\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{mathrsfs} \AtBeginDocument{{\noindent\small Two nonlinear days in Urbino 2017,\newline \emph{Electronic Journal of Differential Equations}, Conference 25 (2018), pp. 221--233.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2018 Texas State University.} \vspace{8mm}} \begin{document} \setcounter{page}{221} \title[\hfilneg EJDE-2018/conf/25\hfil Ground state solutions] {Ground state solutions for Bessel fractional equations with irregular nonlinearities} \author[S. Secchi \hfil EJDE-2018/conf/25\hfilneg] {Simone Secchi} \address{Simone Secchi \newline Dipartimento di Matematica e Applicazioni, Universit\`a degli Studi di Milano Bicocca, Italy} \email{simone.secchi@unimib.it} \thanks{Published September 15, 2018} \subjclass{35KJ60, 35Q55, 35S05} \keywords{Bessel fractional operator, fractional Laplacian} \begin{abstract} We consider the semilinear fractional equation $(I-\Delta)^s u = a(x) |u|^{p-2}u \quad\text{in }\mathbb{R}^N,$ where $$N \geq 3$$, $$02$$, $$00 we introduce the \emph{Bessel function space} $L^{s,2}(\mathbb{R}^N) = \{ f \in L^2(\mathbb{R}^N): f=G_s \star g \ \text{for some g \in L^2(\mathbb{R}^N)} \},$ where the Bessel convolution kernel is defined by $G_s (x) = \frac{1}{(4 \pi )^{s /2}\Gamma(s/2)} \int_0^\infty \exp \Big( -\frac{\pi}{t} |x|^2 \Big) \exp \Big( -\frac{t}{4\pi} \Big) t^{\frac{s - N}{2}-1} \, dt.$ The Bessel space is endowed with the norm~\|f\| = \|g\|_2 if f=G_s \star g. The operator (I-\Delta)^{-s} u = G_{2s}\star u is usually called Bessel operator of order s. In Fourier variables the same operator reads $G_s = \mathcal{F}^{-1} \circ \Big( \big(1+|\xi|^2 \big)^{-s /2} \circ \mathcal{F} \Big),$ so that $\|f\| = \| (I-\Delta)^{s /2} f \|_2.$ For more detailed information, see \cite{Adams, Stein} and the references therein. In \cite{Fall} the pointwise formula $(I-\Delta)^s u(x) = c_{N,s} \operatorname{P.V.} \int_{\mathbb{R}^N} \frac{u(x)-u(y)}{|x-y|^{\frac{N+2s}{2}}} K_{\frac{N+2s}{2}}(|x-y|) \, dy + u(x)$ was derived for functions u \in C_c^2(\mathbb{R}^N). Here c_{N,s} is a positive constant depending only on N and s, P.V. denotes the principal value of the singular integral, and K_\nu is the modified Bessel function of the second kind with order \nu (see \cite[Remark 7.3]{Fall} for more details). However a closed formula for K_\nu is not known. We summarize the main properties of Bessel spaces. For the proofs we refer to \cite[Theorem 3.1]{Felmer}, \cite[Chapter V, Section 3]{Stein}. \begin{theorem} \label{th:sobolev} \begin{enumerate} \item L^{s,2}(\mathbb{R}^N) = W^{s,2}(\mathbb{R}^N) = H^s (\mathbb{R}^N), where the sign of equality must be understood in the sense of an isomorphism. \item If s \geq 0 and 2 \leq q \leq 2_s^*=2N/(N-2s), then L^{s,2}(\mathbb{R}^N) is continuously embedded into L^q(\mathbb{R}^N); if 2 \leq q < 2_s^* then the embedding is locally compact. \item Assume that 0 \leq s \leq 2 and s > N/2. If s -N/2 >1 and 0< \mu \leq s - N/2-1, then L^{s,2}(\mathbb{R}^N) is continuously embedded into C^{1,\mu}(\mathbb{R}^N). If s -N/2 <1 and 0 < \mu \leq s -N/2, then L^{s,2}(\mathbb{R}^N) is continuously embedded into C^{0,\mu}(\mathbb{R}^N). \end{enumerate} \end{theorem} \begin{remark} \rm According to Theorem \ref{th:sobolev}, the Bessel space L^{s,2}(\mathbb{R}^N) is topologically undistinguishable from the Sobolev fractional space H^s(\mathbb{R}^N). Since our equation involves the Bessel norm, we will not exploit this characterization. \end{remark} Going back to \eqref{eq:1.1}, it must be said that in the case s \in (0,1) less is known than in the \emph{local} case s=1. Equation~\eqref{eq:1.1} arises from the more general Schr\"{o}dinger-Klein-Gordon equation $\mathrm{i}\frac{\partial \psi}{\partial t} = (I-\Delta)^s \psi -\psi -f(x,\psi)$ describing the the behaviour of bosons, spin-0 particles in relativistic fields. We refer to \cite{FelmerVergara, Secchi17,Secchi17-1,Secchi17-2} for very recent results about the existence of variational solutions. When s=1/2, the operator (I-\Delta)^{1/2}=\sqrt{\strut I-\Delta} is also called \emph{pseudorelativistic} or \emph{semirelativistic}, and it is very important in the study of several physical phenomena. The interested reader can refer to \cite{CingolaniSecchi15,CingolaniSecchi18} and to the references therein for more information. \begin{remark} \rm The identity operator I is often replaced by a multiple m^2 I, for some real number~m \neq 0. The operator reads then (-\Delta+m^2)^s, but for our purposes this generality does not give any advantage. \end{remark} A common feature in the current literature is that the existence of solutions to \eqref{eq:1.1} is related to the behavior of the potential function a at infinity. This is a very useful tool for applying concentration-compactness methods or for working in weighted Lebesgue spaces. In the present paper, following \cite{AckerChagoya}, we investigate \eqref{eq:1.1} under much weaker assumptions on a, see Section 2. The first existence results for semilinear elliptic equations with \emph{irregular} potentials appeared, as far as we know, in \cite{CeramiMolle}. \section{Variational setting} We introduce some tools that will be used systematically in the rest of this article. \begin{definition} \rm \begin{itemize} \item For any y \in \mathbb{R}^N, we define the translation operator \tau_y acting on a (suitably regular) function f as \tau_y f\colon x \mapsto f(x-y). \item In a normed space X, we denote by B(x,r) the ball centered at x \in X with radius~r >0, and by \overline{B}(x,r) its closure. The boundary of~B(0,1) will be denoted by S(X). \item For any a \in L^\infty(\mathbb{R}^N), we define $\mathscr{P} = \overline{B}(0,|a|_\infty) \subset L^\infty(\mathbb{R}^N).