Proof. We start proving (3.1). If it is false, there exist $\delta \in (0,1)$, a sequence $\mu_n \to 0$ and a sequence $(u_n) \subset S_{\mu_n}$ such that, for every n,

(3.3)\begin{equation} \frac{E(u_n,\mathcal G)}{E(\phi_{\mu_n,1},{\mathbb R})} \le 1-\delta. \end{equation}

Each u_n solves (1.1) for some $\lambda_n \geq 0$. In view of lemma 2.5(d), we have that $\lambda_n \to +\infty$ as $n \to \infty$. To proceed, consider a point q_n in some edge e_n where u_n attains a local maximum. Since u_n solves (1.1), it must be $u_n(q_n) \ge \lambda_n^{\frac{1}{p-2}}$. If equality holds, u_n is constant on e_n by uniqueness in the Cauchy problem, and e_n is a bounded edge. But then

\begin{equation*} \mu_n = \int_\mathcal G |u_n|^2{\,dx} \ge \int_{e_n} |u_n|^2{\,dx} = \lambda_n^{\frac{2}{p-2}}|e_n| \ge \lambda_n^{\frac{2}{p-2}}|e_0|, \end{equation*}

which is impossible since $\lambda_n \to +\infty$ and $\mu_n \to 0$. Therefore, u_n is strictly larger than $\lambda_n^{\frac{1}{p-2}}$ at all its local maximum points, which are, therefore, non-degenerate. Since $u_n \in H_{\mu_n}^1(\mathcal G)$, outside a compact set $K_n \subset \mathcal G$, u_n is smaller than some number, say 1 (certainly smaller than $\lambda_n^{\frac{1}{p-2}})$, and thus all local maximum points of u_n are in K_n. Since by the preceding argument, they are also isolated, it then follows that each u_n can have at most a finite number of maximum points.

Since $\mu_n \to 0$, we can also assume that $\min_e u_n \lt 1$ for every $e \in \mathbb{E}_\mathcal G$ and every n. We then set

\begin{equation*} A_n = \{x \in \mathcal G \mid u_n(x) \gt 1\} \end{equation*}

and we note that no local minimum point of u_n can be in A_n. Each A_n is the disjoint union of a finite number of connected open subsets $B_1^n,\ldots, B_{k_n}^n$ of $\mathcal G$ such that

1. (i) $|B_i^n| \to 0$ for every $i=1,\ldots,k_n$ as $n\to \infty$;

2. (ii) each $B_i^n$ contains at least one local maximum point of u_n;

3. (iii) each $B_i^n$ contains at most one vertex of $\mathcal G$;

4. (iv) if $B_i^n$ contains more than one local maximum point of u_n, then it contains exactly one vertex of $\mathcal G$.

Indeed, (i) is obvious as $u_n \ge 1$ on each $B_i^n$ and $\mu_n \to 0$. Moreover, since each $B_i^n$ is non-empty, bounded, contained in A_n and u = 1 on $\partial B_i^n$, it contains at least one local maximum point of u_n, i.e., property (ii) holds. Property (iii) follows by the fact that, if some $B_i^n$ contains at least two vertices of $\mathcal G$, then, being connected, it contains at least a whole edge of $\mathcal G$, which is impossible because the minimum of u_n on each edge of $\mathcal G$ is strictly smaller than 1. Finally, to prove (iv), observe that, if two local maximum points of u_n belong to the same edge of $\mathcal G$, then the fact that u_n solves (1.1) implies that u_n has a local minimum point halfway between the two local maximum points. Moreover, by phase plane analysis, it is easy to see that the value attained by u_n at this local minimum point coincides with the minimum value of u_n on the whole edge. Hence, u_n is smaller than or equal to 1 at least at one point between two consecutive local maximum points belonging to the same edge, in turn implying that these local maximum points do not belong to the same $B_i^n$. Therefore, if $B_i^n$ contains at least two local maximum points, these must belong to different edges. By connectedness, then, $B_i^n$ must contain a vertex, and by (iii) this vertex is unique.

Now, since $0 \lt u_n\le 1$ on $\mathcal G\setminus A_n$, it follows that

(3.4)\begin{equation} E(u_n,\mathcal G\setminus A_n) \ge -\frac1p \|u_n\|_{L^p(\mathcal G\setminus A_n)}^p \ge - \frac1p \|u_n\|_{L^2(\mathcal G\setminus A_n)}^2 \ge -\frac1p \mu_n = o(1) \end{equation}

as $n \to \infty$.

For every n, we denote shortly by B_n a set $B_{i_n}^n$ such that

(3.5)\begin{equation} E(u_n, B_{i_n}^n) = \min_i E(u_n, B_i^n), \end{equation}

which exists since for every n the number of $B_i^n$’s is finite. Up to subsequences, we can assume that either every B_n contains exactly one vertex $\textrm{v}_n$ or that no B_n contains a vertex, and we treat the two cases separately.

Let us first suppose that B_n contains no vertex of $\mathcal G$, for every n. In this case, for each n, $B_i^n$ is an interval $(-a_n,a_n)$ contained in an edge e_n and the restriction of u_n to $[-a_n,a_n]$ is strictly positive, symmetric with respect to 0 and decreasing, say, on $[0,a_n]$. It satisfies

\begin{equation*} \begin{cases} u_n''+u_n^{p-1}=\lambda_n u_n & \text{on }[-a_n,a_n]\,,\\ u_n(0)= \|u_n\|_{L^\infty(-a_n,a_n)} \gt \lambda_n^{\frac{1}{p-2}},& \\ u_n(\pm a_n) =1 & \end{cases} \end{equation*}

and

\begin{equation*} \frac12 |u_n'(x)|^2 + \frac1p |u_n(x)|^p - \frac{\lambda_n}2 |u_n(x)|^2 = C_n \qquad\text{for every } x \in [-a_n,a_n] \end{equation*}

for some constant C_n. By lemma 2.6, we see that, as $n \to \infty$,

(3.6)\begin{equation} C_n = \frac12 |u_n(0)|^2\left(\frac2p |u_n(0)|^{p-2} -\lambda _n \right) \le \frac12 |u_n(0)|^2 \left(\lambda_n + \frac{\pi^2}{e_o^2} -\lambda _n \right) = O\left(\lambda_n^{\frac{2}{p-2}}\right). \end{equation}

We now set

\begin{equation*} v_n(x):=\lambda_n^{-\frac1{p-2}}u_n\left(x/\sqrt{\lambda_n}\right), \end{equation*}

so that v_n is a strictly positive function defined on the interval $\left[-\sqrt{\lambda_n}a_n,\sqrt{\lambda_n}a_n\right]$, symmetric with respect to 0 and decreasing on $\left[0,\sqrt{\lambda_n}a_n\right]$. It satisfies

(3.7)\begin{equation} \begin{cases} v_n''+v_n^{p-1}=v_n & \text{on }\left[-\sqrt{\lambda_n}a_n,\sqrt{\lambda_n}a_n\right]\,,\\ v_n(0)= \|v_n\|_{L^\infty\left(-\sqrt{\lambda_n} a_n,\sqrt{\lambda_n} a_n\right)}\ge 1, &\\ v_n\left(\pm \sqrt{\lambda_n} a_n\right) =\lambda_n^{-\frac1{p-2}} & \end{cases} \end{equation}

and

\begin{equation*} \frac12 |v_n'(x)|^2 + \frac1p |v_n(x)|^p - \frac12 |v_n(x)|^2 = \lambda_n^{-\frac{p}{p-2}} C_n \qquad\text{for every } x \in \left[-\sqrt{\lambda_n}a_n,\sqrt{\lambda_n}a_n\right]. \end{equation*}

Then, recalling (3.6), we deduce that, as $n\to \infty$,

\begin{align*} \left|v'\left(\pm \sqrt{\lambda_n}a_n\right)\right|^2 &= 2\lambda_n^{-\frac{p}{p-2}} C_n + \left|v\left(\pm \sqrt{\lambda_n}a_n\right)\right|^2 - \frac2p \left|v\left(\pm \sqrt{\lambda_n}a_n\right)\right|^p \\ &= O\left(\lambda_n^{-1}\right) + o(1) = o(1). \end{align*}

Summing up, v_n satisfies (3.7) and $\left(v_n(\pm(\sqrt{\lambda_n}a_n)),v_n'(\pm\sqrt{\lambda_n}a_n)\right) \to (0,0)$ as $n\to+\infty$. By continuity in the phase plane, it then follows that $\sqrt{\lambda_n}a_n \to \infty$ and that

(3.8)\begin{equation} E\left(v_n,[-\sqrt{\lambda_n}a_n,\sqrt{\lambda_n}a_n]\right)=E(\phi,{\mathbb R})+o(1)\qquad\text{as }n\to+\infty\,, \end{equation}

where ϕ is the unique solution in $H^1({\mathbb R})$ of

(3.9)\begin{equation} \begin{cases} \phi''+\phi^{p-1}=\phi & \text{on }{\mathbb R}\\ \phi(0)=\|\phi\|_{L^\infty({\mathbb R})} \ge 1. \end{cases} \end{equation}

