TITLE: How should logical quantifiers be understood if the domain of discourse is not specified?
QUESTION [5 upvotes]: Let $P$ be a predicate. The expression $\forall x:P(x)$ seems to have two different meanings, depending on the context:

If the domain of discourse $D$ is clear from context, then the statement "$\forall x:P(x)$" should actually be read as an abbreviation of "$\forall x\in D:P(x)$". For instance, in real analysis the statement $\forall x:x^2\ge0$ would be understood as an abbreviation of $\forall x\in\mathbb R:x^2\ge0$.
In set theory, it seems that $\forall x:P(x)$ means "for any set $x$, $P(x)$ is true". For instance, the axiom of extensionality is written as $\forall x:\forall y:\forall z:(z\in x\iff z\in y)\implies x=y$.

I don't fully understand the meaning of $\forall x:P(x)$ in the context of set theory. Naïvely, the statement $\forall x:P(x)$ seems to be an abbreviation of $\forall x\in \mathbf{U}:P(x)$, where $\mathbf{U}$ is the universal set. However, the "universal set" is not actually a meaningful concept in standard formulations of set theory, and so this interpretation is clearly wrong. So how should the statement $\forall x:P(x)$ actually be interpreted?

REPLY [3 votes]: Actually, there is nothing wrong with understanding "$∀x ( Q(x) )$" in a foundational set theory to mean "$∀x{∈}U\ ( \ Q(x) \ )$" where $U$ is the type of all objects (i.e. $U$ denotes the entire intended domain). $U$ does not have to be a set. In pure ZFC, $U$ is not an object, because otherwise we get a contradiction. $U$ is commonly called a class. In extensions of ZFC that can reason about classes as objects, such as NBG or MK, we have two sorts, one for "sets" and one for "classes", and $U$ is still not a set, even though it is now an object.
Typically "$∀x{\in}S\ ( \ Q(x) \ )$" can be treated as a short-form for "$∀x\ ( \ x∈S ⇒ Q(x) \ )$", and "$x∈U$" simply reduces to "true". Note that the "$\in$" here does not require any set theory in a strict sense, because $S$ can just be a sort (in many-sorted FOL). However, it is convenient to use the same symbol as for set membership because there is no danger of ambiguity. After all, in NBG and MK we literally have $E ∈ \{ t : Q(t) \} ⇔ Q(E)$ where "$\{ t : Q(t) \}$" is the class notation.
In fact, this usage of quantifying over classes is extremely common in higher set theory, such as "$∀k{∈}ORD$" where $ORD$ is the class of ordinals, which is (as explained above) perfectly fine even in pure ZFC.


TITLE: What are the current obstacles to experimentally testing quantum pseudo telepathy?
QUESTION [2 upvotes]: Quantum pseudo-telepathy refers to how, in some specific coordination games, isolated players can do better when they have pre-shared some entangled qubits. I understand how it works in theory and wrote about it, but I'm curious if it's something we can do in practice.
How close are we to being able to do real life runs of these games, where a quantum strategy is used to win more often than possible with classical strategies? I know we don't have quantum computers with more than a few qubits now, but the game in the wikipedia article seems to only require two per player.
What if we relax some of the constraints, like just saying "you've been isolated" instead of actually separating the players/machines by large distances? Or allowing the entangled qubits to be generated on demand instead of ahead of time?
If we can't do it yet, or in the near future, what are the obstacles?

REPLY [4 votes]: So far, scientists have been able to test spatially separated entanglement between 2 parties, and recently even between 3 parties. The latter experiment (link)(without paywall) states the following challenges:

No attempts have been made to close locality loopholes in Bell
  experiments involving three or more particles. The primary reason
  for this is source brightness. While entangled photon pairs have been
  generated and detected at rates in excess of 1 MHz,
  entangled photon triplets have only been observed at rates on the
  order of, necessitating long measurement times. In
  addition, further experimental challenges include high sensitivity to
  loss, causality relations requiring a complex experimental set-up and
  demanding stability requirements.

However, these type of experiments are interested in measuring the violation of classical bounds (e.g., proving that no classical hidden variable theories are possible, thus confirming Quantum Mechanics), such as Bell's or Mermin's inequalities. Your question is why games have not yet been experimentally demonstrated, but it turns out that testing inequalities is exactly the same.
Edit: I realized only later that quantum games and Bell-type inequalities are actually the exact same. Let me try to explain the equivalence below, and then give an example of what an experiment would look like in the "game-language".
A bell-type inequality says something about expectation values of measurements in pre-defined bases.
The equivalent game dictates that players receive a random question, and depending on the question, they measure in a particular basis.
Now, you will find that the number of "expectation values" and the "number of questions"  are the exact same. Moreover, the corresponding measurements turn out to be along the same basis (assuming we start from the same shared state)! Thus, when performing an optimal stategy multiple times, the players of a pseudo-telepathy game basically sample the expectation values.
The tricky part is that the expectation values in Bell-inequalities do not correspond to winning probablities $\omega$ (a number between 0 and 1), but to so-called "biasses" $\varepsilon$, which is simply $\omega$ rescaled to lie between (-1,1) (indeed, the values of a quantum measurement outcome!).
$$\varepsilon = 2 \omega  - 1$$
Take for example the CHSH game, where the optimal winning probabilitiy $\omega$ equals $\cos^2(\pi/8) = 1/2 + 1/\sqrt{8}$ (on average, per question). Turning these into biasses, and adding all 4 questions together, gives the well-known bound $2\sqrt{2}$, which is indeed the maximal violation of the Bell-inequality.
Now, you can simply look at any Bell-inequality experiment, turn it into the "game-language", and see how well the "players" scored, on average. For example, in the 3-qubit Mermin inequalities paper, take a look at figure 2. It shows the "biasses" per question (a,b,c). The "worst" expectation  found was −0.655 where they should've measured -1, which means a bias of +0.655. This corresponds to a "winning probability" of 0.8275 on that question. Considering that this was the question on which they performed the worst, they clearly did better than the 0.75 classical bound.


TITLE: Normal distribution: how to find the mean $µ$ and standard deviation $σ$?
QUESTION [1 upvotes]: A random variable $X$ is normally distributed with mean $µ$ and standard deviation $σ$.
lf $P(X < 25)$ is $0.1082$ and $P(X> 42)$ is $0.1303$ determine:

(i) $P(25 < X < 42)$
(ii) the values of mean $µ$ and standard deviation $σ$
(iii) $P(X>22)$

For part (i), I have done $P(25 < X < 42) = P(X<42) - P(X<25)$ which gives $(1-0.1303)-0.1082 = 0.7615$.
I'm unsure how to do part (ii). I'm thinking it has something to do with the empirical rule, so $68.3\%$ lies within one standard deviation of the mean, etc. But I'm not sure how to apply this, as there doesn't seem to be any concrete "percent rule" that the $0.7615$ applies to. This is from a high school maths textbook. Any hints would be appreciated.

REPLY [0 votes]: In normal distribution we use the notation $P(X\le a)=\phi(a)$
$P\bigg(\dfrac{x-\mu}{\sigma} < \dfrac {25-\mu}{\sigma}\bigg)=P\bigg(Z< \dfrac {25-\mu}{\sigma}\bigg)=\phi\bigg(\dfrac{25-\mu}{\sigma}\bigg)=0.1082$
Now look at normal table and find $a$ such that $\phi(a)=0.1082$ then,
$a=\dfrac{25-\mu}{\sigma}$
similarly, you have $P(X>42)=.1303$
So you will get two equations two unknows$(\mu$ and $\sigma)$ solve them you will get $
\mu,\sigma$


TITLE: Using the fine structure constant to measure atomic and molecular sizes
QUESTION [3 upvotes]: This is kind of a coursework question but it bring up some really interesting things about the fine structure constant $\alpha$ so I wanted to post it to not only make sure I understood something but to get into some other ideas as well.
So, the question: we're asked to show that the ration of the wavelength of a photon emitted by an atom and its size is related in turn to $\alpha$. I came up with the following:
for a hydrogen-like atom (using the Bohr model) $$r = \frac{n^2 \hbar^2} {Zm_e k e^2}$$ or, in terms of the Bohr radius, $r= \frac{n^2r_0}{Z}$ where $k=\frac{1}{4\pi\epsilon_0}$
Now, the energy of an atom is $E=\frac{kZe^2}{2r}$ (I can get this from Virial theorem) and r in this case is the radius we have from earlier.
Plugging one into the other we have $$E=\frac{kZe^2}{2 \frac{n^2 \hbar^2} {Zm_e k e^2}}=\frac{k^2Z^2e^4 m_e}{2 n^2 \hbar^2}$$
since $E=h \nu = hc/ \lambda$ we have
$$\frac{hc}{\lambda}=\frac{k^2Z^2e^4 m_e}{2 n^2 \hbar^2}$$
Now, the fine structure constant $\alpha = \frac{ke^2}{\hbar c}$ and that turns this into
$$\frac{1}{\lambda}=\frac{k^2Z^2e^4 m_e}{2 n^2 \hbar^2 hc}=\alpha^2 \frac{\pi Z^2 m_ec}{ n^2 \hbar} $$
and multiplying the original r by that: $$\frac{n^2 \hbar^2} {Zm_e k e^2}\alpha^2 \frac{\pi Z^2 m_ec}{ n^2 \hbar}=\alpha \pi Z = \frac{r}{\lambda}$$
which tells me that the ratio of size to wavelength depends entirely on Z and $\alpha$.
I also plugged this into the change-of-energy equation,
$$\Delta E = -Z^2 \frac{ke^2}{2r_0} \left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)$$
and plugging in what we have for $r_0$:
$r_0=\frac{\hbar^2}{m_eke^2}\rightarrow-Z^2 \frac{ke^2}{2(\frac{\hbar^2}{m_eke^2})} \left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)\rightarrow -Z^2\frac{k^2e^4m_e}{2\hbar^2}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)\rightarrow -Z^2\frac{\alpha^2c^2m_e}{2}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)=\Delta E$
Now assuming I did this correctly, would the same apply to a molecule? That is, given a wavelength, I would think you'd just plug that back into the energy equation (using $\Delta E = \frac{hc}{\lambda} $ and get an estimate of the size. But I was checking if my logic was correct.
The other interesting thing to me is how one might use $\alpha$ in other intersting ways for problems like this.
Anyhow, if anyone can say that I did something dumb that's most appreciated. I just want to see if my logic is correct.

REPLY [3 votes]: would the same apply to a molecule? That is, given a wavelength, I would think you'd just plug that back into the energy equation...and get an estimate of the size

No.
In the hydrogen atom, size is the distance of the electron from the proton.
In a molecule, size is primarily the relative locations of the nuclei. Especially for large molecules, absorption wavelengths are reflective of local structure, not the overall structure.  For example a C=C bond will absorb at the about the same wavelength from one molecule to another.
To take an extreme example, consider a protein molecule.
The protein may contain 100s or 1000s of amino acid residues. But all protein molecule have strong absorbance in the 280nm range regardless of size.
An exception to the wavelength being indicative of only part of the molecule, would be if the entire molecule was a conjugated system have delocalized electrons.
In a system with an alternating series of double and single carbon-carbon bonds:
$$-C=C-C=C-C=C-C=C-$$ the wavelength of a transition will increase with the length of the series.
In the case of conjucated double bonds a particle in a box model is sometimes applied.  See Conjugated Bonding in Cyanine Dyes: A "Particle In A Box" Model.


text:
\begin{document}

\title
[Unbounded supersolutions with generalized Orlicz growth]
{The weak Harnack inequality for unbounded supersolutions of equations with generalized Orlicz growth}
\author[Benyaiche, Harjulehto, Hästö, and Karppinen]{Allami Benyaiche, Petteri Harjulehto, Peter Hästö, and Arttu Karppinen}

\subjclass[2010]{35J60 (35B65, 31C45)}
\keywords{Generalized Laplacian, supersolution, generalized Orlicz space, Musielak--Orlicz spaces, nonstandard growth, Harnack's inequality, variable exponent, double phase}

\begin{abstract}
We study unbounded weak supersolutions of elliptic partial differential equations with generalized Orlicz
(Musielak--Orlicz) growth.
We show that they satisfy the weak Harnack inequality with optimal exponent provided that they belong to
a suitable Lebesgue or Sobolev space.
Furthermore, we establish the sharpness of our central assumptions.
\end{abstract}

\maketitle

\section{Introduction}

We prove the weak Harnack inequality for unbounded supersolutions of partial differential equations
with generalized Orlicz growth (also known as Musielak--Orlicz growth).
A general principle states that only intrinsic Harnack inequalities
are possible for PDEs which are not scaling invariant, and we do indeed find that the constant
in the weak Harnack inequality depends on the norm of the supersolution.
With dependence on the $L^s$- or $W^{1,s}$-norms, our result requires that
$s$ be sufficiently large, namely, $s\ge \max\{s_1,s_2\}$, where $s_1$ depends continuity of the
generalized Orlicz functional
and is shown to be sharp and $s_2$ depends on the global growth of the functional and
does not occur in any previously known special cases.
The result is new even for the special case of double phase functionals \cite{BarCM18}.
Our framework includes also the following special cases in which the weak Harnack inequality
was not known before, even for bounded solutions:
perturbed variable exponent \cite{GP13,LiaSZ18,Ok0,Ok18},
Orlicz variable exponent \cite{CapCF18, GPRT17},
degenerate double phase \cite{BCM2,BOh2},
Orlicz double phase \cite{BOh3},
variable exponent double phase \cite{MizNOS20,RagT20},
triple phase \cite{DeFO19}, and
double variable exponent \cite{CenRR18, RadRSZ_pp,ZhaR18}.

