Proof. Let us first observe that the minimization problem 12 is decoupled and can be reduced to an algebraic problem, in particular \[\mathcal{F}^{\infty}(\mu)=\inf\{\mathcal{E}_{p_1}(\mu_1) + \mathcal{E}_{p_2}(\mu_2):\mu_1 + \mu_2 =\mu\}.\] It is well known that \(\mathcal{E}_p(\mu)<0\) for every \(\mu>0\) and \(p>2\) and \(\mathcal{E}_p(\mu)>-\infty\) for every \(\mu>0\) and \(p\) belonging to the \(L^2\)-subcritical regime, i.e. \(2<p<4\). Moreover, by scaling properties of the energy, it follows that \(\mathcal{E}_p(\mu)=-\rho_p \mu^{\frac{2}{4-p}}\), with \(\rho_p=-\mathcal{E}_p(1)\), thus \[\mathcal{F}^{\infty} (\mu) = \inf_{m\in[0,\mu]} g(m),\] where \[g(m)= -\rho_{p_1}\,m^{\frac{2}{4-p_1}} - \rho_{p_2}\,(\mu-m)^{\frac{2}{4-p_2}}.\] The function \(g\) is continuous in \([0,\mu]\), of class \(C^2\) in \((0,\mu)\) and \[\begin{align} g''(m)&= -\frac{2(p_1-2)}{(4-p_1)^2}\,\rho_{p_1} m^{\frac{2(p_1-3)}{4-p_1}} - \frac{2(p_2-2)}{(4-p_2)^2} \,\rho_{p_2}(\mu-m)^{\frac{2(p_2-3)}{4-p_2}} <0 \quad \forall\, m\in (0,\mu), \end{align}\] hence \(g\) attains the minimum either at \(0\) or at \(\mu\), entailing the thesis. ◻

Proof. Let us introduce for every \(\mu>0\) the function \[h(\mu):=\mathcal{E}_{p_1}(\mu,\mathbb{R}^2) -\mathcal{E}_{p_2}(\mu,\mathbb{R}^2)=-\rho_{p_1}\mu^{\frac{2}{4-{p_1}}} +\rho_{p_2}\mu^{\frac{2}{4-p_2}},\] and observe that \[\lim_{\mu\to 0}h(\mu)=0,\quad \lim_{\mu\to+\infty}h(\mu)=\begin{cases} -\infty, \quad \text{if}\quad p_1>p_2,\\ +\infty,\quad \text{if}\quad p_1<p_2, \end{cases}\] By direct computation, \[h'(\mu)=-\frac{2\rho_{p_1}}{4-p_1}\mu^{\frac{p_1-2}{4-p_1}} + \frac{2\rho_{p_2}}{4-p_2}\mu^{\frac{p_2-2}{4-{p_2}}},\] hence there exists \(\widetilde{\mu}>0\) such that \(h'(\mu)<0\) if and only if \((p_1-p_2)(\mu-\widetilde{\mu})<0\). This entails that \(h(\mu)<0\) if and only if \((p_1-p_2)(\mu-\mu^\star)>0\), with \(\mu^\star\) being the only value for which \(h(\mu)=0\): the thesis follows from the definition of \(h\). ◻

Proof of Theorem 1. Let \(U_{n}=(u_{1,n},u_{2,n})\) be a minimizing sequence for 4 at mass \(\mu\), i.e. \[\|U_n\|_{2}^2=\mu\quad \forall\,n\quad\text{and}\quad F(U_n)\to \mathcal{F}(\mu)\quad \text{as}\quad n\to+\infty,\] and \(q_{i,n}\) be the charge associated with \(u_{i,n}\) for \(i=1,2\). Let \(\sigma=\min\{\sigma_1,\sigma_2\}\), fix \(\lambda=4e^{4\pi(2\beta-\sigma)-2\gamma}\) and consider the decomposition \(u_{i,n}=\phi_{i,n}+q_{i,n}\mathcal{G}_\lambda\) for \(i=1,2\): when not necessary, we omit the parameter \(\lambda\) to lighten the notation.

We divide the proof in three steps.

