2. Variational framework and preliminary facts

Since in the present work we focus on (E) in the case $\gamma,b,\inf V > 0$, to simplify the notation, we assume

\[ \gamma = 2\pi,\quad b=1,\ \inf V =1. \]

We use the following notation:

• • $\int =\int _{\mathbb {R}^2}$.

• • $B_r(y)$ denotes the open ball of radius $r>0$ and centre $y\in \mathbb {R}^2$.

• • $L^p$, $1\leq p<+\infty$, and $H^1$ are the usual Lebesgue and Sobolev spaces in $\mathbb {R}^2$, with norm $|u|_{p}=(\int |u|^p{\rm d}x)^{1/p}$ and $\|u\|=(\int (|\nabla u|^2+u^2)\,{\rm d}x)^{1/2}$, respectively.

• • We denote by $C, c_1, c_2,\ldots$ various constants that can also vary from one line to another.

First, in Lemma 2.1, we recall from [Reference Cingolani and Weth10, Reference Stubbe27] that the nonlocal term is well defined and regular in $\widetilde X$. To state this result, let us introduce the following bilinear forms:

\begin{align*} & (u, v) \mapsto B_1(u, v):= \iint \log(1+ |x-y|)u(x)v(y)\, {\rm d}x\,{\rm d}y,\\ & (u, v)\mapsto B_2(u,v):= \iint \log\left(1+\frac{1}{|x-y|}\right)u(x) v(y)\, {\rm d}x\,{\rm d}y,\\ & (u, v)\mapsto B_0(u, v):= B_1(u, v) - B_2(u, v) = \iint \log(|x-y|)u(x) v(y)\, {\rm d}x\, {\rm d}y, \end{align*}

where $u, v:\mathbb {R}^2 \to \mathbb {R}$ are measurable functions such that the integrals are well defined. Correspondingly, let

\[ V_1(u) := B_1(u^2, u^2),\quad V_2(u):= B_2(u^2, u^2),\quad V_0(u):= B_0(u^2, u^2) = V_1(u) - V_2(u). \]

With this notation, the functional related to problem (E) takes the form

\[ I(u) = \frac12 \int (|\nabla u|^2 + V(x)u^2)\ {\rm d}x + \frac14 V_0(u) - \frac 1p |u|_p^p. \]

Lemma 2.1 The functional $I$ is a well-defined $C^2$ functional on the Hilbert space X. Moreover, $V_2$ is of class $C^1$ on $L^\frac 83 (\mathbb {R}^2)$ and the following formulas hold

(2.1)\begin{equation} V'_i(u)[v] = 4B_i(u^2, uv), \quad \forall u, v \in X,\ i=0, 1, 2. \end{equation}

In particular,

\[ V_i'(u)[u]=4V_i(u), \quad \forall u \in X,\ i=0, 1, 2. \]

For the proof, see [Reference Cingolani and Weth10, Lemma 2.2] and [Reference Sardilli26, Lemma 3.1.5].

Moreover, let us recall the Hardy–Littlewood–Sobolev inequality (see [Reference Lieb and Loss18, Theorem 4.3]):

(2.2)\begin{equation} \left|\iint_{\mathbb{R}^{N}\times \mathbb{R}^N} \displaystyle\frac {f(x) h(y)}{|x-y|^\lambda}\, {\rm d}x{\rm d}y\right|\leq C(\lambda, p )|f|_p|h|_{r}, \quad \forall f \in L^{p}(\mathbb{R}^N),\ \forall g \in L^r(\mathbb{R}^N), \end{equation}

where $p,r>1$ and $0 <\lambda < N$ are such that $\frac {1}{p} + \frac {\lambda }{N} + \frac {1}{r} = 2$. Then, by (2.2) and $\log (1+r)\le r$, $\forall r>0$, it is readily seen that

(2.3)\begin{equation} V_2(u)\le \iint \displaystyle\frac{u^2(x)u^2(y)}{|x-y|} \, {\rm d}x\, {\rm d}y\le c\, |u^2|_{4/3} |u^2|_{4/3}=c|u|^4_{8/3}. \end{equation}

Next, we introduce the Cerami compactness condition.