Let us introduce the context of this paper.
Minimizers and (weak) solutions of
\begin{align*}
\inf \int_\Omega F(x, \nabla u) \, dx \quad \text{and}\quad
-\div(f(x,\nabla u)) = 0
\end{align*}
have been actively studied during recent years when $F$ or $f$ has generalized Orlicz growth. For instance,
solutions with given boudary values exist \cite{ChlGZ19,GwiSZ18, HarHK16},
minimizers or solutions with given boundary values are locally bounded, satisfy Harnack's inequality and belong to $C^{0, \alpha}_{\loc}$ \cite{BenK_pp, HarHT17, WanLZ19} or $C^{1, \alpha}_{\loc}$ \cite{HasO_pp},
quasiminimizers satisfy a reverse H\"older inequality \cite{HarHK18},
$\omega$-minimizers are locally Hölder continuous \cite{HarHL_pp},
minimizers for the obstacle problem are continuous \cite{Kar19} and the boundary Harnack inequality
holds for harmonic functions \cite{ChlZ_pp}.
In most cases the assumptions in these results coincides with optimal assumptions in well-known special cases.
The important special cases are the variable exponent case $F(x,\xi) := |\xi|^{p(x)}$
\cite{Alk97, HarHL09, HarKL07, Jul15, Toi12}, the Orlicz case $F(x,\xi) := \phi(|\xi|)$
\cite{ArrH18, Lie87, Lie91},
and  the double phase case $F(x,\xi) := |\xi|^{p} + a(x) |\xi|^q$ \cite{BarCM15, ColM15a, ColM15b}.

The surveys \cite{Chl18, Mar20} include more references of variational problems and partial differential equations of generalized Orlicz growth, while the recent monographs \cite{HarH19, LanM19} present the theory of the
underlying function spaces.

In \cite{BenK_pp}, the weak Harnack inequality for bounded supersolutions and Harnack's inequality for solutions were proved by Moser's iteration in the generalized Orlicz case.
In those results the constants depend on $L^\infty_\loc$-norm of the function.
In \cite{BarCM15}, Harnack's inequality has been proved in the double phase case,
and the constant depends on $L^\infty_\loc$-norm of the function.

We want to study unbounded supersolutions, so we need the constants not to depend on the $L^\infty$-norm.
In the variable exponent case, the constant in the weak Harnack inequality depends
on $L^t_\loc$-norm of the function and $t>0$ can be chosen arbitrarily small; furthermore,
an example shows that the constant in  Harnack's inequality cannot be independent of the function, in contrast to the constant exponent case \cite{HarKL07}. We extend these results to the generalized Orlicz case with sharp assumptions on the continuity of $\phi$.

We assume that $f:\Omega\times  \Rn \to \Rn$ satisfies the following $\phi$-growth conditions:
\begin{equation}\label{eq:growth}
\nu \phi(x,|\xi|) \le f(x,\xi)\cdot \xi
\quad\text{and}\quad
|f(x,\xi)|\, |\xi| \le \Lambda \phi(x,|\xi|)
\end{equation}
for all $x\in \Omega$ and $\xi\in\Rn$, and fixed but arbitrary constants $0<\nu \leq \Lambda$.
We are interested in local  (weak) supersolutions:

\begin{defn}
A function $u \in W^{1, \phi}_\loc(\Omega)$ is a \emph{supersolution} if
\begin{align}
\label{eq:A-eq}
\int_{\Omega} f(x,\nabla u) \cdot \nabla h \, dx \ge 0,
\end{align}
for all non-negative $h \in W^{1, \phi}(\Omega)$ with compact support in $\Omega$.
\end{defn}

We define the limiting exponent
\[
\ell(p):= \frac{p^*}{p'} =
\begin{cases}
\frac n{n-p}(p-1) & \quad\text{if }p<n,\\
\infty & \quad\text{if }p\ge n;
\end{cases}
\]
the ratio of the Sobolev exponent $p^*$ and the H\"older exponent $p'$.
Since $u_p(x):=|x|^{-\frac{n-p}{p-1}}$ is a supersolution of the $p$-Laplace equation
$-\div(|\nabla u|^{p-2}\nabla u) = 0$ in the unit ball and since $u_p\not\in L^{\ell(p)}(B_1)$,
we see that $\ell(p)$ is an upper bound on the exponent of integrability of $p$-supersolutions.

The following is a special case of our main result, Theorem~\ref{thm:weakHarnackGeneral}
together with Propositions \ref{prop:vBound} and \ref{prop:betaBound}.
Theorem~\ref{thm:weakHarnackGeneral} contains also
the cases $\|u\|_{L^\omega(B_{2R})}\le d$ and $\|u\|_{W^{1,\omega}(B_{2R})}\le d$ for general
$\omega\in \Phiw(\Omega)$.
The last statement in the next theorem follows from the example in Section~\ref{sect:example}.
Note that with the choice $s=\infty$ we recover as a special case previously known results for bounded
solutions \cite{BenK_pp} with the correct, \aones{n} assumption.

\begin{thm}[The weak Harnack inequality]\label{thm:weakHarnack}
Suppose $\phi$ satisfies \azero{}, \ainc{p} and \adec{q}, $1<p\le q<\infty$.
Let $u$ be a non-negative supersolution to \eqref{eq:A-eq} in $B_{2R}$.
Assume that one of the following holds:
\begin{enumerate}
\item
$\phi$ satisfies \aones{s_*} and $\|u\|_{L^s(B_{2R})}\le d$, where $s_* := \frac{ns}{n+s}$ and $s\in [q-p,\infty]$.
\item
$\phi$ satisfies \aone{} and $\|u\|_{W^{1,\phi}(B_{2R})}\le d$.
\end{enumerate}
Then there exist  positive constants $\ell_0$ and $C$

such that the weak Harnack inequality holds:
\begin{align*}
\bigg (\fint_{B_{2R}} (u +R)^{\ell_0} \, dx \bigg)^{\frac{1}{\ell_0}}  \le C (\essinf_{B_R} u + R).
\end{align*}
If (1) holds with $s>\max\{\frac np,1\}(q-p)$ or if (2) holds with $p^*>q$, then the weak Harnack inequality
holds for any $\ell_0<\ell(p)$.

The \aones{s_*} assumption is sharp, since
for any $s'<s_*$ if, instead of (1), $\phi$ satisfies \aones{s'} and $\|u\|_{L^s(B_{2R})}\le d$,
then the weak Harnack inequality need not hold.
\end{thm}

The next result follows by Corollary~\ref{cor:doublePhase-2} and the counter-example given in
Section~\ref{sect:example}. Corresponding corollaries could be formulated in the other cases of
double phase type functionals listed in the beginning of this section.

\begin{cor}\label{cor:doublePhase}
Let $\phi(x,t):=t^p + a(x)t^q$ be the double phase functional with
$a^\lambda\in C^{0,\alpha\lambda}(\Omega)$ for some $\lambda>0$.
Let $u$ be a non-negative supersolution to \eqref{eq:A-eq} in $B_{2R}$.
If $u\in L^s(\Omega)$ with
\[
\alpha \ge (\tfrac{n}s+1)(q-p)
\quad\text{and}\quad
s\in [q-p,\infty],
\]
then there exist  positive constants $\ell_0$ and $C$

such that
\begin{align*}
\bigg (\fint_{B_{2R}} (u +R)^{\ell_0} \, dx \bigg)^{\frac{1}{\ell_0}}  \le C (\essinf_{B_R} u + R).
\end{align*}
If additionally $s>\max\{\frac np,1\}(q-p)$, then the inequality holds for any $\ell_0<\ell(p)$.

The bound on $\alpha$ is sharp, since
for every $\alpha < (\tfrac{n}s+1)(q-p)$, there exists $a\in C^{0,\alpha}(\Omega)$ for which
the weak Harnack inequality does not hold.
\end{cor}

As the example in Section~\ref{sect:example} shows, the assumption $\alpha \ge (\tfrac{n}s+1)(q-p)$ is sharp
and this restriction has been previously encountered in \cite{Ok20}, see also \cite{BarCM18}.
The assumption $s\ge q-p$ is a consequence of considering supersolutions as it can be omitted
if $u$ is a solution. It is especially interesting to note that such a restriction does not
occur in the variable exponent case and it is another example that the double phase functional is
more subtle than the variable exponent case \cite{ColM15a}.

\begin{rem}\label{rem:plusR}
Compared to the classical Harnack inequality, our estimate contains an extra $+R$-term. It is not know
whether this is necessary,
but the same phenomenon occurs when $\phi(x,t)= t^{p(x)}$ \cite{Alk97, HarKL07, HarKLMP08}, unless the
exponent $p$ is assumed to belong to $C^1$ and slightly different Harnack's inequality is used \cite{Jul15}.
In the Orlicz and the double phase cases the extra $+R$ term is not needed \cite{ArrH18, BarCM15, Lie91}.
\end{rem}

\begin{rem}\label{rem:japaneseCase}
In the case of the basic double phase functional, namely, when $\lambda=1$ in Corollary~\ref{cor:doublePhase},
the assumption $\alpha \ge (\frac{n}s+1)(q-p)$ implies $s\ge n(q-p) \ge q-p$ and $s>\frac np(q-p)$,
since $\alpha\le 1$ and $q\ge 1$.
Mizuta, Ohno and Shimomura \cite{MizOS20} (see also \cite{HarH_pp}) have considered
the functional $\phi(x,t):=t^p + (a(x)t)^q$,
which corresponds to the case $\lambda=\frac 1q$ above. Also in this case, the first assumption
implies the latter two. However, if $\lambda<\frac1q$, then the first condition can hold while
the others do not.

Note that the parameter $\lambda$ does not impact the restrictions in
Corollary~\ref{cor:doublePhase}. The reason for this is that only the growth of $a$
at $a=0$ is important, see \cite[Proposition~7.2.2]{HarH19}.
\end{rem}

\section{Preliminaries}

We briefly introduce our assumptions. More information about $L^\phi$-spaces can be found in \cite{HarH19}.
\textit{Almost increasing} means that a function satisfies $f(s) \le L f(t)$ for all $s<t$ and some constant $L\ge 1$.
If there exists a constant $C$ such that $f(x) \leq C g(x)$ for almost every $x$, then we write $f \lesssim g$. If $f\lesssim g\lesssim f$, then we write $f \approx g$.

\begin{defn}
We say that $\phi: \Omega\times [0, \infty) \to [0, \infty]$ is a
\textit{weak $\Phi$-function}, and write $\phi \in \Phiw(\Omega)$, if
the following conditions hold:
\begin{itemize}
\item
For every measurable function $f:\Omega\to \R$ the function $x \mapsto \phi(x, f(x))$ is measurable
and for every $x \in \Omega$ the function $t \mapsto \phi(x, t)$ is non-decreasing.
\item
$\displaystyle \phi(x, 0) = \lim_{t \to 0^+} \phi(x,t) =0$  and $\displaystyle \lim_{t \to \infty}\phi(x,t)=\infty$ for every $x\in \Omega$.
\item
The function $t \mapsto \frac{\phi(x, t)}t$ is $L$-almost increasing on $(0,\infty)$ with $L$ independent of $x$.
\end{itemize}
If $\phi\in\Phiw(\Omega)$ is additionally convex and left-continuous, then $\phi$ is a
\textit{convex $\Phi$-function}, and we write $\phi \in \Phic(\Omega)$. If $\phi$ does not depend on $x$, then we omit the set  and write $\phi \in \Phiw$ or $\phi \in \Phic$.
\end{defn}

We denote $\phi^+_{A}(t) = \esssup_{x \in A\cap \Omega} \phi(x,t)$ and
$\phi^-_{A}(t) = \essinf_{x \in A\cap \Omega} \phi(x,t)$.

We define several conditions. See Table~\ref{tab:examples} for an intuition of their meaning
in special cases.
Let $p,q,s>0$ and let $\omega:\Omega\times [0,\infty)\to [0,\infty)$ be almost increasing.
We say that $\phi:\Omega\times [0,\infty)\to [0,\infty)$ satisfies
\begin{itemize}
\item[(A0)]\label{def:a0}
if there exists $\beta \in(0, 1]$ such that $ \beta \le \phi^{-1}(x,
1) \le \frac1{\beta}$ for a.e.\ $x \in \Omega$,

\item[(A1-$\omega$)] \label{def:a1s}
if there exists $\beta \in (0,1]$ such that,  for every ball $B$ and a.e.\ $x,y\in B \cap \Omega$,
\[
\phi(x,\beta t) \le \phi(y,t) \quad\text{when}\quad  \omega_B^-(t) \in \bigg[1, \frac{1}{|B|}\bigg];
\]

\item[(A1-$s$)]
if it satisfies \aones{\omega} for $\omega(x,t):=t^s$;

\item[(A1)]\label{def:a1}
if it satisfies \aones{\phi};

\item[(aInc)$_p$] \label{def:aInc} if
$t \mapsto \frac{\phi(x,t)}{t^{p}}$ is $L_p$-almost
increasing in $(0,\infty)$ for some $L_p\ge 1$ and a.e.\ $x\in\Omega$;

\item[(aDec)$_q$] \label{def:aDec}
if
$t \mapsto \frac{\phi(x,t)}{t^{q}}$ is $L_q$-almost
decreasing in $(0,\infty)$ for some $L_q\ge 1$ and a.e.\ $x\in\Omega$.
\end{itemize}
We say that \ainc{} holds if \ainc{p} holds for some $p>1$, and similarly for \adec{}.

