text:
\begin{document}

\begin{abstract}

In this work we study the existence of solutions $u \in W^{1,p}_0(\Omega)$ to the implicit elliptic problem $ f(x, u, \nabla u, \Delta_p u)= 0$ in $ \Omega $, where $ \Omega $ is  a bounded domain in $ \R^N $, $ N \ge 2 $, with smooth boundary $ \partial \Omega $, $ 1< p< +\infty $, and $ f\colon \Omega \times \R \times \R^N \times \R \to \R $. We choose the particular case when the function $ f $ can be expressed in the form $ f(x, z, w, y)= \varphi(x, z, w)- \psi(y) $, where the function $ \psi $ depends only on the $p$-Laplacian $ \Delta_p u $. We also show some applications of our results.

\end{abstract}

\maketitle

\begin{center}
	\begin{minipage}{12cm}
		\small
		\tableofcontents
	\end{minipage}
\end{center}

\medskip

\section{Introduction and main results}

Let $ \Omega $ be a bounded domain in $ \R^N $, $ N \ge 2 $, with smooth boundary $ \partial \Omega $, let $ 1< p< +\infty $ and let $ f\colon \Omega \times \R \times \R^N \times \R \to \R $. In this paper, we shall consider the following implicit elliptic problem
\begin{equation}
 \label{equazione}
u \in W^{1,p}_0(\Omega), \qquad f(x, u, \nabla u, \Delta_p u)= 0 \quad \text{in} \, \, \, \Omega,

\end{equation}
where $ \Delta_p $ denotes the $p$-Laplace operator, namely
\[
\Delta_p u:= \text{div}(|\nabla u|^{p-2} \nabla u) \quad \forall \, u \in W^{1,p}(\Omega).
\]
We focus on the particular case when the function $ f $ can be expressed in the form $ f(x, z, w, y)= \varphi(x, z, w)- \psi(y) $, where $ \varphi $ is a real-valued function defined on $ \Omega \times \R \times \R^N $, and  $ \psi $ is a real-valued function defined on $ Y $, where $ Y $ is a nonempty interval of $ \R $ (which will be specified later). We require that $ \psi $ depends only on the $p$-Laplacian $ \Delta_p u $.  We further distinguish among the case where $ \varphi $ is a Carath\'eodory function depends on $ x, u $, and $ \nabla u $, and the case where $ \varphi $ is allowed to be highly discontinuous in each variable, for which  the dependance on the gradient is not permitted.

In both cases we first reduce problem  \eqref{equazione} to an elliptic differential inclusion, but  methods used are different,  depending on the regularity of the function $ \varphi $ and on the structure of the problem.

More precisely, in the first case we make use of a result in \cite{ricceri1} to obtain the inclusion
\begin{equation}
\label{inclusione}
-\Delta_p u \in F(x, u, \nabla u),
\end{equation}
where $ F $ is a lower semicontinuous selection of the multifunction
\[
(x, z, w) \mapsto \{y \in Y: \varphi(x, z, w)- \psi(y)= 0 \}.
\]

A function $ u \in W^{1,p}_0(\Omega) $ is called a (weak) solution to \eqref{inclusione} provided  there exists $ v \in L^{p'}(\Omega) $,  $ p' $ being the conjugate exponent of $ p $, such that $ v(x) \in F(x, u(x), \nabla u(x)) $ for almost every $ x \in \Omega $ and
\[
\int_{\Omega} |\nabla u|^{p-2} \nabla u \cdot \nabla w dx= \int_{\Omega} v w dx \quad \forall w \in W^{1,p}_0(\Omega).
\]

We start with the general case when $ Y $ coincides with the whole space $ \R $, and after we deduce, as a particular case, the existence result when $ Y $ is a closed interval of $ \R $.

The main tool to obtain existence of solutions to \eqref{inclusione}  is the result below, \cite[Theorem 3.1]{marmos}, which deals with the existence of solutions for elliptic differential inclusions with lower semicontinuous right-hand side and is based on a selection theorem for decomposable-valued multifunctions (see \cite{bart} and \cite{marano1}).

\begin{theorem}
\label{mosconi}
Let $ F \colon \Omega \times \R \times \R^N \to 2^{\R} $ be a closed-valued multifunction. Suppose that:
\begin{enumerate}
\item[(h1)] $ F $ is $ \mathcal L(\Omega) \otimes \mathcal B(\R \times \R^N)$-measurable;

\item[(h2)] for almost every $  x \in \Omega $, the multifunction $ (z, w) \mapsto F(x, z, w) $ turns out to be lower semicontinuous;

\item[(h3)] there exist $ a \in L^{p'}(\Omega, \R^+_0), b, c \ge 0 $, with $ \frac{b}{\lambda_{1, p}}+ \frac{c}{\lambda_{1, p}^{1/p}}< 1$,  complying with
\[
\inf_{y \in F(x, z, w)} |y|< a(x)+ b|z|^{p-1}+ c|w|^{p-1} \quad \text{in} \quad \Omega \times \R \times \R^N.
\]

\end{enumerate}

Then, \eqref{inclusione} has a solution $ u \in W^{1,p}_0(\Omega) $.

\end{theorem}

Here, $ \lambda_{1, p} $ is the first eigenvalue of the $p$-Laplacian in the space $ W^{1,p}_0 (\Omega) $.

The following is our main result, which extends \cite[Theorem 3.2]{marano1} to the case $ p \ne 2 $.

\begin{theorem}

Let  $ \varphi \colon \Omega \times \R \times \R^N \to \R $ be a Carath\'eodory function and let $ \psi \colon \R \to \R $ be continuous. Suppose that:
\begin{enumerate}

\item[(i)] for all $ (x, z, w)  \in \Omega \times \R \times \R^N $, the set $ \{y \in \R : \varphi(x, z, w)- \psi(y)= 0\} $ has empty interior;

\item[(ii)] for all $ (x, z, w)  \in \Omega \times \R \times \R^N $, the function $ y \mapsto \varphi(x, z,w)- \psi(y) $ changes sign;

\item[(iii)] there exist $ a \in L^{p'}(\Omega, \R^+_0), b, c \ge 0 $, with $ \frac{b}{\lambda_{1,p}}+ \frac{c}{\lambda_{1,p}^{1/p}}< 1 $, such that
\[
\sup \{\vert y \vert: y \in \psi^{-1}(\varphi(x, z, w)) \} < a(x)+ b|z|^{p-1}+ c|w|^{p-1},
\]
for all $ (x, z, w) \in \Omega \times \R \times \R^N $.

\end{enumerate}

Then, there exists $ u \in W^{1,p}_0(\Omega) $ such that
\begin{equation}
\label{equa}
\psi(-\Delta_p u)= \varphi(x, u, \nabla u) \qquad in \quad \Omega .
\end{equation}

\end{theorem}

When $ \varphi $ is discontinuous we essentially follow \cite[Theorem 3.1]{marano4} to construct an appropriate upper semicontinuous multifunction $ F $ related with $\psi^{-1}$ and $\varphi$, and then we solve the elliptic differential inclusion $-\Delta_p u \in F(x,u)$ using the following \cite[Theorem 2.2]{marano2}

\begin{theorem}
\label{plap}
Let $ U $ be a nonempty set, $ \Phi\colon U \to W^{1,p}_0(\Omega), \Psi\colon U \to L^{p'}(\Omega) $ two operators and $ F\colon \Omega \times \R \to 2^{\R} $ a convex closed-valued multifunction. Suppose that the following conditions hold true:
\begin{enumerate}

\item[($i_1$)] $ \Psi $ is bijective and for any $ v_h \rightharpoonup v$ in $ L^{p'}(\Omega) $ there is a subsequence of $ \{\Phi(\Psi^{-1}(v_h))\} $  which converges to $ \Phi(\Psi^{-1}(v)) $ almost everywhere in $ \Omega $. Furthermore, a non-decreasing function $ g\colon \R^+_0 \to \R^+_0 \cup \{+\infty\} $ can be defined in such a way that
\[
\|\Phi(u)\|_{\infty} \le g(\|\Psi(u)\|_{p'}) \qquad \forall \, u \in U;
\]

\item[($i_2$)] $ F(\cdot, z) $ is measurable for all $ z \in \R $;

\item[($i_3$)] $ F(x, \cdot) $ has a closed graph for almost every $ x \in \Omega $;

\item[($i_4$)] There exists $ r> 0 $ such that the function
\[
\rho(x):= \sup_{|z| \le g(r)} d(0, F(x,z)), \quad  x \in \Omega,
\]
belongs to $ L^{p'}(\Omega) $ and $ \|\rho\|_{p'} \le r $.

\end{enumerate}

Then, the problem $ \Psi(u) \in F(x, \Phi(u)) $ has at least one solution $ u \in U $ satisfying $ |\Psi(u)(x)| \le \rho(x) $ for almost every $ x \in \Omega $.

\end{theorem}

Extending \cite[Theorem 3.1]{marano4} to the case  $ p \ne 2 $, we obtain the following result. We denote by $ \pi_0 $ and $ \pi_1 $ the projections of $ \Omega \times \R $ on $ \Omega $ and $ \R $, respectively.

\begin{theorem}
\label{disc}
Let $ \mathcal{F} = \{ A \subseteq \Omega \times \R: A \, \text{is measurable and there exists} \, \, \,  i \in \{0,1\} \text{ such} \\ \text{that } m(\pi_i(A))=0\} $, $(\alpha, \beta) \subseteq \R $ be an interval which does not contain $ 0 $,  $\psi$ a continuous real-valued function defined on $(\alpha, \beta)$, $\varphi$ a real-valued function defined on $ \Omega \times \R $, and $p>N$. Suppose that

\begin{enumerate}

\item[$(i)$] $\varphi$ is $ \mathcal L (\Omega \times \R)$-measurable and essentially bounded;

\item[$(ii)$] the set $D_{\varphi} = \{ (x,z) \in \Omega \times \R: \varphi$ is discontinuous at $(x,z) \}$  belongs to $\mathcal{F} $;

\item[$(iii)$] $\varphi^{-1}(r) \setminus \text{ int} (\varphi^{-1}(r)) \in \mathcal{F}$ for every  $r \in \psi((\alpha, \beta))$;

\item[$(iv)$] $\overline{\varphi (S \setminus D_{\varphi})} \subseteq \psi((\alpha, \beta))$.

\end{enumerate}

Then, there exists $u \in W^{1,p}_0(\Omega) $ such that
\[
\psi(-\Delta_p u) = \varphi (x,u) \qquad in \quad \Omega.
\]

\end{theorem}

\subsection{Structure of the paper} In Section 2 we will introduce the functional analytic setting we will use throughout the work. Section 3 is devoted to the case $ \varphi(x, \cdot, \cdot) $ continuous. Here we will distinguish some cases, which depend on the growth conditions on $ \varphi $ or  on the choice of the set $ Y $. We also will give some examples where these situations apply.  In Section 4 we will consider the discontinuous framework.

\section{Preliminaries}

Let $ X $ be a topological space and let $ V \subseteq X $. We denote by $ \text{int}(V) $ the interior of $ V $ and by $ \bar V $ the closure of $ V $. The symbol $ \mathcal B(X) $ is used to denote the Borel $ \sigma$-algebra of $ X $.

If $ (X, d) $ is a metric space, for every $ x \in X, r \ge 0 $ and every nonempty set $ V \subseteq X $, we define
\[
B(x, r)= \{z \in X: d(x, z) \le r\} \quad \text{and } \quad d(x, V)= \inf_{z \in V} d(x, z).
\]

Let $ X $ and $ Z $ be two nonempty sets. A multifunction $ \Phi $ from $ X $ into $ Z $ (symbolically $ \Phi \colon X \to 2^Z $) is a function from $ X $ into the family of all subsets of $ Z $. A function $ \varphi\colon X \to Z $ is said to be a selection of $ \Phi $ if $ \varphi(x) \in \Phi(x) $ for all $ x \in X $. For every set $ W \subseteq Z $ we define $ \Phi^-(W)= \{x \in X: \Phi(x) \cap W \ne \emptyset \} $.

When $ (X, \mathcal A) $ is a measurable space, $ Z $ is a topological space and for every open set $ W \subseteq Z $ we have $ \Phi^-(W) \in \mathcal A $, we say that the multifunction $ \Phi $ is measurable. If $ X $ and $ Z $ are two topological spaces and, for every open (resp. closed) set $ W \subseteq Z $, the set $ \Phi^-(W) $ is open (resp. closed) in $ X $, we say that $ \Phi $ is lower semicontinuous (resp. upper semicontinuous). When $ (Z, \delta) $ is a metric space, the multifunction $ \Phi $ is lower semicontinuous if and only if, for every $ z \in Z $, the real-valued function $ x \mapsto \delta(z, \Phi(x)) $, $ x \in X $, is upper semicontinuous (see \cite[Theorem 1.1]{ricceri3}). If, moreover, $ X $ is first countable, then the multifunction $ \Phi $ is lower semicontinuous if and only if, for every $ x \in X $, every sequence $ \{x_k\} $ in $ X $ converging to $ x $ and every $ z \in \Phi(x) $, there exists a sequence $ \{z_k\} $ in $ Z $ converging to $ z $ and such that $ z_k \in \Phi(x_k) $, for all $ k \in \N $ (see \cite[Theorem 7.1.7]{klein}).