Note that $\phi=\phi_{\nu,1}$ for $\displaystyle\nu:=\lim_{n\to+\infty}\|v_n\|_{L^2\left(-\sqrt{\lambda_n} a_n,\sqrt{\lambda_n} a_n\right)}^2 \gt 0$. Since by direct computations

\begin{equation*} E(u_n,B_n)=\lambda_n^{\frac{p+2}{2(p-2)}}E\left(v_n,[-\sqrt{\lambda_n}a_n,\sqrt{\lambda_n}a_n]\right), \end{equation*}

by (3.8) it follows

\begin{equation*} \begin{aligned} E(u_n,B_n) \ge \lambda_n^{\frac{p+2}{2(p-2)}}E(\phi_{\nu,1}, {\mathbb R}) + o\left( \lambda_n^{\frac{p+2}{2(p-2)}}\right) = &\,E(\phi_{\nu_n,1},{\mathbb R}) + o( E(\phi_{\nu_n,1},{\mathbb R})) \\ \ge&\, E(\phi_{\mu_n,1},{\mathbb R}) + o( E(\phi_{\mu_n,1},{\mathbb R})), \end{aligned} \end{equation*}

the last inequality coming by the fact that $\nu_n:=\lambda_n^{\frac{6-p}{2(p-2)}}\nu\leq\mu_n$ by definition of ν and

\begin{equation*} \|v_n\|_{L^2\left(-\sqrt{\lambda_n} a_n,\sqrt{\lambda_n} a_n\right)}^2=\lambda_n^{\frac{p-6}{2(p-2)}}\|u_n\|_{L^2(B_n)}^2\leq\lambda_n^{\frac{p-6}{2(p-2)}}\mu_n\qquad\forall n\,. \end{equation*}

Hence, by (3.4),

\begin{align*} E(u_n, \mathcal G) = E(u_n, A_n) + E(u_n, \mathcal G\setminus A_n) &\ge E(u_n, B_n) +o(1) \\ & \ge E(\phi_{\mu_n,1},{\mathbb R}) + o( E(\phi_{\mu_n,1},{\mathbb R})), \end{align*}

so that

\begin{equation*} \frac{E(u_n, \mathcal G) }{E(\phi_{\mu_n,1},{\mathbb R}) } \ge 1 + o(1), \end{equation*}

which violates (3.3) when $n\to \infty$ and completes the proof in case B_n contains no vertex of $\mathcal G$.

We now assume (keeping in mind (3.5)) that every B_n contains a (unique) vertex $\textrm{v}_n$. Thus, each B_n is the union of $k_n \ge 3$ line segments glued together at this vertex.

Hence, letting $k_n:=\text{deg}(\textrm{v}_n)$, there exist $a_1^n,\ldots,a_{k_n}^n \gt 0$ such that B_n can be identified with the intervals $[0,a_1^n),\ldots,[0,a_{k_n}^n)$ glued together at the origin (identified with $\textrm{v}_n$). Since by construction B_n contains at least one local maximum point of u_n on $\mathcal G$, there are $i_n \ge 0$ intervals $[0,a_1^n),\ldots,[0,a_{i_n}^n)$ on which u_n is increasing from 0 to a local maximum point and then decreasing from this maximum point to the end of the interval, whereas on the intervals $[0,a_{i_n+1}^n),\ldots,[0,a_{k_n}^n)$ u_n is decreasing from 0 to the end of the interval. Note that $i_n \lt k_n$, because otherwise u_n would have a local minimum point at $\textrm{v}_n$, and by the Kirchhoff condition this would imply $u_n(\textrm{v}_n)\le 1$, contradicting $\textrm{v}_n\in B_n$.

Observe that, if

(3.10)\begin{equation} \limsup_{n\to+\infty} \lambda_n^{-\frac1{p-2}}u_n(\textrm{v}_n)=0, \end{equation}

arguing as in the previous part of the proof on each interval (if any) $[0,a_i^n]$, $i = 1, \ldots, i_n$, yields, as $n\to+\infty$,

\begin{equation*} E\left(u_n,[0,a_i^n]\right) = \lambda_n^{\frac{p+2}{2(p-2)}}E(\phi_{\nu,1},{\mathbb R})+o\left(\lambda_n^{\frac{p+2}{2(p-2)}}\right) \end{equation*}

for $\nu\leq\lambda_n^{\frac{p-6}{2(p-2)}}\mu_n$. On the other hand, on each interval $[0,a_i^n]$ with $i = i_n+1, \ldots, k_n$, if $a_i^n = o\left(1/\sqrt{\lambda_n}\right)$, then

\begin{equation*} E(u_n,[0,a_i^n]) \ge -\int_0^{a_n}|u_n|^p{\,dx} \ge o \left(\lambda_n^{\frac{p}{p-2}}\right)a_i^n = o \left(\lambda_n^{\frac{p+2}{2(p-2)}}\right). \end{equation*}

If, on the contrary, $\sqrt{\lambda_n}a_i^n$ is bounded away from 0, then $E(u_n, [0,a_i^n]) \gt 0$ (this can be seen, e.g., combining the scaling procedure used in the previous part of the proof with [Reference Dovetta and Tentarelli25, lemma 4.5]). All in all, this yields, as above,

\begin{equation*} E(u_n, B_n) \ge E(\phi_{\mu_n,1},{\mathbb R}) + o( E(\phi_{\mu_n,1},{\mathbb R})). \end{equation*}

Conversely, if (3.10) does not hold, there exists K > 0 such that along a subsequence (not relabelled) we have

\begin{equation*} \lambda_n^{-\frac1{p-2}}u_n(\textrm{v}_n) \geq K\qquad\forall n\,. \end{equation*}

Considering the restriction of u_n to $\overline{B_n}$, set then $v_n(x):=\lambda_n^{-\frac1{p-2}}u_n(x/\sqrt{\lambda_n})$. The function v_n is defined on a star graph with k_n bounded edges, identified with the intervals $\left[0,\sqrt{\lambda_n}a_i^n\right]$, $i= 1, \ldots, k_n$. Moreover, on each edge, v_n is either increasing from 0 to a local maximum point and then decreasing to the end of the edge or decreasing, and it satisfies

\begin{equation*} \begin{cases} v_n''+v_n^{p-1}=v_n & \text{on each } \left[0,\sqrt{\lambda_n}a_i^n\right]\\ v_n(0) \ge K & \\ \sum_{i=1}^{k_n}\frac{dv_n}{dx_i}(0)=0, \end{cases} \end{equation*}

where $\displaystyle\frac{dv_n}{dx_i}(0)$ denotes the outgoing derivative of v_n at 0 along $\left[0,\sqrt{\lambda_n}a_i^n\right]$. Furthermore, $v_n\left(\sqrt{\lambda_n}a_i^n\right)\to0$ as $n\to+\infty$ for every i.

Arguing exactly as in the previous part of the proof, we obtain again that $|v_n'(\sqrt{\lambda_n}a_j^n)|\to0$ and $\sqrt{\lambda_n}a_i^n\to+\infty$ as $n\to+\infty$ for every i. By continuity in the phase plane, this implies that the restriction of v_n to each interval $[0,a_i^n)$ converges to the restriction of the function ϕ of (3.9) to a suitable interval of the form $[b_i,+\infty)$. Together with the fact that v_n satisfies the Kirchhoff condition at the vertex of the star graph, this implies that

\begin{equation*} E\left(v_n,\sqrt{\lambda_n}B_n\right)=E(w_n,S_{k_n})+o(1)\qquad\text{as }n\to+\infty\,, \end{equation*}

where $S_{k_n}$ is the infinite star graph with k_n half-lines and w_n is a $H^1(S_{k_n})$ positive solution of

\begin{equation*} \begin{cases} w_n''+w_n^{p-1}=w_n & \text{on }S_{k_n}\\ w_n(0) \ge K & \\ \sum_{i=1}^{k_n}\frac{dw_n}{dx_i}(0)=0. \end{cases} \end{equation*}

Since $k_n\geq 3$ for every n, it is well known that $E(w_n,S_{k_n}) \ge E(\phi_{\nu_n,1},{\mathbb R})$ (see, e.g., [Reference Adami, Cacciapuoti, Finco and Noja1]), where $\nu_n \le \mu_n$ is the mass of w_n. Hence, recalling that $\displaystyle E(u_n,B_n)=\lambda_n^{\frac{p+2}{2(p-2)}}E\left(v_n,\sqrt{\lambda_n}B_n\right)$, we conclude as in the previous case. This completes the proof of (3.1).

The proof of (3.2) is completely analogous to that of (3.1). The only difference is that if u_n ‘concentrates’ on a pendant, then one must use the half-soliton $\phi_{2\mu_n,1}$ and its level $E(\phi_{2\mu_n,1}, {\mathbb R}^+) \lt E(\phi_{\mu_n,1},{\mathbb R})$ in the asymptotic estimates.

Proof. The proof works exactly as that of [Reference Borthwick, Chang, Jeanjean and Soave13, lemma 3.1], additionally recalling here that, for every $\mu,k \gt 0$, the set

\begin{equation*} A_k:=\left\{u\in H_\mu^1(\mathcal G)\,:\,\|u'\|_2^2\leq k\right\} \end{equation*}

is non-empty whenever $\mathcal G\in\mathbf{G}$ is a non-compact graph with finitely many edges or a periodic graph because for all such graphs we already pointed out in (2.8) that $\lambda_\mathcal G=0$.