If in the definition of \ainc{p} we have $L_p=1$, then we say that $\phi$ satisfies \inc{p},
similarly for \dec{q}.

\begin{table}[h]
\centerline{
\begin{tabular}{l|ccccc}
$\phi(x,t):=$ & \azero{} & \aone{}& \aones{s}& \ainc{} & \adec{} \\
\hline
$\phi(t)$ & $\text{true}$ & $\text{true}$ & $\text{true}$ & $\nabla_2$ & $\Delta_2$ \\
$t^{p(x)}a(x)$ & $a\approx 1$ & $p\in C^{\log}$ & $p\in C^{\log}$ & $p^->1$ & $p^+<\infty$ \\
$t^{p(x)}\log(e+t)$ & $\text{true}$ & $p\in C^{\log}$ & $p\in C^{\log}$ & $p^->1$ & $p^+<\infty$ \\
$t^p + a(x) t^q$ & $a\in L^\infty$ & $a\in C^{0, \frac np (q-p)}$ & $a\in C^{0, \frac ns(q-p)}$ & $p>1$ & $q<\infty$ \\
\end{tabular}
}
\caption{Assumptions in special cases}\label{tab:examples}
\end{table}

We note that \aones{\omega} is a new condition introduced in this article
to combine \aone{} and \aones{n} as well as other cases. Basically, \aones{\omega} is the
appropriate assumption if we have \textit{a priori} information that the
solution is in space $W^{1,\omega}$ or the corresponding Lebesgue or H\"older space.
The most important cases are $\omega=\phi$ and $\omega(x,t)=t^s$, but we may, for instance,
consider $\omega=\phi^{1+\epsilon}$ if we have some higher integrability information.

By \cite[Section~4.1]{HarH19}, \azero{} can be stated equivalently as the existence of $\beta \in(0,1]$ such that
$\phi(x,\beta) \le 1\le \phi(x,1/\beta)$ for almost every $x \in \Omega$. If $\phi$ satisfies \azero{}, then
\aone{} is equivalent to the condition that  there exists $\beta\in (0,1)$ such that
\[
\beta \phi^{-1}(x, t) \le \phi^{-1} (y, t)
\quad\text{when}\quad
t \in \bigg[1, \frac{1}{|B|}\bigg]
\]
for every ball $B$ and a.e.\ $x,y\in B \cap \Omega$ \cite[Section~4.2]{HarH19}.

\begin{rem}
\label{rem:incdec}
Assume that the derivative $\phi'$ with respect to the second variable exists. Then
\[
\frac d{dt}\frac{\phi(x,t)}{t^p} = \frac{\phi'(x, t) t^p- p t^{p-1}\phi(x,t)}{t^{2p}}.
\]
If $\phi$ satisfies \inc{p}, then the derivative is non-negative and so
$\frac{t\phi'(x,t)}{\phi(x,t)} \ge p$.
Similarly, $\frac{t\phi'(x,t)}{\phi(x,t)} \le q$ if $\phi$ satisfies \dec{q}.
If, on the other hand,
\[
p\le \frac{t\phi'(x,t)}{\phi(x,t)} \le q
\]
then $\frac d{dt}\frac{\phi(x,t)}{t^p}$ is non-negative and
$\frac d{dt}\frac{\phi(x,t)}{t^q}$ is non-positive and hence \inc{p} and \dec{q} hold.
Moreover,  the double inequality implies also that $\phi'$ satisfies \ainc{p-1} and \adec{q-1}
if $\phi$ satisfies \inc{p} and \dec{q}.
\end{rem}

\begin{defn}\label{def:Lphi}
Let $\phi \in \Phiw(\Omega)$ and define the \textit{modular}
$\varrho_\phi$ for $u\in L^0(\Omega)$, the set of measurable functions in $\Omega$, by
  \begin{align*}
    \varrho_\phi(u) &:= \int_\Omega \phi(x, |u(x)|)\,dx.
  \end{align*}
The \emph{generalized Orlicz space}, also called Musielak--Orlicz space, is defined as the set
  \begin{align*}
    L^\phi(\Omega) &:= \big\{u \in L^0(\Omega) \colon
      \lim_{\lambda \to 0^+} \varrho_\phi(\lambda u) = 0\big\}
  \end{align*}
equipped with the (Luxemburg) norm
  \begin{align*}
    \|u\|_{L^\phi(\Omega)} &:= \inf \Big\{ \lambda>0 \colon \varrho_\phi\Big(
      \frac{u}{\lambda}  \Big) \leq 1\Big\}.
  \end{align*}
If the set is clear from context we abbreviate $\|u\|_{L^\phi(\Omega)}$ by $\|u\|_{\phi}$.
\end{defn}

We denote by $\phi^*$ the conjugate $\Phi$-function, defined by
\[
\phi^*(x,t):= \sup_{s\ge 0} (st-\phi(x,s)).
\]
By this definition, we have Young's inequality $st \le \phi(x,s) + \phi^*(x,t)$.
H\"older's inequality holds in generalized Orlicz spaces for $\phi\in\Phiw(\Omega)$ with constant $2$
\cite[Lemma~3.2.13]{HarH19}:
\[
\int_\Omega |u|\, |v|\, dx \le 2 \|u\|_\phi \|v\|_{\phi^*(\cdot)}.
\]

\begin{defn}
A function $u \in L^\phi(\Omega)$ belongs to the
\textit{Orlicz--Sobolev space $W^{1, \phi}(\Omega)$} if its weak partial derivatives $\partial_1 u, \ldots, \partial_n u$ exist and belong to the space $L^{\phi}(\Omega)$.
For $u \in W^{1,\phi}(\Omega)$, we define the norm
\[
\| u \|_{W^{1,\phi}(\Omega)} := \| u \|_{L^\phi(\Omega)} + \| \nabla u \|_{L^\phi(\Omega)}.
\]
Here $\| \nabla u \|_{\phi}$ is a abbreviation of $\big{\|} | \nabla u | \big{\|}_{\phi}$.
Again, if $\Omega$ is clear from context, we abbreviate $\| u \|_{W^{1,\phi}(\Omega)}$ by $\| u \|_{1,\phi}$.
\end{defn}

We conclude the section by proving an appropriate version of the Caccioppoli inequality.
We denote by $\eta$ a cut-off function in $B_R$, more precisely,
$\eta \in C_0^{\infty}(B_R)$, $\chi_{B_{\sigma R}} \leq \eta \leq \chi_{B_R}$
and $|\nabla \eta| \leq \frac{2}{(1-\sigma)R}$, where $\sigma \in (0, 1)$. Note that the auxiliary function
$\psi$ is independent of $x$ in the next lemma. This avoids assumptions
regarding the differentiability of $\psi$ in the space variable, but it does mean that
the application of the lemma later on is more complicated compared to classical, standard growth
cases where one simply choses $\psi=\phi$.

\begin{lem}[Caccioppoli inequality]
\label{lem:caccioppoli}
Suppose $\phi \in \Phiw(\Omega)$ satisfies \azero{} and \adec{q},
and let $\psi\in \Phiw$ be differentiable and satisfy \azero{},
\inc{p_1} and \dec{q_1}, $p_1, q_1\ge 1$. Let $u$ be a non-negative supersolution
to equation \eqref{eq:A-eq} and $\eta$ be a cut-off function in $B_R\subset \Omega$.
For any $\ell > \frac1{p_1}$ and $s\ge q$,
\begin{align*}
\int_{B_R} \phi(x, |\nabla u|) \psi( \tfrac{u+R}{R} )^{-\ell} \eta^s \, dx
\leq
\big(\tfrac{c(L_q) s\Lambda}{(1-\sigma)(p_1\ell-1)\nu}\big)^q \int_{B_R} \psi (\tfrac{u+R}{R} )^{-\ell} \phi(x,\tfrac{u+R}{R} ) \eta^{s-q} \, dx.
\end{align*}
\end{lem}

\begin{proof}
Let us simplify the notation by denoting $\tilde u := u + R$ and $v := \frac{\tilde u}{R}$.
Since $\nabla u = \nabla \tilde u$, we see that $\tilde u$ is still a supersolution.
Since $v\ge 1$, $\psi(v)^{-\ell} \eta^{s}\le c_1$ by \azero{} and \inc{p_1}.

We would like to test equation \eqref{eq:A-eq} with $h := \psi(v)^{-\ell} \eta^{s} \tilde u$. Let us first check that $h$ is a valid test function, that is $h \in W^{1,\phi}(B_R)$ and has compact support in $\Omega$. As $\tilde u \in L^{\phi}(B_R)$ and $|h|\le c_1 \tilde u$ it is immediate that $h \in L^{\phi}(B_R)$. By a direct calculation we have
\begin{align*}

\begin{split}
\nabla h
&= -\ell \psi(v)^{-\ell-1} \eta^{s} \tilde u \psi'(v) \nabla v +
s \psi(v)^{-\ell} \eta^{s-1} \tilde u \nabla \eta + \psi(v)^{-\ell} \eta^s \nabla \tilde u.

\end{split}
\end{align*}
Note that $\tilde u \nabla v =  v \nabla \tilde u$. We use Remark~\ref{rem:incdec} and get
\begin{align*}
\big|\ell \psi(v)^{-\ell-1} \eta^{s}  \psi'(v) v \nabla \tilde u \big|
\leq
q_1\ell c_1 |\nabla \tilde u| \in L^{\phi}(B_R).
\end{align*}
For the third term, we obtain $|\psi(v)^{-\ell} \eta^{s} \nabla \tilde u |
\leq  c_1 |\nabla \tilde u| \in L^{\phi}(B_R)$.
The term with $\nabla \eta$ is treated as $h$ itself. Thus $h \in W^{1, \phi}(B_R)$.
Since $s>0$ and $\eta\in C^\infty_0(B_R)$, $h$ has compact support in
$\Omega$ and so it is a valid test-function.

We next calculate
\begin{align*}
f(x,\nabla \tilde u) \cdot \nabla h
&=\psi(v)^{-\ell-1} \eta^{s} [-\ell \psi'(v) v + \psi(v)] f(x,\nabla \tilde u) \cdot \nabla \tilde u\\
&\qquad+ s\psi(v)^{-\ell} \eta^{s-1} \tilde u \, f(x,\nabla \tilde u) \cdot \nabla \eta .
\end{align*}
Since $\tilde u$ is a supersolution, we have
$\int_{B_R} f(x, \nabla \tilde u) \cdot \nabla h \, dx \ge 0$,
which implies with the growth conditions \eqref{eq:growth} that
\begin{align*}
\nu \int_{B_R} \phi(x, |\nabla \tilde u|) \psi(v)^{-\ell-1} \eta^{s}
[\ell  \psi'(v) v - \psi(v)]
\, dx \le
s \Lambda \int_{B_R} \tfrac{\phi(x, |\nabla \tilde u|)}{|\nabla \tilde u|} |\nabla \eta| \, \eta^{s-1}\psi(v)^{-\ell}\tilde u \, dx.
\end{align*}
By Remark~\ref{rem:incdec}, $p_1 \psi(t) \le \psi'(t) t$ so we conclude that
\begin{align*}
& [p_1\ell- 1]\nu \int_{B_R} \phi(x, |\nabla \tilde u|) \psi(v)^{-\ell} \eta^{s}\, dx
\le
s\Lambda \int_{B_R} \tfrac{\phi(x, |\nabla \tilde u|)}{|\nabla \tilde u|}|\nabla \eta |\, \eta^{s-1}\psi(v)^{-\ell}\tilde u
\, dx.
\end{align*}
Here $p_1\ell- 1$ is positive, since $\ell> \frac1{p_1}$.

Recalling that $|\nabla \eta| \tilde u \leq \frac{2}{(1-\sigma)R}\tilde u
=\frac{2}{1-\sigma} v$, we arrive at
\begin{align*}
\int_{B_R} \phi(x, |\nabla \tilde u|)  \psi(v)^{-\ell} \eta^{s} \, dx
\leq
\frac{C}{1-\sigma} \int_{B_R} \tfrac{\phi(x, |\nabla \tilde u|)}{|\nabla \tilde u|} v \psi(v)^{-\ell} \eta^{s-1}\, dx.
\end{align*}
By Young's inequality \cite[(2.4.2)]{HarH19}
\begin{equation*}
\tfrac{\phi(x, |\nabla \tilde u|)}{|\nabla \tilde u|} v
\le
\phi\big(x, \epsilon^{-\frac1{q'}} v\big)
+  \phi^*\big(x, \epsilon^{\frac1{q'}} \tfrac{\phi(x, |\nabla \tilde u|)}{|\nabla \tilde u|}\big).

\end{equation*}
For the first term on the right hand side we use \adec{q} of $\phi$.
For the second term we first  use
\ainc{q'} of $\phi^*$ \cite[Proposition 2.4.9]{HarH19} and then
$\phi^*(\frac{\phi(t)}t) \le \phi(t)$ (comment after \cite[Theorem 2.4.10]{HarH19}), and obtain
\begin{equation*}
\tfrac{\phi(x, |\nabla \tilde u|)}{|\nabla \tilde u|} v
\lesssim
\epsilon^{1-q} \phi(x,v) + \epsilon \phi(x,|\nabla u|) .
\end{equation*}

Finally, we choose $\epsilon:=\frac{1-\sigma}{2C}\eta(x)$ and so
\begin{align*}
&\int_{B_R} \phi(x, |\nabla \tilde u|)  \psi(v)^{-\ell} \eta^{s} \, dx\\
 &\quad\leq \frac{1}{2} \int_{B_R} \phi(x, |\nabla \tilde u|) \psi(v)^{-\ell} \eta^{s} \, dx
 + \frac{C}{(1-\sigma)^q} \int_{B_R} \psi(v)^{-\ell} \phi(x,v) \eta^{s-q} \, dx.
\end{align*}
The first term on the right-hand side can be absorbed in the left-hand side. This gives the claim.
\end{proof}

\section{The weak Harnack inequality}

The plan of the proof follows the usual scheme of Moser iteration.
We first show that the infimum is bounded below by an integral-mean with negative power.
We then prove a reverse H\"older-type inequality for positive exponents below a certain
threshold. These steps are proved by iteration. Jumping over zero is the final piece and it is
accomplished by the John--Nirenberg lemma.