A general result on the lower semicontinuity of a multifunction is the following \cite[Theorem 1.1]{ricceri1}

\begin{theorem}
\label{sci}
Let $ C, D $ be two topological spaces, with $ D $ connected and locally connected, and $ f $ be a real-valued function defined on $ C \times D $. For all $ x \in C $ we set
\[
\begin{split}
& V(x):= \{y \in D : f(x, y)= 0\},  \\
& M(x):= \{y \in D: y \, \, \text{is a local extremum point for} \, \, f(x, \cdot)\}, \quad \text{and} \\
& Q(x):= V(x) \setminus M(x).
\end{split}
\]

Suppose that
\begin{enumerate}

\item[(a)] for all $ x \in C, f(x, \cdot) $ is continuous, and $ 0 \in \text{int}(f(x, D)) $;

\item[(b)] for all $ x \in C $ and for all $ A $ open subset of $ D $, there exists $ \bar y \in A $ such that $ f(x, \bar y) \ne 0 $;

\item[(c)] the set $ \{(y', y'') \in D \times D : \{x \in C: f(x, y')< 0< f(x, y'')\} \, \, \text{is open} \} $ is dense in $ D \times D $.

\end{enumerate}

Then, the multifunction $ Q $ is lower semicontinuous, with nonempty closed values.

\end{theorem}

From now on, $ \Omega $ is a bounded domain in $ \R^N $, $ N \ge 2 $, with a smooth boundary $ \partial \Omega $, the symbol $ \mathcal L(\Omega) $ (respectively, $ m(\Omega) $) denotes the Lebesgue $ \sigma$-algebra (respectively, measure) of $ \Omega $, while $ W^{1,p}_0(\Omega) $ stands for the closure of $ C^{\infty}_0(\Omega) $ in $ W^{1, p}(\Omega) $. On $ W^{1,p}_0(\Omega) $ we introduce the norm
\[
\|u\|:= \biggl(\int_{\Omega} |\nabla u(x)|^p dx \biggr)^{1/p}, \qquad u \in W^{1,p}_0(\Omega).\
\]

Let $ p^*$ be the critical exponent for the Sobolev embedding $ W^{1,p}_0(\Omega) \subseteq L^r(\Omega) $. Recall that
\[
p^*=
\begin{cases}
\frac{Np}{N-p} \quad & \, \text{if} \, \, \, p< N, \\
+\infty & \text{otherwise}
\end{cases}
\]

If $ p \ne N $, then to each $ r \in [1, p^*] $ there corresponds a constant $ c_{rp}> 0 $ satisfying
\[
\|u\|_{L^r(\Omega)} \le c_{rp} \|u\|, \qquad \forall \, u \in W^{1,p}_0(\Omega),
\]
whereas, when $ p= N $, for every $ r \in [1, +\infty) $ we have
\[
\|u\|_{L^r(\Omega)} \le c_{rN} \|u\|, \qquad \forall \, u \in W^{1,N}_0(\Omega).
\]

Finally, the embedding $ W^{1,p}_0(\Omega) \hookrightarrow L^r(\Omega) $ is compact, provided $ 1 \le r< p^* $. When $ p> N $, we get  $W_0^{1,p}(\Omega) \subseteq L^{\infty}(\Omega)$ and
\begin{equation}
\label{1dis}
\Vert u\Vert_{\infty} \leq a\Vert u\Vert, \ \ \ u \in W_0^{1,p}(\Omega),
\end{equation}

for suitable $a>0$; see \cite[Ch. IX]{brezis}.

Given $ p \in  \, ]1, +\infty[ $, the symbol $ p' $ will denote the conjugate exponent of $ p $ while $ W^{-1,p'}(\Omega) $ stands for the dual space of $ W^{1,p}(\Omega) $.  Through \cite[Theorem 6.4]{brezis}, we see that $ L^{p'}(\Omega) $ compactly embeds in $ W^{-1,p'}(\Omega) $. So, there exists $ b> 0 $ satisfying
\begin{equation}
\label{2dis}
\|v\|_{W^{-1, p'}(\Omega)} \le b \|v\|_{L^{p'}(\Omega)}, \quad \forall \, v \in L^{p'}(\Omega).
\end{equation}

Let $ A_p\colon W^{1,p}_0(\Omega) \to W^{-1, p'}(\Omega) $ be the nonlinear operator stemming from the negative $p$-Laplacian, i.e.,
\begin{equation}
\label{operator}
\langle A_p(u), v \rangle := \int_{\Omega} |\nabla u(x)|^{p-2} \nabla u(x) \cdot \nabla v(x) dx, \qquad u,v \in W^{1,p}_0(\Omega),
\end{equation}
and let $ \lambda_{1,p} $ be its first eigenvalue in $ W^{1,p}_0(\Omega) $. The following facts are well known  (see, e.g., \cite{peral}, Appendix A):

\begin{enumerate}

\item[(p$_1$)] $ A_p $ is bijective and uniformly continuous on bounded sets;

\item[(p$_2$)] the inverse operator $ A_p^{-1} $ is $ (W^{-1, p'}(\Omega), W^{1,p}_0(\Omega))$-continuous;

\item[(p$_3$)] $ \|A_p(u)\|_{W^{-1,p'}(\Omega)}= \|u\|_p^{p-1} $ in $ W^{1,p}_0(\Omega) $;

\item[(p$_4$)] $ \|u\|_{L^p(\Omega)}^p \le \frac{1}{\lambda_{1,p}} \|u\|_p^p $, for all $ u \in W^{1,p}_0(\Omega) $.

\end{enumerate}

\section{The case when $ \varphi $ is a Carath\'eodory function}

This section deals with the existence of solutions to the equation
\begin{equation}
\label{equa}
\psi(-\Delta_p u)= \varphi(x, u, \nabla u).
\end{equation}

We first consider the case $ Y= \R $. Throughout the section,  $ p  \in \, ]1, +\infty[ $ and the following assumptions will be posited:

\begin{enumerate}

\item[(i)] for every $ (x, z, w)  \in \Omega \times \R \times \R^N $, the set $ \{y \in \R : \varphi(x, z, w)- \psi(y)= 0\} $ has empty interior;

\item[(ii)] for all $ (x, z, w)  \in \Omega \times \R \times \R^N $, the function $ y \mapsto \varphi(x, z,w)- \psi(y) $ changes sign.

\end{enumerate}

\begin{theorem}
\label{uno}

Let $ \varphi \colon \Omega \times \R \times \R^N \to \R $ be a Carath\'eodory function and let $ \psi \colon \R \to \R $ be continuous. Suppose that (i)-(ii) hold true and, moreover,
\begin{enumerate}

\item[(iii)] there exist $ a \in L^{p'}(\Omega, \R^+_0), b, c \ge 0 $, with $ \frac{b}{\lambda_{1,p}}+ \frac{c}{\lambda_{1,p}^{1/p}}< 1 $, such that
\[
\sup \{\vert y \vert: y \in \psi^{-1}(\varphi(x, z, w)) \} < a(x)+ b|z|^{p-1}+ c|w|^{p-1},
\]
for all $ (x, z, w) \in \Omega \times \R \times \R^N $.

\end{enumerate}

Then, there exists a solution $ u \in W^{1,p}_0(\Omega) $ to equation \eqref{equa}.

\end{theorem}

\begin{proof}

Fix any $ x \in \Omega $. We want to apply Theorem \ref{sci}. Choose $ C= \R \times \R^N $, $ D= \R $, $ f(z,w,y)= \varphi(x, z, w)- \psi(y) $, and for every $ (z, w) \in \R \times \R^N $ set
\[
\begin{split}
F(x, z, w):= \{y \in \R : \varphi(x, z, & w)- \psi(y)= 0,  \, \, \\
& \text{$ y $ is not a local extremum point of} \, \, \psi(\cdot)\}.
\end{split}
\]

Hypothesis (ii) directly yields (a). Moreover, in our context, (b) is equivalent to say that, for all $ (z, w) \in \R \times \R^N $, the set $ U:= \{y \in \R: \varphi(x, z, w)- \psi(y) \ne 0\} $ is dense in $ \R $. Since, by (i), the set $ \R \setminus U $ has empty interior, it follows that $ U $ is dense in $ \R $, as desired.

Let us next analyze the set
\begin{equation}
\label{set}
\begin{split}
\bigl\{(y',y'') \in \R \times \R : \{(z, w) \in \R \times \R^N : \varphi(x, &z, w)- \psi(y')< 0 \\
&< \varphi(x, z, w)- \psi(y'')\} \, \, \text{is open}\bigr\}.
\end{split}
\end{equation}

If one can find $ y', y'' \in \R $ such that
\[ \varphi(x, z, w)- \psi(y')< 0< \varphi(x, z, w)- \psi(y''), \] then $ \varphi(x, z, w) \in \, ]\psi(y''), \psi(y')[ $. So, the set
\[
\{(z, w) \in \R \times \R^N : \varphi(x, z, w)- \psi(y')< 0< \varphi(x, z, w)- \psi(y'')\},
\]
turns out to be open, because $ \varphi(x, \cdot, \cdot) $ is continuous. Otherwise it is empty. So, the set \eqref{set} is the whole space $ \R \times \R $, and (c) follows.

Therefore, thanks to Theorem \ref{sci}, the multifunction $ F(x, \cdot, \cdot) $ is lower semicontinuous, with nonempty closed values.

Moreover, for all $ y', y'' \in \R $ we have
\[
\begin{split}
\{(x, z, w) \in \Omega \times \R \times & \R^N: \varphi(x, z, w)- \psi(y')< 0< \varphi(x, z, w)- \psi(y'')\}= \\
& \{(x, z, w) \in \Omega \times \R \times \R^N: \varphi(x, z, w) \in \, ]\psi(y''), \psi(y')[ \, \} \in \mathcal L(\Omega) \otimes \mathcal B(\R \times \R^N),
\end{split}
\]
cf. \cite[Lemma III.14]{castaing}. Therefore, condition (iii) of \cite[Theorem 3.2]{marano1}, with $ \Lambda^*= \R \times \R $, is satisfied. Arguing as in that theorem we see that, if $ A \subseteq \R  $ is open, then
\[
\begin{split}
F^-(A)= \bigcup_{(y', y'') \in A \times A} \{(x, z, w) & \in \Omega \times \R \times \R^N: \\
& \varphi(x, z, w)- \psi(y')< 0< \varphi(x, z, w)- \psi(y'')\}.
\end{split}
\]

This actually implies that $ F^-(A) \in \mathcal L(\Omega) \otimes \mathcal B(\R \times \R^N)$, i.e. $ F $ is measurable.

Finally, fix any $ y \in F(x, z, w) $. In other words, $ y \in \psi^{-1}(\varphi(x, z, w)) $, therefore hypothesis (iii) implies that
\[
\inf_{y \in F(x, z, w)} |y|< a(x)+ b|z|^{p-1}+ c|w|^{p-1} \quad \text{in} \, \, \, \Omega \times \R \times \R^N.
\]

So all the hypotheses of Theorem \ref{mosconi} are fulfilled, and we get a solution $ u \in W^{1,p}_0(\Omega) $ to equation \eqref{inclusione}. Taking into account the definition of $F$, we have $\psi(-\Delta_pu)=\varphi(x,u,\nabla u)$, that is the thesis.

\end{proof}

\begin{remark}

A very simple situation when hypothesis (iii) occurs is the following.

Suppose that  $ \varphi(\Omega \times \R \times \R^N) \subseteq [\alpha, \beta] $ and $ \psi $ is such that $ \psi^{-1}(B) $ is bounded, for every $ B $ bounded subset of $ \R $. If $ (x, z, w) \in \Omega \times \R \times \R^N $, we get $ \varphi(x, z, w) \in [\alpha, \beta] $, and so $ \psi^{-1}(\varphi(x, z, w)) \subseteq \psi^{-1}([\alpha, \beta]) $. Then, if we choose $ a \in L^{p'}(\Omega, \R^+_0) $ such that $ a(x) > \sup \{|y|: y \in \psi^{-1}([\alpha, \beta])\} $ for all $ x \in \Omega $, we have
\[
|\psi^{-1}(\varphi(x,z,w))| < a(x) \le a(x)+ b|z|^{p-1}+ c|w|^{p-1} \quad \text{in} \, \, \, \Omega \times \R \times \R^N,
\]
that is hypothesis (iii).