Definition 2.2 Let $X$ be a Banach space, $I\in C^1(X)$ and $c \in \mathbb {R}$. A sequence $\{x_n\}_n$ is called a $(C)$ sequence for $I$ at the level $c$ if

(C)\begin{equation} {I(x_n) \to c\ \text{as}\ n\to\infty\ \text{and} \ } {\|I'(x_n)\|_{\mathcal{L}}(1 + \|x_n\|_X) \to 0,\ \text{as}\ n\to\infty.} \end{equation}

We say that $I$ satisfies the Palais–Smale–Cerami condition (at the level $c$), $(C)$ condition for short, if any $(C)$ sequence (at the level $c$) possesses a converging subsequence.

To analyse the compactness condition, we report some known facts: Lemma 2.3 is Lemma 2.6 in [Reference Cingolani and Weth10] and states a continuity property of the bilinear form $B_1$, Lemma 2.4 provides the compactness of the embedding of $X$ in the Lebesgue spaces.

Since bounded sets in $X$ are also bounded sets in $\widetilde X$ (see (1.2), (1.3)), to verify Lemma 2.4 it is sufficient to exploit the coercivity of the map $|x|\mapsto \log (1+|x|)$ and apply, for example, Theorem XIII.65 in [Reference Reed and Simon24] (see also [Reference Rabinowitz23]).

Finally, let us state the topological tools we will use in the proof of Theorem 1.1.

Lemma 2.5 see [Reference Li and Wang17, Lemma 2.6]

Let $X$ be a Banach space, $I \in C^{1}(X),$ $c \in \mathbb {R},$ $\varepsilon >0$ and suppose that there exists $\alpha >0$ such that

(2.4)\begin{equation} (1 + \|x\|_X) \|I'(x)\|_{\mathcal{L}} \geq \alpha > 0 \quad \forall x \in I^{c+2\epsilon}_{c-2\epsilon}. \end{equation}

Then there exists a continuous map $\eta :X \to X$, such that:

1. 1. $\eta (I^{c+\epsilon }) \subset (I^{c-\epsilon });$

2. 2. $\eta (x) = x,$ for all $x \not \in I^{c+2\epsilon }_{c-2\epsilon },$

where $I^a=\{x\in X\ :\ I(x)\le a\}$ and $I^b_a=\{x\in X\ :\ a\le I(x)\le b\}$, for all $a< b$.

From Lemma 2.5 there follows the following variant of the Mountain Pass Theorem.

Theorem 2.6 Let $I \in C^{1}(X)$, X a Banach space. We suppose that $I(0) = 0$ and there exists a closed subspace $S \subset X$ that disconnects $X$ in two path-connected components $X_1$ and $X_2$. We assume that $0 \in X_1$ and there exists $A>0$, $\bar u \in X_2$ such that

• • $I\big |_S\geq A;$

• • $I(\bar u)\leq 0$.

Set

(2.5)\begin{equation} c = \inf_{\gamma \in \Gamma} \max_{t \in [0,1]} I(\gamma(t)) \end{equation}

where $\Gamma :=\{\gamma : [0,1] \to X: \gamma\mbox{ is continuous,}\ \gamma (0)=0,\ \gamma (1) \in X_2,\ I(\gamma (1))\leq 0 \}$. If $I$ verifies the (C) condition at $c,$ then $c$ is a critical value for $I$.

The proof of Theorem 2.6 is standard so we give here only a sketch of it, for the sake of completeness.