$ Looking at L^\infty(\mathbb{R}^N) as the dual space of L^1(\mathbb{R}^N), the set \mathscr{P} will be endowed with the weak* topology. It is well-known that \mathscr{P} becomes a compact metrizable space, see \cite[Theorems 3.15 and 3.16]{Rudin}. \item For any a \in L^\infty(\mathbb{R}^N), we define the subset \mathscr{A} = \left\{ \tau_y a: y \in \mathbb{R}^N \right\} of \mathscr{P}, endowed with the relative topology. Finally, we introduce \mathscr{B} = \overline{\mathscr{A}} \setminus \mathscr{A}. \item For any a \in L^\infty(\mathbb{R}^N), we define \begin{equation} \label{eq:a-bar} \bar{a} = \sup \left\{ \operatorname{ess\,sup} u: u \in \mathscr{B} \right\}. \end{equation} If \mathscr{B}=\emptyset, we agree that \bar{a} = -\infty. \end{itemize} \end{definition} The following is the main assumption of this article. \begin{itemize} \item[(A1)] The function a \in L^\infty(\mathbb{R}^N) is such that a^{+} = \max\{a,0\} is not identically zero, and either (i) \bar{a} \leq 0 or (ii) \bar{a} \leq a. \end{itemize} Weak solutions to \eqref{eq:1.1} are critical points of the functional I_a \colon L^{s,2}(\mathbb{R}^N) \to \mathbb{R}^N defined by $I_a(u) = \frac{1}{2} \| u \|_{L^{s,2}}^2 - \frac{1}{p} \int_{\mathbb{R}^N} a |u|^p.$ \begin{definition} \rm A solution u \in L^{s,2}(\mathbb{R}^N) is called a ground-state solution to \eqref{eq:1.1} if I_a attains at u the infimum over the set of all solutions to \eqref{eq:1.1}, namely $I_a(u) = \min \big\{ I_a(v): \text{v \in L^{s,2}(\mathbb{R}^N) solves \eqref{eq:1.1}} \big\}.$ \end{definition} We now state the main result of our paper. \begin{theorem} \label{th:main} Equation \eqref{eq:1.1} has (at least) a positive ground state provided that 2 b$$, then for some $$\delta>0$$ we can say that the set $$\Omega = \{ x \in \mathbb{R}^N: a(x) \geq b + \delta \}$$ has positive measure. Let us define $$\varphi = \chi_\Omega / \mathcal{L}^N(\Omega)$$, so that $\int_{\mathbb{R}^N} a \varphi = \frac{1}{\mathcal{L}^N(\Omega)} \int_\Omega a \geq b+\delta,$ contrary to \eqref{eq:3.2}. This completes the proof. \end{proof} Recall from assumption (A1) that $a^{+} \neq 0$ as an element of $L^\infty(\mathbb{R}^N)$. Therefore Lemma \ref{lem:3.1} yields a function $\varphi \in S(L^1(\mathbb{R}^N))$ such that $\varphi \geq 0$ and $\int_{\mathbb{R}^N} a \varphi >0$. By a standard mollification argument, we can assume without loss of generality that $\varphi \in C_c^\infty(\mathbb{R}^N)$. Since $L^{s,2}(\mathbb{R}^N)$ is continuously embedded into $L^p(\mathbb{R}^N)$ for every $20 \Big\} \] and $\mathscr{S}_a^{+} = \mathscr{B}_a^{+} \cap S(L^{s,2}(\mathbb{R}^N)).$ \begin{lemma} The set$\mathscr{B}_a^{+}$is non-empty and open in$L^{s,2}(\mathbb{R}^N)$. \end{lemma} \begin{proof} We already know that$\varphi \in \mathscr{B}_a^{+}$. Furthermore, the map$u \mapsto \int_{\mathbb{R}^N} a|u|^p$is continuous from$L^{s,2}(\mathbb{R}^N)$to$\mathbb{R}$, since$a \in L^\infty(\mathbb{R}^N)$and$20$of$h$. By direct computation,$ t u \in \mathscr{N}_a$if, and only if,$t=\bar{t}(u)$. Explicitly, $\bar{t}(u) = \frac{\|u\|_{L^{s,2}}^2}{\int_{\mathbb{R}^N} a|u|^p}.