\end{remark}

As an application of the previous result, we consider the following example.

\begin{example}
\label{ese-uno}

Let $ g \in L^{2}(\Omega) $ and $ \gamma \in \, ]0, 1[ $. Then, for every $ \lambda \ne 0, \mu \in \R $, there exists a solution $ u \in W^{1,2}_0(\Omega) $ to the equation
\begin{equation}
\label{eqex1}
-\Delta u= g(x) + \mu(|u|+ |\nabla u|)^{\gamma}+ \lambda \sin(-\Delta u).
\end{equation}

\end{example}

\begin{proof}

Fix $ \lambda, \mu \in \R $.  For every $ (x,z,w) \in \Omega \times \R \times \R^N  $ and every $ y \in \R $, set
\[
\varphi(x,z,w):= g(x)+ \mu(|z|+ |w|)^{\gamma}, \quad \quad \psi(y):= y- \lambda \sin y.
\]

Since $ \lim_{y \to \pm \infty} (y- \lambda \sin y)= \pm \infty $,  the function $ y \mapsto \varphi(x, z, w)- \psi(y) $ surely changes sign. Moreover, since it vanishes only at points of $ \R $ and not in intervals,   the set
\[
\{y \in \R : \varphi(x, z, w)- \psi(y)= 0\}
\]
has empty interior in $ \R $. Hence, hypotheses (i) and (ii) are fulfilled.

Fix now $ (x, z, w) \in \Omega \times \R \times \R^N $. In order to verify hypothesis (iii), we want to find  $ b, c\ge 0 $, with $ \frac{b}{\lambda_{1,2}}+ \frac{c}{\lambda_{1,2}^{1/2}}< 1 $, and $ a \in L^2(\Omega, \R^+_0) $ such that
\begin{equation}
\label{max}
\max\{\vert y \vert: y \in \psi^{-1}(\varphi(x, z, w))\}< a(x)+ b\vert z \vert+ c\vert w \vert.
\end{equation}
Notice that we can consider the maximum in \eqref{max} instead of the supremum, since the set  $\psi^{-1}(\varphi(x, z, w))$ is compact. Of course, \eqref{max} is equivalent to prove that $ \vert y \vert< a(x)+ b\vert z \vert+ c\vert w \vert $, for every $ y $ solution  of the equation
\begin{equation}
\label{eqesempio}
\psi(y)= \varphi(x, z, w).
\end{equation}

Thanks to Young's inequality with exponents $ 1/{\gamma} $ and $ 1/(1- \gamma) $, we have
\begin{equation}
\label{young}
\begin{split}
|\varphi(x, z,w)| &= |g(x)+ \mu(|z|+ |w|)^{\gamma}| \le |g(x)|+  |\mu| |z|^{\gamma}+  |\mu| |w|^{\gamma} \\
& \le |g(x)|+ \eps |z|+ \eps |w|+ C_{\gamma, \eps, \mu} \\
& \le \tilde g(x)+ \eps |z|+ \eps |w|,
\end{split}
\end{equation}
where $ \tilde g(x):= |g(x)| + C_{\gamma, \eps, \mu} $ for every $ x \in \Omega $. Then, if $ \tilde y $ is a solution to \eqref{eqesempio}, from \eqref{young} it follows that
\[
\vert \psi(\tilde y) \vert= \vert \varphi(x, z, w) \vert \le \tilde g(x)+ \eps |z|+ \eps |w|.
\]

On the other hand, by the definition of $ \psi $, we have
\[
\vert \psi(\tilde y) \vert= \vert \tilde y- \lambda \sin \tilde y \vert \ge \vert \tilde y \vert- \vert \lambda \vert,
\]
which implies that
\[
\begin{split}
\vert \tilde y \vert & \le \vert \psi(\tilde y) \vert+ \vert \lambda \vert \le \tilde g(x)+ \vert \lambda \vert+ \eps |z|+ \eps |w|< \\
& < \bar g(x)+ \eps \vert z \vert+ \eps \vert w \vert,
\end{split}
\]
where $ \bar g(x):=  \tilde g(x) + 2 \vert \lambda \vert $, for every $ x \in \Omega $. Observe that $ \bar g \in L^2(\Omega, \R^+_0) $. \\
Then, if we choose $ \eps $ in such a way that
\[
\frac{\eps}{\lambda_{1,2}}+ \frac{\eps}{\lambda_{1,2}^{1/2}}< 1,
\]
 we have hypothesis (iii)  with $ a:= \bar g $ and $ b:= c:= \eps $. Thanks to Theorem \ref{uno}, there exists a solution $ u \in W^{1,2}_0(\Omega) $ to  equation \eqref{eqex1}.

\end{proof}

In the following example the function $ \psi $ exhibits a behavior very different from the previous one.

\begin{example}

Let $ p \in [2, +\infty[ $, $ f \in L^{p'}(\Omega) $ and  $ \gamma \in \, ]0, p-1[ $. Then, for every $ \mu \in \R $ and $ \lambda \in \R^+ $, there exists a solution $ u \in W^{1,p}_0(\Omega) $ to the equation
\begin{equation}
\label{eq}
-\Delta_p u= f(x) + \mu (|u|+ |\nabla u|)^{\gamma}- \lambda e^{-\Delta_p u}.
\end{equation}

\end{example}

\begin{proof}

Fix $ \mu \in \R $ and $ \lambda \in \R^+ $. As before, for every $ (x, z, w) \in \Omega \times \R \times \R^N $ and $ y \in \R $, we set
\[
\varphi(x, z, w):= f(x)+ \mu (|z|+ |w|)^{\gamma},  \quad \quad \psi(y):= y+ \lambda e^y.
\]

Since $ \lim_{y \to \pm \infty} (y+ \lambda e^y)= \pm \infty $, one immediately gets that (i) and (ii)  are fulfilled. In order to verify hypothesis (iii), we argue as in Example \ref{ese-uno}.  First of all, applying Young's inequality  with exponents $ \frac{p-1}{\gamma}, \frac{p-1}{p-1- \gamma} > 1 $, we have
\[
\begin{split}
|\varphi(x, z, w)| &= |f(x)+ \mu (|z|+ |w|)^{\gamma}|\le |f(x)|+ 2^{p-1}(|\mu| |z|^{\gamma}+ |\mu| |w|^{\gamma}) \\
& \le |f(x)|+ \eps |z|^{p-1}+ \eps  |w|^{p-1}+ C_{\gamma, \eps, \mu} \\
&= \tilde f(x)+ \eps |z|^{p-1}+ \eps  |w|^{p-1},
\end{split}
\]
where $ \tilde f(x):= |f(x)| + C_{\gamma, \eps, \mu} $ for every $ x \in \Omega $.
Let now $ \tilde y $ be a solution to the equation $ \varphi(x, z, w)- \psi(y)= 0 $. Then, from the previous inequality, we have
\[
\vert \psi(\tilde y) \vert= \vert \varphi(x, z, w) \vert \le \tilde f(x)+ \eps |z|^{p-1}+ \eps  |w|^{p-1}.
\]

On the other hand, for every $ y \in \R $, and in particular for $ \tilde y $, we have
\begin{equation}
\label{xi}
\vert \psi(\tilde y) \vert= \vert \tilde y+ \lambda e^{\tilde y} \vert \ge \vert \tilde y \vert- \vert \xi \vert,
\end{equation}
$ \xi $ being the only solution to the equation $ y+ \lambda e^{y}= 0 $. Indeed, fix $ y \in \R $. If $ y \ge \xi $, we have
\[
\begin{split}
\vert y+ \lambda e^y \vert &= \vert y+ \lambda e^y- \xi- \lambda e^{\xi} \vert= \vert y- \xi+ \lambda(e^y- e^{\xi})\vert \\
& \ge \vert y- \xi \vert \ge \vert y \vert- \vert \xi \vert,
\end{split}
\]
which is \eqref{xi}. Suppose now that $ y< \xi $, then
\[
\begin{split}
\vert y+ \lambda e^y \vert&= \vert y-	\xi+ \lambda (e^y- e^{\xi}) \vert= \vert \xi- y+  \lambda(e^{\xi}- e^y) \vert \\
& \ge \vert \xi- y \vert \ge \vert y \vert- \vert \xi \vert,
\end{split}
\]
which again gives \eqref{xi}. This implies that
\[
\begin{split}
\vert \tilde y \vert & \le \vert \psi(\tilde y) \vert+ \vert \xi \vert \le \tilde f(x)+ \eps |z|^{p-1}+ \eps  |w|^{p-1} + \vert \xi \vert \\
& < \bar f(x)+ \eps |z|^{p-1}+ \eps  |w|^{p-1},
\end{split}
\]
$ \bar f(x):=  \tilde f(x) + 2 \vert \xi \vert $, for every $ x \in \Omega $ (note that one cannot have $ \xi= 0 $). Observe that $ \bar f \in L^{p'}(\Omega, \R^+_0) $. Then, if we choose $ \eps $ in such a way that
\[
\frac{\eps}{\lambda_{1,p}}+ \frac{\eps}{\lambda_{1,p}^{1/p}}< 1,
\]
we have hypothesis (iii) with $ a:= \bar f $ and $ b:= c:= \eps $. Thanks to Theorem \ref{uno}, there exists a solution $ u \in W^{1,p}_0(\Omega) $ to equation \eqref{eq}.
\end{proof}

To state the next result, we set $ \delta_{\Omega}:= \text{diam}(\Omega) $ and denote by $ \hat C $ the constant given by \cite[Proposition 3.3]{marmos}.

\begin{theorem}
\label{due}

Let $ \varphi $ and $ \psi $ as in Theorem \ref{uno}. Suppose that hypotheses (i)-(ii) hold true and, moreover,
\begin{enumerate}
\item[(iii)$'$]  there exist $ a \in L^q(\Omega, \R^+_0) $, $ q> N $, $ g\colon \R^+_0 \times \R^+_0 \to \R^+_0 $ nondecreasing with respect to each variable separately, such that
\[
\sup \{\vert y \vert: y \in \psi^{-1}(\varphi(x, z, w)) \} < a(x)+ g(|z|, |w|),
\]
for all $ (x, z, w) \in \Omega \times \R \times \R^N $;

\item[(iv)] there exists $ R> 0 $ such that
\[
\|a\|_{L^q(\Omega)}+ m(\Omega)^{1/q} g(\delta_{\Omega} \hat C R^{1/(p-1)}, \hat C R^{1/(p-1)}) \le R.
\]

\end{enumerate}

Then, equation \eqref{equa} has a solution $ u \in W^{1,p}_0(\Omega) $.

\end{theorem}

\begin{proof}

As before, fix $ x \in \Omega $, and for all $ (z, w) \in \R \times \R^N $, define
\[
\begin{split}
F(x, z, w):= \{y \in \R : & \, \, \varphi(x, z,w)-  \psi(y)= 0, \\
& y \, \, \text{is not a local extremum point of} \, \ \psi(\cdot)\}.
\end{split}
\]

Reasoning like in the previous theorem,  the multifunction $ F $ actually takes nonempty  closed values, is lower semicontinuous w.r.t. $ (z, w) $ and $ \mathcal L(\Omega) \otimes \mathcal B(\R \times \R^N) $-measurable.

Fix now  $ y \in F(x, z, w) $. Since, in other words,  $ y \in \psi^{-1}(\varphi(x, z, w)) $, hypothesis (iii)$'$ implies that
\[
\inf_{y \in F(x, z, w)} |y|< a(x)+ g(|z|, |w|) \quad \text{in} \quad \Omega \times \R \times \R^N.
\]

Taking into account (iv), we see that all the hypotheses of \cite[Theorem 3.4]{marmos} are fulfilled. Therefore, there exists $ u \in W^{1,p}_0(\Omega) $ such that $ -\Delta_p u \in F(x, u, \nabla u) $. Exploiting the definition of $ F $, this means that $ u $ is a solution to equation \eqref{equa}.

\end{proof}

As an application of the previous result, we consider the following example, which has been inspired by \cite[Corollary 1]{kacz}. Observe that, unlike \cite{kacz}, here we consider a function $ \varphi $ which is not necessarily continuous w.r.t. the variable $ x $, but only in a suitable $ L^q(\Omega) $. Moreover, here we deal with partial differential equations.

\begin{example}

Let $ h \in L^q(\Omega) $, with $ q> N $. Then, for every $ k \ne 0 $ and any sufficiently small $ \Vert h \Vert_q $ there exists a solution $ u \in W^{1,2}_0(\Omega) $ to the equation
\begin{equation*}
\label{laplacian}
-\Delta u= h(x) + u^3+ \vert \nabla u \vert^2+ k \sin (-\Delta u).
\end{equation*}

\end{example}

\begin{proof}

Fix $ k \in \R $ and for all $ (x, z, w) \in \Omega \times \R \times \R^N$ and all $ y \in \R $ define
\[
\varphi(x, z, w):= h(x)+ z^3+ \vert w \vert^2, \quad \quad \psi(y):= y- k \sin y.
\]

Reasoning like in Example \ref{ese-uno}, we have that hypotheses (i)-(ii) are fulfilled.