Proof. We argue in three steps.

Step 1: proof of $c_\rho(\mathcal G)\leq c_\rho({\mathbb R})+o(c_\rho({\mathbb R}))$. Let $\mathcal G\in\mathbf{G}$ and p > 6 be fixed. Let $e\in\mathbb{E}_\mathcal G$ be any given edge of $\mathcal G$, and let $4\ell:=|e|$ be its length. To prove the first part of proposition 4.3, we will now show that, as soon as µ → 0, there exists a path $\gamma=(\gamma_t)_{t\in[0,1]}\subset H_\mu^1(\mathcal G)$ such that γ_t is compactly supported on e for every $t\in[0,1]$ and $\displaystyle\max_{t\in[0,1]}E_\rho(\gamma_t,\mathcal G)\leq c_\rho({\mathbb R})+o(c_\rho({\mathbb R}))$ for every $\rho\in\left[1/2,1\right]$.

To this end, let $\overline{\gamma}\subset H_\mu^1({\mathbb R})$ be the path given in remark 4.2 and define $v_t:e\to{\mathbb R}$ as

\begin{equation*} v_t(x):=\begin{cases} \overline{\gamma}_t(x) & \text{if }x\in\left[-\ell,\ell\right]\\ \overline{\gamma}_t(\ell)(2-x/\ell) & \text{if }x\in\left(\ell,2\ell\right]\\ \overline{\gamma}_t(-\ell)(2+x/\ell) & \text{if }x\in\left[-2\ell,-\ell\right), \end{cases} \end{equation*}

where we tacitly identified the edge e with the interval $[-2\ell,2\ell]$. Extending then v_t to be identically zero on $\mathcal G\setminus e$, we define $\gamma_t\in H_\mu^1(\mathcal G)$ as

\begin{equation*} \gamma_t:=\frac{\sqrt{\mu}}{\|v_t\|_2}\,v_t\,. \end{equation*}

Note that, by definition of v_t, remark 4.2 and (2.3),

(4.1)\begin{equation} \begin{aligned} \|v_t\|_2^2&=\|\overline{\gamma}_t\|_{L^2(-\ell,\ell)}^2+\frac{2\ell}3\overline{\gamma}_t(\ell)^2\,=\mu+\frac{2\ell}3\overline{\gamma}_t(\ell)^2-2\|\overline{\gamma}_t\|_{L^2(\ell,+\infty)}^2\\ &\,=\mu+C\mu^{2\alpha}e^{-2\tau c\mu^\beta\ell}+o\left(\mu^{2\alpha}e^{-2\tau c\mu^\beta\ell}\right)\qquad\text{as }\mu\to0\,, \end{aligned} \end{equation}

for constants $C,c \gt 0$ depending on $p,\rho,t$ only and both bounded from above and away from zero uniformly for $\rho\in\left[1/2,1\right]$ and $t\in[0,1]$, and $\tau:=\alpha/\beta$ (with $\alpha,\beta$ as in (2.2)). Analogous computations lead to

\begin{equation*} \begin{aligned} \|v_t'\|_2^2&\,=\|\overline{\gamma}_t'\|_2^2+C'\mu^{2\alpha+\beta}e^{-2\tau c\mu^\beta\ell}+o\left(\mu^{2\alpha+\beta}e^{-2\tau c\mu^\beta\ell}\right)=\|\overline{\gamma}_t'\|_2^2+o\left(\|\overline{\gamma}_t'\|_2^2\right)\\ \|v_t\|_p^p&\,=\|\overline{\gamma}_t\|_p^p+C''\mu^{\alpha p}e^{-p\tau c\mu^\beta\ell}+o\left(\mu^{\alpha p}e^{-p\tau c\mu^\beta\ell}\right)\\ &=\|\overline{\gamma}_t\|_p^p++o\left(\|\overline{\gamma}_t\|_p^p\right)\qquad\text{as }\mu\to0\,, \end{aligned} \end{equation*}

again with $C',C'' \gt 0$ depending on $p,\rho,t$ only and both bounded from above and away from zero uniformly for $\rho\in\left[1/2,1\right]$ and $t\in[0,1]$ by remark 4.2. All in all, as soon as µ is small enough we obtain $\gamma_0\in A_\delta$, $\gamma_1\in B$ and

\begin{equation*} E_\rho(\gamma_t,\mathcal G)=\frac{\mu}{\|v_t\|_2^2}\frac12\|v_t'\|_2^2-\left(\frac\mu{\|v_t\|_2^2}\right)^\frac p2 \frac1p\|v_t\|_p^p= E_\rho(\overline{\gamma}_t,{\mathbb R})+o(E_\rho(\overline{\gamma}_t,{\mathbb R})) \end{equation*}

uniformly for $\rho\in\left[1/2,1\right]$ and $t\in[0,1]$, in turn yielding

\begin{equation*} c_\rho(\mathcal G)\leq\max_{t\in[0,1]}E_\rho(\gamma_t,\mathcal G)=\max_{t\in[0,1]}\left(E_\rho(\overline{\gamma}_t,{\mathbb R})+o\left(E_\rho(\overline{\gamma}_t,{\mathbb R})\right)\right)=c_\rho({\mathbb R})+o(c_\rho({\mathbb R})) \end{equation*}

for sufficiently small µ.

Step 2: proof of $c_\rho(\mathcal G) \leq c_\rho({\mathbb R}^+)+o(c_\rho({\mathbb R}^+))$ for graphs with a pendant. This step works exactly as Step 1, with the only difference that we now construct a path $\gamma\subset H_\mu^1(\mathcal G)$ such that γ_t is compactly supported on the pendant of $\mathcal G$ for every $t\in [0,1]$ and $\displaystyle\max_{t\in[0,1]}E_\rho(\gamma_t)\leq c_\rho({\mathbb R}^+)+o(c_\rho({\mathbb R}^+)) \lt c_\rho({\mathbb R})$. By remark 4.2, such a path γ can be obtained repeating verbatim the construction of Step 1 starting with the restriction to ${\mathbb R}^+$ of $\overline{\gamma}\subset H_{2\mu}^1({\mathbb R})$.

Step 3: proof of $c_\rho(\mathcal G) \lt c_\rho({\mathbb R})$ for graphs with a signpost. The argument is again analogous to that of Step 1, suitably adapted to functions defined on the tadpole graph, i.e., the graph $\mathcal{T}_\ell$ given by a loop $\mathcal{L}$ of length $2\ell \gt 0$ and a single half-line $\mathcal H$. Note first that, for every µ, ρ, it holds

(4.2)\begin{equation} c_\rho(\mathcal{T}_\ell) \lt c_\rho({\mathbb R})\,. \end{equation}

To see this, take again the path $\overline{\gamma}\subset H_\mu^1({\mathbb R})$ introduced in remark 4.2 and define $\widetilde{\gamma}\subset H_\mu^1(\mathcal{T}_\ell)$ as

\begin{equation*} \widetilde{\gamma}_t(x):=\begin{cases} \overline{\gamma}_t(x) & \text{if }x\in\left[-\ell,\ell\right]\\ \overline{\gamma}_t\left(\frac{x+2\ell}2\right) & \text{if }x\in{\mathbb R}^+ \end{cases} \end{equation*}

where, with a slight abuse of notation, we identified the loop $\mathcal{L}$ with the interval $[-\ell,\ell]$ and the half-line $\mathcal H$ with ${\mathbb R}^+$. Straightforward computations show that, for every $t\in[0,1]$,

(4.3)\begin{equation} \begin{aligned} &\|\widetilde{\gamma}_t\|_2^2=\mu\,,\qquad\qquad\|\widetilde{\gamma}_t\|_{L^p(\mathcal{T}_\ell)}^p=\|\overline{\gamma}_t\|_{L^p({\mathbb R})}^p\,,\\ &\|\widetilde{\gamma}_t'\|_{L^2(\mathcal{T}_\ell)}^2=\|\overline{\gamma}_t'\|_{L^2(-\ell,\ell)}^2+\frac14\|\overline{\gamma}_t'\|_{L^2({\mathbb R}\setminus[-\ell,\ell])}^2=\|\overline{\gamma}_t'\|_2^2-\frac34\|\overline{\gamma}_t'\|_{L^2({\mathbb R}\setminus[-\ell,\ell])}^2\,, \end{aligned} \end{equation}

so that $\widetilde{\gamma}_0\in A_\delta$, $\widetilde{\gamma}_1\in B$ and $E_\rho(\widetilde{\gamma}_t,\mathcal{T}_\ell) \lt E_\rho(\overline{\gamma}_t,{\mathbb R})$, in turn yielding (4.2).

