\frac{\alpha}{2}+1\geq\chi+1-\mu-\chi\sigma$. The
relation is automatically satisfied both in cases $1)$, $2)$ and $3)$ of
(5.90), and in case $4)$ it is equivalent to the growth condition. We have
thus shown (5.89), as required.

To prove $(5)$ and $(6)$, we use $h(t)=t^{\sigma}/\log t$ in (5.95) and we
consider for convenience the $p$-Laplacian ($q=0$) to obtain

$\displaystyle\displaystyle\Delta_{\varphi}u\geq
C_{1}\frac{r^{(\sigma-1)(p-1)-1}}{\log^{p-1}t}\Big{\\{}(p-1)(\sigma-1)\big{(}1+o(1)\big{)}+\zeta
r^{1+\frac{\alpha}{2}}\big{(}1+o(1)\big{)}\Big{\\}},$

as $r\rightarrow\infty$, for some constant $C_{1}(\sigma,p)>0$.

In order to meet the volume conditions in $(5),(6)$ and
$\chi\sigma<\chi+1-\mu$, necessarily $\alpha>-2$ and thus

$\displaystyle\displaystyle\Delta_{\varphi}u\geq
C_{2}\frac{r^{(\sigma-1)(p-1)+\frac{\alpha}{2}}}{\log^{p-1}t}\big{(}1+o(1)\big{)}$

as $r\rightarrow\infty$. On the other hand, by the definition of
$\Delta_{\varphi}$ on $\Omega_{\gamma}$ we have

$(1+r)^{-\mu}\frac{\varphi(|\nabla u|)}{|\nabla u|^{\chi}}\leq
C_{3}\frac{r^{-\mu+(\sigma-1)(p-1)-\chi(\sigma-1)}}{\log^{p-1}r}\log^{\chi}r$

Combining the last two inequalities, we examine two cases.

  * -

In case $(5)$, $\chi>0$ and $u$ solves (5.89) whenever $\alpha$ is big enough
to satisfy

$(\sigma-1)(p-1)+\frac{\alpha}{2}>-\mu+(\sigma-1)(p-1)-\chi(\sigma-1)$

that is,

$1+\frac{\alpha}{2}>\chi+1-\mu-\chi\sigma>0,$ (5.99)

that in view of (5.88) implies

$\lim_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{\chi+1-\mu-\chi\sigma}}=\infty,$

as required.

  * -

In case $(6)$, $\chi=0$ and for $u$ to solve (5.89) it is sufficient the weak
inequality

$1+\frac{\alpha}{2}\geq\chi+1-\mu-\chi\sigma>0,$

and choosing $\alpha$ in order to meet the equality sign, we obtain

$\lim_{r\to\infty}\frac{\log\mathrm{vol}(B_{r})}{r^{\chi+1-\mu-\chi\sigma}}\in(0,\infty),$

as required.

######  Remark 5.25.

_Example 5.94 satisfies (5.3) with $\bar{p}=p-q$. The case_

$\begin{array}[]{ll}3^{\prime})&\quad\chi\sigma=\chi+1-\mu,\qquad\alpha=-2,\qquad\sigma>1\qquad\text{and}\\\\[5.69046pt]
&\displaystyle\lim_{r\rightarrow\infty}\frac{\log\mathrm{vol}(B_{r})}{\log
r}>\bar{p}-\sigma(\bar{p}-1)\end{array}$

_is not covered by our counterexamples._

##  6 Strong maximum principle and Khas’minskii potentials

The aim of this section is to prove Theorem 1.14 in the Introduction. We
observe that the argument is based on the existence of what we call a
“Khas’minskii potential”, according to the following

######  Definition 6.1.

A Khas’minskii potential at $o\in M$ is a function $\bar{w}$ depending on the
parameters $r_{0}$, $r_{1}$, $\eta$, $K$, $\varepsilon>0$ satisfying the next
requirements:

$\left\\{\begin{array}[]{l}\Delta_{\varphi}\bar{w}\leq
Kb(x)l(|\nabla\bar{w}|)\qquad\text{on }\,M\backslash B_{r_{0}}\\\\[5.69046pt]
\bar{w}>0\qquad\text{on }\,M\backslash B_{r_{0}},\\\\[5.69046pt]
\bar{w}\leq\eta\qquad\text{ on }\,B_{r_{1}}\backslash
B_{r_{0}},\\\\[5.69046pt] \bar{w}(x)\rightarrow+\infty\qquad\text{as
}\,r(x)\rightarrow\infty,\\\\[5.69046pt]
|\nabla\bar{w}|\leq\varepsilon\quad\text{on }\,M\backslash
B_{r_{0}}\end{array}\right.$ (6.1)

The strategy to prove Theorem 1.14 is as follows: assuming, by contradiction,
that $(\mathrm{SMP}_{\infty})$ does not hold, the Khas’minskii potential $w$
will be compared to a non-constant, bounded solution $u\in C^{1}(M)$ of