If, by contradiction, $c$ is not a critical value for $I$, then (2.4) has to be verified for suitable $\varepsilon,\alpha >0$, because $I$ verifies the (C) condition at the level $c$. We can choose $\varepsilon < A$. Now, let $\gamma _\varepsilon \in \Gamma$ be such that $\max _{[0,1]}I(\gamma _\varepsilon (t))\le c+\varepsilon$. If we consider $\tilde \gamma _\varepsilon :=\eta \circ \gamma _\varepsilon$, where $\eta$ is the deformation provided by Lemma 2.5, it turns out that

\[ \tilde\gamma_\varepsilon\in\Gamma,\quad \max_{[0,1]}I(\tilde \gamma_\varepsilon(t))\le c-\varepsilon, \]

contrary to the definition of $c$.

3. Compactness

This section is devoted to prove that the (C) condition holds at positive values, that is the range where we are looking for critical levels:

Indeed, by $p\ge 4$ and assumption (V)$(b)$,

\begin{align*} I(u) & =I(u) -\displaystyle\frac{1}{4}I'(u)[u] = \frac{1}{4}\displaystyle\int (|\nabla u|^2{\rm d}x + V(x)u^2)\, {\rm d}x +\left(\frac 14- \frac{1}{p}\right)|u|^p_p \\ & \geq\displaystyle\frac{1}{4} \displaystyle\int ( |\nabla u|^2 + V(x)u^2)\,{\rm d}x \geq C\|u\|^2>0. \end{align*}

A key point to prove the proposition is the following lemma that states the boundedness of the $(C)$ sequences at the positive levels.

Lemma 3.3 Under the assumptions of Proposition 3.2, the sequence $\{u_n\}_n$ is bounded in $X$. Moreover, there exist a ball $B_2(\bar x)$ and a constant $d>0$ such that, up to a subsequence,

(3.1)\begin{equation} \displaystyle\int_{B_2(\bar x) }u_n^2(x)\, {\rm d}x \ge d \quad \forall n\in\mathbb{N}. \end{equation}

Proof. Since $\{u_n\}_n$ is a (C) sequence, it turns out that

(3.2)\begin{equation} I'(u_n)[u_n] = o(1) \end{equation}

and so we have

\begin{align*} c+o(1) & =I(u_n) -\displaystyle\frac{1}{4}I'(u_n)[u_n] \\ & =\displaystyle\frac{1}{4} \displaystyle\int ( |\nabla u_n|^2 + V(x)u_n^2)\,{\rm d}x +\left(\frac 14-\frac 1p\right)\, |u_n|_p^p\\ & \geq C \|u_n\|^2. \end{align*}

Therefore, $\{u_n\}_n$ is bounded in $H^1(\mathbb {R}^2)$, hence in $L^q(\mathbb {R}^2)$ for all $q\in [2, +\infty )$, and

(3.3)\begin{equation} \displaystyle\int (|\nabla u_n|^2 + V(x)u_n^2) \,{\rm d}x\le C\quad\forall n\in\mathbb{N}. \end{equation}

So, we are left to prove that

\[ |u_n|^2_{{\ast}}:=\int \log(1 + |y|)u^2_n(x)\, {\rm d}x< C,\quad \forall n\in\mathbb{N}. \]

Let us define

\[ \delta := \limsup_{n\to\infty} \left(\max_{y\in\mathbb{R}^2} \int_{B_1(y)}|u_n|^2{\rm d}x\right). \]

We claim that $\delta >0$. Assume, by contradiction, that $\delta = 0$. Then by [Reference Lions19, Lemma I.1] there follows

(3.4)\begin{equation} u_n \to 0\text{ in }\ L^{q}(\mathbb{R}^2),\quad\forall q\in (2,+\infty) \end{equation}

and so, by (2.3), we get

(3.5)\begin{equation} V_2(u_n) \to 0 \quad \text{ as }\quad n\to+\infty. \end{equation}