Step 1. There exist \(\phi_i\in H^1(\mathbb{R}^2)\) and \(q_i\in \mathbb{C}\) such that, up to subsequences, \[\phi_{i,n}\rightharpoonup\phi_i \quad \text{in}\quad H^1(\mathbb{R}^2),\quad q_{i,n}\to q_i \quad \text{in}\quad \mathbb{C}.\] By using that \(\|U_n\|_{2}^2=\mu\), 16 and 15 and observing that \[(\sigma+\theta_\lambda)|q_{i,n}|^2\leqslant(\sigma_i+\theta_\lambda)|q_{i,n}|^2\leqslant\mathcal{N}_{\sigma_i}(u_{i,n})^2,\quad \forall\,n\in\mathbb{N},\quad i=1,2,\] one obtains that \[\label{bound-Fun-partial} \begin{align} F(U_n)\geqslant& \frac{1}{2}\mathcal{N}_{\sigma_1}(u_{1,n})^2+ \frac{1}{2}\mathcal{N}_{\sigma_2}(u_{2,n})^2-\frac{\beta}{\sigma+\theta_\lambda} \mathcal{N}_{\sigma_1}(u_{1,n})\mathcal{N}_{\sigma_2}(u_{2,n})\\ &- \frac{\lambda}{2}\mu - \frac{C_{p_1}\,\mu}{p_1}\mathcal{N}_{\sigma_1}(u_{1,n})^{p_1-2} - \frac{C_{p_2}\,\mu}{p_2}\mathcal{N}_{\sigma_2}(u_{2,n})^{p_2-2} \end{align}\tag{18}\] Since \(\sigma+\theta_\lambda=2\beta\) according to our choice of \(\lambda\), 18 entails \[\label{bound-Fun} F(U_n)\geqslant\frac{1}{4}\mathcal{N}_{\sigma_1}(u_{1,n})^2+ \frac{1}{4}\mathcal{N}_{\sigma_2}(u_{2,n})^2-\frac{C_{p_1}\,\mu}{p_1}\mathcal{N}_{\sigma_1}(u_{1,n})^{p_1-2} - \frac{C_{p_2}\,\mu}{p_2}\mathcal{N}_{\sigma_2}(u_{2,n})^{p_2-2} -\frac{\lambda}{2}\mu.\tag{19}\] Since \(F(U_n)<0\) and \(p_i<4\), by 19 it follows that \(\mathcal{N}_{\sigma_i}(u_{i,n})\) are equibounded for \(i=1,2\), in particular \(\phi_{i,n}\) are equibounded in \(H^1(\mathbb{R}^2)\) and \(q_{i,n}\) are equibounded in \(\mathbb{C}\), hence by Banach-Alaoglu-Bourbaki theorem the thesis of Step 1 follows.
Step 2. It holds that \(\mathcal{F}(\mu)<\mathcal{F}^\infty(\mu)\). By lemma 1 and [26], we have that \[\label{eq:Finf62F1F2} \mathcal{F}^{\infty}(\mu)=\min\{\mathcal{E}_{p_1}(\mu),\mathcal{E}_{p_2}(\mu)\}> \min\{\mathcal{F}_{1}(\mu), \mathcal{F}_{2}(\mu)\}.\tag{20}\]

Moreover, let us notice that \[\label{eq:inf-Fmu-moduli} \inf_{U\in\Lambda_{\mu}}F(U)= \inf_{U\in\Lambda_{\mu}}\{F_{1}(u) + F_{2}(v) - \beta \abs{q_{1}}\,\abs{q_{2}}\}\quad\forall\,\mu>0.\tag{21}\] Indeed, on the one hand it holds for every \(U=(u,v)\in \Lambda_\mu\) that \(F(U)\geqslant F_1(u)+F_2(v)-\beta|q_1||q_2|\). On the other hand, if \(u=\phi_1+q_1\mathcal{G}_\lambda\) and \(v=\phi_2+q_2\mathcal{G}_\lambda\), then it is possible to find a proper phase factor \(e^{i\theta}\) such that the function \(\widetilde{U}=(u,e^{i\theta} v)\) satisfies \[\mathrm{Re}\left\{q_1\overline{e^{i\theta}q_2}\right\}=|q_1||q_2|,\] hence passing to the infimum the equality is achieved.