$ This shows that the map$u \mapsto \bar{t}(u)$is continuous from$\mathscr{B}_a^{+}$to$(0,+\infty)$. The rest of the proof follows easily. \end{proof} \begin{lemma} The set$\mathscr{N}_a$is closed in$L^{s,2}(\mathbb{R}^N)$. \end{lemma} \begin{proof} If$u \in \mathscr{N}_a$, then $\|u\|_{L^{s,2}}^2 = \int_{\mathbb{R}^N} a |u|^p \leq \int_{\mathbb{R}^N} a^{+} |u|^p \leq S_p |a^{+}|_\infty \|u\|_{L^{s,2}}^p.$ It follows that \begin{equation} \label{eq:3.3} \inf_{u \in \mathscr{N}_a} \|u\|_{L^{s,2}} \geq \frac{1}{S_p |a^{+}|_\infty^{1/(p-2)}}. \end{equation} As a consequence,$0$is not a cluster point of$\mathscr{N}_a$, which turns out to be closed. \end{proof} It is now standard to invoke the Implicit Function Theorem to prove that$\mathscr{N}_a$is a$C^2$-submanifold of$L^{s,2}(\mathbb{R}^N)$and that \eqref{eq:3.3} implies $\inf_{u \in \mathscr{N}_a} I_a(u) \geq \Big( \frac{1}{2} - \frac{1}{p} \Big) \frac{1}{S_p^2 |a^{+}|_\infty^{2/(p-2)}}.$ More importantly,$\mathscr{N}_a$is a \emph{natural constraint} for$I_a$, i.e. every critical point of the restriction$\bar{I}_a$of$I_a$to$\mathscr{N}_a$is a nontrivial critical point of$I_a$. The following result was proved in \cite[Proposition 3.2]{FelmerVergara}, and allows us to consider only positive ground states. \begin{proposition} Any weak solution to \eqref{eq:1.1} is strictly positive. \end{proposition} \begin{proposition} \label{prop:3.7} Let$\bar{I}_a$be the restriction of the functional$I_a$to the manifold$\mathscr{N}_a$. Every Palais-Smale sequence at level$c$for$\bar{I}_a$is also a Palais-Smale sequence at level$c$for$I_a$. \end{proposition} \begin{proof} Assume that$\{u_n\}_n \subset \mathscr{N}_a$is a Palais-Smale sequence at level$c$for$\bar{I}_a$, namely $\lim_{n \to +\infty} \bar{I}_a(u_n) =c$ and $\lim_{n \to +\infty} D\bar{I}_a(u_n) =0$ in the norm topology. It suffices to show that the sequence$\{ \nabla I_a(u_n)\}_n$converges to zero in$L^{s,2}(\mathbb{R}^N)$. Let us abbreviate$\psi(u) = D I_a(u)[u]$, so that$\mathscr{N}_a = \psi^{-1}(\{0\}) \setminus \{0\}$. From the fact that$u_n \in \mathscr{N}_a$, we deduce that$I_a(u_n) = (1/2-1/p)\|u_n\|_{L^{s,2}}^2$, and hence the sequence$\{u_n\}_n$is bounded. This implies that \begin{equation} \label{eq:3.4} \sup_n \frac{\| \nabla \psi (u_n) \|_{L^{s,2}}}{\|u_n\|_{L^{s,2}}} < +\infty. \end{equation} Explicitly, we have that, for every$n \in \mathbb{N}$, \begin{equation} \label{eq:3.5} \langle \nabla \psi(u_n)\mid u_n \rangle = (2-p) \|u_n\|_{L^{s,2}}^2 <0 \end{equation} and \begin{equation} \label{eq:3.6} \nabla \bar{I}_a(u_n) = \nabla I_a(u_n) - \frac{\langle \nabla I_a(u_n)\mid \nabla \psi (u_n) \rangle}{\|\nabla \psi(u_n)\|_{L^{s,2}}^2} \nabla \psi(u_n). \end{equation} Observe that$\nabla I_a(u_n) \perp u_n$because$u_n \in \mathscr{N}_a$. If we consider the quantity $\| \nabla \psi (u_n)\|_{L^{s,2}}^2 - \Big( \frac{\langle \nabla I_a(u_n)\mid \nabla \psi(u_n) \rangle}{\|\nabla