In order to verify hypothesis (iii)$'$, let $ g(|z|, |w|):= |z|^3+ |w|^2 $, for all $ (z, w) \in \R \times \R^N $. Of course $  g\colon \R^+_0 \times \R^+_0 \to \R^+_0  $ is nondecreasing w.r.t. each variable, separately. Let  $ \tilde y $ be a solution to the equation $ \psi(y)= \varphi(x, z, w) $. It follows that
\begin{equation*}
\label{eseh}
\begin{split}
\vert \psi(\tilde y) \vert &= |\tilde y- k \sin \tilde y|= |\varphi(x, z, w)| \\
& \le |h(x)|+ |z|^3+ |w|^2= \vert h(x) \vert+ g(\vert z \vert, \vert w \vert).
\end{split}
\end{equation*}

On the other hand, we always have
\[
\vert \psi(\tilde y) \vert= \vert \tilde y- k \sin \tilde y \vert \ge \vert \tilde y \vert- \vert k \vert,
\]
which implies that
\[
\begin{split}
\vert \tilde y \vert & \le \vert \psi(\tilde y) \vert+ \vert k \vert \le \vert h(x) \vert+ g(\vert z \vert, \vert w \vert)+ \vert k \vert \\
& < \bar h(x)+ g(\vert z \vert, \vert w \vert),
\end{split}
\]
where $ \bar h(x):= |h(x)|+ 2 \vert k \vert $, for every $ x \in \Omega $. Of course, $ \bar h \in L^q(\Omega, \R^+_0) $.  Hence we get hypothesis (iii)$'$.

Moreover, in order to verify hypothesis (iv), we have to check  if there exists $ R> 0 $ such that
\[
\|\bar h\|_{L^q(\Omega)}+ m(\Omega)^{1/q} g(\delta_{\Omega} \hat C R, \hat C R) \le R,
\]
that is
\begin{equation}
\label{diseg}
\|\bar h\|_{L^q(\Omega)}+ m(\Omega)^{1/q} \delta_{\Omega}^3 \hat C^3 R^3+ m(\Omega)^{1/q} \hat C^2 R^2 \le R.
\end{equation}

If $ 0< R <<1 $, choosing $ \bar h $ in such a way that $ \| \bar h\|_{L^q(\Omega)}< \frac{R}{2} $, we have that  \eqref{diseg} is immediately satisfied, since the terms containing $ R^2 $ and $ R^3 $ are negligible with respect to $ R $.  So all the hypotheses of Theorem \ref{due} are fulfilled, and we get the thesis.

\end{proof}

The next result provides solutions to equation \eqref{equa} when the function $ \psi $ is of the form $ y \mapsto y- h(y)$, with $ h $ continuous and bounded. Note that here we have to require a specific growth condition on $ \varphi $.

\begin{theorem}

Let $ \varphi\colon \Omega \times \R \times \R^N \to \R $ be a Carath\'eodory function and let $ h \in L^{\infty}(\R) $ be continuous. Suppose that (i)-(ii) hold true and, moreover,
\begin{enumerate}

\item[(iii)$''$] there exist $ f \in L^{p'}(\Omega, \R^+_0)$, with $ f(x) \ge \|h\|_{\infty} $ for all $ x \in \Omega $, $ \mu> 0 $, $ \gamma \in \, ]0, p-1[ $ such that
\[
\sup_{(x, z, w) \in \Omega \times \R \times \R^N} |\varphi(x, z, w)| < f(x)+ \mu(|z|+ |w|)^{\gamma}.
\]

\end{enumerate}

Then, there exists a solution $ u \in W^{1,p}_0(\Omega) $ to the equation
\begin{equation}
\label{h}
-\Delta_p u- h(-\Delta_p u)= \varphi(x, u, \nabla u).
\end{equation}

\end{theorem}

\begin{proof}

Fix $ x \in \Omega $ and define, for all $ (z, w) \in \R \times \R^N $,
\[
\begin{split}
F(x, z, w):= \{y \in \R : \, \, & \varphi(x, z,w)- (y- h(y))= 0, \\
& y \, \, \text{is not a local extremum point of} \,  \, y \mapsto y- h(y)\}.
\end{split}
\]

Reasoning as in the above proofs ensures that $ F $ is lower semicontinuous w.r.t. $ (z, w) $, with nonempty  closed values, and  $ \mathcal L(\Omega) \otimes \mathcal B(\R \times \R^N) $-measurable.

Fix  $ (x, z, w) \in \Omega \times \R \times \R^N $. If $ y \in F(x, z, w) $, then it solves the equation $ \varphi(x, z, w)= y- h(y) $. Two cases occur. First, $ \gamma \in  [1, p-1[ $. Applying Young's inequality with exponents $ \frac{p-1}{\gamma}, \frac{p-1}{p-1-\gamma}> 1 $, we have
\[
\begin{split}
|y|= |y- h(y)+ h(y)| & \le |y- h(y)|+ |h(y)| \le |\varphi(x, z, w)|+ \|h\|_{\infty} \\
& < f(x)+ \mu(|z|+ |w|)^{\gamma}+ \|h\|_{\infty} \\
& \le 2 f(x) +  2^{\gamma-1} \mu (|z|^{\gamma}+ |w|^{\gamma}) \\
& \le  2 f(x)+  2^{\gamma-1} \mu (\eps|z|^{p-1}+ \eps |w|^{p-1}+ K_{\eps})  \\
& \le 2 f(x)+ C_{\eps}+   2^{\gamma-1} \mu \eps(|z|^{p-1}+ |w|^{p-1}),
\end{split}
\]
that is $ \vert y \vert < 2 f(x)+ C_{\eps}+ 2^{\gamma-1} \mu \eps(|z|^{p-1}+ |w|^{p-1}) $, where $ C_{\eps}:=   2^{\gamma-1} \mu K_{\eps} $. Hence
\[
\inf_{y \in F(x, z, w)} |y|< 2 f(x)+ C_{\eps}+ 2^{\gamma-1} \mu \eps(|z|^{p-1}+ |w|^{p-1}).
\]
If we choose $ \eps $ in such a way that
\[
\frac{2^{\gamma-1} \mu \eps}{\lambda_{1,p}}+ \frac{2^{\gamma-1} \mu \eps}{\lambda_{1,p}^{1/p}}< 1,
\]
hypothesis (h$3$) of Theorem \ref{mosconi} is fulfilled, with $ a:= 2 f+ C_{\eps} \in L^{p'}(\Omega, \R^+_0) $ and $ b:= c:= 2^{\gamma-1} \mu  \eps $.

Suppose now  $ \gamma \in \, ]0, 1[ $. Since, for every $ a, b \ge 0 $ we have $ (a+ b)^{\gamma} \le a^{\gamma}+ b^{\gamma} $, reasoning as before yields
\[
\vert y \vert < 2 f(x)+ \tilde C_{\eps}+ \mu \eps(\vert z \vert^{p-1}+ \vert w \vert^{p-1}),
\]
where $ \tilde C_{\eps}:= \mu K_{\eps} $. If we now choose $ \eps $ in such a way that
\[
\frac{\mu \eps}{\lambda_{1,p}}+ \frac{\mu \eps}{\lambda_{1,p}^{1/p}}< 1,
\]
hypothesis (h$3$) of Theorem \ref{mosconi} is again fulfilled, with $ a:= 2 f+ \tilde C_{\eps} \in L^{p'}(\Omega, \R^+_0) $ and $ b:= c:= \mu  \eps $.

In both cases, there exists $ u \in W^{1,p}_0(\Omega)$ such that $ -\Delta_p u \in F(x, u, \nabla u) $. Through a familiar argument, this entails that $ u $ is a solution to equation \eqref{h}.

\end{proof}

We conclude this section considering the case when $ Y $ is a closed interval of $ \R $. Observe that here no growth conditions on $ \varphi $ are required.

\begin{theorem}
\label{intervallo}

Let $ \varphi \colon \Omega \times \R \times \R^N \to \R $ be a Carath\'eodory function and let $ \psi\colon [\alpha, \beta] \to \R $ be continuous. Suppose that:
\begin{enumerate}

\item  for every $ (x, z, w)  \in \Omega \times \R \times \R^N $, the set $ \{y \in [\alpha, \beta] : \varphi(x, z, w)- \psi(y)= 0\} $ has empty interior;

\item for every $ (x, z, w)  \in \Omega \times \R \times \R^N $,  the function $ y \mapsto \varphi(x, z,w)- \psi(y) $ changes sign in $ [\alpha, \beta] $.

\end{enumerate}

Then, there exists a solution $ u \in W^{1,p}_0 (\Omega) $  to equation \eqref{equa}.

\end{theorem}

\begin{proof}

As before, fix $ x \in \Omega $, and for all $ (z, w) \in \R \times \R^N $ define
\[
\begin{split}
F(x, z, w):= \{y \in [\alpha, \beta] : & \, \,  \varphi(x, z,w)- \psi(y)= 0, \\
& y \, \, \text{is not a local extremum point of} \, \, \psi(\cdot)\}.
\end{split}
\]

A familiar argument ensures that   $ F $ takes nonempty  closed values, is lower semicontinuous w.r.t. $ (z, w)$ and $ \mathcal L(\Omega) \otimes \mathcal B(\R \times \R^N) $-measurable.

Fix now $ y \in F(x,z,w) $.  In particular we have  $ |y| \le \max \{ \vert \alpha \vert, \vert \beta \vert \} $. Then, hypothesis (h$3$) of Theorem \ref{mosconi} is immediately satisfied with $ a(x):= 2 \max \{ \vert \alpha \vert, \vert \beta \vert \} $ for every $ x \in \Omega $  and $ b:= c:= 0 $.  Therefore, there exists $ u \in W^{1,p}_0 (\Omega) $  such that $ -\Delta_p u \in F(x, u, \nabla u ) $, i.e. $ u $ is a solution to  \eqref{equa}.

\end{proof}

As application of the previous theorem, we consider two examples, which differ by the behavior of the function  $ \psi $. In both  cases, the condition which permits to get a solution is the boundedness of $ \varphi $.

\begin{example}

Let $ f \in L^{\infty}(\Omega) $, $ k \in \N $, $ k $ even and such that $ k \pi> \Vert f \Vert_{\infty} $, and let $ \psi\colon [-k \pi, k \pi] \to \R $ be defined by $ \psi(y)= y \cos y $. Then, there exists a solution $ u \in W^{1,p}_0(\Omega) $ to the equation
\begin{equation}
\label{appl}
\psi(-\Delta_p u)= f \qquad \text{in } \, \Omega.
\end{equation}

\end{example}

\begin{proof}

Observe that assumption (1) is clearly satisfied. Moreover, for every $ x \in \Omega $, we have
\[
\begin{split}
& f(x)- \psi(k \pi)= f(x)- k \pi \, \cos(k \pi)= f(x)- k \pi \, (-1)^k= f(x)- k \pi< 0 \\
 \text{and} \qquad & f(x)- \psi(-k \pi)= f(x)+ k \pi \cos(-k \pi)= f(x)+ k \pi> 0.
\end{split}
\]
Therefore, hypothesis (2) is also satisfied. Thanks to Theorem \ref{intervallo}, there exists at least a solution $ u \in W^{1,p}_0(\Omega) $ to equation \eqref{appl}.

\end{proof}

Note that the interval $ [\alpha, \beta] $ could  be  unbounded, as  the following example shows.

\begin{example}

Let $ p \in \, ]1, +\infty[ $, $ f \in L^{p'}(\Omega)$,  $ \gamma> 0 $ and $ \varphi\colon \Omega \times \R \times \R^N \to \R $. Suppose that there exists $ \lambda \in \R^+ $ such that
\begin{equation}
\label{iplambda}
\sup_{(x, z, w) \in \Omega \times \R \times \R^N} \vert \varphi(x, z, w) \vert < \lambda.
\end{equation}
Then, there exists a solution $ u \in W^{1, p}_0(\Omega) $ to equation
\begin{equation*}
\label{eqex}
\varphi(x, u, \nabla u)- \lambda e^{\Delta_p u}- \Delta_p u= 0.
\end{equation*}

\end{example}

\begin{proof}

Define $ \psi(y):= \lambda e^{-y}- y $ for every $ y \in [0, +\infty[ $. Observe that hypothesis (1) is immediately satisfied. Moreover, thanks to \eqref{iplambda}, for every $ (x, z, w) \in \Omega \times \R \times \R^N $ we have
\[
\begin{split}
& \varphi(x, z, w)- \psi(0)= \varphi(x, z, w)- \lambda< 0 \\
\text{and} \quad & \lim_{y \to +\infty} (\varphi(x, z, w)- \psi(y))= +\infty.
\end{split}
\]

Therefore,  hypothesis (2) holds true too, and the conclusion follows from Theorem \ref{intervallo}.

\end{proof}

\section{The discontinuous framework}

This section is  devoted to the proof of Theorem \ref{disc}, which we rewrite here, for the reader's convenience. Given $(x,z)\in S:= \Omega \times \R  $, set $\pi_0(x,z)=x$ and $\pi_1(x,z)=z $. Moreover,  fix $ p> N $ and define
\[
\mathcal{F}= \{ A \subseteq S: A \, \, \text{is measurable and there exists } i \in \{0,1\} \text{ such that } m(\pi_i(A))=0\}.
\]

\begin{theorem}

\label{ultimo}

Let $ (\alpha, \beta) \subseteq \R $ be such that $ 0 \notin (\alpha, \beta) $,  $\psi\colon (\alpha, \beta) \to \R $ be  continuous and $\varphi \colon \Omega \times \R \to \R $. Suppose that the following conditions hold true:

\begin{enumerate}

\item[$(i)$] $\varphi$ is $ \mathcal L(\Omega \times \R)$-measurable and essentially bounded;

\item[$(ii)$] the set $D_{\varphi} = \{ (x,z) \in S: \varphi$ is discontinuous at $(x,z) \}$  belongs to $\mathcal{F} $;

\item[$(iii)$] $\varphi^{-1}(r) \ \setminus \text{ int} (\varphi^{-1}(r)) \in \mathcal{F}$ for every  $r \in \psi((\alpha, \beta))$;

\item[$(iv)$] $\overline{\varphi (S \backslash D_{\varphi})} \subseteq \psi((\alpha, \beta))$.