To show then that the same inequality holds on $\mathcal G$ when it has a signpost, it is enough to adapt the construction in Step 1 to the path $\widetilde{\gamma}$. Denote by $\mathcal{P}$ the signpost of $\mathcal G$ and let $2\ell$ be the length of its loop and 2a be that of its bounded edge. For every $t\in[0,1]$, we define $v_t:\mathcal{P}\to{\mathbb R}$ as

\begin{equation*} v_t(x):=\begin{cases} \widetilde{\gamma}_t(x) & \text{if }x\in\mathcal{L}\\ \widetilde{\gamma}_t(x) & \text{if }x\in \left[0,a\right]\cap\mathcal H\\ \widetilde{\gamma}_t(a)\left(2-\frac xa\right) & \text{if }x\in[a,2a]\,, \end{cases} \end{equation*}

again with the slight abuse of notation to identify the loop of $\mathcal{P}$ with the loop $\mathcal{L}$ of the tadpole $\mathcal{T}_\ell$ and its bounded edge with the interval $[0,2a]$ on the half-line $\mathcal H$ of $\mathcal{T}_\ell$. Extending v_t to zero on the rest of $\mathcal G$, we then define $\gamma\subset H_\mu^1(\mathcal G)$ as $\displaystyle\gamma_t:=\frac{\sqrt{\mu}}{\|v_t\|_2}\,v_t$, for every $t\in[0,1]$. Arguing as in Step 1, by (2.3), there results

(4.4)\begin{align} \|v_t\|_2^2&=\|\widetilde{\gamma}_t\|_{L^2(\mathcal{L}\cup(\mathcal H\cap[0,a]))}^2 +\frac{\overline{\gamma}_t\left(\frac{a+2\ell}2\right)^2a}{3}=\mu-\|\widetilde{\gamma}_t\|_{L^2(\mathcal H\cap(a,+\infty))}^2+\frac{\overline{\gamma}_t\left(\frac{a+2\ell}2\right)^2a}{3} \nonumber\\ &=\mu-\int_{a}^{+\infty}\left|\overline{\gamma}_t\left(\frac{x+2\ell}{2}\right)\right|^2\,dx+\frac{\overline{\gamma}_t\left(\frac{a+2\ell}2\right)^2a}{3}\nonumber\\ &=\mu-2\int_{\frac{a+2\ell}{2}}^\infty\left|\overline{\gamma}_t(y)\right|^2\,dy+\frac{\overline{\gamma}_t\left(\frac{a+2\ell}2\right)^2a}{3}\nonumber\\ &=\mu+K\mu^{2\alpha}e^{-2\tau c\mu^\beta\left(\frac{a+2\ell}2\right)}+o\left(\mu^{2\alpha}e^{-2\tau c\mu^\beta\left(\frac{a+2\ell}2\right)}\right)\qquad\text{as }\mu\to0\,, \end{align}

where $c,\tau$ are the same as in (4.1) and K > 0 is a constant depending on $p,\rho,t$ only, bounded from above and away from zero uniformly for $\rho\in\left[1/2,1\right]$ and $t\in[0,1]$. Similarly, recalling also (4.3), we have

(4.5)\begin{align} \|v_t'\|_2^2&=\|\widetilde{\gamma}_t'\|_{L^2\left(\mathcal{L}\cup(\mathcal H\cap[0,a])\right)}^2+\frac{\overline{\gamma}_t\left(\frac{a+2\ell}{2}\right)^2}{a} \nonumber\\ &=\|\widetilde{\gamma}_t\|_{L^2(\mathcal{T}_\ell)}^2-\|\widetilde{\gamma}_t'\|_{L^2(\mathcal H\cap(a,+\infty))}^2+\frac{\overline{\gamma}_t\left(\frac{a+2\ell}{2}\right)^2}{a} \nonumber\\ &=\|\overline{\gamma}_t'\|_2^2-\frac34\|\overline{\gamma}_t'\|_{L^2({\mathbb R}\setminus[-\ell,\ell])}^2-\|\widetilde{\gamma}_t'\|_{L^2(\mathcal H\cap(a,+\infty))}^2+\frac{\overline{\gamma}_t\left(\frac{a+2\ell}{2}\right)^2}{a} \nonumber\\ &=\|\overline{\gamma}_t'\|_{2}^2-K'\mu^{2\alpha+\beta}e^{-2\tau c\mu^\beta\ell}+o\left(\mu^{2\alpha+\beta}e^{-2\tau c\mu^\beta\ell}\right) \end{align}

and

(4.6)\begin{align} \|v_t\|_p^p&=\|\widetilde{\gamma}_t\|_{L^p(\mathcal{L}\cup(\mathcal H\cap[0,a]))}^p+\frac{\left|\overline{\gamma}_t\left(\frac{a+2\ell}{2}\right)\right|^pa}{p+1} \nonumber\\ &= \|\widetilde{\gamma}_t\|_{L^p(\mathcal{T}_\ell)}^p-\|\widetilde{\gamma}_t\|_{L^p(\mathcal H\cap(a,+\infty))}^p+\frac{\left|\overline{\gamma}_t\left(\frac{a+2\ell}{2}\right)\right|^pa}{p+1} \nonumber\\ &=\|\overline{\gamma}_t\|_{L^p({\mathbb R})}^p+K''\mu^{\alpha p}e^{-p\tau c\mu^\beta\left(\frac{a+2\ell}{2}\right)}+o\left(e^{-p\tau c\mu^\beta\left(\frac{a+2\ell}{2}\right)}\right)\,, \end{align}

for constants $K',K'' \gt 0$ depending on $p,\rho,t$ only and bounded from above and away from zero uniformly for $\rho\in\left[1/2,1\right]$ and $t\in[0,1]$. Coupling (4.5) and (4.6) yields, since p > 2 and a > 0,

\begin{equation*} E_\rho(v_t,\mathcal G)=E_\rho(\overline{\gamma}_t,{\mathbb R})-K'\mu^{2\alpha+\beta}e^{-2\tau c\mu^\beta\ell}+o\left(\mu^{2\alpha+\beta}e^{-2\tau c\mu^\beta\ell}\right) \end{equation*}

as µ → 0, which together with (4.4) (and recalling again remark 4.2 and (2.6)) gives

\begin{align*} E_\rho(\gamma_t,\mathcal G)&=\frac{\mu}{\|v_t\|_2^2}\frac12\|v_t'\|_2^2-\left(\frac\mu{\|v_t\|_2^2}\right)^\frac p2 \frac1p\|v_t\|_p^p \nonumber\\ &=\left(1-O\left(\mu^{2\alpha}e^{-2\tau c\mu^\beta\left(\frac{a+2\ell}2\right)}\right)\right)E_\rho(v_t,\mathcal G) \nonumber\\ &= E_\rho(\overline{\gamma}_t,{\mathbb R})-K'\mu^{2\alpha+\beta}e^{-2\tau c\mu^\beta\ell}E_\rho(\overline{\gamma}_t,{\mathbb R})\nonumber\\ &\quad +o\left(\mu^{2\alpha+\beta}e^{-2\tau c\mu^\beta\ell}E_\rho(\overline{\gamma}_t,{\mathbb R})\right) \end{align*}

as µ → 0. Since K ^ʹ is bounded away from zero uniformly for $\rho\in\left[1/2,1\right]$ and $t\in[0,1]$, by remark 4.2 the previous identity entails $\displaystyle c_\rho(\mathcal G)\leq \max_{t\in[0,1]}E_\rho(\gamma_t,\mathcal G) \lt \max_{t\in[0,1]}E_\rho(\overline{\gamma}_t,{\mathbb R})=c_\rho({\mathbb R})$ as soon as µ is small enough, and we conclude.

Proof. The validity of (4.7) is evident as soon as $\mathcal G$ has at least one half-line. To show that $c_\rho(\mathcal G)=c_\rho({\mathbb R})$ whenever $\mathcal G$ satisfies assumption (H) recall that, by standard properties of rearrangements on graphs (see, e.g., [Reference Adami, Serra and Tilli3, Section 3]), if $u\in H^1(\mathcal G)$ is a positive function on $\mathcal G$ satisfying assumption (H), then its symmetric decreasing rearrangement on the line $\widehat{u}\in H^1({\mathbb R})$ is such that

(4.8)\begin{equation} \|u\|_r=\|\widehat{u}\|_r,\quad\forall r\geq1,\qquad\|u'\|_2\geq\|\widehat{u}'\|_2\,. \end{equation}

By (4.8) it follows that, if $\gamma\in C\left([0,1],H_\mu^1(\mathcal G)\right)$ is a path in $H_\mu^1(\mathcal G)$ such that $\gamma_0\in A_\alpha$ and $\gamma_1\in B$ on $\mathcal G$, then the symmetric decreasing rearrangement $\widehat{\gamma}_t$ of γ_t provides a path $\widehat{\gamma}\in C\left([0,1], H_\mu^1({\mathbb R})\right)$ (the strong continuity in $H^1({\mathbb R})$ of the map $t\mapsto\widehat{\gamma}_t$ can be proved, e.g., as in [Reference Coron19]) such that $\widehat{\gamma}_0\in A_\alpha$ and $\widehat{\gamma}_1\in B$ on ${\mathbb R}$, and $E_\rho(\widehat{\gamma}_t,{\mathbb R})\leq E_\rho(\gamma_t,\mathcal G)$ for every t. As this ensures that $c_\rho(\mathcal G)\geq c_\rho({\mathbb R})$, coupling with (4.7) yields the desired identity.