I_a(u_n)\|_{L^{s,2}}^2} \Big)^2,$ we immediately see that it equals the square of the norm of the projection of the vector$\nabla \psi(u_n)$onto the subspace of$L^{s,2}(\mathbb{R}^N)$orthogonal to the unit vector$\nabla I_a(u_n)/\|\nabla I_a(u_n)\|$. Since this subspace contains in particular the vector$u_n/\|u_n\|_{L^{s,2}}, it follows from the Pythagorean Theorem that \begin{equation} \| \nabla \psi (u_n) \|_{L^{s,2}}^2 - \Big( \frac{\langle \nabla I_a(u_n)\mid \nabla \psi(u_n) \rangle}{\|\nabla I_a(u_n)\|_{L^{s,2}}^2} \Big)^2 \geq \Big( \frac{\langle \nabla \psi(u_n)\mid u_n \rangle}{\|u_n\|_{L^{s,2}}} \Big)^2. \end{equation} This yields, recalling \eqref{eq:3.6}, \eqref{eq:3.5} and \eqref{eq:3.4}, \begin{align*} &\| \nabla \bar{I}_a(u_n) \|_{L^{s,2}} \| \nabla I_a(u_n)\|_{L^{s,2}} \\ &\geq \langle \nabla \bar{I}_a(u_n)\mid \nabla I_a(u_n) \rangle \\ &= \frac{\|\nabla I_a(u_n)\|_{L^{s,2}}^2}{\|\nabla \psi(u_n)\|_{L^{s,2}}^2} \bigg( \| \nabla \psi(u_n) \|_{L^{s,2}}^2 - \Big( \frac{\langle \nabla I_a(u_n)\mid \nabla \psi(u_n)\rangle}{\|\nabla I_a(u_n)\|_{L^{s,2}}^2} \Big)^2 \bigg)^2 \\ &\geq \frac{\|\nabla I_a(u_n)\|_{L^{s,2}}^2}{\|\nabla \psi(u_n)\|_{L^{s,2}}^2} \Big( \frac{\langle \nabla \psi(u_n)\mid u_n \rangle}{\|u_n\|_{L^{s,2}}} \Big)^2 \\ &= \frac{\|\nabla I_a(u_n)\|_{L^{s,2}}^2}{\|\nabla \psi(u_n)\|_{L^{s,2}}^2} (2-p)^2 \|u_n\|_{L^{s,2}}^2 \\ &\geq C \|\nabla I_a(u_n)\|_{L^{s,2}}^2. \end{align*} This argument proves that\lim_{n \to +\infty} \|\nabla I_a(u_n)\|_{L^{s,2}}=0$, and we complete the proof. \end{proof} \section{Splitting and vanishing sequences} The analysis of Palais-Smale sequences can be harder than in the more familiar case of a potential function$a$that has a precise asymptotic behavior at infinity. For this reason, we recall a language taken from \cite{AckerChagoya}. \begin{definition} \rm A map$F \colon X \to Y$between two Banach spaces splits in the BL sense (BL stands for Brezis and Lieb.) if for any sequence$\{u_n\}_n \subset X$such that$u_n \rightharpoonup u$in$X$there results $F(u_n-u) = F(u_n)-F(u)+o(1)$ in the norm topology of~$Y$. \end{definition} \begin{lemma} \label{lem:4.2} Suppose that$\{u_n\}_n \subset L^{s,2}(\mathbb{R}^N)$and$\{y_n\}_n \subset \mathbb{R}^N$are such that$\tau_{-y_n}u_n \rightharpoonup u_0$in$L^{s,2}(\mathbb{R}^N)$. Then $I_{\tau_{-y_n}a}(\tau_{-y_n}u_n) - I_{\tau_{-y_n}a} (\tau_{-y_n}u_n-u_0) -I_{\tau_{-y_n}a}(u_0)=o(1)$ and $DI_{\tau_{-y_n}a}(\tau_{-y_n}u_n) - DI_{\tau_{-y_n}a} (\tau_{-y_n}u_n-u_0) - DI_{\tau_{-y_n}a}(u_0) =o(1).