From (3.2), (3.4) and (3.5), we infer

\[ \int \left(|\nabla u_n|^2 + V(x)u_n^2\right){\rm d}x + V_1(u_n) = I'(u_n)[u_n] + V_2(u_n) + |u_n|^p_p=o(1), \]

that, by the positivity of $V_1$, implies

(3.6)\begin{equation} \displaystyle\int (|\nabla u_n|^2 + V(x)u_n^2)\,{\rm d}x=o(1)\quad \mbox{ and }\quad V_1(u_n)=o(1). \end{equation}

Summarizing, from (3.4), (3.5) and (3.6), we infer

\[ I(u_n) = \frac{1}{2}\int (|\nabla u_n|^2 + V(x)u^2_n)\,{\rm d}x + \frac{1}{4}V_1(u_n) - \frac{1}{4} V_2(u_n) - \frac{1}{p}|u_n|^p_p =o(1), \]

contrary to $I(u_n)\to c>0$, and the claim follows.

Then, $\delta > 0$ and there exists a sequence $\{x_n\}_n \subset \mathbb {R}^2$ such that

(3.7)\begin{equation} \displaystyle\int_{B_1(x_n)} u^2_n(x)\, {\rm d}x > \displaystyle\frac{\delta}{2}, \end{equation}

for n large enough.

The task is now to prove that $\{x_n\}$ is bounded. Assume by contradiction that $|x_n| \to \infty$, up to a subsequence.

Let us fix an arbitrary $M>0$ and call

\[ A_n := \{x \in B_1(x_n)\ : \ V(x)\leq M\}, \quad B_n := \{x \in B_1(x_n)\ : \ V(x)>M\}. \]

By (V) $(c)$ and $|x_n| \to \infty$, we obtain

(3.8)\begin{equation} \left|A_n\right| \leq \left|\{x \in \mathbb{R}^2\ :\ |x|\geq|x_n|-2, \ V(x)\leq M\}\right| \mathop{\longrightarrow} 0,\quad\mbox{ as }n\to\infty. \end{equation}

Observe that from (3.8) we infer, for any fixed $q>2$,

\[ \int_{A_n}u_n^2(x)\, {\rm d}x\le |u_n|_q^2|A_n|^{1-\frac 2q}=o(1). \]

As a consequence

\[ \int_{B_n}u_n^2(x)\, {\rm d}x =\int_{B_1(x_n)}u_n^2(x)\,{\rm d}x-\int_{A_n}u_n^2(x)\,{\rm d}x\ge \frac\delta 2+o(1). \]

Then, taking into account (2.3), we have

(3.9)\begin{equation} \begin{aligned} c+o(1)= I(u_n) & \geq \displaystyle\frac{1}{2} \displaystyle\int_{B_n}V(x)u_n^2{\rm d}x - \frac{1}{4}V_2(u_n) - C_1\frac{1}{p}\|u_n\|^p \\ & \geq \frac\delta 4\, M - C_2, \end{aligned} \end{equation}

where $C_2$ is a constant independent of $M$. Letting $M\to \infty$ in (3.9), we get a contradiction, hence $\{x_n\}$ has to be bounded.

Therefore, there exists $\bar {x}\in \mathbb {R}^2$ such that $x_n\to \bar {x}$, up to a subsequence. Taking into account (3.7), we can also assume that

(3.10)\begin{equation} \displaystyle\int_{B_2(\bar{x})}u_n^2{\rm d}x \geq \displaystyle\frac{\delta}{2} > 0, \end{equation}

that proves (3.1).