\end{enumerate}

Then, there exists $u \in W^{1,p}_0(\Omega)$ such  that
\[
\psi(-\Delta_p u) = \varphi (x,u) \qquad \text{in} \quad \Omega.
\]

\end{theorem}

\begin{proof}

The first part essentially follows the proof of  \cite[Theorem 3.1]{marano4}. Thanks to assumption (i), there exists a constant $c > 0$ such that
\[
S \setminus D_{\varphi} \subseteq \{ (x,z) \in S: \vert \varphi(x,z) \vert \leq c \}.
\]
Set
\[
a = \min{ \overline{ \varphi(S \backslash D_{\varphi}) }} \ \ \ \text{and} \ \ \
b = \max{\overline{\varphi(S \backslash D_{\varphi})}} .
\]
Hypothesis (iv) allows us to choose $y', y'' \in (\alpha, \beta) $ in such a way that $\psi(y') = a $ and $  \psi(y'') = b$. Pick a continuous function $\lambda\colon [0,1] \rightarrow (\alpha, \beta) $ complying with $\lambda(0)=y'$, $\lambda(1) = y''$, and define $ \tilde \psi (t) = \psi(\lambda(t))$, $t \in [0,1] $.

If $ \tilde \psi $ is constant, then $a=b$ and, consequently, $\varphi(S \backslash D_{\varphi}) = \{ a \}$. Let $ u \in W_0^{1,p} (\Omega) $ be such that $-\Delta_p u= y'$. Since $\psi( -\Delta_p u)=\psi(y')=a$,  the conclusion will be achieved by showing that the set $\Omega_{\varphi}= \{ x \in \Omega: (x, u(x)) \in D_{\varphi} \}$ has measure zero.

First of all, observe that an elementary computation gives
\begin{equation}
\label{Omegaphi}
\Omega_{\varphi} \subseteq \pi_0(D_{\varphi}) \cap u^{-1} (\pi_1(D_{\varphi}))
\end{equation}
and, due to (ii), $ m(\pi_i(D_{\varphi})) = 0 $, for some $i \in \{ 0,1\}$. Suppose $ i= 0 $. From \eqref{Omegaphi} we obtain
\[
m(\Omega_{\varphi}) \le m(\pi_0(D_{\varphi}) \cap u^{-1} (\pi_1(D_{\varphi}))) \le m(\pi_0(D_{\varphi}))= 0,
\]
whence $ m(\Omega_{\varphi})= 0 $. Let now $ i= 1 $. Lemma 1 in \cite{buttazzo} ensures that $ \nabla u(x)= 0 $ a.e. in $ u^{-1}(\pi_1(D_{\varphi})) $. Thanks to \cite[Theorem 1.1]{lou}, we have  $ y'= 0 $ on $ \{x \in \Omega: \, \nabla u(x)= 0 \} $, and, in particular, on $ u^{-1}(\pi_1(D_{\varphi})) $ (notice that our calculation showed that $ u^{-1}(\pi_1(D_{\varphi})) \subseteq \{x \in \Omega: \, \nabla u(x)= 0\} $) . Since $ y' \in (\alpha, \beta) $, this is possible if and only if $ m(u^{-1}(\pi_1(D_{\varphi})))= 0 $. From \eqref{Omegaphi} again we get
\[
m(\Omega_{\varphi}) \le m(\pi_0(D_{\varphi}) \cap u^{-1} (\pi_1(D_{\varphi}))) \le m(u^{-1} (\pi_1(D_{\varphi}))),
\]
which implies $ m(\Omega_{\varphi})= 0 $.

Suppose now that $ \tilde \psi $ is non constant and choose $t_1,t_2 \in [0,1]$ fulfilling
\[
\tilde \psi (t_1) = \min_{t \in [0,1]}  \tilde \psi (t), \qquad \tilde \psi (t_2) = \max_{t \in [0,1]} \tilde \psi (t).
\]
Obviously, $t_1 \neq t_2$ and there is no loss of generality in assuming $t_1 < t_2$. Let $h\colon \tilde \psi ([0,1]) \to [0,1]$ be defined by $h(r) = \min{(\tilde \psi^{-1}(r) \cap [t_1,t_2])}$, for every $r \in \tilde \psi ([0,1])$.

We claim that $h$ is strictly increasing. Indeed, pick $r_1, r_2 \in \tilde \psi ([0,1])$, with $r_1 < r_2$. Then, $h(r_1) \neq h(r_2)$ and $t_1 < h(r_2)$. From $ \tilde \psi (h(r_2))=r_2 > r_1$, $ \tilde \psi (t_1) \le r_1$, taking into account the continuity of $ \tilde \psi $, we immediately infer $h(r_1) < h(r_2)$.

Therefore, the family $D_k$ of all discontinuity points of the function $k\colon \mathbb{R} \rightarrow (\alpha, \beta)$ given by

\[
k(r)=\begin{cases}
 \ \lambda (h(\tilde{\psi}(t_1))) \quad & \, \text{if} \, \, \, r \in \, ]-\infty, \tilde{\psi}(t_1)[ \\
 \ \lambda (h(r)) \quad & \, \text{if} \, \, \, r \in \tilde{\psi}([0,1]) \\
 \ \lambda (h(\tilde{\psi}(t_2))) \quad & \, \text{if} \, \, \, r \in \, ]\tilde{\psi}(t_2),+\infty [
\end{cases}
\]
is at most countable. Owing to hypotheses (ii) and (iii), this implies that the set

\begin{equation}
\label{D}
D = D_{\varphi} \cup \{ \bigcup_{r \in D_k} [\varphi^{-1}(r) \backslash \text{ int}(\varphi^{-1}(r)) ] \}
\end{equation}
has measure zero.

Define now $ f(x, z):= k(\varphi(x,z))$, $(x, z) \in S $. Of course, the function $f\colon S \to \R $ is bounded, since $f(S) \subseteq \lambda([0,1])$. Moreover,  as in \cite[Theorem 3.1]{marano4},  we see that $ f $ is continuous.  Put
\begin{equation*}
F(x,z) = \overline{\text{co}} \left( \bigcap_{\delta > 0 } \bigcap_{ E \in \mathcal{E} } \overline{f(B_{\delta}(x,z) \setminus E)} \right),
\end{equation*}
where
\[
\begin{split}
\mathcal{E} &= \{ E \subseteq S : m(E)= 0 \} \\
\text{and} \qquad B_{\delta}(x,z) &= \{ (x',z') \in S: \vert x-x' \vert + \vert z - z' \vert \leq \delta \}.
\end{split}
\]
A standard argument (see, e.g,  \cite[Theorem 3.1]{marano4}), ensures that $F$ is upper semicontinuous, and nonempty, convex and closed-valued. Further, $ F(\cdot, z) $ is measurable for every $ z \in \R $, $ F(x, \cdot) $ has a closed graph for almost all $ x \in \Omega $, and
\begin{equation*}
F(x,z) = \{ f(x,z) \}, \quad \text{as soon as} \quad (x,z) \in S \setminus D.
\end{equation*}
Consider now the problem
\[
-\Delta_p u \in F(x, u)  \, \,  \, \text{in} \, \, \Omega, \, \, u \in W^{1,p}_0(\Omega).
\]

A solution will be obtained via Theorem \ref{plap}. So, let us verify its hypotheses. Choose $U:=A_p^{-1}(L^{p'}(\Omega))$, where $ A_p $ is given in \eqref{operator}, $\Phi(u):=u$ and $\Psi(u):=A_p(u)$ for every $u \in U$. Observe that the operator $ A_p\colon U \rightarrow L^{p'}(\Omega)$ is bijective.

Let $v_h \rightharpoonup v$ in $L^{p'}(\Omega)$. Since $ \{v_h\} $ is bounded in $ L^{p'}(\Omega) $, and $ L^{p'}(\Omega) $ compactly embeds in $ W^{-1,p'}(\Omega)$, there exists a subsequence, still denoted by $ \{v_h\} $, such that $ v_h \to v $ in $ W^{-1,p'}(\Omega) $. Since, from property ($p_2$), $ A_p^{-1} $ is strongly continuous, it follows that  $\{A_p^{-1}(v_h) \}$ converges to $ A_p^{-1}(v)$ almost everywhere in $\Omega$.

Let now $ g\colon \R^+_0 \to \R^+_0 $ be defined by
\[
g(t) := a(bt)^{1/(p-1)} \qquad \forall \, t \in \R^+_0,
\]
where the constants $ a $ and $ b $ derive from  inequalities \eqref{1dis}-\eqref{2dis}. Clearly, $ g $ is monotone increasing in $ \R^+_0 $. Moreover, taking into account property ($p_3$), if $ u \in U $ then
\[
\Vert u \Vert_{\infty} \leq a \Vert u \Vert = a \Vert A_p(u)\Vert_{W^{-1,p'}(\Omega)}^{1/(p-1)}\leq a(b\Vert A_p(u)\Vert_{p'})^{1/(p-1)} = g (\Vert A_p(u) \Vert_{p'}).
\]

This shows ($i_1$). Since hypotheses ($i_2$) and ($i_3$) are already satisfied, we have only to check ($i_4$). Define, for every $ x \in \Omega $,
\[
\rho(x):= \sup_{|z| \le g(r)} d(0,F(x, z)).
\]
Reasoning as in  \cite[Theorem 3.1]{marano3}, we  see that $ \Vert \rho \Vert_{p'} \le r $ once the same property holds true for the function $ x \mapsto j(x):= \sup_{|z| \le g(r)} \vert f(x, z) \vert $.

If $ \vert z \vert \le g(r) $, then
\[
\int_{\Omega} \vert f(x, z) \vert^{p'} dx \le \int_{\Omega} \Vert f(\cdot, z) \Vert_{\infty}^{p'} dx= \Vert f(\cdot, z) \Vert_{\infty}^{p'} \, m(\Omega),
\]
whence
\[
\begin{split}
\int_{\Omega} \vert j(x) \vert^{p'} dx&= \int_{\Omega}  \sup_{|z| \le g(r)} \vert f(x, z) \vert^{p'} dx \le \sup_{|z| \le g(r)} \int_{\Omega} |f(x, z)|^{p'} dx  \\
& \le \sup_{|z| \le g(r)} \Vert f(\cdot, z) \Vert_{\infty}^{p'} \, m(\Omega)= \Vert f(\cdot, z) \Vert_{\infty}^{p'} \, m(\Omega).
\end{split}
\]

Choosing  $ r \ge \Vert f(\cdot, z) \Vert_{\infty} m(\Omega)^{1/p'} $, we get $ j \in L^{p'}(\Omega) $ and $ \Vert j \Vert_{p'} \le r $.

Now, thanks to Theorem \ref{plap}, there exists $u \in U \subseteq W_0^{1,p}(\Omega)$ such that
\begin{equation}
\label{incl}
-\Delta_p u(x) \in F(x, u(x)) \qquad \text{a.e. in} \, \, \Omega
\end{equation}
and $ \vert \Delta_p u(x) \vert \le \rho(x) $, for almost every $ x \in \Omega $.
Define $\Omega_f = \{ x \in \Omega: (x,u(x)) \in D \}$. From \eqref{D}, it follows that
\[
\begin{split}
\Omega_f \subseteq & \{\pi_0(D_{\varphi}) \cap u^{-1}(\pi_1(D_{\varphi}))\} \\
& \cup \biggr \{\bigcup_{r \in D_k} \bigl[\pi_0(\varphi^{-1}(r) \setminus \text{int} (\varphi^{-1}(r)))\cap u^{-1} (\pi_1(\varphi^{-1}(r) \setminus \text{int} (\varphi^{-1}(r)))) \bigr] \biggr\},
\end{split}
\]
which, in particular, implies
\[
\begin{split} m(\Omega_f) & \le m \bigl(\pi_0(D_{\varphi}) \cap u^{-1}(\pi_1(D_{\varphi}))\bigr) \\
&+ m\bigl(\bigcup_{r \in D_k} [\pi_0(\varphi^{-1}(r) \setminus \text{int} (\varphi^{-1}(r)))\cap u^{-1} (\pi_1(\varphi^{-1}(r) \setminus \text{int} (\varphi^{-1}(r))))]\bigr) \\
& \le m \bigl(\pi_0(D_{\varphi}) \cap u^{-1}(\pi_1(D_{\varphi}))\bigr) \\
&+ \bigcup_{r \in D_k} m\bigl([\pi_0(\varphi^{-1}(r) \setminus \text{int} (\varphi^{-1}(r)))\cap u^{-1} (\pi_1(\varphi^{-1}(r) \setminus \text{int} (\varphi^{-1}(r))))]\bigr).
\end{split}
\]

Assumption (ii) entails $ m(\pi_i(D_{\varphi}))= 0 $ for some $ i \in \{0, 1\} $. Likewise, due to (iii), for each $ r \in D_k $, there exists $ i_r \in \{0, 1\} $ such that $ m(\pi_{i_r}(\varphi^{-1}(r) \setminus \text{int}(\varphi^{-1}(r))))= 0 $. Therefore, reasoning as in the case when $ \tilde \psi $ is constant, we obtain $ m(\Omega_f)= 0 $. This implies $F(x,u(x)) = \{ f(x,u(x)) \}$ and so,  on account  of \eqref{incl},
\[
-\Delta_p u(x)= f(x,u(x)) \qquad \text{a.e. in } \, \, \Omega.
\]

Therefore,
\[
\psi(-\Delta_p u(x))= \psi (f(x, u(x))) = \psi (k (\varphi (x, u(x)))) = \varphi (x, u(x)),
\]
which completes the proof.