$ \end{lemma} \begin{proof} Since both$F(u)=p^{-1}|u|^p$and$F'(u)=|u|^{p-2}u$split from$L^{s,2}(\mathbb{R}^N)$into$L^1(\mathbb{R}^N), see \cite[Lemma 4.4]{Secchi17}, we can write \begin{align*} &\int_{\mathbb{R}^N} | ( \tau_{-y_n} a ) ( F( \tau_{-y_n}u_n) - F(\tau_{-y_n}u_n-u_0)-F(u_0) ) | \\ & \leq |a|_\infty \int_{\mathbb{R}^N} \left| F( \tau_{-y_n}u_n) - F(\tau_{-y_n}u_n-u_0)-F(u_0) \right| =o(1) \end{align*} and \begin{align*} &\int_{\mathbb{R}^N} \left| \left( \tau_{-y_n} a \right) \left( F'(\tau_{-y_n} u_n) - F'(\tau_{-y_n} u_n-u_0) - F'(u_0) \right) \right|^{p/(p-1)} \\ &\leq |a|_\infty^{p/(p-1)} \int_{\mathbb{R}^N} \left| F'(\tau_{-y_n} u_n) - F'(\tau_{-y_n} u_n-u_0) - F'(u_0) \right|^{p/(p-1)}. \end{align*} Recalling that the squared norm splits in the BL sense, the proof is complete. \end{proof} \begin{definition} \label{def:4.3} \rm A sequence\{u_n\}_n \subset L^{s,2}(\mathbb{R}^N)$vanishes if$\tau_{x_n} u_n \rightharpoonup 0$in$L^{s,2}(\mathbb{R}^N)$for any sequence$\{x_n\}_n$of points in$\mathbb{R}^N$. \end{definition} \begin{remark} \label{rem:4.4} \rm Any vanishing sequence is necessarily bounded in$L^{s,2}(\mathbb{R}^N)$, and by the Rellich-Kondratchev theorem (see \cite[Corollary 7.2]{DiNezza})$\tau_{x_n} u_n \to 0$strongly in$L_{\mathrm{loc}}^2(\mathbb{R}^N)$for every sequence$\{x_n\}_n \subset \mathbb{R}^N$. This yields that, for every$R>0$, $\lim_{n \to +\infty} \sup \Big\{ \int_{B(x,R)} |u_n|^2: x \in \mathbb{R}^N \Big\} =0.$ By the fractional version of Lions' vanishing lemma \cite[Proposition II.4]{Secchi12}, we deduce that$u_n \to 0$strongly in$L^q(\mathbb{R}^N)$for every$20$, independent of$i$, such that$\|u^i\|_{L^{s,2}} \geq C$. For every$j$we also have $0 \leq \|u_{n}^{j+1}\|_{L^{s,2}}^2 = \|u_n\|^2_{L^{s,2}} - \sum_{i=1}^j \|u^i\|_{L^{s,2}}^2 + o(1),$ which implies that the iteration must stop after finitely many steps. Therefore there exists a positive integer$k$such that$\{u_n^{k+1}\}_n$vanishes,$\{u_n^{k+1}\}_n$converges to zero strongly in$L^p(\mathbb{R}^N)$and \eqref{eq:4.7} holds true. Similarly, $-\int_{\mathbb{R}^N} a | u_n^{k+1} |^p \leq I_a(u_n^{k+1}) = I_a(u_n) - \sum_{i=1}^k I_{a^i} (u^i) + o(1),$ and also \eqref{eq:4.8} follows from$c = \lim_{n \to +\infty} I_a(u_n)$. The proof is complete. \end{proof} \section{Existence of a ground state} The proof of the following comparison lemma is probably known, but we reproduce here for the reader's convenience. \begin{lemma} \label{lem:5.1} Suppose that$a_1$,$a_2 \in L^\infty(\mathbb{R}^N)$. If$a_1 \geq a_2$, then$c_{a_1} \leq c_{a_2}$. If, in addition,$a_1 \neq a_2$and$I_{a_2}$possesses a ground state, then$c_{a_1}0. \] We can therefore define \begin{equation} \label{eq:5.1} t = \Big( \frac{\int_{\mathbb{R}^N} a_2 |u|^p}{\int_{\mathbb{R}^N} a_1 |u|^p} \Big)^{1/(p-2)} \leq 1. \end{equation} Then we have $DI_{a_1}(tu)[tu] = t^2 \Big( \|u\|_{L^{s,2}}^2 - t^{p-2} \int_{\mathbb{R}^N} a_1 |u|^p \Big) = t^2 D I_{a_2}(u)[u]=0,$ and hence $tu \in \mathscr{N}_{a_1}$. Since \begin{align*} I_{a_2}(u) &= \frac{1}{2} \|u\|_{L^{s,2}}^2 - \frac{1}{p} \int_{\mathbb{R}^N} a_2 |u|^p = \Big( \frac{1}{2} - \frac{1}{p} \Big) \|u\|_{L^{s,2}}^2 \\ &\geq \Big( \frac{1}{2} - \frac{1}{p} \Big) \|tu\|_{L^{s,2}}^2 = J_{a_1}(u) \geq c_{a_1}, \end{align*} we conclude that $c_{a_2} = \inf_{u \in \mathscr{N}_{a_2}}I_{a_2}(u) \geq c_{a_1}$. Furthermore, if $a_1 \neq a_2$ (as elements of $L^\infty(\mathbb{R}^N)$) and $u$ is a ground state of $I_{a_2}$, then $|u|>0$. In \eqref{eq:5.1} we then have $t<1$, and it follows that $c_{a_2}=I_{a_2}(u) > I_{a_1}(tu) \geq c_{a_1}$. \end{proof} Recall the definition \eqref{eq:a-bar} of $$\bar{a}$$. Then we have the following result. \begin{proposition} \label{prop:5.3} It results \begin{align*} c_{a} < c_{\bar{a}}. \end{align*} \end{proposition} \begin{proof} We first consider (i) of assumption (A1). Since $$\bar{a}\leq 0$$, we have $$c_{\bar{a}}=\infty$$. But $$c_{a} \in \mathbb{R}$$ because $$a^{+} \neq 0$$, and there is nothing more to prove. We can assume that $$\bar{a}>0$$ in the rest of the proof. If (ii) of assumption (A1) holds, recalling that $$\bar{a}>-\infty$$ entails $$\mathscr{B}\neq \emptyset$$ we can conclude that $$a \neq \bar{a}$$. Now Lemma \ref{lem:5.1} implies that $$c_a < c_{\bar{a}}$$, since $$I_{\bar{a}}$$ has a ground state by the arguments of \cite[Theorem 1.1]{Ambrosio}. \end{proof} We are now ready to prove our main existence result. \begin{proof}[Proof of Theorem \ref{th:main}] We have $$\mathscr{N}_{a} \neq \emptyset$$ and $$c_{a}<\infty$$ because $$a^{+} \neq 0$$. From \eqref{eq:3.3} we get $$c_{a}>0$$. An application of Ekeland's Principle yields in a standard way a mimnimizing sequence $$\{u_n\}_n \subset \mathscr{N}_{a}$$ for the functional $$\bar{I}_{a}$$ defined as the restriction of $$I_{a}$$ to $$\mathscr{N}_{a}$$. This sequence is also a (PS)-sequence for $$\bar{I}_{a}$$ at the level $$c_{a}$$. By Proposition \ref{prop:3.7} $$\{u_n\}_n$$ is a (PS)-sequence for $$I_{a}$$ at the level $$c_{a}$$. The strong convergence of $$\{u_n\}_n$$ to zero is easily ruled out, since $$I_{a}(u_n) \to c_{a}>0$$. Proposition \ref{prop:4.8} yields then a number $$k \in \mathbb{N}$$, functions $$a^i \in \overline{\mathscr{A}}$$ and non-trivial critical points $$u^i$$ of $$I_{a^i}$$ such that \begin{align*} c_{a} \geq \sum_{i=1}^k I_{a^i}(u^i). \end{align*} From the knowledge that each $$u^i$$ is a non-trivial critical point of $$I_{a^i}$$ we deduce $$(a^i)^{+}\neq 0$$ for every $$i=1,\ldots,k$$. Again by \eqref{eq:3.3} we get $$I_{a^i}(u^i)>0$$ for every $$i=1,\ldots,k$$. Suppose that for \emph{some} index $$i$$ there results $$a^i \in \mathscr{B}$$. Then $$a^i \leq \bar{a}$$, and Lemma \ref{lem:5.1} together with Proposition \ref{prop:5.3} yield $$I_{a^i}(u^i) \geq c_{a^i} \geq c_{\bar{a}}>c_{a}$$. This is a contrdiction. Therefore each $$a^i$$ is a translation of $$a$$, and $$I_{a^i}(u^i) \geq c_{a}$$ for every $$i=1,\ldots,k$$. This forces $$k=1$$, and a translation of $$u^1$$ is a ground state of $$I_a$$. \end{proof} \section{An example} Assumption (A1) can be rephrased in a more familiar way for continuous bounded potentials. \begin{proposition} For any $a \in L^\infty(\mathbb{R}^N)$, define \begin{align*} \hat{a} = \lim_{R \to +\infty} \operatorname*{ess\,sup}_{x \in \mathbb{R}^N \setminus