Proof. Since $(u_n)_n$ is bounded in $H^1(\mathcal G)$ by (c), so is $(\lambda_n)_n \subset {\mathbb R}$. Hence, up to subsequences, $\lambda_n\to\lambda$ as $n\to+\infty$, for some $\lambda\in{\mathbb R}$.

By weak lower semicontinuity, $m:=\|u\|_2^2\in[0,\mu]$. If m = 0, then $u\equiv0$. Conversely, if $m=\mu$, then $u\in H_\mu^1(\mathcal G)$, $u_n\to u$ in $L^2(\mathcal G)$ and thus in $L^p(\mathcal G)$, so that by (b) and $\lambda_n\to\lambda$ it follows

\begin{equation*} \|u_n'-u'\|_2^2+\lambda\|u_n-u\|_2^2=o(1)\qquad\text{as }n\to+\infty\,, \end{equation*}

that is $u_n\to u$ in $H^1(\mathcal G)$, and u is a critical point of E_ρ constrained to $H_\mu^1(\mathcal G)$ at level $c_\rho(\mathcal G)$. Hence, to prove the lemma, it is enough to rule out the case $m\in(0,\mu)$.

To this end, we argue by contradiction. Assume thus $0 \lt m \lt \mu$, so that $u\not\equiv0$ on $\mathcal G$ satisfies

(5.1)\begin{equation} \begin{cases} u''+\rho u^{p-1}=\lambda u & \text{on every }e\in\mathbb{E}_\mathcal G\\ u\geq 0 &\text{on }\mathcal G\\ \sum_{e\succ \textrm{v}}\frac{du}{dx_e}(\textrm{v})=0 & \text{for every }\textrm{v}\in\mathbb{V}_\mathcal G\,. \end{cases} \end{equation}

By standard arguments, u > 0 on $\mathcal G$, that is $u\in S_{m,\rho}$ (where $S_{m,\rho}$ is the set defined in remark 3.2). Thus we have from lemma 2.5(c) that λ > 0. By remark 3.2, see also corollary 3.3, there exist $\overline{\mu} \gt 0$ (depending on p and $\mathcal G$ only), such that for every $\mu\in\left(0,\overline{\mu}\right)$ and every $\displaystyle\rho\in\left[1/2,1\right]$, there results

(5.2)\begin{equation} E_\rho(u,\mathcal G) \ge \frac{1}{2} E_\rho(\phi_{2m, \rho}, {\mathbb R}^+)\,. \end{equation}

Set then $v_n:=u_n-u$. By weak convergence of u_n to u in $H^1(\mathcal G)$, as $n \to \infty$, we have

(5.3)\begin{equation} \|v_n\|_2^2=\mu-m+o(1),\qquad\|v_n'\|_2^2=\|u_n'\|_2^2-\|u'\|_2^2+o(1)\,, \end{equation}

so that by Brezis–Lieb lemma [Reference Brezis and Lieb15] and (a)

(5.4)\begin{equation} E_\rho(v_n,\mathcal G)=E_\rho(u_n,\mathcal G)-E_\rho(u,\mathcal G)+o(1)=c_\rho(\mathcal G)-E_\rho(u,\mathcal G)+o(1)\qquad\text{as }n\to+\infty\,. \end{equation}

Moreover, by (b), (c), (5.1), and $\lambda_n\to\lambda$, it is easy to see that $v_n\to0$ in $H^1(\mathcal K)$, where $\mathcal K$ denotes as usual the compact core of $\mathcal G$. This can be easily seen for instance by taking the difference between (b) and (5.1) both tested with $\varphi_n:=(u_n-u)\eta$, where η is compactly supported on a fixed neighbourhood of $\mathcal K$ and $\eta\equiv1$ on $\mathcal K$. Hence, up to negligible corrections in $H^1(\mathcal G)$ that do not affect the validity of (5.3) and (5.4), we can take v_n to be identically zero on $\mathcal K$ for every n. Since $\displaystyle\mathcal G\setminus\mathcal K=\bigcup_{i=1}^N\mathcal H_i$, where $\mathcal H_i,\,i=1,\ldots,N$, are all the half-lines of $\mathcal G$, this means that we can think of v_n as the sum of its restrictions $v_{n,i}$ to the half-lines $\mathcal H_i$, $i=1,\ldots,N$. As $v_{n,i}$ vanishes at the vertex of $\mathcal H_i$, we can embed $\mathcal H_i$ in ${\mathbb R}$ and extend each $v_{n,i}$ by 0. Up to translations, there is no loss of generality in further assuming that each $v_{n,i}$ satisfies $\|v_{n,i}\|_\infty=v_{n,i}(0)$. Moreover, standard arguments (see, e.g., [Reference Coti Zelati and Rabinowitz20]) show that, for every $i=1,\ldots, N$, as $n\to+\infty$,

(5.5)\begin{equation} \int_{\mathbb R} v_{n,i}'\varphi'\,dx-\rho\int_{\mathbb R} v_{n,i}^{p-1}\varphi\,dx+\lambda\int_{\mathbb R} v_{n,i}\varphi\,dx=o(1)\|\varphi\|_{H^1({\mathbb R})}\qquad\forall \varphi\in H^1({\mathbb R})\,. \end{equation}

Observe that, together with (5.3) and λ > 0, this entails that (up to subsequences) $\|v_{n,i}\|_\infty\not\to0$ as $n\to+\infty$ at least for one value of i. Indeed, otherwise $\|v_{n,i}\|_p^p \to0$ as $n\to+\infty$ contradicting (5.5).

All in all, (5.3) and (5.5) imply that there exists $k\in{\mathbb N}$, $k\geq1$, such that

\begin{equation*} E_\rho(v_n,\mathcal G)=\sum_{i=1}^NE_\rho(v_{n,i},{\mathbb R})=k E_\rho\left(\phi_{\frac{\mu-m}{k},\rho},{\mathbb R}\right)+o(1) \end{equation*}

as soon as n is large enough. Coupling with (5.2) and (5.4) then yields (recalling (2.6))

\begin{align*} c_\rho(\mathcal G) &\geq k E_\rho\left(\phi_{\frac{\mu-m}{k},\rho},{\mathbb R}\right)+ E_\rho(u,\mathcal G) + o(1) \geq E_\rho\left(\phi_{\mu-m,\rho},{\mathbb R}\right)\\ &\quad + \frac{1}{2} E_\rho(\phi_{2m,\rho},{\mathbb R}^+) +o(1) \\ &= \theta_p \rho^{\frac{4}{6-p}}\left((\mu-m)^{2\beta +1} + 2^{2\beta -1} m^{2\beta +1}\right) +o(1)\\ & \ge (1-\tau)^{2\beta+1}E_\rho(\phi_{\mu,\rho},{\mathbb R})+ o(1) \end{align*}

where $(1-\tau)^{2\beta+1} \gt 1$ since $2\beta + 1 \lt 0$, and $\tau:=\left(2 (\frac{1}{2})^{-2\beta}+1\right)^{-1}$ is such that τµ is the minimum point of the function $m\mapsto (\mu-m)^{2\beta +1} + 2^{2\beta -1} m^{2\beta +1}$ on $(0,\mu)$. Note that τ depends on p only.

However, by proposition 4.3 (and recalling that τ < 1 and $2\beta+1 \lt 0$), up to a possible reduction of the threshold $\overline{\mu} \gt 0$, there results $c_\rho(\mathcal G)\leq (1-\tau/2)^{2\beta+1}c_\rho({\mathbb R})=(1-\tau/2)^{2\beta+1} E_\rho(\phi_{\mu,\rho},{\mathbb R})$ for every $\mu\in(0,\overline{\mu}]$ and every $\displaystyle\rho\in\left[1/2,1\right]$, contradicting the preceding inequality and completing the proof.

Proof. Proof of theorem 1.1

The proof is divided in two steps: in the first one, we prove existence of positive critical points of E_ρ constrained to $H_\mu^1(\mathcal G)$ for almost every $\displaystyle\rho\in\left[1/2,1\right]$, in the second one we prove that (a subsequence of) these critical points converges, as $\rho\to1^-$, to a positive critical point of E in $H_\mu^1(\mathcal G)$.