\end{proof}

Hypothesis (iv) and the assumption $ 0 \notin (\alpha, \beta) $ are \emph{essential} to obtain the existence of a solution for equations as in the previous theorem. Below we list two examples, apparently very similar, and such that one admits a solution while the other one doesn't.

\begin{example}

Let $ \varphi \colon \R \to \R $ be defined by
\[
\varphi(z)=
\begin{cases}
0 \, \,  & \text{if } \, z \ne 0 \\
1 & \text{if } \, z= 0.
\end{cases}
\]
and let $ \psi\colon [1, +\infty[ \to \R $ be such that $ \psi(y)= y $. Consider the following equation
\begin{equation}
\label{ex}
-\Delta_p u= \varphi(u).
\end{equation}

Note that equation \eqref{ex} doesn't have any solution $ u \in W^{1,p}_0(\Omega) $. Suppose on the contrary  that $ u $ is such a solution. Since $ \varphi(u) \ge 0 $, then from equation \eqref{ex} we have  $ -\Delta_p u \ge 0 $, and so the Strong Maximum Principle implies that $ u \equiv 0 $ or $ u> 0 $. Suppose that $ u \equiv 0 $, then this would imply that $ -\Delta_p u \equiv 0 $, which is in contrast with \eqref{ex}. Suppose now that $ u> 0 $. Then, taking into account the definition of $ \varphi $ and equation \eqref{ex}, we have $ -\Delta_p u= 0 $. This fact, jointly with the boundary condition $ u  |_{\partial \Omega}= 0 $, implies $ u \equiv 0 $ which is again impossible.

It is also evident from the definition of $ \varphi $ that hypothesis (iv) and $ 0 \notin (\alpha, \beta) $ cannot be verified simultaneously.

Fix now $ \lambda \in \, ]0, 1[ $ and consider the function $ \tilde \varphi \colon \R \to \R $ be defined by
\[
\tilde \varphi(z)=
\begin{cases}
1 \, \,  & \text{if } \, z \ne 0 \\
\lambda & \text{if } \, z= 0.
\end{cases}
\]

In this case $ 0 \notin [1, +\infty[ $ and hypothesis (iv) is now verified, since
\[
\{1\}= \overline{\tilde \varphi(\R \setminus \{0\})} \subseteq \psi([1, +\infty[)= [1, +\infty[,
\]
and so we get a solution to the equation $ -\Delta_p u= \tilde \varphi (u) $.

\end{example}


TITLE: A zero-dimensional Hausdorff space is totally disconnected
QUESTION [13 upvotes]: The full question: A space is zero-dimensional if the clopen subsets form a basis for the topology. Show that a zero-dimensional Hausdorff space is totally disconnected. Recall a space is totally disconnected if the only connected subsets are singletons (one-point subsets).
Let X = {X1, X2, ...} be the set of clopen subsets of the space. We know that X is a basis for the topology T, so any open set in T can be written as a union or finite intersection of elements in X.
In a topological space, we know the union/finite intersection of open sets are also open (by definition) and the union/finite intersection of closed sets are closed, so any union/finite intersection of clopen sets is also clopen.
Since X is a basis, then any open set in T is also closed, since it will be the union/finite intersection of clopen sets. Does this mean our space is discrete? If it is discrete, then the only connected subsets are singletons, and then our space is totally disconnected.
I have a strong feeling I've gone in circles and my argument is incorrect (especially the discrete part...) Any help would be appreciated.

REPLY [7 votes]: Suppose $X$ is a zero dimensional Hausdorff space, and $\mathcal{B}$ a clopen basis for it.
Suppose $A$ is a set containing points $x$ and $y$. If they are distinct we can find an open set containing $x$ not containing $y$, hence we can find a basis element $B \in \mathcal{B}$ containing $x$ but not $y$. The sets $A\cap B$ and $A\cap B^c$ are both open in $A$ and disjoint, and their union is $A$, hence $A$ is disconnected.
Thus the space doesn't even need to be Hausdorff - the $T_1$ axiom is sufficient.


TITLE: groups generated by two elements of order 3
QUESTION [1 upvotes]: I'd like to know whether it is possible to find a characterization (cardinality?) of the set of finite, non-abelian groups generated by two elements $a$ and $b$ whose order is $3$? Is it the same task than finding all possible relations between $a$ and $b$ (for instance $aabba=bab$)?
Thanks by advance for any comment

REPLY [0 votes]: Maybe interesting to know, this touches the famous Burnside problem, posed by William Burnside in 1902 and one of the oldest and most influential questions in group theory. It asks whether a finitely generated group in which every element has finite order must necessarily be finite.


text:
\section{Positive Integer is Well-Defined}
Tags: Integers

\begin{theorem}
"[[Definition:Positive Integer|Positive]]" as applied to an [[Definition:Integer|integer]] is [[Definition:Well-Defined Relation|well-defined]].
\end{theorem}

\begin{proof}
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]] under the [[Definition:Congruence Relation|congruence relation]] $\boxminus$.
$\boxminus$ is the [[Definition:Congruence Relation|congruence relation]] defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, [[Definition:Integer/Formal Definition/Notation|as suggested]].
Thus, what we are trying to prove is:
:$\eqclass {a, b} {} = \eqclass {c, d} {} \land b < a \implies d < c$
By definition:
:$\eqclass {a, b} {} = \eqclass {c, d} {} \iff a + d = b + c$
So:
{{begin-eqn}}
{{eqn | l = b
      | o = <
      | r = a
}}
{{eqn | ll= \leadsto
      | l = \exists p \in \N: a
      | r = b + p
}}
{{eqn | ll= \leadsto
      | l = b + p + d
      | r = b + c
}}
{{eqn | ll= \leadsto
      | l = p + d
      | r = c
}}
{{eqn | ll= \leadsto
      | l = d
      | o = <
      | r = c
}}
{{end-eqn}}
{{qed}}
[[Category:Integers]]
tfoqxrhyoxppc1d4o3kos1sbvkcow8n
\end{proof}

config: wiki


TITLE: Laurent series for $\frac{1}{z(z-1)}$ in the domain $1 < |z-2| < 2$
QUESTION [1 upvotes]: Expand $f(z) = \cfrac{1}{z(z-1)}$ in a Laurent series valid for $1 < |z-2| < 2$

First of all I wrote the fraction in partial fraction form:
$\cfrac{1}{z(z-1)} = \cfrac{-1}{z} + \cfrac{1}{z-1}$
Then, I tried manipulating the fraction to get (z-2) part:
$\cfrac{-1}{z} = \cfrac{-1}{(z-2)+2} = \cfrac{1}{2}\cfrac{-1}{1+(\cfrac{z-2}{2})}$
Now, $|\cfrac{z-2}{2}| < 1$
Thus, I can apply the series expansion of:
$\cfrac{1}{1+x} = 1 - x + x^2 - ...$, when $|x| < 1$
Thus:
$\cfrac{-1}{z} = \cfrac{-1}{2}[1 - (\cfrac{z-2}{2})^2 + ...]$
$\implies \cfrac{-1}{z} = \cfrac{-1}{2}\sum_{n = 0}^{\infty}(-1)^n(\cfrac{z-2}{2})^n$
$\cfrac{-1}{z} = \sum_{n=0}^{\infty}(-1)^{n+1}\cfrac{(z-2)^n}{2^{n+1}}\tag{1}$
But, how can I manipulate the second fraction?
Thanks

REPLY [2 votes]: $$\frac{-1}{z} = \frac{1}{-2-(z-2)} = -\frac12\;\; \frac{1}{1 - (\frac{2-z}{2})}$$
$$= -\sum_{n=0}^{\infty}(-1)^n\frac{(z-2)^n}{2^{n+1}} $$
$$\frac{1}{z-1} = \frac{1}{1-(2-z)} =\frac{1}{z-2}\;\; \frac{1}{1 - ({2-z})^{-1}}$$


TITLE: Prove IVT from clopen sets in $\mathbb{R}$.
QUESTION [0 upvotes]: I am attempting to prove the Intermediate Value Theorem from the proof of clopen sets in $\mathbb{R}$. I have several lemmas and a corollary inbetween which do most of the heavy work (I've excluded their proofs to save space).
Have I missed anything, and or are the assumptions/proofs valid?
$\textbf{Proposition.}$ The only clopen subsets of $\mathbb{R}$ are the $\emptyset$ and $\mathbb{R}$ itself.
$\textit{Proof.}$
Let $Y$ be the subset $Y \subset \mathbb{R}$ and $\chi_{Y}$ a characteristic function $\chi_{Y} : \mathbb{R} \to \{0, 1\}, y \mapsto
\begin{cases}
      1 & y \in Y \\
      0 & y \notin Y
\end{cases}
$.
Let $A \subset \mathbb{R}$ be an open subset (or subsets). Then the characteristic function $\chi_{Y}$ is continuous iff the preimage $I^{-1} \equiv \chi_{y}^{-1}(A)$ is open. If elements $0,~1 \in A$ then $I^{-1} = \emptyset$, and is open. If $0,~1 \notin A$ then $I^{-1} = \mathbb{R}$, which is also open. If $0 \notin A$ and $1 \in A$ then $I^{-1} = Y$, and must be open. Similarly, If $0 \in A$ and $1 \notin A$ then $I^{-1} = \mathbb{R} \setminus Y$, and must also be open. But then $Y$ must be closed by definition. Therefore, the only clopen subsets of $\mathbb{R}$ are the $\emptyset$ and the set $\mathbb{R}$ itself.
$\textbf{Corollary (Connectedness).} ~\textit{The set $W$ is connected iff there are subsets $X \subset W$ and $Y \subset W$} \\
\textit{such that $X$ and $Y$ are clopen.}$
Now for IVT...
$\textbf{Lemma 1.}~\textit{Let $a$ and $b$ be any elements in $\mathbb{R}$. Intervals of form} \\
\textit{$[a,b]$, $(a,b]$, $[a,b)$, $(a,b)$ are the only subsets in $\mathbb{R}$ that are connected.}$
$\textbf{Lemma 2.}~\textit{If a connected space is continuous then so is its image.}$
$\textbf{Intermediate Value Theorem.} ~\textit{If $f : \mathbb{R} \to \mathbb{R}$ is a function which is continuous at every point}$ $\textit{of the interval $[a,b]$ and $f(a) < 0, f(b) < 0$ then $f(x) = 0$ at some point $x \in (a,b)$.}$
$\textit{Proof.}$
By Lemma 1, the interval $[a,b]$ is connected. Then by Lemma 2, the image $f([a,b])$ is also connected. Therefore, there exists some point $x \in (a,b)$.

REPLY [1 votes]: In the corollary are you saying that the only connected subset of $\mathbb{R}$ are $\mathbb{R}$ and $\emptyset$? But in lemma 1 you say $[a,b]$ is connected.
Intervals of the form $(a,b), [a,b), (a,b]$ are also connected.


TITLE: If the limit of $1-(1-1/x)^x$ equals $1-1/e$, does this mean that only $63.21\%$ of possible things ever happen?
QUESTION [0 upvotes]: $1-(1-\frac1x)^x$ is the analytic function modelling an ever-growing lottery where the number of attempts at the lottery correspond to the chance of winning said lottery; eg playing a $100$-ticket lottery $100$ times only gives the chance of winning at least once to be $1-(1-\frac1{100})^{100} = 63.4\%$, not $100\%$. As $x$ increases, the chance goes to $1-1/e$. At the limit, eg: infinitely many darts whose summed width corresponds to the area of the dartboard, this means that in an infinitely large universe, not everything possible actually gets to happen since the distribution is not uniform.
Follow up question, what does this distribution look like? Some things happen more than once. How would you calculate how many times the most-frequent outcome happens based on the size of $x$, and at the limit?

REPLY [0 votes]: Your different attempts have to be independent for this theory to apply.
For instance you play 100 different 100-ticket lotteries.
Or you play a lottery where they pick the numbers out for you at random.
Then yes, playing 100 independent times with a 1/100 chance of winning gives you a probability of winning of 1-1/e$ \approx 63 $ %.
But you can go further. Play 200 times. Then the probability of winning is $1-e^{-2} \approx 90$ %. And so on.