By 21 , it follows that \[\label{eq:Fmu-beta0} \begin{align} \mathcal{F}(\mu)=\inf_{U\in\Lambda_{\mu}}F(U)&= \inf_{U\in\Lambda_{\mu}}\{F_{1}(u) + F_{2}(v) - \beta \abs{q_{1}}\,\abs{q_{2}}\}\\ &\leqslant\inf_{U\in\Lambda_{\mu}}\{F_{1}(u) + F_{2}(v)\}\\ &=\inf_{t\in[0,\mu]}\{\mathcal{F}_1(t)+\mathcal{F}_2(\mu-t)\}. \end{align}\tag{22}\] Since \(\mathcal{F}_{1}\) and \(\mathcal{F}_{2}\) are strictly concave functions of the mass by [1], then also the function \(\mathcal{F}_1(t)+\mathcal{F}_2(\mu-t)\) turns out to be a strictly concave function of \(t\) and attains its minimum at \(t=0\) or \(t=\mu\), so that \[\label{eq:inf-t-F1F2} \inf_{t\in[0,\mu]}\{\mathcal{F}_1(t)+\mathcal{F}_2(\mu-t)\}=\min\{\mathcal{F}_{1}(\mu), \mathcal{F}_{2}(\mu)\}.\tag{23}\] By combining 20 , 22 and 23 , we conclude the proof of Step 2.
Step 3. The function \(U=(u_1,u_2)=(\phi_1+q_1\mathcal{G}_\lambda, \phi_2+q_2\mathcal{G}_\lambda)\) is a ground state of 4 at mass \(\mu\). By Step 1, we deduce that \(u_{i,n}\rightharpoonup u_i\) weakly in \(L^2(\mathbb{R}^2)\) since \(\phi_{i,n}\rightharpoonup\phi_i\) weakly in \(L^2(\mathbb{R}^2)\) and \(q_{i,n}\mathcal{G}_\lambda\to q_i\mathcal{G}_\lambda\) in \(L^2(\mathbb{R}^2)\).

Set now \(m=\|U\|_{L^{2}(\mathcal{I})}^{2}:=\|u_1\|_{L^{2}(\mathbb{R}^2)}^{2}+\|u_2\|_{L^{2}(\mathbb{R}^2)}^{2}\). First, we observe that by weak lower semicontinuity of the norms \[\|u_1\|_{L^{2}(\mathbb{R}^2)}^{2}+\|u_2\|_{L^{2}(\mathbb{R}^2)}^{2}\leqslant\liminf_{n\to+\infty}\|u_{1,n}\|_{L^{2}(\mathbb{R}^2)}^{2}+\liminf_{n\to+\infty}\|u_{2,n}\|_{L^{2}(\mathbb{R}^2)}^{2}\leqslant\liminf_{n\to+\infty}\|U_{n}\|_{L^{2}(\mathcal{I})}^{2},\] thus \(m\leqslant\mu\).

Suppose first that \(m=0\), i.e. \(u\equiv 0\) and \(v\equiv 0\). In particular, this entails that \(q_{1,n}\to 0\) and \(q_{2,n}\to 0\) as \(n\to +\infty\). Therefore, relying on Step 2 we deduce that for \(n\) sufficiently large \[\begin{align} \mathcal{F}^{\infty}(\mu)>\mathcal{F}(\mu)=& \lim_{n\to +\infty}\left\{ F_{1}(u_{1,n}) + F_{2}(u_{2,n}) -\beta \,\abs{q_{1,n}}\,\abs{q_{2,n}}\right\}\\ =& \lim_{n\to +\infty} \left\{E_{p_1}(\phi_{1,n}) + E_{p_2}(\phi_{2,n})\right\}\\ \geqslant& \lim_{n\to +\infty}\left\{\mathcal{E}_{p_1}({\left\Vert\phi_{1,n}\right\Vert}^2_2) + \mathcal{E}_{p_2}({\left\Vert\phi_{2,n}\right\Vert}_2^2)\right\}\\ =& \lim_{n\to +\infty}\left\{\mathcal{E}_{p_1}({\left\Vert u_{1,n}\right\Vert}^2_2) + \mathcal{E}_{p_2}({\left\Vert u_{2,n}\right\Vert}_2^2)\right\}\\ \geqslant&\, \inf\{\mathcal{E}_{p_1}(\mu_1) \,+\, \mathcal{E}_{p_2}(\mu_2):\mu_1 +\mu_2=\mu\}=\mathcal{F}^{\infty}(\mu), \end{align}\] thus \(m\neq 0\).