Now, observe that for every $R>0$

\[ 1 + |x - y|\geq 1 +\frac{|y|}{2}\geq \sqrt{1 + |y|} \quad \forall y\in \mathbb{R}^2\setminus B_{2R}(0),\ \forall x\in B_R(0). \]

Then, fixing $\bar R>|\bar x|+2$ and taking into account (3.10), we get

(3.11)\begin{equation} \begin{aligned} V_1(u_n) =B_1(u_n^2,u_n^2) & \ge \displaystyle\int_{\mathbb{R}^2\setminus B_{2\bar R}(0)}\left(\int_{B_{\bar R}(0)} \log(1 + |x - y|)u^2_n(x)\, {\rm d}x\right)u^2_n(y)\, {\rm d}y \\ & \geq \left(\displaystyle\int_{B_2(\bar x)}u_n^2(x){\rm d}x\right)\cdot \left(\int_{\mathbb{R}^2\setminus B_{2\bar R}(0)} \log\left(1 + \displaystyle\frac{|y|}{2}\right)u^2_n(y)\, {\rm d}y\right) \\ & \geq \frac{\delta}{4}\int_{\mathbb{R}^2\setminus B_{2\bar R}(0)} \log(1 + |y|)u^2_n(y)\, {\rm d}y\\ & = \frac{\delta }{4}\left(|u_n|^2_{{\ast}} - \int_{B_{2\bar R}(0)} \log(1 + |y|)\, u^2_n(y)\, {\rm d}y\right) \\ & \geq \frac{\delta}{4}(|u_n|^2_{{\ast}} - \log (1+2\bar R) |u_n|^2_2). \end{aligned} \end{equation}

Thus, we conclude that

(3.12)\begin{equation} |u_n|^2_{{\ast}} \leq \displaystyle\frac{4}{\delta}V_1(u_n) + C |u_n|^2_2. \end{equation}

Since $\{|u_n|_2\}_n$ is bounded, if we prove that $\{V_1(u_n)\}_n$ is bounded, we are done.

Observe that

(3.13)\begin{equation} {-}\displaystyle\frac{1}{4}V_1(u_n) + \frac{1}{4} V_2(u_n) + \left(\frac{1}{2} - \frac{1}{p}\right)|u|^p_p = I(u_n) -\frac{1}{2}I'(u_n)[u_n] = c + o(1), \end{equation}

from which we infer, taking into account (2.3),

(3.14)\begin{equation} \displaystyle\frac{1}{4}V_1(u_n) = \frac{1}{4}V_2(u_n) + \left(\frac{1}{2} - \frac{1}{p}\right)|u_n|^p_p -c +o(1) \leq C|u_n|^4_{ {8}/{3}} + C|u_n|^p_p -c+o(1)\leq C, \end{equation}

that is the desired result.

Proof of Proposition 3.2. By Lemma 3.3, $\{u_n\}_n$ is bounded in the Hilbert space $X$, so, taking also into account Lemma 2.4, there exists $\bar u\in X$ such that, up to a subsequence,

(3.15)\begin{equation} u_n \to \bar{u}\text{ weakly in } X,\mbox{ in }\ L^s(\mathbb{R}^2), \text{ for every }s\in[2,+\infty), \mbox{ and a.e. in }\mathbb{R}^2. \end{equation}

Clearly, the sequence $\{u_n\}$ is bounded also in the weighted Hilbert space

\[ H_V:=\left\{u\in H^1(\mathbb{R}^2)\ :\ \int V(x)u^2(x)\, {\rm d}x<\infty\right\},\quad\|u\|_V^2:=\int(|\nabla u|^2+V(x)u^2)\, {\rm d}x. \]

Hence, we can assume that $\{u_n\}$ weakly converges in $H_V$ to a function $\tilde u\in H_V$. Observe that $\tilde u=\bar u$. Indeed for every fixed $\varphi \in {\mathcal {C}}^\infty _0(\mathbb {R}^2)$ the map

\[ u\mapsto \int u\varphi\, {\rm d}x \]

is a continuous linear form on $X$ and on $H_V$, so that

\[ \lim_{n\to\infty}\int u_n\varphi\, {\rm d}x=\int\bar u\varphi\, {\rm d}x \quad\mbox{ and }\quad \lim_{n\to\infty}\int u_n\varphi\, {\rm d}x=\int\tilde u\varphi\, {\rm d}x. \]

Then we can conclude that

\[ \int (\bar u-\tilde u)\varphi\, {\rm d}x =0\quad\forall \varphi\in{\mathcal{C}}^\infty_0(\mathbb{R}^2) \]

and the assertion follows by the fundamental lemma of Calculus of Variations (see e.g. [Reference Brezis5, Corollary 4.24]).