$\Delta_{\varphi}u\geq Kb(x)l(|\nabla u|)$

on an appropriate subset of $\Omega_{\eta,\varepsilon}$ to reach the
conclusion. This approach is very old and we can trace it back, for instance,
to Phrágmen-Lindelöff in the realm of classical complex analysis. In more
recent times, it has been used by Redheffer [130, 129], and later refined in
[110, Sect. 6], [111, Thm. 18] and [2, Ch. 3]. A similar approach, although
with a somehow different point of view, was systematically used by Serrin
[143] and recently by Pucci-Serrin [123, Thm. 8.1.1]. In all of the quoted
results, the Khas’minskii potential (or variants thereof) is either a data of
the problem or it is constructed “ad-hoc”. These methods seem difficult to be
applied when a nontrivial gradient term $l$ is present. Thus, we provide the
existence of an appropriate potential via the solution of an associated ODE
problem, see Lemma 6.2 below, that will be coupled with a condition on the
Ricci tensor to transplant the function on the original manifold.

###  6.1 Ricci curvature and $(\mathrm{SMP}_{\infty})$

To investigate the ODE problem mentioned above, we assume the following:

$\left\\{\begin{array}[]{l}\displaystyle\varphi\in C(\mathbb{R}^{+}_{0})\cap
C^{1}(\mathbb{R}^{+}),\qquad\varphi(0)=0,\qquad\varphi^{\prime}>0\ \text{ on
}\,\mathbb{R}^{+},\\\\[5.69046pt] \displaystyle f\in C(\mathbb{R}),\qquad
f(0)=0,\qquad f>0\ \text{ on }\,\mathbb{R}^{+},\\\\[5.69046pt] \displaystyle
l\in C(\mathbb{R}^{+}_{0}),\qquad l>0\ \text{ on
}\,\mathbb{R}^{+},\\\\[5.69046pt] \beta\in
C(\mathbb{R}^{+}_{0}),\qquad\beta>0\ \text{ on
}\,\mathbb{R}^{+}_{0},\\\\[2.84544pt] \end{array}\right.$ (6.2)

and the next growth conditions:

$\left\\{\begin{array}[]{ll}\displaystyle\frac{t\varphi^{\prime}(t)}{l(t)}\in
L^{1}(0^{+}),&\\\\[11.38092pt] \displaystyle l(t)\geq
C_{1}\frac{\varphi(t)}{t^{\chi}}&\quad\text{on }\,(0,1]\ \text{ for some
}\,C_{1}>0,\ \chi\geq 0\\\\[11.38092pt] \displaystyle\varphi(t)\leq
Ct^{p-1}&\quad\text{on }\,[0,1],\ \text{ for some }\,C>0,\
p>1.\end{array}\right.$ (6.3)

Fix a “volume” function $v\in C^{2}(\mathbb{R}^{+})$ such that

$v>0\ \text{ on }\mathbb{R}^{+},\qquad v^{\prime}\geq 0\qquad\text{on
}\,\mathbb{R}^{+}.$ (6.4)

We are now ready to prove

######  Lemma 6.2.

Assume the validity of (6.2), (6.3), (6.4) and furthermore suppose that

$\begin{array}[]{l}\displaystyle\beta(r)\geq C(1+r)^{-\mu}\qquad\text{on
$\,\mathbb{R}^{+}_{0}$, for some $\,\mu\leq\chi+1$,}\\\\[8.5359pt]
\displaystyle\limsup_{r\rightarrow\infty}\frac{1}{v(r)}\int_{r_{0}}^{r}\frac{v(s)}{(1+s)^{\mu}}\mathrm{d}s<\infty,\end{array}$
(6.5)

for some (hence any) $r_{0}>0$, and that either

$\begin{array}[]{l}\displaystyle\mu<\chi+1\quad\text{ and
}\quad\displaystyle\liminf_{t\rightarrow\infty}\frac{\log\int_{r_{0}}^{t}v}{t^{\chi+1-\mu}}<\infty\quad(=0\
\text{ if }\,\chi=0),\ \ \text{ or}\\\\[11.38092pt]
\displaystyle\mu=\chi+1\quad\text{ and
}\quad\displaystyle\liminf_{t\rightarrow\infty}\frac{\log\int_{t_{0}}^{t}v}{\log
t}<\infty\quad(\leq p\ \text{ if }\,\chi=0).\\\\[11.38092pt] \end{array}$
(6.6)

Then, for each $0<r_{0}<r_{1}$, $\eta>0$ and $\varepsilon>0$ there exists a
function $w\in C^{1}([r_{0},\infty))$ satisfying:

$\left\\{\begin{array}[]{l}\big{[}v\varphi(w^{\prime})\big{]}^{\prime}\leq
v\beta f(w)l(|w^{\prime}|)\qquad\text{on }\,[r_{0},\infty)\\\\[5.69046pt]
w>0,\ w^{\prime}>0\qquad\text{on }\,[r_{0},\infty),\\\\[5.69046pt]
w\leq\eta\qquad\text{ on }\,[r_{0},r_{1}],\\\\[5.69046pt]
w(t)\rightarrow\infty\qquad\text{as }\,t\rightarrow\infty,\\\\[5.69046pt]
|w^{\prime}|\leq\varepsilon\quad\text{on }\,[r_{0},\infty).\end{array}\right.$
(6.7)

###### Proof.

Clearly, it is enough to prove the result for $\beta(r)=C(1+r)^{-\mu}$ and for
$\eta$ and $\varepsilon$ small enough. First of all, we modify $f,l,\varphi$
and choose a suitable $\eta$. These adjustments will be essential to prove the
$L^{\infty}$-gradient bound for $w$. Fix $\xi\in(0,1]$ and choose $\bar{l}(t)$
satisfying

$\begin{array}[]{l}\displaystyle\bar{l}\in
C(\mathbb{R}^{+}_{0}),\qquad\bar{l}>0\ \text{ on
}\,\mathbb{R}^{+},\qquad\bar{l}\leq 2\|l\|_{L^{\infty}([0,\xi])}\ \text{ on
}\,\mathbb{R}^{+},\\\\[8.5359pt] \bar{l}(t)=l(t)\ \text{ if
}\,t\in[0,\xi),\qquad\bar{l}(t)t^{\chi}\geq\bar{C}_{2}\ \text{ on
}\,[\xi,\infty),\end{array}$ (6.8)

for some constant $\bar{C}_{2}$ (here, we use $\chi\geq 0$). Regarding
$\varphi$, we choose $\bar{\varphi}\in C^{1}(\mathbb{R}^{+}_{0})$ such that

$\bar{\varphi}^{\prime}>0\ \text{ on
}\mathbb{R}^{+},\quad\bar{\varphi}=\varphi\ \text{ on
}\,(0,\xi),\quad\bar{\varphi}\leq\bar{C}_{1}t^{\chi}\bar{l}(t)\ \text{ on
}\,[\xi,\infty)$

note that this is possible if $\bar{C}_{1}$ is sufficiently large, by the last
of (6.8). By construction, since $\varphi(0)=0$,

$\bar{\varphi}(t)\leq Ct^{p-1}\ \text{ on }\,[0,1],\quad\bar{\varphi}(t)\leq
C_{3}t^{\chi}\ \text{ on }\,[\xi,\infty),\quad\bar{l}(t)\geq
C_{4}\frac{\bar{\varphi}(t)}{t^{\chi}}\qquad\text{on }\,\mathbb{R}^{+},$ (6.9)

for some constants $C_{3},C_{4}>0$. For $\eta>0$ and $\xi>0$, we introduce the
notation

$\begin{array}[]{ll}\beta_{0}=\min_{[r_{0},r_{1}]}\beta,&\quad\beta_{1}=\max_{[r_{0},r_{1}]}\beta;\\\\[5.69046pt]
v_{0}=\min_{[r_{0},r_{1}]}v,&\quad v_{1}=\max_{[r_{0},r_{1}]}v;\\\\[5.69046pt]
f_{2\eta}=\max_{[0,2\eta]}f,&\quad l_{\xi}=\max_{[0,\xi]}\bar{l}.\end{array}$
(6.10)

Given $\sigma\in(0,\xi)$ to be specified later, we choose
$\eta_{\sigma}\in[0,\sigma)$ small enough in order to satisfy

$\frac{v_{1}}{v_{0}}\left[(r_{1}-r_{0})\beta_{1}f_{2\eta_{\sigma}}l_{\xi}+\bar{\varphi}\left(\frac{\eta_{\sigma}}{r_{1}-r_{0}}\right)\right]<\bar{\varphi}(\sigma).$
(6.11)

This is possible because $\bar{\varphi}(0)=0$ and
$f_{2\eta_{\sigma}}\rightarrow 0$ as $\eta_{\sigma}\rightarrow 0$ (since
$f(0)=0$). We next choose $f_{\sigma}\in C(\mathbb{R}_{0}^{+})$ satisfying

$\begin{array}[]{ll}0\leq
f_{\sigma}(t)\leq\min\big{\\{}f(t),1\big{\\}},&\qquad f_{\sigma}(t)=0\ \
\text{ if }\,t\leq\eta_{\sigma},\\\\[5.69046pt] f_{\sigma}>0\quad\text{if
}\,t>\eta_{\sigma},&\qquad f_{\sigma}(\eta_{\sigma}+t)\leq K^{\prime}(t)\
\text{ for }\,t\in[0,\xi],\end{array}$ (6.12)

where $K(t)$ is the function defined in (1.19). The last condition can be
satisfied because of the positivity of $\varphi^{\prime}$ and $l$ on
$\mathbb{R}^{+}$, hence of $K^{\prime}$. We consider the Dirichlet problem