Suppose now that \(0<m<\mu\). Note that, since \(u_{i,n}\rightharpoonup u_i\) in \(L^{2}(\mathbb{R}^2)\) for \(i=1,2\), then \[\label{eq:mass-m-mu} \|U_{n}-U\|_{L^{2}(\mathcal{I})}^{2}={\left\Vert U_n\right\Vert}_2^2 - {\left\Vert U\right\Vert}_2^2 + o(1) =\mu-m+o(1),\quad\text{as}\quad n\to+\infty.\tag{24}\]

On the one hand, since \(p_i>2\) for \(i=1,2\), \(\frac{\mu}{\|U_{n}-U\|_{2}^{2}}=\frac{\mu}{\mu-m}+o(1)>1\) as \(n\to+\infty\) by 24 and \({\left\Vert\sqrt{\frac{\mu}{\|U_{n}-U\|_{2}^{2}}} (U_{n}-U)\right\Vert}_2^2 =\mu\), there results that \[\begin{align} \mathcal{F}(\mu)&\leqslant F\left(\sqrt{\frac{\mu}{\|U_{n}-U\|_{2}^{2}}} (U_{n}-U)\right)\\ &=\frac{1}{2}\frac{\mu}{\|U_{n}-U\|_{2}^{2}}Q(U_n-U)-\sum_{i=1,2}\frac{1}{p_i} \left(\frac{\mu}{\|U_{n}-U\|_{2}^{2}}\right)^{\frac{p_i}{2}}{\left\Vert u_{i,n} -u_i\right\Vert}_{p_i}^{p_i}\\ &<\frac{\mu}{\|U_{n}-U\|_{L^{2}(\mathcal{I})}^{2}}F(U_{n}-U), \end{align}\] hence \[\label{eq:FUn-U} \liminf_{n\to+\infty} F(U_{n}-U)\geqslant\frac{\mu-m}{\mu}\mathcal{F}(\mu).\tag{25}\] On the other hand, by similar computations \[\mathcal{F}(\mu)\leqslant F\left(\sqrt{\frac{\mu}{\|U\|_{2}^{2}}}U\right)<\frac{\mu}{\|U\|_{2}^{2}}F(U)\] thus \[\label{eq:FU62mmu} F(U)>\frac{m}{\mu}\mathcal{F}(\mu).\tag{26}\]

In addition, since \(u_{i,n}\rightharpoonup u_i\) in \(L^2(\mathbb{R}^2)\), \(\phi_{i,n}\rightharpoonup\phi_{i}\) in \(H^{1}(\mathbb{R}^2)\) and \(q_{i,n}\to q_i\) in \(\mathbb{C}\), then \[\begin{align} Q(U_{n}-U)&=Q(U_{n})-Q(U)+o(1),\quad\text{as}\quad n\to +\infty, \end{align}\] while by Brezis-Lieb lemma [33] applied to \(u_{i,n}\) we get \[\|u_{i,n}\|_{L^{p_i}(\mathbb{R}^2)}^{p_i}=\|u_{i,n}-u\|_{L^{p_i}(\mathbb{R}^2)}^{p_i}+\|u\|_{L^{p_i}(\mathbb{R}^2)}^{p_i}+o(1),\quad\text{as}\quad n\to+\infty.\] Thus \[\label{eq:BrezisLieb} F(U_{n})= F(U_{n}-U)+F(U)+o(1),\quad\text{as}\quad n\to+\infty.\tag{27}\] Combining 25 , 26 and 27 , it follows that \[\mathcal{F}(\mu)=\liminf_{n\to+\infty} F(U_{n})=\liminf_{n\to+\infty} F(U_{n}-U)+F(U)>\frac{\mu-m}{\mu}\mathcal{F}(\mu)\,+\,\frac{m}{\mu}\mathcal{F}(\mu)=\mathcal{F}(\mu),\] which, being a contradiction, forces \(m=\mu\). In particular, \(U\in \Lambda_{\mu}\) and thus \(u_{i,n}\to u_i\) in \(L^{2}(\mathbb{R}^2)\) and \(\phi_{i,n}\to \phi_{i}\) in \(L^{2}(\mathbb{R}^2)\).