B(0,R)} a(x). \end{align*} If {\rm (A1)} holds with $\bar{a}$ replaced by $\hat{a}$, then {\rm (A1)} holds with $\bar{a}$. \end{proposition} \begin{proof} If $$\mathscr{B} = \emptyset$$, then $\bar{a}=-\infty$ and (A1) holds. We may assume that $$\mathscr{B} \neq \emptyset$$, so that $$a$$ cannot be constant. Let us prove that \begin{align} \bar{a} \leq \hat{a}. \label{eq:6.1} \end{align} Pick $b \in \mathscr{B}$. There is a sequence $\{x_n\}_n \subset \mathbb{R}^N$ such that $\tau_{x_n}a \rightharpoonup^\star b$. Translations are continuous in the weak$^\star$ topology of $L^\infty(\mathbb{R}^N)$, since they are continuous in $L^1(\mathbb{R}^N)$. For the sake of contradiction, suppose that $\{x_n\}_n$ contains a bounded subsequence. Up to a further subsequence, there must exist a point $\xi\in\mathbb{R}^N$ such that $x_n \to \xi$ and $\tau_{x_n} a \rightharpoonup^\star \tau_{\xi}a$. Since $\mathscr{P}$ is metrizable, $\tau_{\xi} a = b \notin \mathscr{A}$, a contradiction. Therefore $\lim_{n \to +\infty} |x_n|=+\infty$. Let $\varepsilon>0$ be given, and apply Lemma \ref{lem:3.1}: there exists $\varphi \in L^1(\mathbb{R}^N)$ with $\varphi \geq 0$ and $\|\varphi\|_{L^1}=1$ such that $\int_{\mathbb{R}^N} b \varphi \geq \operatorname{ess\,sup}b - \frac{\varepsilon}{2}.$ Choose $\tilde{\psi} \in C_c^\infty(\mathbb{R}^N)$ such that $\tilde{\psi} \geq 0$ and $\| \varphi - \tilde{\psi} \|_{L^1} \leq \frac{\varepsilon}{4 \|b\|_{L^\infty}}.$ Now $\psi = \tilde{\psi}/\|\tilde{\psi}\|_{L^1} \in C_c^\infty(\mathbb{R}^N)$ satisfies $\| \varphi - \psi \|_{L^1} \leq \frac{\varepsilon}{2 \|b\|_{L^\infty}},$ $\psi \geq 0$ and $\|\psi\|_{L^1} = 1$. This implies $\int_{\mathbb{R}^N} b \psi = \int_{\mathbb{R}^N} b \varphi - \int_{\mathbb{R}^N} b (\varphi-\psi) \geq \int_{\mathbb{R}^N} b \varphi - \|b\|_{L^\infty} \|\psi - \varphi \|_{L^1} \geq \operatorname{ess\,sup}b - \varepsilon.$ Suppose that $\operatorname{supp}\psi \subset B(0,R)$: then \begin{align*} \operatorname{ess\,sup}b - \varepsilon &\leq \int_{\mathbb{R}^N} b \psi = \lim_{n \to +\infty} \int_{\mathbb{R}^N} (\tau_{x_n}a)\psi \\ &\leq \lim_{n \to +\infty} \operatorname*{ess\,sup}_{x \in B(-x_n,R)} a(x) \int_{\mathbb{R}^N} \psi \\ &\leq \lim_{n \to +\infty} \operatorname*{ess\,sup}_{x \in \mathbb{R}^N \setminus B(0,|x_n|-R)} a(x) = \hat{a}. \end{align*} Since $\varepsilon>0$ is arbitrary, we conclude that $\operatorname{ess\,sup}b \leq \hat{a}$. If (i) of assumption (A1) holds, then \eqref{eq:6.1} yields $\bar{a} \leq \hat{a} \leq 0$. If (ii) holds, then \eqref{eq:6.1} yields $\bar{a} \leq \hat{a} \leq a$, and the proof is complete. \end{proof} The following corollary is an immediate consequence of Theorem \ref{th:main}. \begin{corollary} If $$a$$ is a bounded continuous function such that either $\limsup_{|x| \to +\infty} a(x) \leq 0$ or $\limsup_{|x| \to +\infty} a(x) \leq a,$ then equation \eqref{eq:1.1} has (at least) a positive ground state as soon as \$2