Step 1. For every fixed µ > 0, by lemma 4.1 and theorem 2.3, it follows that, for almost every $\displaystyle\rho\in\left[1/2,1\right]$, there exists a bounded sequence $(u_n)_n\subset H_\mu^1(\mathcal G)$ satisfying properties (i)–(iv) of theorem 2.3. Furthermore, as pointed out in [Reference Borthwick, Chang, Jeanjean and Soave14, remark 1.4], it is not restrictive to assume $u_n\geq0$ on $\mathcal G$ for every n. Set

(5.6)\begin{equation} \lambda_n:=-\frac{E_\rho'(u_n,\mathcal G)[u_n]}{\|u_n\|_2^2}=\frac{\rho\|u_n\|_p^p-\|u_n'\|_2^2}{\mu}\,. \end{equation}

Since $(u_n)_n$ is bounded in $H^1(\mathcal G)$ by property (iii) of theorem 2.3, $(\lambda_n)_n$ is bounded too, so that as $n\to+\infty$ up to subsequences $\lambda_n\to\lambda_\rho$, for some $\lambda_\rho\in{\mathbb R}$, and $u_n\rightharpoonup u_\rho$ in $H^1(\mathcal G)$, for some $u_\rho\in H^1(\mathcal G)$ satisfying

(5.7)\begin{equation} \begin{cases} u_\rho''+u_\rho^{p-1}=\lambda_\rho u_\rho & \text{on every }e\in\mathbb{E}_\mathcal G\\ u_\rho\geq0 &\text{on }\mathcal G\\ \sum_{e\succ \textrm{v}}\frac{du_\rho}{dx_e}(\textrm{v})=0 & \text{for every }\textrm{v}\in\mathbb{V}_\mathcal G\,. \end{cases} \end{equation}

Since $\mathcal G$ is a graph with at least one half-line and u_n satisfies properties (iv) of theorem 2.3, by lemma 2.4 and [Reference Borthwick, Chang, Jeanjean and Soave13, lemma 3.2] we obtain $\lambda_\rho\geq0$.

We now show that $u_\rho\not\equiv0$ on $\mathcal G$. Assume by contradiction that this is the case, so that $u_n\rightharpoonup 0$ in $H^1(\mathcal G)$ and $u_n\to0$ in $L_{loc}^\infty(\mathcal G)$ as $n\to+\infty$. Note that the convergence of u_n to 0 cannot be in $L^\infty(\mathcal G)$, because then $u_n\in H_\mu^1(\mathcal G)$ would imply $u_n\to0$ in $L^p(\mathcal G)$, which together with (5.6) and $\liminf \lambda_n\geq0$ would yield $E_\rho(u_n,\mathcal G)\to0$, that is impossible since $E_\rho(u_n,\mathcal G)\to c_\rho(\mathcal G) \gt 0$. Hence, $u_n\to0$ in $L_{loc}^\infty(\mathcal G)$ but not in $L^\infty(\mathcal G)$. Whenever the mass is smaller than a threshold independent of ρ, arguing as in the proof of lemma 5.1 it is then easy to see that, denoting by $\mathcal H_i$, $i=1,\ldots,N$, the half-lines of $\mathcal G$, as $n\to+\infty$ there results

\begin{equation*} \begin{aligned} &\|u_n\|_{L^2(\bigcup_{i=1}^N\mathcal H_i)}^2=\mu+o(1)\\ &\int_{\mathcal H_i}u_n'\varphi'\,dx-\int_{\mathcal H_i}u_n^{p-1}\varphi\,dx+\lambda\int_{\mathcal H_i}u_n\varphi\,dx=o(1)\|\varphi\|_{H^1(\mathcal H_i)},\qquad\forall i=1,\ldots,N\,, \end{aligned} \end{equation*}

and, up to negligible corrections in $H^1(\mathcal G)$, u_n can be taken to be identically zero on $\mathcal K$. Since $\|u_n\|_\infty$ is uniformly bounded away from 0, it then follows that there exists $k\in{\mathbb N}$, $k\geq1$, such that

(5.8)\begin{align} E_\rho(u_n,\mathcal G)&=k E_\rho\left(\phi_{\frac\mu k,\rho},{\mathbb R}\right)+o(1)\geq E_\rho(\phi_{\mu,\rho},{\mathbb R})+o(1) \nonumber\\ &=c_\rho({\mathbb R})+o(1)\qquad\text{as }n\to+\infty\,, \end{align}

the first inequality following by (2.6). However, since by assumption $\mathcal G$ contains either a pendant or a signpost, proposition 4.3 ensures that $E_\rho(u_n,\mathcal G)=c_\rho(\mathcal G)+o(1) \lt c_\rho({\mathbb R})$ as µ → 0 (uniformly in ρ), contradicting (5.8). Hence, $u_{\rho}\not\equiv0$ on $\mathcal G$.

Up to possibly lowering the threshold on the mass, by lemma 5.1, we then have that u_ρ is a critical point of E_ρ constrained to $H_\mu^1(\mathcal G)$ at level $c_\rho(\mathcal G)$. Moreover, since u_ρ satisfies (5.7) and $\mathcal G$ is a non-compact graph with finitely many edges, we have $\lambda_\rho \gt 0$ and $u_\rho \gt 0$ on $\mathcal G$.

Step 2. Since the previous argument works for every $\mu\in(0,\mu_{p,\mathcal G}]$ and almost every $\displaystyle\rho\in\left[1/2,1\right]$, for some value $\mu_{p,\mathcal G} \gt 0$ depending on p and $\mathcal G$ only, by Step 1 there exists $u_\rho \gt 0$ critical point of E_ρ constrained to $H_\mu^1(\mathcal G)$ at level $c_\rho(\mathcal G)$, for every $\mu\in(0,\mu_{p,\mathcal G}]$ and for almost every $\displaystyle\rho\in\left[1/2,1\right]$. We then fix any µ in this interval and consider the limit $\rho\to1^-$ (along a subsequence of ρ’s, not relabelled). Recall first that u_ρ satisfies (5.7) with $\lambda_\rho \gt 0$. Since $\|u_\rho\|_2^2=\mu$ for every ρ, arguing exactly as in the proof of proposition 3.1 proves that $(\lambda_\rho)_\rho$ is bounded as $\rho\to1^-$. By (5.7), the boundedness of λ_ρ and that of $E_\rho(u_\rho,\mathcal G)=c_\rho(\mathcal G)$, we obtain that $(u_\rho)_\rho$ is bounded in $H_\mu^1(\mathcal G)$ as $\rho\to1^-$. All in all, this implies that, as ρ → 1, $(u_\rho)_\rho\subset H_\mu^1(\mathcal G)$ satisfies $E(u_\rho)\to c(\mathcal G)+o(1)$. Indeed, the continuity of the map $\rho\mapsto c_\rho(\mathcal G)$ as $\rho\to1^-$ follows repeating the argument of [Reference Jeanjean27, lemma 2.3]. Furthermore,

\begin{equation*} E'(u_\rho,\mathcal G)+\lambda_{\rho} u_\rho\to0\qquad\text{in the dual of }H^1(\mathcal G) \end{equation*}

where, up to subsequences, $\lambda_\rho\to\lambda\geq0$ as $\rho\to1^-$. Therefore, repeating verbatim the argument of Step 1 yields a positive critical point of E in $H_\mu^1(\mathcal G)$ at level $c(\mathcal G)$, i.e., a solution of (1.1) as desired.

Proof. Since the proof is very similar to that of lemma 5.1, we only highlight the main differences.

Let us consider first the case of a periodic graph $\mathcal G$ with no pendant. If we assume $m:=\|u\|_2^2\in(0,\mu)$, arguing exactly as above we obtain that $u\in S_{m,\rho}$ solves (5.1) for some $\lambda \in {\mathbb R}$, where up to subsequences $\lambda_n\to\lambda$ as $n\to+\infty$. Note that by lemma 2.5(b), $\lambda \geq 0$ and also, assuming µ > 0 sufficiently small, λ > 0 by lemma 2.5(c). Moreover, exploiting the periodicity of $\mathcal G$, we can consider $u_n-u$ and suitably translate it on $\mathcal G$ to obtain a function v_n such that, as $n\to+\infty$,

\begin{equation*} \begin{aligned} &\|v_n\|_2^2=\mu-m+o(1)\\ &E_\rho(v_n,\mathcal G)=c_\rho(\mathcal G)-E_\rho(u,\mathcal G)+o(1)\\ &E_\rho'(v_n,\mathcal G)+\lambda v_n\to0\,\text{in the dual of }H^1(\mathcal G) \end{aligned} \end{equation*}

and, for every n, v_n attains its $L^\infty$ norm in a given compact subset of $\mathcal G$ independent of n. Since $(v_n)_n$ is bounded in $H^1(\mathcal G)$ by definition, the argument used for u_n implies that $v_n\rightharpoonup w$ in $H^1(\mathcal G)$, for some w satisfying again (5.1) with the same λ as u. Iterating this procedure, we eventually end up with finitely many functions w_k, $k=1,\ldots, N$, each one solving (5.1) with the same λ > 0 (here $w_1=u$, $w_2=w$ as above) and such that

\begin{equation*} \mu=\|u_n\|_2^2=\sum_{k=1}^N\|w_k\|_2^2,\qquad c_\rho(\mathcal G)=E_\rho(u_n,\mathcal G)=\sum_{k=1}^NE_\rho(w_k,\mathcal G)+o(1) \end{equation*}

as $n \to +\infty$. Note however that $N\geq2$ since by assumption $\|w_1\|_2^2=m\in(0,\mu)$, so that in particular there exists $\overline{k}\in\left\{1,\ldots,N\right\}$ for which $\displaystyle\|w_{\overline{k}}\|_2^2\leq\frac\mu2$. Remark 3.2 and proposition 4.3 then ensure the existence of a value $\overline{\mu} \gt 0$, depending only on p and $\mathcal G$, such that for every $\displaystyle\rho\in\left[1/2,1\right]$ and $\mu \in (0, \overline{\mu}]$,

\begin{align*} & E_\rho(\phi_{\mu,\rho},{\mathbb R})+o\left(E_\rho(\phi_{\mu,\rho},{\mathbb R})\right)\geq c_\rho(\mathcal G) \gt E_\rho(w_{\overline{k}},\mathcal G)\\ &\geq E_\rho\left(\phi_{\frac\mu2,\rho},{\mathbb R}\right)+o\left(E_\rho\left(\phi_{\frac\mu2,\rho},{\mathbb R}\right)\right)\,, \end{align*}

i.e., a contradiction in view of (2.6).