TITLE: Magnetic analog for how charge distributions can be approximated as point charges?
QUESTION [1 upvotes]: I know that far away from a distribution of charges, the fields can be approximated using Gauss' Law, and the fields become very similar to that of a point charge. Is there an analog for this in magnetics?
For example, how would this work for a loop of current, could it be approximated as a "point magnet"? That doesn't seem right to me, so I'm guessing it has no analog.
But if a ball was magnetized, could it be approximated in such a fashion? I guess I am asking, at a sufficient distance from this magnetized ball, what do the nature of the fields caused by the equivalent currents due to its magnetization become "similar to"?

REPLY [1 votes]: The loop of current can indeed be considered as a small bar magnet with the same magnetic moment as the current loop. Unlike in electrostatics, magnetic monopoles do not exist. The magnetized ball is a magnetic dipole, and hence it will have the same effects as any other magnetic dipole including a current carrying loop, at a large distance.


TITLE: Find all $x,y\in \mathbb{N}$, such that $x^{y}=y^{y-x}$
QUESTION [7 upvotes]: Find all $x,y\in \mathbb{N}$, such that $ x^{y}=y^{y-x} $.
I have found the solutions $(1,1)$ and $(2,4)$. I've been trying to prove that there aren't any more. I tried splitting the cases by parity, but nothing interesting seemed to come out.
I also tried rearranging it into:
$x=y^{\frac{y-x}{y}}$.
Is there any way to proceed with this problem?
Thank you very much

REPLY [8 votes]: Actually, the function can be written as
$$ \left(\frac{y}{x}\right)^{y} = y^{x} $$
If you let $t = \frac{y}{x}$ then we can get
$$ t^{t} = y $$
So if you apply $y = t^{t}$ and $x = t^{t - 1}$ with $t$ an integer you get infinite pairs of solutions.
Furthermore, if $t$ is not an integer then $t = q/p$ which is irreducible.
Then $y = (q/p)^{q/p}$ is an integer so $y^{p} = (q/p)^q$ is an integer, which is not possible.
So $(x, y)$ is a solution to the equation iff $(x, y) = (t^{t-1}, t^t)$ with $t$ an integer, and there are infinitely many pairs.


TITLE: Showing that a ring is a field as well for one of the provided choices.
QUESTION [1 upvotes]: Let $\mathbb M$ be one of the following rings: $\mathbb{R}, \mathbb{Q},  \mathbb{F_9}, \mathbb{C} $. Let $I$ be the ideal generated by $x^4+2x-2$. Is the ring $\mathbb M[x]/I$ a field for some of the above choices for $R$?

I have been trying to work my way around some problems involving ideals, but I get really confused when they get mixed with rings and fields. I'm stuck and don't know how to proceed.

REPLY [5 votes]: In general, given a ring $R$ and an ideal $J$ of $R,$ we have that $R/J$ is a field if and only if $J$ is a maximal ideal of $R.$ Now, since each $\Bbb M$ is a field, here, then $\Bbb M[x]$ is a principal ideal domain, and so $I$ is maximal if and only if $I=\langle p(x)\rangle$ for some irreducible polynomial $p(x)\in\Bbb M[x].$ So, what you must determine is whether $x^4+2x-2$ is irreducible over any of the given fields $\Bbb M.$
Added: As a hint for how to address the case that $\Bbb M=\Bbb F_9,$ note that $\Bbb F_9$ has characteristic $3,$ and so $x^4+2x-2=x^4+2x+1,$ as far as $\Bbb F_9$ is concerned.


TITLE: Vector Calculus Divergence
QUESTION [0 upvotes]: Prove that $$\nabla.\left(\vec{r}\,\nabla\left(\frac{1}{r^3}\right)\right)= \frac3{r^4}.$$

My doubt is that here should we consider  the term
$\vec{r}$ and $\nabla\left(\frac{1}{r^3}\right)$ as the ones to be multiplied or as dot product?
Also please give the solution.
Thanks in advance!

REPLY [0 votes]: $\nabla\big(\frac{1}{r^3}\big) = \frac{-3}{r^4}\nabla(r) = \frac{-3}{r^5}\vec{r} $
$\bigg[r^2 = x^2 +y^2 + z^2$
$\frac{\partial r}{\partial x} = \frac{x}{r}$, similarly for y and z.
$\nabla r = \Sigma(\hat i \frac{\partial r}{\partial x}) = \frac{1}{r}\Sigma(x \hat i) = \frac{\vec{r}}{r}\bigg] $
$y = \nabla.\bigg( r.\nabla\big(\frac{1}{r^3}\big)\bigg) = \nabla.\bigg( r.\frac{-3}{r^5}\vec{r}\bigg) = -3\nabla.\bigg( \frac{\vec{r}}{r^4}\bigg) = -3\bigg[\frac{1}{r^4}\nabla.\bigg( \vec{r}\bigg) + \vec{r}.\nabla(\frac{1}{r^4})\bigg]$
$y =-3\bigg[\frac{1}{r^4}3 + \vec{r}.\frac{-4}{r^6}\vec{r}\bigg] = -3\bigg[\frac{1}{r^4}3 + {r^2}\frac{-4}{r^6}\bigg] = \frac{3}{r^4}  $
[Formulae used: $grad(r) = \frac{\vec{r}}{r} \ , \ div(\vec{r}) = 3 \ , \ \vec{r}.\vec{r} = r^2 $]


TITLE: A "dense" extension of the set of primitive recursive functions
QUESTION [6 upvotes]: Let $\mathcal{PR}$ be the set of primitive recursive functions. Let $\mathcal{PR}(f)$ be $\mathcal{PR}$ which we have amplified by adding (a recursive) $f$ the in the set of initial functions. To make this a true extension, assume $f$ outgrows all primitive recursive functions.
Now, $\mathcal{PR}$ is "dense" in the sense that it has the Ritchie-Cobham property, i.e. every function computable in primitive recursive time is primitive recursive and vice versa.

Does $\mathcal{PR}(f)$ have the Ritchie-Cobham property?

To me it seems natural the extension preserves the property and I would guess there is some "direct" argument for this, which I just fail to see.

REPLY [5 votes]: The answer is yes.
Suppose a function $g$ is computable by a procedure $p$ whose
computation running time is bounded by a function
$h\in\newcommand\PR{\text{PR}}\PR(f)$. I claim that $g\in\PR(f)$.
To see this, let's first argue that $\PR(f)$ is closed under the
bounded search operator $(x,z)\mapsto\mu y<z[R(x,y,z)]$, where $R$
is a relation whose characteristic function is in $\PR(f)$ and
where the $\mu$ operator returns the least $y<z$ such that
$R(x,y,z)$, if there is one, and otherwise returns $z$ as a signal
that the search failed. The proof is just the usual proof that the
primitive recursive functions are closed under bounded search,
since one can easily define this instance of the $\mu$ operator by
recursion, by using $R$. In other words, the usual argument is not
sensitive to the particular functions one starts with in the $\PR$
hierarchy.
It follows now that $g$ is in $\PR(f)$, since we can say that
$g(x)=y$ just in case there is a code for a computation of length
$h(x)$ showing that the output of $g$ on $x$ is $y$ according to
the fixed program $p$. This code can be checked using bounded
search, since we can bound how big the code will be if we know how
long the computation will be.


TITLE: Adding two complex numbers
QUESTION [0 upvotes]: I have two complex numbers: $Z_1=e^{-j45°}$ and $Z_2=e^{j135°}$.
I need to calculate $2*Z_1+Z_2$.
I know how to do it when I convert with help of Eulers formula n form $\cos(x)+j\sin(x)$, but that is just too complicated to calculate.
Is there any easier way to do the given operation.
Thanks!!!

REPLY [1 votes]: Hint #1: $2z_1+z_2=z_1+(z_1+z_2)$.
Hint #2: Plot $z_1$ and $z_2$ on the complex plane. Does their sum have an obvious "nice" value?

REPLY [1 votes]: $e^{-i45^\circ}$ is a complex number with angle $-45^\circ = 315^\circ$
$e^{i 135^\circ}$ is a complex number with angle $135^\circ$
What happens when you add two numbers whose angles differ by $180^\circ$?


TITLE: What parameters affect the range and flight path of playing card? How?
QUESTION [1 upvotes]: An ordinary playing card can fly quite a long distance if you spin it during a throw.  What parameters affect the range and flight path?  How?

REPLY [0 votes]: Very interesting question ! I don't know exact answer, but surely it has something to do with playing card angular rotation speed. The more it rotates faster - the less it will experience air resistance, aka. drag force. If rotation speed would be low or zero, then card will experience huge drag, because of tumbling which increases drag because of increased card's cross-section. Thus, it will fall down sooner. Of course card has no such parameter as angle of attack, but you can understand it's flight dynamics by studying boomerang flight laws


TITLE: Show that $S=\frac{1}{a_1}+\cdots+\frac{1}{a_n}$ is not an integer.
QUESTION [4 upvotes]: So I have a bunch of integers $a_1,...,a_n$ and a prime number $p$ that divides only one of the numbers in this sequence, say $a_k$
I want to show, that $S=\dfrac{1}{a_1}+\cdots+\dfrac{1}{a_n}$ is not an integer.
Well we can express the sum in another way, by putting them under a commond denominator, then the sum is
$$\sum_{i=1}^n\dfrac{\prod_{j=1,j \neq i}^n a_j}{\prod_{j=1}^n a_j}$$
All of the members of this sum are divisible by $p$ expect the $k$'th one. But is this helpful in any way?

REPLY [2 votes]: The pedantic way: The $p$-adic absolute value on $\mathbf Q$ satisfies the ultrametric inequality $|x+y|_p \leq \max\{|x|_p, |y|_p\}$. Let $S$ be your sum. Then
$$|S-1/a_k| \leq \max_{i \neq k}\{|1/a_i|_p\} \leq  1$$
because the $a_i$'s are $p$-adic units for $i \neq k$.
On the other hand, $$p\leq  |1/a_k|_p \leq \max\{|S-1/a_k|_p, |S|_p\}\leq \max\{1, |S|_p\}$$
implies $|S|_p\geq p$, so $S$ is not an integer.


TITLE: Number of triangles in the figure
QUESTION [1 upvotes]: How many triangles are in the figure show here:

I tried using casework, but it got really messy really fast. I also think that you could easily miscount and get something wrong.

REPLY [1 votes]: There are 36 triangles in that square.


TITLE: Is there a non-locally compact Hausdorff space in which all infinite compact sets (of which there is at least one) have uncountable interiors?
QUESTION [4 upvotes]: Here is the background material from which I am working:

The Cantor set is an uncountable compact Hausdorff subspace of $\mathbb{R}$ with empty interior.
In a locally compact Hausdorff space with no isolated points, each countable set has empty interior.
The rational numbers with the subspace topology is a non-locally compact Hausdorff space in which all compact sets have empty interior.

I am trying to find a non-locally compact Hausdorff space in which there are infinite compact sets, and where all of the infinite compact sets have uncountable interiors. I am guessing the example will be an exotic function space.
I first posed this question without specifying that there should be at least one infinite compact set, and this was solved by Stefan H. on this site. I then posed this updated question, which was solved by Dejan Govc.