We next define a
differentiable approximation of $\phi_{B_R}^-$ with nice growth properties.
We assume that $\phi\in \Phiw(\Omega)$ satisfies \adec{} and \ainc{p}, $p\ge 1$.

\begin{defn}\label{def:psi}
We define an auxiliary weak $\Phi$-function $\psi_r$ by setting $\psi_r(0):=0$ and
\[
\psi_r(t):= \int_0^t \tau^{p-1}\sup_{s\in (0, \tau]} \frac{\phi_{B_r}^-(s)}{s^p} \, d\tau
\quad\text{for }t>0.
\]
\end{defn}

Then $\frac{\psi'_r(t)}{t^{p-1}} = \sup_{s\in (0, t]} \frac{\phi_{B_r}^-(s)}{s^p}$ is increasing and positive
so that $\psi_r'$ satisfies \inc{p-1} and $\psi_r$ is convex, strictly increasing and satisfies \inc{p}.

Since $\phi^-_{B_r}$ satisfies \ainc{p} and \adec{}, the integrand is finite in every point, and so
$\psi_r$ is continuous. Thus $\psi_r \in \Phic$.
Further, by \adec{},
\[
\psi_r(t) \ge  \int_{t/2}^t \tau^{p-1}\sup_{s\in (0, \tau]} \frac{\phi_{B_r}^-(s)}{s^p} \, d\tau
\gtrsim \frac{\phi_{B_r}^-(t/2)}{(t/2)^p} \Big( \frac{t}2 \Big)^p \approx \phi_{B_r}^-(t),
\]
and since $\phi^-_{B_r}$ satisfies \ainc{p} we obtain
\[
\psi_r(t) \lesssim \int_0^t \tau^{p-1} \frac{\phi_{B_R}^-(t)}{t^p} \, d\tau = \tfrac1p \phi_{B_r}^-(t).
\]
Thus $\psi_r(t)\approx \phi_{B_r}^-(t)$.
It follows that $\psi_r$ satisfies \adec{q}, and since it is convex,
it satisfies \dec{} \cite[Lemma~2.2.6]{HarH19}.
Therefore, $\psi_r$ satisfies the assumptions of Lemma~\ref{lem:caccioppoli} with
$p_1=p$; $q_1$ is a function of $q$, but it does not affect the constant.
See Proposition~\ref{prop:vBound} for a simpler, sufficient condition for \eqref{eq:vBound}.

\begin{thm}\label{thm:essinf}
Suppose $\phi\in \Phiw(\Omega)$ satisfies \azero{}, \ainc{p} and \adec{q}, $1\le p\le q$.
Let $u$ be a nonnegative supersolution to \eqref{eq:A-eq} in $B_R$.
Let $\omega \in \Phi_w(\Omega)$ satisfy \azero{} and \adec{}.
Assume that $\phi$ satisfies \aones{\omega}, and
\begin{equation}\label{eq:vBound}
\omega_{B_R}^-\bigg(\fint_{B_R} \frac{u+R}R\, dx\bigg) \le \frac d{|B_R|}
\end{equation}
for some $d>0$. For any $\ell > 0$, there exists a constant $C_{\ell,d} = C(p, q, L_p, L_q, n, \ell,d)>0$
such that
\begin{align*}
\essinf_{B_{R/2}} u+R
\geq
C_{\ell,d}\Big(\fint_{B_R} (u+R)^{-\ell} \, dx  \Big)^{-\frac{1}{\ell}}.
\end{align*}
\end{thm}

\begin{proof}
Let us assume that $r\in [\frac{R}{2},R]$ and
denote $ \tilde u := u+R, v:= \frac{\tilde u}{r}$ and $n' := \frac{n}{n-1}$.
Let $\psi_r \in \Phic$ be as in Definition~\ref{def:psi} in the ball $B_r$,
and abbreviate $\psi := \psi_r$.
Let  $s>0$ be a constant that will be fixed later.
We use the $W^{1,1}$-Sobolev--Poincar\'e inequality for the function $\psi(v)^{-\ell} \eta^{s}$,
where $\eta \in C_0^\infty (B_r)$ is a cut-off function as before.
We see that $\psi(v)^{-\ell}\eta^{s} \in W^{1,1}(\Omega)$  and it has  a compact support in $\Omega$ by the same arguments as in the proof of the Caccioppoli inequality (Lemma~\ref{lem:caccioppoli}).
The Sobolev--Poincar\'e inequality gives us
\begin{equation*}
\begin{split}
\Big (\int_{B_r} \psi(v)^{-\ell n'} \eta^{sn'} \, dx \Big )^{\frac{1}{n'}}
&\lesssim \int_{B_r} |\nabla(\psi(v)^{-\ell} \eta^{s})| \, dx \\
&\le \int_{B_r} \psi(v)^{-\ell-1} \eta^{s-1} [ s\psi(v) |\nabla \eta|  + \ell \eta |\nabla \psi(v)| ]\, dx.
\end{split}
\end{equation*}
By the definition of $\psi$ and \ainc{p} of $\phi$,
$\psi'(t) \approx \frac1t \phi^-_{B_r} (t) \le \frac1t \phi(x,t)$ for $x \in B_r$. Thus
\[
|\nabla \psi(v)|
=
|\psi'(v) \tfrac{1}{r} \nabla \tilde u|
\lesssim
\tfrac{1}{rv} \phi(x,v) |\nabla \tilde u|
\]
almost everywhere in $B_r$. We use this and the estimate $|\nabla \eta| \leq \frac{2}{(1-\sigma)r}$:
\begin{align}
\label{eq:after-sob}
 \Big(\int_{B_r} \psi(v)^{-\ell n'} \eta^{sn'} \, dx  \Big)^{\frac{1}{n'}}
 &\leq
\frac {C(s+\ell)}{r}
\int_{B_r} \psi(v)^{-\ell-1} \eta^{s-1}
\big[\tfrac{\psi(v)}{1-\sigma}  + \eta \tfrac{1}{v}  \phi(x,v) |\nabla \tilde u| \big]\, dx .
\end{align}

By Young's inequality \cite[(2.4.1)]{HarH19} and $\phi^*(x,\frac{1}{t}\phi(x,t))\le \phi(x,t)$
\cite[p.~35]{HarH19}, we have
\[
\tfrac{1}{v} \phi(x,v) |\nabla \tilde u|
\le \phi^*\big(x,\tfrac{1}{v} \phi(x,v)\big) + \phi(x,|\nabla \tilde u|)
\le
\phi(x,v) + \phi(x,|\nabla \tilde u|).
\]
This and the Caccioppoli inequality (Lemma \ref{lem:caccioppoli}) for $u+R-r$ in $B_r$ yield
\[
\begin{split}
\int_{B_r} \psi(v)^{-\ell-1} \eta^{s} \tfrac{1}{v}\phi(x,v) |\nabla \tilde u| \, dx
&\le   \int_{B_r} \psi(v)^{-\ell-1} \eta^{s} \big(\phi(x,v) + \phi(x,|\nabla \tilde u|)\big) \, dx\\
&\le
\frac C{(1-\sigma)^q} \int_{B_r} \psi(v)^{-\ell-1} \eta^{s-q}  \phi(x,v)\, dx,
\end{split}
\]
where we assumed that $(\ell+1)p >1$ and $s\ge q$ and used $\eta^{s} \le \eta^{s-q}$.
We next divide \eqref{eq:after-sob} by $r^{n-1}$, use this estimate as well as
$\eta^{s-1} \le \eta^{s-q}$ and $\psi(t) \lesssim \phi(x,t)$:
\begin{align*}
\Big(\fint_{B_r} \psi(v)^{-\ell n'} \eta^{sn'} \, dx  \Big)^{\frac{1}{n'}}
&\leq
\frac{C(s+\ell)}{(1-\sigma)^q}
\fint_{B_r} \psi(v)^{-\ell-1} \eta^{s-1}\psi(v)  + \psi(v)^{-\ell-1} \eta^{s-q}  \phi(x,v)\, dx\\
&\lesssim
\frac {s+\ell}{(1-\sigma)^q}
\fint_{B_r} \psi(v)^{-\ell-1} \eta^{s-q}  \phi(x,v)\, dx.
\end{align*}

Let us denote $V_R:= (\omega_{B_R}^-)^{-1}(d/|B_R|)$ and $E:= \{v(x) < V_R\}$.
Since $v\ge 1$, we find by \azero{}, \adec{} and \aones{\omega} that $\phi(x, v) \approx \psi(v)$
in $E$. Hence
\[
\frac{1}{|B_r|}\int_{B_r \cap E} \psi(v)^{-\ell-1} \eta^{s-q} \phi(x,v)\, dx
\approx
\frac{1}{|B_r|} \int_{B_r \cap E} \psi(v)^{-\ell} \eta^{s-q} \, dx.
\]
Since $\phi$ satisfies \adec{q} and $\psi$ satisfies \ainc{p}, we see that
$t\mapsto \psi(t)^{-\ell-1}\phi(x,t)$ is almost decreasing when $(\ell+1)p\ge q$.
Then when $v\ge V_R$ we obtain that
\[
\begin{split}
\frac{1}{|B_r|}\int_{B_r \setminus E} \psi(v)^{-\ell-1} \eta^{s-q} \phi(x,v)\, dx
&\lesssim
\frac{1}{|B_r|}\int_{B_r \setminus E} \psi(V_R)^{-\ell-1} \eta^{s-q}  \phi(x, V_R)\, dx\\
&\lesssim
\fint_{B_r} \psi(V_R)^{-\ell} \eta^{s-q}  \, dx
\lesssim  \psi(V_R)^{-\ell},
\end{split}
\]
again by $\phi(x, V_R) \approx \psi(V_R)$ (from \aones{\omega}). Furthermore, by \eqref{eq:vBound}
we obtain
\[
\begin{split}
\fint_{B_{r/2}} v \, dx
\lesssim
\fint_{B_{R}} \frac{u+R}{R} \, dx
\lesssim
(\omega^-_R)^{-1}\Big(\frac{d}{|B_R|} \Big) = V_R.
\end{split}
\]
This and Jensen's inequality for the convex function $t\mapsto \psi(t)^{-\ell}$ yield
\[
\psi(V_R)^{-\ell}
\lesssim
\psi\bigg( \fint_{B_{r/2}} v\, dx\bigg)^{-\ell}
\lesssim
\fint_{B_{r/2}} \psi(v)^{-\ell} \, dx
\lesssim
\fint_{B_r} \psi(v)^{-\ell} \eta^{s-q} \, dx.
\]
Combining the estimates in $E$ and $B_r\setminus E$ and the previous inequality, we have established that
\begin{equation}\label{eq:splitEstimate}
\Big(\fint_{B_r} \psi(v)^{-\ell n'} \eta^{sn'} \, dx  \Big)^{\frac{1}{n'}}
\leq
\frac {C(s+\ell)}{(1-\sigma)^{q}}
\fint_{B_r} \psi(v)^{-\ell } \eta^{s-q} \, dx  .
\end{equation}

Let us choose $s:=\ell-(n-1)q$, and suppose that $\ell$ is so large that $s \ge q$,
$(\ell+1)p\ge q$ and $\ell\ge nq$.
Raising both sides of the previous inequality to the power $-\frac1\ell$ gives
\begin{align*}
\underbrace{\Big(\fint_{B_r} \psi(v)^{-\ell n'} \eta^{\ell n' - nq} \, dx  \Big)^{-\frac{1}{n' \ell}}}
_{=:\Psi(n'\ell)}
\ge
\Big(\frac {C\ell}{(1-\sigma)^{q}} \Big)^{-\frac1\ell}
\underbrace{\Big(\fint_{B_r} \psi(v)^{-\ell} \eta^{\ell - nq} \, dx \Big)^{-\frac1\ell}}
_{=\Psi(\ell)}.

\end{align*}
Let us then set $\ell=n_k:=(n')^k$. For $k\ge k_0$ (so that the required
lower bounds on $\ell$ hold)
we use the standard iteration technique. By induction, we obtain that
\[
\Psi(n_{K})
\ge
\exp\Big( -\sum_{k=k_0}^{K-1} \frac{\ln n_k}{n_k}\Big)(C(1-\sigma))^{-\sum_{k=k_0}^{K-1}\frac{q}{n_k}}
\Psi(n_{k_0}).
\]
Denote $\gamma:=n_{k_0}$.
Since $\sum_{k=k_0}^\infty \frac{q}{n_k} = \frac{nq}\gamma =:\beta<\infty$ as a geometric series and
$\sum_{k=k_0}^\infty \frac{\ln n_k}{n_k} < \infty$ by comparison with a geometric series, we get
\begin{align*}
\essinf_{B_{\sigma r}} \psi(v)
&\ge
\essinf_{B_{r}} \frac{\psi(v)}\eta
= \lim_{K\to \infty} \Psi(n_{K}) \\
&\gtrsim
(1-\sigma)^{-\beta} \Psi(\gamma)
=
(1-\sigma)^{-\beta} \bigg(\fint_{B_r} \psi(v)^{-\gamma} \eta^{\gamma - nq}\, dx \bigg)^{-\frac1\gamma} \\
&\ge
(1-\sigma)^{-\beta} \bigg(\fint_{B_{r}} \psi(v)^{-\gamma} \, dx \bigg)^{-\frac1\gamma}
=
 (1-\sigma)^{-\beta} \psi\bigg(\xi^{-1}\Big(\fint_{B_{r}} \xi(v)\, dx \Big)\bigg)\\
&\gtrsim  \psi\bigg( (1-\sigma)^{-\frac\beta p} \xi^{-1}\Big(\fint_{B_{r}} \xi(v)\, dx \Big)\bigg),
\end{align*}
where $\xi(t):=\psi(t)^{-\gamma}$, and \ainc{p} of $\psi$ was used in the last inequality.
As $\psi$ is strictly increasing, this implies
\[
\essinf_{B_{\sigma r}} v \gtrsim (1-\sigma)^{-\frac\beta p} \xi^{-1}\Big(\fint_{B_{r}} \xi(v)\, dx \Big).
\]
Since $t\mapsto \xi(t^{-1/(\gamma q)})$ satisfies \adec{1}, it is equivalent to
a concave function \cite[Lemma~2.2.1]{HarH19}, and so by Jensen's inequality
\begin{align*}
\essinf_{B_{\sigma r}} v
\gtrsim (1-\sigma)^{-\frac\beta p}
\xi^{-1}\Big(\fint_{B_{r}} \xi(v)\, dx \Big)
&\gtrsim(1-\sigma)^{-\frac\beta p}
\Big( \fint_{B_{r}} v^{-\gamma q} \, dx  \Big)^{- \frac{1}{\gamma q}}.
\end{align*}
Then we recall that  $v= \frac{u+R}{r}$ and multiply both sides by $r$:
\begin{align}\label{equ:suuret-arvot}
\essinf_{B_{\sigma r}} u + R
&\gtrsim
(1-\sigma)^{-\frac\beta p} \Big( \fint_{B_r} (u + R)^{-\gamma q} \, dx  \Big)^{- \frac{1}{\gamma q}},
\end{align}
where $\frac{R}{2}\le r \le R$ and the constant depends only on $p$, $q$, $L_p$, $L_q$, $n$, $\ell$ and $d$.
This is the claim for $\ell=\gamma q$.
For exponents larger than $\gamma q$ the claim follows by Hölder's inequality.
We have thus established the claim for large exponents.