If $\mathcal G$ has a pendant, the proof is exactly the same, making use of the corresponding estimates in remark 3.2 and proposition 4.3.

Proof. It is a direct consequence of the fact that $\lambda_\mathcal G=0$ (with $\lambda_\mathcal G$ given by (2.8)) for every periodic graph $\mathcal G$. Indeed, from this it follows that, for every λ < 0, there exists $w_0\in H^1(\mathcal G)$ with compact support such that

\begin{equation*} \|w_0'\|_2^2\leq \frac{-\lambda}{2-\lambda}\|w_0\|_2^2\,, \end{equation*}

i.e., w ₀ satisfies (6.1). Exploiting the periodicity of the graph, it is then straightforward to take $w_1,w_2\in H^1(\mathcal G)$ to be two different copies of w ₀, suitably translated on $\mathcal G$ so that $w_0, w_1, w_2$ have disjoint supports. The proof is then completed simply taking $Y=\text{span}\left\{w_0,w_1,w_2\right\}$.

Proof of theorem 1.2

The proof follows exactly the same line as that of theorem 1.1, replacing whenever needed lemma 5.1 with lemma 6.1 and [Reference Borthwick, Chang, Jeanjean and Soave13, lemma 3.2] with lemma 6.2. The only significant difference arises when one shows that the weak limit u_ρ of the Palais–Smale sequence $(u_n)_n\subset H_\mu^1(\mathcal G)$ for E_ρ at level $c_\rho(\mathcal G)$ is not identically equal to 0 on $\mathcal G$. In the case of a periodic graph, it is even easier to perform this passage. Indeed, exploiting the periodicity of $\mathcal G$, there is no loss of generality in taking from the very beginning each u_n to attain its $L^\infty$ norm in a given compact subset of $\mathcal G$ independent of n. Like this, if we assume by contradiction that $u_n\rightharpoonup0$ in $H^1(\mathcal G)$ as $n\to+\infty$, then $u_n\to0$ in $L^p(\mathcal G)$, in turn entailing (just as in the proof of theorem 1.1) $\|u_n'\|_2^2\to0$ and $E_\rho(u_n,\mathcal G)\to0$ as $n\to+\infty$, that is impossible since $E_\rho(u_n,\mathcal G)\to c_\rho(\mathcal G) \gt 0$ by assumption.

Proof. Recall first that, by [Reference De Coster, Dovetta, Galant, Serra and Troestler22], if $\mathcal G\in\mathbf{G}$ is a non-compact graph with finitely many edges, for every p > 2 and λ > 0 the minimization problem $\displaystyle \inf_{v\in\mathcal N_{\lambda,p}(\mathcal G)}J_{\lambda,p}(v,\mathcal G)$ admits a solution if

(7.4)\begin{equation} \inf_{v\in\mathcal N_{\lambda,p}(\mathcal G)}J_{\lambda,p}(v,\mathcal G) \lt \inf_{v\in\mathcal N_{\lambda,p}({\mathbb R})}J_{\lambda,p}(v,{\mathbb R})\,. \end{equation}

Hence, to prove the lemma, it is enough to show that (7.4) holds at a certain value of λ for every p sufficiently close to 6, whenever $\mathcal G$ has finitely many edges, exactly one half-line and no pendant. Moreover, since at fixed λ > 0 the map $\displaystyle p\mapsto\inf_{v\in\mathcal N_{\lambda,p}(\mathcal G)}J_{\lambda,p}(v,\mathcal G)$ is easily proved to be continuous, we are left to prove the validity of (7.4) at p = 6.

Let then p = 6. In this case, by the aforementioned result of [Reference Adami, Serra and Tilli5], there exists $\overline{u}\in H_{\mu_{\mathbb R}}^1(\mathcal G)$ such that

(7.5)\begin{equation} E(\overline{u},\mathcal G)=\inf_{u\in H_{\mu_{\mathbb R}}^1(\mathcal G)}E(u,\mathcal G) \lt 0\,. \end{equation}

By [Reference Dovetta, Serra and Tilli24, theorem 1.3], it then follows that

\begin{equation*} J_{\overline{\lambda},6}(\overline{u},\mathcal G)=\inf_{v\in\mathcal N_{\overline{\lambda},6}(\mathcal G)}J_{\overline{\lambda},6}(v,\mathcal G), \quad \text{with} \quad \overline{\lambda}:=\frac{\|\overline{u}\|_6^6-\|\overline{u}'\|_2^2}{\mu_{\mathbb R}}\,. \end{equation*}

By (7.1) and (7.5), we then have

\begin{equation*} \inf_{v\in\mathcal N_{\overline{\lambda},6}(\mathcal G)}J_{\overline{\lambda},6}(v,\mathcal G)=E(\overline{u},\mathcal G)+\frac{\overline{\lambda}}{2}\|\overline{u}\|_2^2 \lt \frac{\overline{\lambda}}{2}\mu_{\mathbb R}=\inf_{v\in\mathcal N_{\overline{\lambda},6}({\mathbb R})}J_{\overline{\lambda},6}(v,{\mathbb R})\,, \end{equation*}

and we conclude.

Proof of theorem 1.3

Since it is a standard fact that solutions of the problem $\displaystyle\inf_{v\in\mathcal N_\lambda(\mathcal G)}J_{\lambda,p}(v,\mathcal G)$ satisfy (1.1), in view of lemma 7.1 to complete the proof of theorem 1.3, it is enough to show that, as soon as p is sufficiently close to 6, functions $u\in\mathcal N_{\overline{\lambda},p}(\mathcal G)$ for which $\displaystyle J_{\overline{\lambda},p}(u,\mathcal G)=\inf_{v\in\mathcal N_{\overline{\lambda},p}(\mathcal G)}J_{\overline{\lambda},p}(v,\mathcal G)$ satisfy $E(u,\mathcal G) \lt 0$ (where $\overline{\lambda}$ is the number given by lemma 7.1).

To this end, we argue by contradiction. Assume that there exist exponents $p_n\to 6^+$ and functions $u_n\in\mathcal N_{\overline{\lambda},p_n}(\mathcal G)$ such that $\displaystyle J_{\overline{\lambda},p_n}(u_n,\mathcal G)=\inf_{v\in\mathcal N_{\overline{\lambda},p_n}(\mathcal G)}J_{\overline{\lambda},p_n}(v,\mathcal G)$ and $E(u_n,\mathcal G)\geq0$. Since p_n is bounded and the proof of lemma 7.1 shows that $\displaystyle \inf_{v\in\mathcal N_{\overline{\lambda},p_n}(\mathcal G)}J_{\overline{\lambda},p_n}(v,\mathcal G) \lt \inf_{v\in\mathcal N_{\overline{\lambda},p_n}({\mathbb R})}J_{\overline{\lambda},p_n}(v,{\mathbb R})$, it follows that $(u_n)_n$ is bounded in $H^1(\mathcal G)$. Hence, up to subsequences $u_n\rightharpoonup u$ in $H^1(\mathcal G)$ and $u_n\to u$ in $L_{loc}^\infty(\mathcal G)$ as $n\to+\infty$, for some $u\in H^1(\mathcal G)$.

Note first that $u\not\equiv0$ on $\mathcal G$. Indeed, if it were $u\equiv0$ on $\mathcal G$, we would have $u_n\to0$ in $L^\infty(\mathcal K)$. Therefore, denoting by $\mathcal H$ the unique half-line of $\mathcal G$, it would follow

(7.6)\begin{align} \inf_{v\in\mathcal N_{\overline{\lambda},p_n}(\mathcal G)}J_{\overline{\lambda},p_n}(v,\mathcal G)&=J_{\overline{\lambda},p_n}(u_n,\mathcal G)=J_{\overline{\lambda},p_n}(u_n,\mathcal H)+o(1) \nonumber\\ &\geq \inf_{v\in \mathcal N_{\overline{\lambda},p_n}({\mathbb R})} J_{\overline{\lambda},p_n}(v,{\mathbb R})+o(1) \end{align}

which is impossible by the proof of lemma 7.1.