In order to conclude that \(U\) is a ground state, we are left to prove that \[\label{Fsigma-lsc} F(U)\leqslant\liminf_{n\to+\infty} F(U_{n})=\mathcal{F}(\mu),\tag{28}\] which reduces to prove that \(u_{i,n}\to u_i\) in \(L^{p_i}(\mathbb{R}^2)\) for \(i=1,2\). By 16 , we obtain that \[\|u_{i,n}-u_i\|_{L^{p_i}(\mathbb{R}^2)}^{p_i}\leqslant C_{p_i}\mathcal{N}_{\sigma_i}(u_{i,n}-u_i)^{p_i-2}\|u_{i,n}-u_i\|_{L^{2}(\mathbb{R}^2)}^{2}\to 0,\quad \text{as}\quad n \to +\infty,\] which concludes the proof since \(\mathcal{N}_{\sigma_i}(u_{i,n}-u_i)\) is equibounded and \(u_{i,n}\to u_i\) in \(L^{2}(\mathbb{R}^2)\). ◻

Proof. Let \(U=(u_1,u_2)\) be a ground state of \(F\) at mass \(\mu\) and \(\lambda_i>0\) for \(i=1,2\), so that \(u_i=\phi_i+q_i\mathcal{G}_{\lambda_i}\). Since \(U\) solves 47 , there results that

\[\begin{cases} \phi_{1}(0)&= (\sigma_1 + \theta_{\lambda_1})q_1-\beta\,q_2,\\ \phi_{2}(0)&=-\beta\,q_1 + (\sigma_2 + \theta_{\lambda_2})q_2. \end{cases}\]

Step 1. \(u_i\not\equiv 0\) for \(i=1,2\). Let us first prove that \(u_1\not\equiv 0\). Suppose by contradiction that \(u_1\equiv 0\). Therefore, since \(\beta>0\), it follows that \(q_2=0\), hence \(u_2\in H^1(\mathbb{R}^2)\). Therefore, since \(U\) is a ground state of \(F\) at mass \(\mu\), by Step 2 in Theorem 1 it holds that \[\mathcal{F}(\mu)=F(U)= F_{2}(u_2)= \mathcal{E}_{p_2}(\mu)\geqslant\mathcal{F}^{\infty}(\mu)>\mathcal{F}(\mu),\] which is a contradiction. Analogously, one can prove that \(u_2\not\equiv 0\).
Step 2. \(q_i\neq 0\) for \(i=1,2\). Let us first prove that \(q_1\neq 0.\) If we suppose by contradiction that \(q_1=0\), then \[F(U)= E_{p_1}(u_1) + F_{2}(u_2)=\inf_{(u_1,u_2)\in \Lambda_\mu} \{E_{p_1}(u_1)+F_2(u_2)\}.\] In particular, since the problems on the two planes are decoupled in this case, one can argue as in the proof of Lemma 1 and prove that \[\label{eq:inf-F-m-mu} F(U)=\inf_{(u_1,u_2)\in \Lambda_\mu} \{E_{p_1}(u_1)+F_2(u_2)\}= \inf_{m\in[0,\mu]} \{\mathcal{E}_{p_1} (m) + \mathcal{F}_{2}(\mu-m) \}.\tag{29}\]