Finally we show the claim for small exponents. We use the $+R$ to simplify the proof, but this
is not essential here. So let $\ell\in (0,\gamma q)$. We observe that
\begin{align*}
\essinf_{B_{\sigma r}} u + R
&\ge
C (1-\sigma)^{-\frac\beta p} \Big( \fint_{B_r} (u + R)^{-\gamma q} \, dx  \Big)^{- \frac{1}{\gamma q}} \\
& \ge
C (1-\sigma)^{-\frac\beta p} \Big( \fint_{B_r} (u + R)^{-\ell} \, dx  \Big)^{- \frac{1}{\gamma q}}
(\essinf_{B_r} u+R)^{\frac{\gamma q-\ell}{\gamma q}}.
\end{align*}
We denote $Q:=\frac{\gamma q-\ell}{\gamma q}$, $\Lambda(r):=\essinf_{B_r} u+R$ and
$A:= ( \fint_{B_R} (u + R)^{-\ell} \, dx )^{-1/\ell}$. Thus, for $r\in [\frac R{2}, R]$,
\[
\Lambda(\sigma r) \ge C (1-\sigma)^{-\frac\beta p} A^{\frac \ell{\gamma q}} \Lambda(r)^Q.
\]
Then we set $r:=(1-2^{-(k+1)})R$ and $\sigma r := (1-2^{-k})R$ so that $1-\sigma \approx 2^{-k}$.
With $\Lambda_k:=\Lambda((1-2^{-k})R)$ and iteration we obtain
\[
\begin{split}
\Lambda_1
&\ge
C 2^{\frac\beta p} A^{\frac \ell{\gamma q}} \Lambda_2^{Q}
\ge
2^{\frac\beta p(1+2Q)} (CA^{\frac \ell{\gamma q}})^{1+Q} \Lambda_3^{Q^2}
\ge \ldots \\
&\ge
2^{\frac\beta p \sum_{k=1}^\infty kQ^{k-1}} (CA^{\frac \ell{\gamma q}})^{\sum_{k=0}^\infty Q^k} \liminf_{k \to \infty} \Lambda_k^{Q^k}.
\end{split}
\]
Since $\Lambda_k\ge R$ and $Q\in (0,1)$, $ \liminf_{k \to \infty} \Lambda_k^{Q^k}\ge 1$. Furthermore,
$\sum_{k=1}^\infty kQ^{k-1} < \infty$ and $\sum_{k=0}^\infty Q^k = \frac 1{1-Q} = \frac{\gamma q}\ell$.
Hence $\Lambda_1 \gtrsim A$, which is the claim for $\ell$.
\end{proof}

The previous proof only works for negative $\ell$. In fact, the largeness of $-\ell$ and the condition
\aones{\omega}  were only used in the paragraph with \eqref{eq:splitEstimate}.
With some modifications, we can iterate also for some positive exponents.
We use the limiting exponent $\ell(p)$ defined in the introduction.
See Proposition~\ref{prop:betaBound} for a simpler, sufficient condition for \eqref{eq:betaBound}.

\begin{prop}\label{prop:betaCondition}
Suppose $\phi\in \Phiw(\Omega)$ satisfies \azero{}, \ainc{p} and \adec{q}, $p,q>1$.
Let $u$ be a nonnegative supersolution to \eqref{eq:A-eq} in $B_R$.
Assume that
\begin{equation}\label{eq:betaBound}
\fint_{B_r} \Big(\frac{\phi(x,v)}{\phi_{B_r}^-(v)}\Big)^{\beta}\, dx \le d,
\end{equation}
for some $\beta>\max\{\frac np,1\}$ and all $B_r\subset B_R$, where $v:=\frac{u+r}r$.
For any $\ell_0>0$ and $\ell< \ell(p)$,
there exists a constant $C = C(p,q, L_p,L_q,n, \ell_0, \ell, d)>0$ such that
\begin{align*}
\Big(\fint_{B_{2R}} (u+R)^{\ell_0} \, dx  \Big)^{\frac{1}{\ell_0}}
\ge
C \psi^{-1}\bigg(\Big(\fint_{B_R} \psi(v)^{\frac \ell p} \, dx  \Big)^{\frac{p}{\ell}} \bigg)
\ge
C\Big(\fint_{B_R} (u+R)^{\ell} \, dx  \Big)^{\frac{1}{\ell}}.
\end{align*}
\end{prop}

\begin{proof}
We proceed as in the previous proof, but use the $W^{1,\gamma}$-Sobolev inequality
instead of the $W^{1,1}$-version. Here we will eventually take $\gamma \nearrow \min\{p,n\}$.
In place of \eqref{eq:after-sob} we obtain
\[
\begin{split}
\Big(\fint_{B_r} \psi(v)^{-\ell \gamma^*} \eta^{s \gamma^*} \, dx  \Big)^{\frac{\gamma}{\gamma^*}}
& \lesssim
\fint_{B_r} \big(\psi(v)^{-\ell-1} \eta^{s-1}
\big[\tfrac{\psi(v)}{1-\sigma}  + \eta  \psi'(v) |\nabla \tilde u| \big]\big)^\gamma\, dx \\

&\lesssim \fint_{B_r} \psi(v)^{-\ell \gamma}  \tfrac{\eta^{\gamma(s-1)}}{(1-\sigma)^\gamma}  + \eta^{s \gamma}  \psi(v)^{-\ell \gamma -1} \tfrac{\psi(v)}{v^\gamma} |\nabla \tilde u|^\gamma\, dx
\end{split}
\]
where we already divided by $r^{n-1}$, and used Remark~\ref{rem:incdec} as well as
$(a+b)^\gamma \approx a^\gamma+b^\gamma$.

We estimate $\frac{\psi(v)}{v^\gamma} |\nabla \tilde u|^\gamma$ with Young's inequality.
Define $\xi(t):=\psi(t^{1/\gamma})$. Then $\xi^{-1}(t) = \psi^{-1}(t)^\gamma$ and
\[
(\xi^*)^{-1}(t) \approx \frac{t}{\xi^{-1}(t)} = \frac{t}{\psi^{-1}(t)^\gamma}
\]
since $\xi^{-1}(t)(\xi^*)^{-1}(t)\approx t$ by \cite[Theorem~2.4.8]{HarH19}. Hence
\[
\xi^*\Big(\frac{\psi(t)}{t^\gamma} \Big) \approx \psi(t)
\]
and so by Young's inequality
\[
\frac{\psi(v)}{v^\gamma} |\nabla \tilde u|^\gamma
\le
\xi(|\nabla \tilde u|^\gamma) + \xi^*\Big( \frac{\psi(v)}{v^\gamma} \Big)
\approx
\psi(|\nabla \tilde u|) + \psi(v)
\lesssim
\phi(x,|\nabla \tilde u|) + \psi(v).
\]

Then we estimate with the Caccioppoli inequality (Lemma~\ref{lem:caccioppoli})
\[
\fint_{B_r} \psi(v)^{-\ell \gamma -1} \eta^{s \gamma} \phi(x,|\nabla \tilde u|)\, dx
\lesssim
\tfrac{1}{(1-\sigma)^q} \fint_{B_r} \psi(v)^{-\ell \gamma-1} \eta^{s\gamma -q} \phi(x,v) \, dx
\]
provided $\ell \gamma +1> \frac 1p$, which means that $\ell$ can also be negative.
Thus we have
\[
\begin{split}
\Big(\fint_{B_r} \psi(v)^{-\ell \gamma^*} \eta^{s \gamma^*} \, dx  \Big)^{\frac{\gamma}{\gamma^*}}
&\lesssim
\fint_{B_r}  \psi(v)^{-\ell \gamma}  \tfrac{\eta^{\gamma(s-1)}}{(1-\sigma)^\gamma}
+ \psi(v)^{-\ell \gamma-1} \tfrac{\eta^{s\gamma-q}}{(1-\sigma)^q} \phi(x,v)
+ \eta^{s \gamma}  \psi(v)^{-\ell \gamma} \, dx \\

&\le  \tfrac{1}{(1-\sigma)^q} \fint_{B_r}  \psi(v)^{-\ell \gamma-1} \eta^{s-q} \phi(x,v)  \, dx,
\end{split}
\]
where $\psi(v) \lesssim \phi(x, v)$,  $\eta^{\gamma (s-1)} \le \eta^{s-q}$ and $\eta^{\gamma s} \le \eta^{s-q}$ were used.
In contrast to the previous proof, we next use H\"older's inequality
\[
\fint_{B_r} \psi(v)^{-\ell\gamma-1} \eta^{s-q}  \phi(x,v)\, dx
\le
\bigg(\fint_{B_r} \psi(v)^{-\ell \gamma \lambda} \eta^{(s-q)\lambda}  \, dx \bigg)^\frac1\lambda
\bigg(\fint_{B_r} \Big(\frac{\phi(x,v)}{\psi(v)}\Big)^{\lambda'}\, dx \bigg)^\frac1{\lambda'}.
\]
Since the second factor on the right-hand side is bounded by \eqref{eq:betaBound} when
$\lambda' \in (\frac{n}{\gamma}, \beta]$, we obtain that
\[
\Big(\fint_{B_r} \psi(v)^{-\ell \gamma^*} \eta^{s\gamma^*} \, dx  \Big)^{\frac{1}{\gamma^*}}
\lesssim
\bigg(\fint_{B_r} \psi(v)^{-\ell \gamma \lambda} \eta^{(s-q)\lambda} \, dx\bigg)^\frac1{\gamma\lambda}.
\]
Since $\lambda < (\frac{n}{\gamma})' = \frac n{n-\gamma}$, we conclude that
$\gamma \lambda < \gamma^*$.
Therefore, we have obtained a reverse H\"older
type inequality, which can be iterated (as in the previous proof) to show that
\[
\bigg(\fint_{B_{2r}} v^{\ell_0} \, dx\bigg)^\frac1{\ell_0}
\gtrsim
\psi^{-1}\Bigg(\Big(\fint_{B_r} \psi(v)^{\gamma_2} \, dx  \Big)^{\frac{1}{\gamma_2}} \Bigg)
\ge
\bigg(\fint_{B_{r}} v^{p\gamma_2} \, dx\bigg)^\frac1{p\gamma_2}
\]
for $\gamma_2\le -\ell \gamma^*$ and any $\ell_0>0$; the last step is just Jensen's inequality.
Since we can choose any value of $\ell$ with $-\ell < \frac1{\gamma p'}$, we have the inequality
for any $\gamma_2 < \frac{\gamma^*}{\gamma p'} = \frac n{n-\gamma} \frac{p-1}p$.
Thus letting $\gamma \nearrow \min\{p,n\}$ we can obtain the weak Harnack inequality for exponent up to, but not including, $\ell(p)$.
\end{proof}

We are now ready for the proof of the main result, the weak Harnack inequality.
Note that here we need to add the \ainc{} assumption for $\phi$ compared to Theorem~\ref{thm:essinf}.

\begin{thm}\label{thm:weakHarnackGeneral}
Suppose $\phi$ satisfies \azero{}, \ainc{p} and \adec{q}, $1<p\le q<\infty$.
Let $u$ be a non-negative solution to \eqref{eq:A-eq} on $B_{2R}$.
Assume that there exists $\omega \in \Phiw(\Omega)$ which satisfies \azero{}  and \adec{} such that
one of the following holds:
\begin{enumerate}
\item
$\phi$ satisfies \aones{\omega} and \eqref{eq:vBound} and \eqref{eq:betaBound}, with $\beta=1$, hold.