Now, from the boundedness in $X$ of the (C) sequence $\{u_n\}_n$, we get

(3.16)\begin{equation} \begin{aligned} o(1) & = I'(u_n)[u_n - \bar{u}]= \displaystyle\int (|\nabla u_n|^2 + V(x)u_n^2)\, {\rm d}x - \int ( \nabla u_n \cdot \nabla \bar{u}+ V(x)u_n\bar{u})\,{\rm d}x \\ & \quad+ V'_1(u_n)(u_n - \bar{u}) - V'_2(u_n)(u_n - \bar{u}) - \displaystyle\int |u_n|^{p-2}u_n(u_n - \bar{u})\, {\rm d}x.\\ \end{aligned} \end{equation}

The weak convergence in $H_V$ yields

(3.17)\begin{equation} \liminf_{n\to\infty} \displaystyle\int (|\nabla u_n|^2 + V(x)u_n^2)\, {\rm d}x \geq \int (|\nabla \bar{u}|^2 + V(x)\bar{u}^2)\, {\rm d}x. \end{equation}

By Lemmas 2.1 and 2.3,

(3.18)\begin{equation} \begin{aligned} V'_1(u_n)[u_n - \bar{u}] & = 4B_1(u_n^2, u_n(u_n - \bar{u})) = 4B_1(u_n^2, (u_n - \bar{u})^2) + 4B_1(u_n^2, \bar{u}(u_n - \bar{u}))\\ & =4B_1(u_n^2, (u_n - \bar{u})^2)+o(1). \end{aligned} \end{equation}

By (2.2) and (3.15)

(3.19)\begin{equation} \begin{aligned} |V'_2(u_n)[u_n - \bar{u}]| & = |4B_2(u_n^2, u_n(u_n - \bar{u}))| \\ & = 4\left| \iint \log\left(1 + \displaystyle\frac{1}{|x - y|}\right)u^2_n(x)u_n(y)(u_n - \bar{u})(y)\,{\rm d}x\,{\rm d}y\right|\\ & \leq 4\iint \frac{u_n^2(x)\,|u_n(y)|\, |(u_n - \bar{u})(y)|}{|x-y|}\, {\rm d}x\,{\rm d}y \\ & \leq 4 |u_n|^3_{\frac{8}{3}}|u_n - \bar{u}|_{\frac{8}{3}}=o(1). \end{aligned} \end{equation}

By (3.15),

(3.20)\begin{equation} \left|\displaystyle\int |u_n|^{p-2}u_n(u_n - \bar{u})\, {\rm d}x\right|\le |u_n|_p^{p-1}|u_n-u|_p=o(1). \end{equation}

From (3.16), taking into account (3.17), (3.18), (3.19) and (3.20), we infer

\begin{align*} o(1) & = \int \left(|\nabla u_n|^2 + V(x)u_n^2\right){\rm d}x - \int \left(|\nabla \bar u |^2 + V(x)\bar u^2\right){\rm d}x + 4B_1(u_n^2, (u_n - \bar{u})^2) + o(1)\\ & \geq o(1). \end{align*}

Thus, it follows that

(3.21)\begin{equation} \displaystyle\int(|\nabla u_n|^2 + V(x)u_n^2)\,{\rm d}x \to \int(|\nabla \bar{u}|^2 + V(x)\bar{u}^2)\,{\rm d}x, \end{equation}

and

(3.22)\begin{equation} B_1(u_n^2, (u_n - \bar{u})^2) \to 0, \text{ as } n\to+\infty. \end{equation}

Taking into account (3.1) and arguing exactly as in (3.11), we get

(3.23)\begin{equation} B_1(u_n^2, (u_n - \bar{u})^2) \ge\displaystyle\frac{d}{2}(|u_n - \bar{u}|^2_*-C|u_n - \bar{u}|_2^2). \end{equation}

From (3.23), (3.22) and $u_n\to \bar u$ in $L^2$, there follows

\[ |u_n-\bar u|_*\to 0, \]

that, together with (3.21), gives $\|u_n-\bar u\|_X\to 0$.