REPLY [3 votes]: It seems the following.
At last you obtain a positive answer, because of the following
Lemma. Each infinite locally compact Hausdorff space $X$ contains an infinite closed subspace with a countable interior.
So if you have an infinite compact subspace $X$ of a Hausdorff space $Z$ then  $X$ contains an infinite closed subspace with a countable interior not only in $Z$, but even in $X$.
Proof of Lemma. Suppose the converse. Let $Y=\{y_n:n\in\omega\}$ be an arbitrary countable infinite subset of $X$. Then the set $\overline Y$ is uncountable. Let $x\in \overline Y\backslash Y$ be an arbitrary point. Since the space $X$ is Hausdorff, for each number $n\in\omega$ there exist disjoint open neighborhoods $U_n$ and $V_n$ of the points $y_n$ and $x$ respectively. The set $\overline Y\backslash\bigcup U_n $ is a closed set with an empty interior and therefore it is finite. Then the set $\overline Y\cap \bigcap V_n\subset \overline Y\backslash\bigcup U_n$ is finite too. Thus the point $x$ has a countable pseudocharacter $\psi(x,\overline{Y})$ in the space $\overline {Y}$. Since the space $X$ is locally compact, there exists an open neighborhood $V$ of the point $x$ such that the set $\overline V$ is compact. Then $\psi(x,\overline{Y\cap V})\le \psi(x,\overline{Y})\le\omega$. Since the space $\overline {V\cap Y}$ is compact, the character $\chi(x,\overline {V\cap Y})$ is countable too. Since the set $Y$ is dense in the space $\overline{V\cap Y}$ and $x\not\in Y$, there exists a sequence $\{x_n\}$ of distinct points of the set $Y$  converging to $x$. Then the set $\{x_n\}\cup\{x\}$ is a countable infinite compact set, and hence an infinite closed subset of the space $X$ with countable interior, a contradiction.$\square$
Update. Moreover, we can improve Lemma to
Lemma 2. Each infinite Hausdorff space $X$ contains an infinite closed subspace with a countable interior.
Proof. Suppose the converse. Let $Y=\{y_n:n\in\omega\}$ be an arbitrary countable infinite subset of $X$. Then the set $\overline Y\backslash Y$ is uncountable. For each $n\in\omega$ we define by induction an open neighborhood $U_n$ of the point $y_n$ such that a set $Z_n=\overline Y\backslash (Y\cup\bigcup_{i\le n} U_i)$ is infinite and a point $z_n\in Z_n$ such that all the points $z_i$ are distinct and $z_i\not\in U_n$ for each $i<n$. If for each neighborhood $U$ of the point $y_0$ the set $\overline Y\backslash (Y\cup U)$ is finite, then for any countable infinite subset $Z$ of the set $\overline Y\backslash Y$ the set $Z\cup \{y_0\}$ is a countable compact set, a contradiction. Therefore we can choose an open neighborhood $U_0$ of the point $y_0$ such that the set $Z_0=\overline Y\backslash (Y\cup U)$ is infinite and a point $z_0\in Z_0$. If for each neighborhood $U$ of the point $y_1$ the set $Z_0\backslash U$ is finite, then for any countable infinite subset $Z$ of the set $Z_0$ the set $Z\cup \{y_1\}$ is a countable compact set, a contradiction. Therefore we can choose an open neighborhood $U_1\not\ni z_0$ of the point $y_1$ such that the set $Z_1=Z_0\backslash U_1$ is infinite and a point $z_1$ in $Z_1\backslash\{z_0\}$, and so forth. Then $\overline{\{z_n:n\in\omega\}}\subset\overline{Y}\backslash\bigcup U_n$ is a nowhere dense closed infinite subset of the space $\overline Y$, a contradiction.$\square$
Update 2. Moreover, we can prove
Proposition. Each infinite Hausdorff space $X$ containing infinitely many non-isolated points contains also an infinite closed subset with empty interior.
Proof. Suppose the converse. Denote  the set of all non-isolated points of $X$ as $X’$. Let $Y=\{y_n:n\in\omega\}\subset X’$ be an arbitrary countable infinite subset. It is clear that $\overline Y\subset X’$.
We shall need the following
Lemma 3. There are no infinite subset $Z$ of the set $X’$ and a point $x\in X$ such that for each neighborhood $U$ of the point $x$ the set $Z\backslash U$ is finite.
Proof. Suppose the converse. Then the set $Z_0=Z\cup \{x\}$ is a compact set. Moreover, let
$y\in Z_0\backslash \{x\}$ be an arbitrary point. There exist disjoint open sets $U\ni x$ and $V\ni y$. Hence $V\cap Z_0$ is a finite neighborhood of the point $y$. Therefore, the space $Z_0\backslash \{x\}$ is discrete. The set $Z_0$ is closed in $X$ as its compact subset. Since the set $Z_0$ is infinite, there exists a non-empty open subset $V$ of $X$ such that $V\subset Z_0$. Since the set $Z_0\backslash \{x\}$ is dense in $Z_0$, there exists a point $y\in V\cap (Z_0\backslash \{x\})$. Since the space $Z_0\backslash \{x\}$ is discrete, there exists a neighborhood $W$ of the point $y$ such that $W\cap (Z_0\backslash \{x\})$ is a singleton. But then a set $W\cap V\subset W\cap (Z_0\backslash \{x\})$ is a singleton too, a contradiction, because all points from the set $Z_0\backslash \{x\}\subset X’$ are non-isolated. $\square$
Now we shall consider two cases.
$1.$ The set $\overline Y\backslash Y$ is infinite. For each $n\in\omega$ we define by induction an open neighborhood $U_n$ of the point $y_n$ such that a set $Z_n=\overline Y\backslash (Y\cup\bigcup_{i\le n} U_i)$ is infinite and a point $z_n\in Z_n$ such that all the points $z_i$ are distinct and $z_i\not\in U_n$ for each $i<n$. By Lemma 3, there exists an open neighborhood $U_0$ of the point $y_0$ such that the set $ Z_0=\overline Y\backslash (Y\cup U)$ is infinite. Choose an arbitrary point $z_0\in Z_0$. Again by Lemma 3, there exists an open neighborhood $U_1\not\ni z_0$ of the point $y_1$ such that the set $Z_1=Z_0\backslash U_1$ is infinite. Choose an arbitrary point $z_1\in Z_1\backslash\{z_0\}$, and so forth. Then $\overline{\{z_n:n\in\omega\}}\subset\overline{Y}\backslash\bigcup U_n$ is a nowhere dense closed infinite subset of the space $\overline Y$, a contradiction.
$2.$ The set $\overline Y\backslash Y$ is finite. Since $\overline Y\subset X’$, without loss of generality, we can assume that
$\overline Y=Y$. For each $n\in\omega$ we define by induction an open neighborhood $U_n$ of the point $y_n$ such that a set $Z_n=Y\backslash \bigcup_{i\le n} U_i$ is infinite and a point $z_n\in Z_n$ such that all the points $z_i$ are distinct. By Lemma 3, there exists an open neighborhood $U_0$ of the point $y_0$ such that the set $Z_0=Y\backslash U_0$ is infinite. Choose an arbitrary point $z_0\in Z_0$. Again by Lemma 3, there exists an open neighborhood $U_1$ of the point $y_1$ such that the set $Z_1=Z_0\backslash U_1$ is infinite. Choose an arbitrary point $z_0\in Z_1\backslash\{z_0\}$, and so forth. Put $Z=\{z_i:i\in\omega\}$. Then $\overline Z\subset\overline Y=Y$. Let $y\in Y$. Then $y=y_n$ for some $n\in\omega$. Then the intersection $U_n\cap Z\subset U_n\cap (Z\backslash Z_n)$ is a finite set. Therefore $Z$ is a discrete (in particular, a closed) subset of the space $X$. Suppose that the set $Z$ has non-empty interior. Then there exists a non-empty open set $U$ such that $U\subset Z$. Then there exists
a number $n$ such that $U_n\cap U$ is a nonempty open set. But $U_n\cap U\subset U_n\cap Z$ is a singleton, a contradiction, because all points from the set $Z$ are non-isolated. $\square$


TITLE: Irreducibility of polynomials over field of rational functions
QUESTION [0 upvotes]: I was reading Dummit and Foote. On page 551 of the book, there is an example mentioned about irreducibile inseparable polynomials. I am facing difficulty in proving the irreducibility of polynomials.
How I can show this : -
The polynomial $p(x) = x^{2^m} - t$ over $F = \mathbb{F}_2(t)$ is irreducible. For small $m$ like 2 or 4 one can enumerate cases and get done but what about a general m?

REPLY [1 votes]: At least two simple ways of getting there:

If $u$ is a zero of your polynomial in an extension field $K$ of $F$ then $u^{2^m}=t$ and, by Freshman's dream,
$$p(x)=(x-u)^{2^m}=x^{2^m}-u^{2^m}=x^{2^m}-t.$$ So, by uniqueness of polynomial factorization, any eventual factor of $p(x)$ in $F[x]$ must be of the form $(x-u)^i$ for some $i,0<i<2^m$. But the constant term $u^i$ is in $F$ only when $i=0$ or $i=2^m$, and we are done.
You can apply Eisenstein's criterion to the prime $t$, irreducible in $\Bbb{F}_2[t]$. All the coefficients of $p(x)$, save the leading $1$, are divisible by $t$, and the constant coefficient is not divisible by $t^2$. Therefore Eisenstein applies.


TITLE: Generating Coprime Integers
QUESTION [4 upvotes]: Is there a formula for generating a set of Coprime integers that every element of this set is coprime to the other elements in this set?
I want to create a collection of this formulas!

REPLY [0 votes]: As shown above, a nice way is to add $1$. I think that this is rather a cool way of visualizing that $a \times b + 1$ will generate coprime integers (that is both $a$ and $b$ will not divide $a \times b + 1$).
Suppose $a = 3, b = 4$ then we have 4 rows of 3 dots so we can draw 3 vertical lines of 4 dots, or 4 horizontal lines of 3 dots
...
...
...
...

Now consider $a \times b + 1$
...
...
...
....


TITLE: $a_n =(\frac{n}{n+1})^n$ find convergence by geometric series
QUESTION [0 upvotes]: $a_n =(\frac{n}{n+1})^n$
I calculated  a few terms:
$a_1 = 1/2$
$a_2 = 4/9$
$a_3 = 27/64$
But not sure how to find $r$ so i can calculate its value as $n$ approaches infinity (if it does converge)

REPLY [3 votes]: Consider $$\frac 1{a_n}=\left(\frac{n+1}{n}\right)^n=\left(1+\frac{1}{n}\right)^n.$$ It is well known that this sequence converges to $e$. Hence, $a_n\to 1/e\approx0\mathord.368$ as $n\to\infty$.

REPLY [1 votes]: Hint:
$$\lim_{n\to{\infty}}{a_n}=\lim_{n\to{\infty}}\frac{1}{\left(1+\frac{1}{n} \right)^n}=\frac{1}{\lim\limits_{n\to{\infty}}\left(1+\frac{1}{n} \right)^n}=\frac{1}{e}.$$


TITLE: Can the Binomial Theorem be proved probabilistically?
QUESTION [7 upvotes]: I believe that the answer is affirmative, as there is a reputable-appearing subscription-hidden link on the web claiming such, and because, although I have not seen that proof, I have (I believe) a proof of my own (which I’m pretty sure is similar to the hidden one).
Here is the link to the subscription-hidden article.
And here’s my take on the question that I have posed:
Using elementary Probability Theory, and an appeal to symmetry and continuity, it is easy to prove the Binomial Theorem.
First, we note that it is easy to establish that $\binom{n}{k}$ = C(n,k), by first (easily) establishing the formula for P(n,k), and then (easily) modifying it to give C(n,k). We can easily show that the binomial coefficients must be C(n,k) for the expansion of ${(x + y)}^n$, for positive integers x and y, by noting the following situation:
If an urn contains x white balls and y black balls, then, given that (exactly) $n$ balls are selected, with replacement, the events:
0. picking n white balls
1. picking n – 1 white balls and 1 black ball
2. picking n – 2 white balls and 2 black balls
…
n – 1. picking 1 white ball and n – 1 black balls
n. picking n black balls
are mutually-exclusive and exhaustive events, and so the sum of their probabilities is unity.
However, it is obvious that the number of ways in which event k can occur is C(n,k). Since all the terms have a denominator of ${(x + y)}^n$, we can multiply both sides of the equation by ${(x + y)}^n$ to get the Binomial Theorem. The case for arbitrary integers x and y follows by symmetry, and the case for arbitrary real numbers x and y follows by continuity. Done.
Notation: $\binom{n}{k}$ is $\frac{n!}{(k!)(n – k)!}$, C(n,k) is the number of combinations of n things taken k at a time, and P(n,k) is the number of permutations of n things taken k at a time.

REPLY [10 votes]: The (almost one-page) paper does have a "probabilistic proof" of the binomial theorem, in the sense that it is a proof of the binomial theorem using tools from the theory of probability, and it follows the basic outline of the argument you give above (which is also correct, by the way), in slightly more advanced terminology. So the answer to your question is yes. However, I don't find the paper particularly evocative. When I think of how the binomial theorem is related to probability, I think of this (I know it's not a proof, but still):

(see the wikipedia article on the Galton box)


TITLE: Line Integral using Green's theorem
QUESTION [0 upvotes]: Problem statement:
Calculate $\int_{\gamma }(3e^{(y-3x)^{2}}-y)\mathrm{dx}+(-e^{(y-3x)^{2}}+2x)\mathrm{dy}$ where $\gamma$ is the curve $y=x^2$ from $(0,0)$ to $(3,9)$.
Progress
First idea was to parametrize the curve: $f(t,t^2), t\in [0,3]$ but using this approach I ended up with an integral whose primitive could not be expressed in elementary functions.
Secondly I noticed that "$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = 3$" so Green's theorem could be suitable. So I enclosed a region using a vertical and a horizontal line segment. However, these couldn't be solved in elementary functions.
What are some other ways?

REPLY [1 votes]: Let $\Omega$ be the domain which is enclosed by $y=3x$ and $y=x^2$, and $\Gamma$ be the line $y=3x$, by the Green's Formula, we have
\begin{eqnarray*}
\int_{\partial\Omega}Pdx+Qdy&=&\int_{\Omega}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\ dxdy=3|\Omega|\\
&=&3\int_0^3 (3x-x^2)\ dx=\frac{27}{2}.
\end{eqnarray*}
On the other hand, we have
\begin{eqnarray*}
\int_{\partial\Omega}Pdx+Qdy&=&\int_{\gamma}Pdx+Qdy-\left(\int_\Gamma Pdx+Qdy\right).
\end{eqnarray*}
Notice that
\begin{eqnarray*}
\int_\Gamma Pdx+Qdy&=&\int_0^3\left[(3-3x)+(-1+2x)\right]\ dx=\int_0^3[2-x]\ dx=\frac{3}{2}.
\end{eqnarray*}
Hence we get
\begin{eqnarray*}
\int_{\gamma}Pdx+Qdy=15.
\end{eqnarray*}