\item
$\phi$ satisfies \aones{\omega} and $\| u\|_{W^{1,\omega}(B_{R})} \le d$.
\end{enumerate}
Then there exist $\ell_0=\ell_0(p,q,L_p,L_q,\beta,d,n)>0$ and  $C = C(p, q, L_p,L_q,\beta, d, n)>0$ such that
\begin{align*}
\bigg (\fint_{B_{2R}} (u +R)^{\ell_0} \, dx \bigg)^{\frac{1}{\ell_0}}  \le C (\essinf_{B_R} u + R).
\end{align*}
If \eqref{eq:betaBound} holds with $\beta>\max\{\frac np,1\}$, then we can choose any $\ell_0<\ell(p)$.
\end{thm}

\begin{proof}
Let $0< r \le \frac12 R$ and denote $v := \frac{u +r}r$ so that $v \approx \frac{u+2r}{2r}$.
By the $W^{1,1}$-Poincar\'e inequality we get
\begin{align*}
\fint_{B_{r}} \Big|\log(u+R) - \fint_{B_r}\log(u+R)\, dy\Big| \, dx
\leq
r \fint_{B_r} |\nabla (\log(u+R))| \, dx
\approx \fint_{B_r} \frac{|\nabla u|}{v} \, dx.
\end{align*}
Considering the cases $ \frac{|\nabla u|}{v}  \le 1$ and $\frac{|\nabla u|}{v} >1$,
we conclude that
\[
\frac{|\nabla u|}{v}
\lesssim
1 + \frac{\phi_{B_{2 r}}^-(|\nabla u|)}{\phi_{B_{2 r}}^-(v)}

\lesssim
1 + \frac{\phi(x,|\nabla u|)}{\psi_{2 r}(v)},
\]
where \ainc{1} was used in the case $\frac{|\nabla u|}{v} >1$, and $\psi_{2 r} \approx \phi^-_{B_{2 r}}$ in the last step.
Note that $\psi_{2 r}$ satisfies \inc{p}.
It follows from the Caccioppoli inequality (Lemma~\ref{lem:caccioppoli})
with $\ell=1> \frac1{p}$ and $s=q$ that
\[
\begin{split}
\int_{B_r} \frac{\phi(x,|\nabla u|)}{\psi_{2 r}(v)} \, dx
&\lesssim
\int_{B_{2r}} \frac{\phi(x,|\nabla u|)}{\psi_{2 r}\big(\frac{u+2r}{2r}\big)} \eta^q \, dx
\lesssim
\int_{B_{2r}}  \frac{\phi\big(x, \frac{u+2r}{2r}\big)}{\psi_{2 r}\big(\frac{u+2r}{2r}\big)} \, dx
\approx
\int_{B_{2r}}  \frac{\phi\big(x, \frac{u+2r}{2r}\big)}{\phi^-_{2 r}\big(\frac{u+2r}{2r}\big)} \, dx.
\end{split}
\]
Then we divide by $|B_{r}|$, note that $|B_{r}|\approx |B_{2r}|$, and combine with the previous inequalities to obtain that
\[
\fint_{B_r} \frac{|\nabla u|}{v} \, dx
\lesssim
1+
\fint_{B_r} \frac{\phi(x,|\nabla u|)}{\psi_{2 r}(v)} \, dx
\lesssim
1+
\fint_{B_{2r}}  \frac{\phi\big(x, \frac{u+2r}{2r}\big)}{\phi^-_{2 r}\big(\frac{u+2r}{2r}\big)} \, dx
\le 1+d,
\]
where \eqref{eq:betaBound} have been used in the last inequality.
Thus we have established $\log(u+R) \in BMO$ under assumption  (1).

Next we consider assumption (2) with $\|\nabla u\|_{L^\omega(B_{2R})} \le d$.
Define
\begin{align*}
E &:= \{x \in B_{2r} : \omega^{-1}(x,\tfrac1{r^n}) < v(x)\}.

\end{align*}
In $B_r\setminus E$, we have $\phi^+_{B_{2r}}(v)\approx \phi^-_{B_{2r}}(v)$ by the \aones{\omega}
condition of $\phi$ and $v \ge 1$. Then we use the Caccioppoli inequality as before, except instead of
$\psi_r$ we use a corresponding function with $\psi_{2r} \approx \phi^+_{B_{2r}}$ and so
do not need \eqref{eq:betaBound}.

In $E$, we use Young's inequality for the $\Phi$-function $\xi:=r^n \omega$:
\[
\frac{|\nabla u|}{v}
\le
\xi(x,\tfrac1\epsilon|\nabla u|) + \xi^*(x, \tfrac{\epsilon}{v})
=
r^n \omega(x,\tfrac1\epsilon|\nabla u|) + r^n\omega^*(x, \tfrac{\epsilon}{r^{n}v}),
\]
since $\xi^* (t) = r^n \omega^*(t/r^n)$  by \cite[Lemma 2.4.3]{HarH19}.
For the first term, $\fint_{B_r} r^n \omega(x,\tfrac1\epsilon|\nabla u|)\, dx \le c$ by the assumption
$\|\nabla u\|_{L^\omega(B_{2R})} \le d$ and \adec{} of $\omega$.
In $E$, $v > \omega^{-1}(x,\tfrac1{r^n}) \approx \frac{1}{r^n (\omega^*)^{-1}(x,\frac1{r^n})}$,
since $\omega^{-1}(t) (\omega^*)^{-1}(t) \approx t$ \cite[Theorem 2.4.8]{HarH19}.
So $\frac1{r^{n}v} \lesssim (\omega^*)^{-1}(x,\frac1{r^n})$, and hence
$\omega^*\big(x, \frac\epsilon{r^{n}v}\big) \le \frac1{r^n}$ for appropriate $\epsilon>0$.
Therefore
\begin{align*}
\frac{1}{|B_{2r}|}\int_{E} r^n\omega^*(x, \tfrac{\epsilon}{r^{n}v}) \, dx
\le
\frac{1}{|B_{2r}|} \int_{E} 1 \, dx \leq 1.
\end{align*}
We have bounded all terms, so $\log(u+R) \in BMO$ also under assumptions (2).

Since $\log(u+R)\in BMO$, the John--Nirenberg lemma says that there exist positive
constants $\ell_0$ and $C$ such that
\begin{align*}
\bigg (\fint_{B_{2R}}  (u+R)^{\ell_0} \, dx  \bigg) \bigg(\fint_{B_{2R}} (u+R)^{-\ell_0} \, dx  \bigg) \leq C.
\end{align*}
Let us note that under assumption (2), condition \eqref{eq:vBound} holds by
the Poincar\'e inequality and Jensen's inequality, see Proposition~\ref{prop:vBound}.
By the previous inequality and Theorem~\ref{thm:essinf} we obtain
\[
\bigg (\fint_{B_{2R}} (u+R)^{\ell_0} \, dx  \bigg)^{\frac{1}{\ell_0}}
\lesssim
\bigg (\fint_{B_{2R}} (u+R)^{-\ell_0} \, dx  \bigg)^{-\frac{1}{\ell_0}}
\lesssim
\essinf_{B_R} u + R .
\]

Assume then that \eqref{eq:betaBound} holds with $\beta>\max\{\frac np,1\}$.
Then by Proposition~\ref{prop:betaCondition} we  have
for any $\ell_0>0$ and $\ell< \ell(p)$,  that
\begin{align*}
\Big(\fint_{B_{4R}} (u+2R)^{\ell} \, dx  \Big)^{\frac{1}{\ell}}
\lesssim
\Big(\fint_{B_{2R}} (u+2R)^{\ell_0} \, dx  \Big)^{\frac{1}{\ell_0}},
\end{align*}
and hence the claim follows when we combine this with the previous inequality.
(With this proof we obtained the ball $B_{4R}$ on the right-hand side. Proving the
intermediate results for $B_{\sqrt 2 R}$ give $B_{2R}$.)
\end{proof}

\section{Special cases}\label{sect:special}

Let us next investigate conditions \eqref{eq:vBound} and \eqref{eq:betaBound}.
We define the ``Sobolev conjugate'' $\omega^\#\in\Phiw$ by the condition
$(\omega^\#)^{-1}(t) := t^{-1/n} \omega^{-1}(t)$ for $t>0$. Note that for $\omega^\#$
to be in $\Phiw$, it must be increasing and so $\omega$ must satisfy \dec{n}, see \cite[Lemma~5.2.3]{HarH19}.
If $\omega(t)=t^s$, then $\omega^\#(t)=t^{s^*}$, where $s^*:=\frac{ns}{n-s}$ is
the Sobolev exponent. Also $s=n$ is included and in this case $\|u\|_{\omega^\#} = \|u\|_\infty$.
Thus we must not assume that $\omega^\#$ satisfies \adec{}, which means that extra care
must be taken in the next proof regarding constants and inverse functions, but tools for this were
established in \cite{HarH19}.

Note that (2) implies (1) if we have a Sobolev inequality, but then we have difficulties
with the case $s\ge n$ in (2), so we provide separate proofs for the two cases.

\begin{prop}\label{prop:vBound}
Let $B_R \subset \Omega$, $R \le 1$ and
let $\omega\in \Phiw(\Omega)$ satisfy \azero{} and \adec{}.
Assume that $\phi\in \Phiw(\Omega)$ satisfies \aones{\omega} and that one of the
following hold:
\begin{enumerate}
\item
$u\in L^{\omega^\#}(B_R)$ with $\|u\|_{\omega^\#}\le d$, where $\omega^\#\in \Phiw(\Omega)$.

\item
$u\in W^{1,\omega}(B_R)$ with $\|u\|_{1,\omega}\le d$.
\end{enumerate}
Then \eqref{eq:vBound} holds, for $v:=\frac{u+R}R$, i.e.\
\[
\omega_{B_R}^-\bigg(\fint_{B_R} v \, dx \bigg) \le \frac {d_2}{|B_R|}.
\]
\end{prop}
\begin{proof}
By an elementary embedding, $\|u\|_{W^{1,\omega_{B}^-}(B)}\lesssim \|u\|_{W^{1,\omega}(\Omega)}$.
Therefore it suffices to prove the result for $\omega\in \Phiw$ (i.e.\ $\omega$ independent of $x$)
and apply this result to $\omega_ {B_R}^-$, and similarly in the case of assumption (1).

Let us first use assumption (1).
Jensen's inequality \cite[Lemma 4.3.1]{HarH19}, and \azero{} of $\omega^\#$ yield that, for sufficiently
small $d'>0$,
\[
\begin{split}
\omega^\#\bigg(2d'\fint_{B_R} \tfrac{u+R}2\, dx\bigg)
&\le \fint_{B_R} \omega^\#(\tfrac{u+\beta_0}2)\, dx
\lesssim
|B_R|^{-1}+1 \lesssim |B_R|^{-1}
\end{split}
\]
with $\beta_0$ from \azero{}.
By \cite[Lemma~2.3.9]{HarH19}, $(\omega^\#)^{-1}\big(\omega^\#(t)\big)\approx t$ or $\omega^\#(t)=0$.
In the former case,
\[
\fint_{B_R} u+R\, dx
\lesssim
(\omega^\#)^{-1}(c \,|B_R|^{-1})
=
c' R \omega^{-1}(c \,|B_R|^{-1})
\approx
R \omega^{-1}(|B_R|^{-1}).
\]
Since $\omega$ satisfies \adec{}, we can move the constants outside.
We divide both sides by $R$ and apply $\omega$ to obtain
$\omega(\fint_{B_R} v\, dx)\lesssim |B_R|^{-1}$, i.e.\ \eqref{eq:vBound}.
In the case $\omega^\#(t)=0$ we conclude from \azero{} and \dec{n} of $\omega$ that
$t\le d''$ and so
\[
\omega\bigg(\fint_{B_R} v\, dx\bigg)
\le
\omega\bigg(\frac{1}{d' R} 2d'\fint_{B_R} \tfrac{u+R}2\, dx\bigg)
\le
\omega\bigg(\frac{d''}{d' R}\bigg)
\lesssim
R^{-n}
\]
also by \dec{n} of $\omega$. Thus we have \eqref{eq:vBound} under assumption (1).

For the other case, we first use  Hölder's inequality to obtain
\[
\fint_{B_R} v\, dx \lesssim R^{-n} \Big(\int_{B_R}  (u+R)^{n'}\, dx\Big)^\frac1{n'}.
\]
Then we use that $B_R$ is a $W^{1,1}$-extension domain,
extend $u+R$ to $\Rn$ and denote this extension by $v$. We obtain by the $W^{1,1}$-Sobolev embedding
\[
\|u+R\|_{L^{n'}(B_R)}
\le
\|v\|_{L^{n'}(\Rn)}
\lesssim \|v\|_{W^{1,1}(\Rn)}
\lesssim \|u+R\|_{W^{1,1}(B_R)}.
\]
Combining the above steps, we have
\[
\fint_{B_R} v\, dx \le \fint_{B_R}  u+R+ |\nabla u| \, dx.
\]
Jensen's inequality, \dec{} of $\omega$, \azero{} of $\omega$ , $R \le 1$, and $\|u\|_{1,\omega}\le d$ yield
\[
\begin{split}
\omega\Big( \fint_{B_R} v\, dx \Big) &\le  \omega\Big( \fint_{B_R} u+R+|\nabla u|\, dx \Big)
\le \fint_{B_R} \omega (u+R+|\nabla u|)\, dx\\
&\lesssim  \fint_{B_R} \omega (u)+\omega(|\nabla u|) + \omega(1)\, dx \lesssim \frac1{|B_R|}.\qedhere
\end{split}
\]
\end{proof}

Note that in the next proposition if we consider case (2) with $\omega=\phi$, then $s=p$ and
$s^*>\max\{\frac np,1\} (q-p)$ is equivalent to $p^* > q$, the condition from Theorem~\ref{thm:weakHarnack}.

\begin{prop}\label{prop:betaBound}
Let $\omega\in \Phiw(\Omega)$ satisfy \azero{}, \adec{}, and \ainc{s}, $s\ge 1$.
Let $\phi\in \Phiw(\Omega)$ satisfy \azero{}, \aones{\omega} and \adec{} and assume that one of the
following hold:
\begin{enumerate}
\item
$u\in L^{\omega^\#}(B_R)$ with $\|u\|_{\omega^\#}\le d$, where $\omega^\#\in \Phiw(\Omega)$.
\item
$u\in W^{1,\omega}(B_R)$ with $\|u\|_{1,\omega}\le d$.
\end{enumerate}
If $s^*\ge\beta (q-p)$, then \eqref{eq:betaBound} holds,
i.e.\
\[
\fint_{B_r} \Big( \frac{\phi(x, v)}{\phi_{B_r}^-(v)}\Big)^\beta\, dx \le C
\]
for and $B_r\subset B_R$ and $v:=\frac{u+r}r$.
\end{prop}
\begin{proof}
Fix $B_r\subset B_R$.
As in the previous proof, it suffices to consider $\omega\in \Phiw$ independent of $x$.
Points where $u(x)\le r$ make only a constant contribution to the integral,
so we may assume that $u>r$ and take $v=\frac ur$ for simplicity.  Here we used \azero{} and \adec{} of $\phi$.