4. Proof of Theorem 1.1

To prove Theorem 1.1, we are going to apply Theorem 2.6. Since condition $(C)$ holds on $(0,+\infty )$ by Proposition 3.2, we are left to show that also the geometric conditions hold.

Observe that, by (2.3) and by the Sobolev inequality, for every $u \in X \setminus \{0\}$

(4.1)\begin{equation} \begin{aligned} I(u) & \ge \displaystyle\frac{1}{2} \displaystyle\int_{\mathbb{R}^2} (|\nabla u|^2 + V(x)u^2)\, {\rm d}x - \frac{1}{4}\iint \log\left(1 + \frac{1}{|x-y|}\right)u^2(x)u^2(y)\, {\rm d}x\,{\rm d}y - \frac{1}{p}|u|^p_p\\ & \geq c_1 \|u\|^2 - c_2|u|^4_{\frac{8}{3}} - c_3\|u\|^p \\ & \geq c_1\|u\|^2 - c_4\|u\|^4 - c_3\|u\|^p. \end{aligned} \end{equation}

Since $p\ge 4$, there exist $A >0$ and $\bar {\rho }>0$ such that $I_{|_{S }}\geq A$, where $S :=\{u \in X: \|u\| = \bar \rho \}$. The closed set $S$ disconnects $X$ in the two path-connected components

\[ X_1:=\{u \in X: \|u\|<\bar{\rho} \}\quad\mbox{ and }\quad X_2:= \{u \in X: \|u\|>\bar{\rho} \}. \]

Clearly, $0 \in X_1$ and $I(0)=0$.

Now, we are going to show that there exists $\bar u\in X_2$ such that $I(\bar u)<0$. If $p>4$, let us fix $\tilde u\in X\setminus \{0\}$ and observe that $\lim \limits _{t\to \infty }I(t\tilde u)=-\infty$, so the claim follows.

If $p=4$, then let us fix $\tilde u\in {\mathcal {C}} (\mathbb {R}^2)$ with compact support and define $u_t:=t^2\tilde u(t\,\cdot )$, for $t>0$. We can assume that $0$ is a Lebesgue point for $V$, then, by standard computations,

(4.2)\begin{equation} \begin{aligned} I(u_t) & =\left[\displaystyle\frac 12 |\nabla \tilde u|_2^2\right]\, t^4+\left[\frac {V(0)+o(1)}{2}\, |\tilde u|_2^2\right]\, t^2 \\ & \quad+\left[\frac 14 \iint \log(|x-y|)\tilde u^2(x)\tilde u^2(y)\,{\rm d}x\,{\rm d}y\right]\, t^4\\ & \quad-\left[\frac 14 |\tilde u|_2^4\right]\, t^4\log t -\left[\frac 14 |\tilde u|_4^4\right]\, t^6, \end{aligned} \end{equation}

where $o(1)\to 0$ as $t\to \infty$. Since (4.2) implies $\lim \limits _{t\to \infty }I( u_t)=-\infty$, and $\|u_t\|\to \infty$ as $t\to \infty$, the existence of $\bar u\in X_2$ such that $I(\bar u)<0$ follows.