Arguing in a similar way, we also observe that u_n converges strongly in $H^1(\mathcal H)$ to u. Indeed, u_n being a solution of (1.1) with $\lambda=\overline{\lambda}$ implies that $u_n(x)=\varphi_{\overline{\lambda},p_n}(x+a_n)$ for every $x\in\mathcal H$, where $\varphi_{\overline{\lambda},p_n}$ is the unique H ¹ solution of (2.1) with $\lambda=\overline{\lambda}$ and $p=p_n$ (and ρ = 1), and $a_n\in{\mathbb R}$ is a suitable number depending on n. It is then easy to see that

(7.7)\begin{equation} \liminf_{n\to+\infty}a_n \gt -\infty\,, \end{equation}

because if this were not the case we would obtain again a contradiction as in (7.6). By (7.7), $p_n\to6^+$, and the properties of $\varphi_{\overline{\lambda},p_n}$, it thus follows that the restriction of u_n to $\mathcal H$ converges strongly in $H^1(\mathcal H)$ to $\varphi_{\overline{\lambda},6}(\cdot+a)$, for a suitable $a\in{\mathbb R}\cup\{+\infty\}$ (if $a=+\infty$, then u_n tends to 0 strongly in $H^1(\mathcal H)$). By uniqueness of the limit, this means that $u(\cdot)\equiv\varphi_{\overline{\lambda},6}(\cdot+a)$ on $\mathcal H$ and, in particular,

(7.8)\begin{equation} \|u_n\|_{L^2(\mathcal H)}\to\|u\|_{L^2(\mathcal H)},\qquad\|u_n\|_{L^{p_n}(\mathcal H)}\to\|u\|_{L^6(\mathcal H)}\qquad\text{as }n\to+\infty\,. \end{equation}

Moreover, since $u_n\to u$ in $L^\infty_{loc}(\mathcal G)$, then $u_n(x)\to u(x)$ almost everywhere on the compact core $\mathcal{K}$ as $n\to+\infty$. Since $\mathcal K$ is a set of finite measure by assumption and u_n is uniformly bounded from above (by the uniform boundedness in $H^1(\mathcal G)$), by dominated convergence, we have

\begin{equation*} \|u_n\|_{L^{p_n}(\mathcal K)}\to\|u\|_{L^6(\mathcal K)}\qquad\text{as }n\to+\infty\,. \end{equation*}

Together with (7.8), the strong convergence of u_n to u in $L^2(\mathcal K)$ and weak lower semicontinuity, this yields

\begin{equation*} \begin{aligned} &\|u_n\|_2\to\|u\|_2,\qquad\|u_n\|_{p_n}\to\|u\|_6\qquad\text{as }n\to+\infty\\ &\liminf_{n\to+\infty}\|u_n'\|_2\geq\|u'\|_2\,, \end{aligned} \end{equation*}

that is

\begin{equation*} \sigma:=\frac{\|u'\|_2^2+\overline{\lambda}\|u\|_2^2}{\|u\|_6^6}\leq\liminf_{n\to+\infty}\frac{\|u_n'\|_2^2+\overline{\lambda}\|u_n\|_2^2}{\|u_n\|_{p_n}^{p_n}}=1\,, \end{equation*}

the last equality being a consequence of $u_n\in\mathcal N_{\overline{\lambda},p_n}(\mathcal G)$ for every n. Noting that $\sigma^{\frac14}u\in\mathcal N_{\overline{\lambda},6}(\mathcal G)$ by definition, we obtain

\begin{align*} &\inf_{v\in\mathcal N_{\overline{\lambda},6}(\mathcal G)}J_{\overline{\lambda},6}(v,\mathcal G)\leq J_{\overline{\lambda},6}\left(\sigma^{\frac14}u\right)\\ &\quad =\left(\frac12-\frac16\right)\sigma^{\frac32}\|u\|_6^6\leq \liminf_{n\to+\infty}\left(\frac12-\frac1{p_n}\right)\|u_n\|_{p_n}^{p_n}\\ &\quad =\liminf_{n\to+\infty}\inf_{v\in\mathcal N_{\overline{\lambda},p_n}(\mathcal G)}J_{\overline{\lambda},p_n}(v,\mathcal G)=\inf_{v\in\mathcal N_{\overline{\lambda},6}(\mathcal G)}J_{\overline{\lambda},6}(v,\mathcal G),\end{align*}

where we exploited also the continuity of the map $p\mapsto\inf_{v\in\mathcal N_{\overline{\lambda},p}(\mathcal G)}J_{\overline{\lambda},p}(v,\mathcal G)$. The previous chain of inequalities shows that σ = 1, $u\in\mathcal N_{\overline{\lambda},6}(\mathcal G)$, $J_{\overline{\lambda},6}(u,\mathcal G)=\inf_{v\in\mathcal N_{\overline{\lambda},6}(\mathcal G)}J_{\overline{\lambda},6}(v,\mathcal G)$ and that the convergence of u_n to u is strong in $H^1(\mathcal G)$. By lemma 7.1, [Reference Dovetta, Serra and Tilli24, theorem 1.3], and the aforementioned results of [Reference Adami, Serra and Tilli5], this implies that $\|u\|_2^2=\mu_{\mathbb R}$ and $E(u,\mathcal G)=\inf_{v\in H_{\mu_{\mathbb R}}^1(\mathcal G)}E(v,\mathcal G) \lt 0$. By strong convergence in $H^1(\mathcal G)$, this entails $E(u_n,\mathcal G) \lt 0$ for sufficiently large n, providing the contradiction we seek and concluding the proof.

Proof of theorem 1.4

Let $\mathcal G$ be a non-compact graph with finitely many edges, at least one of which bounded, and such that every vertex is attached at an even number (possibly zero) of half-lines. Observe first that, on a graph with these properties, for every λ > 0, there exists a positive solution u_λ of (1.1). Indeed, since for every vertex $\textrm{v}_i\in\mathbb{V}_{\mathcal G}$ there exists $k_i\in{\mathbb N}$ such that $\textrm{v}_i$ is attached to $2k_i$ half-lines, we can group these half-lines in pairs and think of them as k_i full lines crossing at the same point $\textrm{v}_i$. Then, identifying $\textrm{v}_i$ with the origin of each of these k_i lines and denoting by φ_λ the unique solution of the NLS equation (2.1) on ${\mathbb R}$ (with ρ = 1), it is readily seen that the function

\begin{equation*} u_\lambda(x):=\begin{cases} \lambda^{\frac1{p-2}} & \text{for }x\in\mathcal K\\ \varphi_\lambda(x+\tau_\lambda) & \text{for }x\,\text{on each of the} k_{\textrm{i}} \text{lines attached to} \textrm{v}_i, \text{for every} \textrm{v}_i\in\mathbb{V}_{\mathcal G}, \end{cases} \end{equation*}

with $\mathcal K$ the compact core of $\mathcal G$ and $\tau_\lambda \gt 0$ such that $\displaystyle\varphi_\lambda(\tau_\lambda)=\lambda^{\frac1{p-2}}$, solves (1.1) on $\mathcal G$. Note that τ_λ exists because

\begin{equation*} \| \varphi_{\lambda} \|_{\infty} = \varphi_{\lambda}(0) = \left( \frac{\lambda p}{2} \right)^{\frac{1}{p-2}} \gt \lambda^{\frac{1}{p-2}}. \end{equation*}

To complete the proof of theorem 1.4, we are thus left to show that, varying λ, u_λ has negative energy as soon as its mass is sufficiently large. Since u_λ is made of a constant function on the compact core of $\mathcal G$ and a certain number, say k, of full copies of φ_λ, we have

\begin{align*} \| u_{\lambda} \|_{2}^{2} &= |\mathcal K| \lambda^{\frac{2}{p-2}} + k \| \varphi_{\lambda} \|_{2}^{2} = |\mathcal K| \lambda^{\frac{2}{p-2}} + k \lambda^{\frac{6-p}{2(p-2)}} \| \varphi_1 \|_{2}^{2}\,,\\ \| u_{\lambda} \|_{p}^{p} &= |\mathcal K| \lambda^{\frac{p}{p-2}} + k \| \varphi_{\lambda} \|_{p}^{p} = |\mathcal K| \lambda^{\frac{p}{p-2}} + k \lambda^{\frac{p+2}{2(p-2)}} \| \varphi_1 \|_{p}^{p}\,,\\ \| u_{\lambda}' \|_{2}^{2} &= k \| \varphi'_{\lambda} \|_{2}^{2} = k \lambda^{\frac{p+2}{2(p-2)}} \| \varphi'_1 \|_{2}^{2}\,, \end{align*}

where we used the well-known relation $\displaystyle \varphi_{\lambda}(x) = \lambda^{\frac{1}{p-2}} \varphi_1\left(\sqrt{\lambda}x\right).$ Therefore,

\begin{equation*} E(u_{\lambda},\mathcal G) = \frac{1}{2}\| u_{\lambda}' \|_{2}^{2} - \frac{1}{p} \| u_{\lambda} \|_{p}^{p} = k \left( \frac{\| \varphi'_1 \|_{2}^{2}}{2} - \frac{\| \varphi_1 \|_{p}^{p}}{p} \right) \lambda^{\frac{p+2}{2(p-2)}} - \frac{|\mathcal K|}{p} \lambda^{\frac{p}{p-2}}. \end{equation*}

Since p > 2, one has $\frac2{p-2} \gt 0$ and $\frac{p}{p-2} \gt \frac{p+2}{2(p-2)}$. Hence, if λ is large enough (depending on k, $|\mathcal K|$ and p), then $\|u_\lambda\|_2$ is large and $E(u_{\lambda},\mathcal G) \lt 0$.