We denote $V_r:=\omega^{-1}(|B_r|^{-1})$
and $E:=\{v<V_r\}$. Then $\omega (V_r) \le \frac1{|B_r|}$ and  \aones{\omega} of $\phi$ yields
\[
\frac1{|B_r|} \int_{B_r\cap E} \Big( \frac{\phi(x, v)}{\phi^-_{B_r}(v)}\Big)^\beta\, dx
\approx
\frac1{|B_r|} \int_{B_r\cap E} \Big( \frac{\phi(x, v)}{\phi(x,v)}\Big)^\beta\, dx
\le 1;
\]
we used $v>1$ and \azero{} for the lower bound of \aones{\omega}.
This holds in both case (1) and (2), since these assumption were not used yet.

In $B_r\setminus E$ we have
\[
\frac{\phi(x, v)}{\phi^-_{B_r}(v)}
\lesssim
\frac{\phi(x, V_r)}{\phi^-_{B_r}(V_r)} (\tfrac v {V_r})^{q-p}
\approx
(\tfrac v {V_r})^{q-p}
\]
by \ainc{p}, \adec{q} and \aones{\omega}.
Consider first the case $\|u\|_{\omega^\#}\le d$.
By the definition of $(w^\#)^{-1}$, we obtain that $(w^\#)^{-1}$ satisfies \adec{\frac1s - \frac1n}
and hence $w^\#$ satisfies \ainc{\frac{sn}{n-s}}
and thus also \ainc{\beta(q-p)}. Therefore
$t\mapsto \omega^\#(t^{1/(\beta(q-p))})$ satisfies \ainc{1}, and hence a Jensen-type inequality \cite[Lemma 4.3.1]{HarH19} yields
\[
\begin{split}
\bigg(\fint_{B_r} (\tfrac v{V_r})^{\beta(q-p)}\, dx\bigg)^\frac1{\beta(q-p)}
&\le
\frac1{r V_r} (\omega^\#)^{-1}\bigg( \fint_{B_r} \omega^\#(u)\, dx \bigg)
\lesssim
\frac{(\omega^\#)^{-1}( |B_r|^{-1} )}{r V_r}\\
&=  \frac{ |B_r|^{1/n} \omega^{-1}( |B_r|^{-1} )}{r V_r} \approx 1
\end{split}
\]
where we used $\rho_{\omega^\#}(u)\le C$ and the definition of $V_r$.
This completes the estimate in case (1).

In the case $\|u\|_{1,\omega}\le d$, the H\"older inequality (with $\beta(q-p)\le s^*$),
the Sobolev inequality and the Jensen inequality (with \ainc{p}) give
\[
\begin{split}
\bigg(\fint_{B_r} (\tfrac v {V_r})^{\beta(q-p)}\, dx\bigg)^\frac1{\beta(q-p)}
&\le
\frac1 {V_r} \bigg(\fint_{B_r} (\tfrac ur)^{s^*}\, dx\bigg)^\frac1{s^*}
\lesssim
\frac1 {V_r} \bigg(\fint_{B_r} u^s+|\nabla u|^s\, dx\bigg)^\frac1{s} \\
&\lesssim
\frac1{V_r} \omega^{-1}\bigg( \fint_{B_r} \omega(u) + \omega(|\nabla u|) \, dx \bigg)
\lesssim
\frac{\omega^{-1}( |B_r|^{-1} )}{V_r}= 1. \qedhere
\end{split}
\]
\end{proof}

We can now state the weak Harnack inequality for the variable exponent case as a corollary.
For notation and terminology we refer to \cite{DieHHR11}. Note that here we
can choose any $s>0$.

\begin{cor}
Let $\phi(x,t):=t^{p(x)}$ be the variable exponent functional
and let $u$ be a non-negative solution to \eqref{eq:A-eq} on $B_{2R}$.
We assume that $p$ is $\log$-H\"older continuous with constant $c_{\log}$, $1<p^-\le p^+<\infty$,
and $\|u\|_s \le d$ for some $s\in (0,\infty]$.
Then for every  $0<\ell_0< \ell(p^-_{B_R})$, there exists
$C = C(p^-, p^+, \ell_0,c_{\log}, d, n)$ and $R_0$ such that
\begin{align*}
\bigg (\fint_{B_{2R}} (u+R)^{\ell_0} \, dx \bigg)^{\frac{1}{\ell_0}}  \le C (\essinf_{B_R} u + R)
\quad\text{for all } r \le R_0.
\end{align*}
\end{cor}
\begin{proof}
We check that the assumptions of Theorem~\ref{thm:weakHarnackGeneral} are fulfilled.
Define $s_*:=\frac{ns}{n+s}$ so that $(s_*)^* = s$ and let $\omega(t):=t^{s_*}$.
Now \aones{s_*} reads as $\beta^{p(x)} t^{p(x) - p(y)} \le 1$ for $t \in [1, |B|^{-\frac{1}{s_*}}]$.
Since $p$ is $\log$-H\"older continuous, this holds (see \cite[Section~7.1]{HarH19} for details).
Thus by  Proposition~\ref{prop:vBound} (1) condition \eqref{eq:vBound} holds.

Let $\beta:=n >\frac{n}{p^-_{B_R}}$. We  choose $R_0$ so small that $n(p_{B_R}^+ - p_{B_R}^-)< s$ for $R \le R_0$. Then $\omega^\#(t) = t^s$ satisfies \ainc{\beta(p_{B_R}^+ - p_{B_R}^-)}, and hence \eqref{eq:betaBound} holds  by Proposition~\ref{prop:betaBound} (1). Thus the exponent can be chosen up to $\ell(p^-_{B_r})$.
\end{proof}

\section{The double phase case and counter-examples}\label{sect:example}

Let us study the double phase case. Note that when $s=\infty$ in the next result we obtain
the special case of bounded supersolutions from \cite{BenK_pp}.

\begin{cor}\label{cor:doublePhase-2}
Let $\phi(x,t):=t^p + a(x)t^q$ be the double phase functional
and let $u$ be a non-negative supersolution to \eqref{eq:A-eq} on $B_{2R}$.
We assume that $a\in C^{0,\alpha}$ and $\|u\|_{s} \le d$, $s\in (0,\infty]$. If
$\alpha \ge (\tfrac{n}s+1)(q-p)$ and $s\ge q-p$, then
there exist  positive constants $\ell_0$ and $C$, such that
\begin{align*}
\bigg (\fint_{B_{2R}} (u +R)^{\ell_0} \, dx \bigg)^{\frac{1}{\ell_0}}  \le C (\essinf_{B_R} u + R).
\end{align*}
Furthermore, if $s>\max\{\frac np,1\}(q-p)$, then the inequality holds for any $\ell_0<\ell(p)$.
\end{cor}

\begin{proof}
Let us show that assumption (1) of Theorem~\ref{thm:weakHarnackGeneral} is fulfilled.
Let $\omega(t):=t^{s_*}$, where $s_*= \frac{ns}{n+s}$.
Let $x,y\in B_R$ and $t\in [1, |B_R|^{-1/s_*}]$. Then
\[
\phi(x,t) = t^p + a(x) t^q
\le
t^p + (a(y) + c R^\alpha) t^q
\le
(1+cR^\alpha t^{q-p} ) \phi(y,t).
\]
Since $t\lesssim R^{-\frac n{s_*}}$, the coefficient is bounded provided $\alpha -\frac n{s_*}(q-p)\ge 0$.
Since $\frac n{s_*} = \frac ns + 1$, this is $\alpha \ge (\tfrac{n}s+1)(q-p)$. We have proved that $\phi$ satisfies \aones{s_*}. Now $(\omega^\#)^{-1}(t) = t^{-1/n} \omega^{-1}(t) = t^{-1/n + 1/s_*}= t^{1/s}$, and thus
$\rho_{\omega^\#}(u) = \rho_{L^s}(u) \le d^s$. Hence \eqref{eq:vBound} holds by Proposition~\ref{prop:vBound}.
Let us then consider \eqref{eq:betaBound}. Since $\omega^\#(t) = t^{s}$, it satisfies \ainc{q-p} provided that $s\ge q-p$ and \ainc{\beta(q-p)} for $\beta>\max\{\frac np,1\}$ provided that $s> \max\{\frac np,1\}(q-p)$.
Thus \eqref{eq:betaBound} follows by Proposition~\ref{prop:betaBound} and the claims follow by
Theorem~\ref{thm:weakHarnackGeneral}.
\end{proof}

We will next give an example that  the weak Harnack inequality need not hold if
$\alpha < (\frac{n}s+1)(q-p)$.
We consider the one-dimensional case and focus on the role of $s$.
In the double phase (or $(p,q)$-growth) case Fonseca, Mal\'y and Mingione \cite{FonMM04}
have given more sophisticated counter-examples for $s=p$,
which is related to the assumption \aone{} in this case. See also \cite{BalDS_pp}.
However, to the best of our knowledge, examples for $s\ne p$ have not been considered before.

Let $\phi\in \Phiw(\R)$ be defined by $\phi(x,0):=0$ and
\[
\phi'(x,t) := \max\{t^{p-1}, a(x) t^{q-1}\},
\]
so that $\phi(x,t) \approx \max\{t^p, a(x) t^q\} \approx t^p + a(x) t^q$, the double phase functional.
Let $u$ be a solution of $\big(\phi'(x,|u'|)\frac {u'}{|u'|}\big)'=0$ on the interval
$(a,b)$. We assume that $\lim_{x \to a^+}u(x)< \lim_{x\to b^-}u(x)$, so $u$ is increasing and $\frac {u'}{|u'|}=1$.
Then the differential equation reduces to $\phi'(x,u')\equiv c$, i.e.\
\[
u'(x) =
\begin{cases}
c^{\frac1{p-1}}, & \text{when } c^{-\frac{q-p}{p-1}} \ge a(x), \\
(c/a(x))^{\frac1{q-1}}, & \text{otherwise.} \\
\end{cases}
\]
We further assume that $a(x):=\max\{-x,0\}^\alpha$. Since $a$ is decreasing,
we obtain that
\[
u'(x) =
\begin{cases}
c^{\frac1{p-1}}, & \text{when } x \ge -x_0, \\
(c |x|^{-\alpha})^{\frac1{q-1}}, & \text{when } x < -x_0, \\
\end{cases}
\]
for $x_0 := c^{-\frac1\alpha \frac{q-p}{p-1}}$. Some solutions are illustrated in Figure~\ref{fig:solutions} for different values of $c$ with zero left boundary values at $x=-1$.

\begin{figure}[ht!]
\definecolor{cqcqcq}{rgb}{0.7529411764705882,0.7529411764705882,0.7529411764705882}
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=4.0cm,y=1.0cm]
\draw [color=cqcqcq,, xstep=1.0cm,ystep=1.0cm] (-1.1,-0.1) grid (1.1,6.3);
\draw[->,color=black] (-1.1,0.) -- (1.1,0.);
\foreach \x in {-1,-0.5,0.5,1}
\draw[shift={(\x,0)},color=black] (0pt,2pt) -- (0pt,-2pt) node[below] {\footnotesize $\x$};
\draw[->,color=black] (0.,-0.5) -- (0.,6.3);
\foreach \y in {1,2,3,4,5,6}
\draw[shift={(0,\y)},color=black] (2pt,0pt) -- (-2pt,0pt) node[left] {\footnotesize $\y$};
\draw[color=black] (0pt,-10pt) node[right] {\footnotesize $0$};
\clip(-1.1,-0.4) rectangle (1.1,6.3);
\draw [line width=1.2pt,domain=-1:-(1.01)^(-18)] plot(\x,{2*1.01 - 2*1.01*sqrt((-\x))}) node{$\bullet$};
\draw [line width=1.2pt,domain=-(1.01)^(-18):1] plot(\x,{2*1.01 - 2*(1.01)^(-8)+(1.01)^(10)*((\x)+(1.01)^(-18)});
\draw [line width=1.2pt,domain=-1:-(1.1)^(-18)] plot(\x,{2*1.1 - 2*1.1*sqrt((-\x))}) node{$\scriptstyle\bullet$};
\draw [line width=1.2pt,domain=-(1.1)^(-18):1] plot(\x,{2*1.1 - 2*(1.1)^(-8)+(1.1)^(10)*((\x)+(1.1)^(-18)});
\draw [line width=1.2pt,domain=-1:-(1.2)^(-18)] plot(\x,{2*1.2 - 2*1.2*sqrt((-\x))}) node{$\scriptstyle\bullet$};
\draw [line width=1.2pt,domain=-(1.2)^(-18):1] plot(\x,{2*1.2 - 2*(1.2)^(-8)+(1.2)^(10)*((\x)+(1.2)^(-18)});
\draw [smooth,samples=50,line width=1.2pt,domain=-1:-(1.3)^(-18)] plot(\x,{2*1.3 - 2*1.3*sqrt((-\x))}) node{$\scriptstyle\bullet$};
\draw [line width=1.2pt,domain=-(1.3)^(-18):0.3] plot(\x,{2*1.3 - 2*(1.3)^(-8)+(1.3)^(10)*((\x)+(1.3)^(-18)});
\draw [smooth,samples=50,line width=1.2pt,domain=-1:-(1.4)^(-18)] plot(\x,{2*1.4 - 2*1.4*sqrt((-\x))}) node{$\scriptstyle\bullet$};
\draw [line width=1.2pt,domain=-(1.4)^(-18)-0.001:0.2] plot(\x,{2*1.4-2*(1.4)^(-8)+(1.4)^(10)*((\x)+(1.4)^(-18)});

\begin{scriptsize}

\end{scriptsize}
\end{tikzpicture}
\caption{Solution for $c\in \{1.01, 1.1, 1.2, 1.3, 1.4\}$ in $[-1, 1]$. The parameters are  $p=1.1$, $q=2$ and $\alpha =0.5$. The right boundary values have been partly cut away but they are in the range $[2, 32]$. The
dots indicate $x_0$.}\label{fig:solutions}
\end{figure}