Then, the value

(4.3)\begin{equation} c:=\inf_{\gamma \in \Gamma}\max_{t\in[0,1]} I(\gamma(t)), \end{equation}

where $\Gamma := \{\gamma :[0,1] \to X: \gamma\mbox{ is continuous,}\ \gamma (0)=0, \gamma (1) \in X_2, I(\gamma (1))\leq 0 \}$, is a critical value for $I$, according to Theorem 2.6, with $c\ge A>0$.

Now, let $p>4$. To verify that $c$ is a ground state level, let us fix any nontrivial solution $\tilde u$ and consider the path $t\mapsto t\tilde u$. A direct computation shows that the real function $t\mapsto I( t\tilde u)$ has a unique critical point $t_{\tilde u}$, that corresponds to a maximum. From $\frac {d\,}{dt}I(t\tilde u)=I'(t\tilde u)[\tilde u]$ we infer $t_{\tilde u}=1$. Then, taking into account that $\lim \limits _{t\to \infty }I(t\tilde u)=-\infty$ and (4.3), it is readily seen that

\[ c\le \max_{t\ge 0}I(t\tilde u)=I(\tilde u), \]

that is our claim.

Our next goal is to show that there exists a constant sign solution. Let $u$ be a critical point for $I$ at the level $c$ and consider the path $t\mapsto t|u|$. Since the functional is even, $c=\max _{t\ge 0}I(t|u|)=I(|u|)$ from which we infer that $\bar u:=|u|$ is a critical point for $I$, too. So, $\bar u$ is the weak solution we are looking for.

Now, we consider the regularity issue. For classical regularity results, we refer the reader to [Reference Lieb and Loss18, §10] (in particular Theorems 10.2 and 10.3), or to [Reference Gilbarg and Trudinger16, §8].

Observe that

(4.4)\begin{equation} -\Delta \bar u=\varphi(x)\quad\mbox{ for }\varphi(x)={-}V(x)\bar u+\phi_u(x)\bar u+\bar u^{p-1}, \end{equation}

(see (1.1)).

Assume $V\in L^q_{\mathop {\rm loc}\nolimits }(\mathbb {R}^2)$ for some $q>1$. Since $\bar u\in L^r(\mathbb {R}^2)$ for every $r\in [2,\infty )$, and $\phi _u\in {\mathcal {C}}^3(\mathbb {R}^2)$ (see [Reference Cingolani and Weth10, Proposition 2.3]), we get $\varphi \in L^s_{\mathop {\rm loc}\nolimits }(\mathbb {R}^2)$ for every $s\in [1,q)$. So we are in position to apply classical regularity results and obtain $\bar u\in {\mathcal {C}}^{0,\alpha }_{\mathop {\rm loc}\nolimits }(\mathbb {R}^2)$. Moreover, we can also write

\begin{align*} -\Delta \bar u=\psi(x)\bar u\quad\mbox{ for }\psi(x)={-}V(x) +\phi_u(x) +\bar u^{p-2} \end{align*}

and then, since $\psi \in L^q_{\mathop {\rm loc}\nolimits }(\mathbb {R}^2)$, by the Harnack inequality, we can conclude $\bar u>0$ (see [Reference Pucci and Serrin22, Theorem 7.2.1]).

If $V\in L^q_{\mathop {\rm loc}\nolimits }(\mathbb {R}^2)$ for some $q>2$, then we can argue as before and conclude $\varphi \in L^s_{\mathop {\rm loc}\nolimits }(\mathbb {R}^2)$ for every $s\in [1,q)$, so that $\bar u\in {\mathcal {C}}_{\mathop {\rm loc}\nolimits }^{1,\alpha }(\mathbb {R}^2)$.

If $V\in {\mathcal {C}}^{0,\alpha }_{\mathop {\rm loc}\nolimits }(\mathbb {R}^2)$, then the RHS in (4.4) is locally Hölder continuous, so $u\in {\mathcal {C}}^2(\mathbb {R}^2)$ by standard elliptic regularity (see, for example, [Reference Lieb and Loss18, §10]).