For $c>0$ and $r>x_0$, we next consider a solution with $u(-x_0-2r)=0$. When
$\frac\alpha{q-1}\ne 1$ we have, for $\rho\in [0,r]$ and $\alpha_2:=1-\frac\alpha{1-q}$,
\[
u(-x_0-\rho)
=
c^{\frac1{q-1}} \int_{-x_0-2r}^{-x_0-\rho} |x|^{-\frac \alpha{q-1}}\, dx
=
c^{\frac1{q-1}} \frac{(x_0+2r)^{\alpha_2} - (x_0+\rho)^{\alpha_2}}{\alpha_2}
\approx c^{\frac1{q-1}} r^{\alpha_2}.
\]
Furthermore, $u(-x_0+\rho) = u(-x_0) + \rho c^{\frac1{p-1}}$ since the derivative is constant on
$(-x_0,\infty)$.
With $c^{\frac1{p-1}-\frac1{q-1}} = c^{\frac{q-p}{(p-1)(q-1)}} = x_0^{-\frac\alpha{q-1}}$, we calculate
\[
\frac{u(-x_0+r)}{u(-x_0)}
=
1 + \alpha_2\frac{r c^{\frac1{p-1}-\frac1{q-1}} x_0^{-\alpha_2}}{(1+2r/x_0)^{\alpha_2} - 1}
=
1 + \alpha_2 \frac{r/x_0}{(1+2r/x_0)^{\alpha_2} - 1}.
\]
Since $\alpha_2<1$, we see that the constant in the Harnack inequality
in $B(-x_0,r)$ blows up if $r/x_0\to \infty$.

We next estimate the $L^s$-norm of $u$ when $s>0$.
For $\rho\in [\frac r2,2r]$, $u(-x_0+\rho) = u(-x_0) + \rho c^{\frac1{p-1}} \approx rc^{\frac1{p-1}}$.
Since $u\ge 0$ is increasing and since only large values of the function are important when $s>0$,
we find that
\[
\|u\|_{L^s(B(-x_0,2r))}
\approx c^{\frac1{p-1}} r^{1+\frac1s}
= x_0^{1+\frac1s-\frac\alpha{q-p}} (\tfrac r{x_0})^{1+\frac1s}
\quad\text{and}\quad
\bigg(\fint_{B(-x_0,r)} u^s\, dx\bigg)^\frac1s \approx c^{\frac1{p-1}} r.
\]
Similarly, if $x\in [-x_0- r, -x_0]$, then
by the earlier formula $u(x) \approx c^{\frac1{q-1}} r^{\alpha_2}$ and,
since only small values of the function are important when $-s<0$, we obtain that

\[
\bigg(\fint_{B(-x_0,r)} u^{-s}\, dx\bigg)^{-\frac1s} \approx c^{\frac1{q-1}} r^{\alpha_2}.
\]

We can therefore say in this example that
\[
\esssup_{B(-x_0,r)} u =u(-x_0+r)\approx \bigg(\fint_{B(-x_0,r)} u^s\, dx\bigg)^\frac1s
\]
and
\[
\essinf_{B(-x_0,r)} u =u(-x_0-r) \approx \bigg(\fint_{B(-x_0,r)} u^{-s}\, dx\bigg)^{-\frac1s}
\]
for every $s>0$, with constant depending on $s$ but not on $c$ or $r$.
It follows that a possible failure in the Harnack inequality is due to the
passing over zero.

As was mentioned before, the failure of the weak Harnack inequality happens if $\tfrac r{x_0}\to\infty$,
for instance if we choose $r:= x_0 \log \frac1{x_0}$. Furthermore,
\[
\|u\|_{L^s(B(-x_0,2r))}
\approx
x_0^{1+\frac1s-\frac\alpha{q-p}} (\tfrac r{x_0})^{1+\frac1s}
=
x_0^{1+\frac1s-\frac\alpha{q-p}} (\log\tfrac1{x_0})^{1+\frac1s}
\]
remains bounded as $x_0\to 0$ if $1+\frac1s-\frac\alpha{q-p} > 0$.
Since $n=1$, this is equivalent to $\alpha<(1+\frac ns)(p-q)$, the complement of the inequality in
Corollary~\ref{cor:doublePhase}. This shows the sharpness of the \aones{s_*} assumption
in Theorem~\ref{thm:weakHarnack}.

In addition, we have $\phi(x,u')\approx u' \phi'(x,u')=c u'$. Thus
\[
\int_{B(-x_0,2r)} \phi(x,u')\, dx = c(u(-x_0+2r) - u(-x_0 -2r))
= c u(-x_0+2r)
\approx c^{\frac p{p-1}}r
= x_0^{1-\frac{p\alpha}{q-p}} \tfrac r{x_0} .
\]
Here we see that the Harnack inequality does not hold with uniform
constant even though the $W^{1,\phi}$-norm of $u$ remains bounded
provided that $\frac{p\alpha}{q-p} < 1$. On the other hand,
for the double phase functional \aone{} is equivalent to $\frac{p\alpha}{q-p}\ge 1$,
and we showed above that the weak Harnack inequality holds in this case.
This proves the sharpness of the \aone{} assumption in Theorem~\ref{thm:weakHarnack}.

config: arxiv


TITLE: Proof: Categorical Product = Topological Product
QUESTION [2 upvotes]: Is there a nice way to prove that the categorical product equals the topological product?
What I mean is the following:
Starting with a given family of topological spaces $X_i$ and any topological space $Z$ together with continuous mappings $f_i:Z\to X_i$ let me construct a categorical product in the category $\mathbf{Top}$.
First, I consider them as merely sets so I can construct a categorical product in the category $\mathbf{Set}$.
That is taking the cartesian product $\prod_{i\in I}X_i$ as object and the cartesian projections $\pi_i:\prod_{i\in I}X_i\to X_i$ as morphisms there is a uniquely determined morphism $f:Z\to \prod_{i\in I}X_i$ s.t. the diagram commutes $f\circ\pi_i=f_i$.
Now, I start to equip that product with some topology and I have to do that in such a way that the projections $\pi_i$ are continuous and that the respective mappings $f$ are continuous as well; so I see that the product topology must neither be to fine nor to coarse.
Since it has to hold for arbitrary objects $Z$ I rather start with the projections $\pi_i$ than with the mappings $f$; thus, I choose the coarsest topology s.t. the projections $\pi_i$ are still continuous and hope that this topology is coarse enough to make the mappings $f$ continuous.
That is I take all sets $\pi_i^{-1}(U_i)$ as generator for my product topology. That is certainly the coarsest topology s.t. the projections $\pi_i$ are still continuous.
At that point, I'm stuck by showing that the mapping $f$ actually become continuous w.r.t. the chosen topology!
Finally, this uniquely defines a categorical product in the category $\mathbf{Top}$ up to isomorphism by category theoretic considerations - so I'm done from the category theory side.
Next, I can check wether this is actually the topological product - but this is a rather easy exercise:
Given any map $f:Z\to\prod_{i\in I}X_i$.
If $f$ is continuous so is $\pi_i\circ f$ continuous as composition of continuous maps.
Conversely, if $\pi_i\circ f$ is continuous so is $f$ continuous by the existence and uniqueness of morphism for a categorical product.
Besides I want to understand:
Apart from the fact that the categorical product is unique up to isomorphism the question now arises wether the topological product is unique since careful considerations show two aspects:

The start off for a topological product is different data:
$X_i$, $X$(new!) plus $\pi_i:X\to X_i$(new!) and $Z$ together with $f:Z\to X$(switched! + merely map!) where $X:=\prod_{i\in I}X_i$
The characteristic property does not claim the existence and uniqueness of an according morphism making the diagram commute:
$f_i:=\pi_i\circ f$(taken not given!) and $f$(given not stated!)

So, going from the characteristic(!) property (I strictly distinguish from universal property here) I cannot claim the uniqueness of the topological product up to isomorphism without referring to the relation to the categorical product and its universal(!) property.
So, does somebody know how to carefully!!!!!! untwine this - again I want to stress carefully!!!!!! and not by some pretty standard blah blah blah which I read already here and there (I'm really sorry for being so rude in here - I just want to pretend by sort of everytime popping out nowhere leading answers)
Thanks everybody alot for hints, insights and ideas! =)

REPLY [0 votes]: @Besides I want to understand:
I think I got the difference now:
While the universal property defines the object completely, the characteristic property defines the topology only!
That is, there is one and only one(!) topological product $\prod_{i\in I}X_i$ uniquely determined by the universal property - but there can be many(!) pairs $(X,\pi_i)$ just as soon as one chooses a specific pair its topology is uniquely determined by the characteristic property.
For example:
Take the real line $\mathbb{R}$ there is one and only one 'topological product' $\mathbb{R}$ however many different 'subspaces' like $[0,1]$ or $(0,1]$ or $[0,1]\cup[2,3]$.
Moreover:
It can happen that some of those realize a product as well as they happen to be homeomorphic - even though they at first glance seem to satisfy the characteristic property only (e.g. $(0,1)\cong\mathbb{R}$).


TITLE: If $\log_510=\log_7x(\log_nm)$ then the values of x,m and n are?
QUESTION [0 upvotes]: I have the question that if $\log_510=\log_7x(\log_nm)$ then values of $x$,$m$ and $n$ are?
This question looks easy but i tried to get the expression down to the form
$$\log_ab=\log_ac\tag{1.}$$
and i failed.
My attempt:-

the expression can be rewritten as :
$$\frac{\log_710}{\log_75}=\log_7{\frac{\log_7m^x}{\log_7n}}$$
After this step i don't get any clue to move forward.

Edit:-

The question has options one of which is the correct answer:-
(a) $10,7,5$
(b) $-1,2,3$
(c) $7,5,3$
(d) $7,5,8$

Help please....

REPLY [1 votes]: HINT.- I hope this answer can help your inquire. If you take the function $$f(X)=\log_7(X)-\log_5 (10)$$ you have a unique and determined point, $a$, such that $f(a)=0$; this point is approximately equal to $16.183108$ so your “true” equation is $\log_n(m^x)=a$
 really.
(Optionally you have $$f(X)=\log_7(X)-\frac{\log_7(10)}{\log_7(5)}=
\log_7(10)-\log_7(10^{\frac{1}{\log_7(5)}})
=\log_7\left(\frac{X}{10^{\frac{1}{\log_7(5)}}}\right)=0$$ so you have $a=10^{\frac{1}{\log_7(5)}}\approx 16.183108$).
You have then to solve the equation $$\log_n(m^x)=a$$ If, for example (arbitrarily) $n=9$, the equation $\log_9(m^x)=a$ has a unique positive value, $b$, for $m^x$ since $a\gt 1$, $\log_9(x)$ is injective and its range is $\mathbb R$ so you have the equation $$m^x=b$$ Take now (again arbitrarily) $m=21$ and you have a unique $x$ such that $21^x=b$.
What conclusion can you have?  See at Kamil 09875’s comment.


TITLE: Converging to moments obeying Carleman's condition
QUESTION [3 upvotes]: I believe that the following is true, and I'd like to make sure that it is and to have a reference.  Suppose that $\mu_N$ are a sequence of measures on $\mathbb{R}$.  Let $m_{N,k}$ be the $k$-th moment of $\mu_N$.  Suppose that, for each $k$, $m_{N,k}$ converges to a limit $m_k$ as $N\rightarrow \infty$; no uniformity assumption is made, in that for each $k$ the convergence may be different.  Suppose further that $m_k$ obeys Carleman's condition:
$\sum_{n=1}^\infty m_{2n}^{-\frac{1}{2n}}=\infty$
Then, I would like to know that the following 3 results are true:
1) There exists a measure $\mu$ such that the moments of $\mu$ are equal to $m_k$.
2) Such a measure $\mu$ is unique.
3) The measures $\mu_N$ weakly converge to $\mu$ as $N \rightarrow \infty$.
Result (2) I think follows straightforwardly from Carleman's, but I want to verify that (1),(3) are correct.

REPLY [1 votes]: Yes, this works. It's convenient to view the $\mu_n$ as measures on the compact space $\mathbb R_{\infty}$, with $\mu_n(\{\infty\})=0$. By a diagonal process, we then find a subsequence (which I'll write as the original sequence) such that $x^{2k}\, d\mu_n(x) \to d\rho_k(x)$ weak $*$ for certain measures $\rho_k$ for all $k$; this follows because $\int_{\mathbb R_{\infty}} x^{2k}\, d\mu_n(x)\le C_k$ by assumption.
From this condition, we also deduce that $\rho_k(\{\infty\})=0$.
By comparing integrals of compactly supported continuous functions, we then find that $d\rho_k(x) = x^{2k}\, d\rho_0(x)$.
This measure $\rho=\rho_0$ has the correct even moments, by construction, and if $n$ is odd, then we write
$$
\int x^n\, d\rho(x) = \int x^{n-1} g(x)\, d\rho(x) + \int x^{n+1} h(x)\, d\rho(x),
$$
with both $g,h\in C(\mathbb R_{\infty})$, and $g$ supported near $x=0$ and $h$ supported away from $x=0$. Again, it follows that $\int x^n\, d\rho(x)= m_n$.
This gives (1). Then (2) is clear because the moment problem with the $m_n$'s is determinate, and then (3) also follows because, as we saw, the only possible limit point of the $\mu_n$ is the unique measure $\rho$ with these moments.