Since both \(\mathcal{E}_{p_1}\) and \(\mathcal{F}_{2}\) are strictly concave functions of the mass, then also the function \(m\mapsto\mathcal{E}_{p_1} (m) + \mathcal{F}_{2}(\mu-m)\) is strictly concave in \([0,\mu]\) and attains its minimum at \(m=0\) or \(m=\mu\). If one considers \(U=(u_1,u_2)\) with \(u_i\not\equiv 0\) for \(i=1,2\), then \(U\) cannot realize the infimum in 29 : indeed, if this is the case, then by strict concavity of the function \(m\mapsto\mathcal{E}_{p_1} (m) + \mathcal{F}_{2}(\mu-m)\) there exists a function \(\widetilde{U}\) with either \(u_1\equiv 0\) or \(u_2\equiv 0\) with energy strictly less than the energy of \(U\), contradicting the hypothesis that \(U\) is a ground state of \(F\). Therefore, it holds that either \(u_1\equiv 0\) or \(u_2\equiv 0\), but this is in contradiction with Step 1. In the same way, it is possible to prove that \(q_2\neq 0\), concluding the proof of Step 2.
Step 3. \(\phi_i\not \equiv 0\) for \(i=1,2\). Let us prove first that \(\phi_{1}\not\equiv 0\). Indeed, if we suppose by contradiction that \(\phi_{1}\equiv 0\), then by the first equation in 47 we have that \[\begin{align} &\omega-\lambda_1=|q_1|^{p_1-2}\mathcal{G}_{\lambda_{1}}^{p_1-2}(x),\qquad \, \forall\, x\in\mathbb{R}^2\setminus\{0\}, \end{align}\] but this is a contradiction since the left-hand side is constant, while the right-hand side is a multiple of \(\mathcal{G}_{\lambda_1}\), hence \(\phi_{1}\not\equiv 0.\) Analogously, one can show that \(\phi_{2}\not\equiv 0.\)
Step 4. \(q_i>0\) for \(i=1,2\), up to a phase factor. First of all, let us observe that \(q_1=|q_1|e^{i\varphi_1}\) and \(q_2=|q_2|e^{i\varphi_2}\) share the same phase, i.e. \(e^{i\varphi_1}=e^{i\varphi_2}\). Indeed, suppose by contradiction that \(e^{i\varphi_1}\neq e^{i\varphi_2}\). Then the function \(U_\varphi=(u_1,e^{i(\varphi_1-\varphi_2)}u_2)\) satisfies \(\|U_\varphi\|_2^2=\|U\|_2^2\) and \[\begin{align} F(U_\varphi) &= F_1(u_1)+F_2\left(u_2 e^{i(\varphi_1-\varphi_2)}\right)-\beta \mathrm{Re}\{q_1\overline{q_2}e^{-i(\varphi_1-\varphi_2)}\}\\ &=F_1(u_1)+F_2(u_2)-\beta|q_1||q_2|\\ &<F_1(u_1)+F_2(u_2)-\beta\mathrm{Re}\{q_1\overline{q_2}\}=F(U), \end{align}\] which contradicts the fact that \(U\) is a ground state of \(F\) at mass \(\mu\). Now, if \(q_i\) are not positive, i.e. \(q_i=|q_i|e^{i\varphi}\) with \(e^{i\varphi}\neq 1\), then it is sufficient to observe that the function \(e^{-i\varphi}U\) has positive charges \(q_i\) and satisfies \(\|e^{-i\varphi}U\|_2^2=\mu\) and \(F\left(e^{-i\varphi}U\right)=F(U)\), thus without loss of generality we can suppose that \(q_i>0\) for \(i=1,2\). ◻

Proof. Let \(U=(u_1,u_2)\) be a ground state of \(F\) at mass \(\mu\) and \(\omega>0\) be the associated Lagrange multiplier, given by 32 . Assume by contradiction that there exists \(V=(v_1,v_2)\in N_{\omega}\) such that \(S_{\omega}(V)<S_{\omega}(U)\) and let \(\tau>0\) be such that \(||\tau V||_{L^2(\mathbb{R}^2)}^2=\mu\). Then \[\begin{align} S_{\omega}(\tau V)= \frac{\tau^2}{2} Q_{\omega}(V) - \frac{\tau^{p_1}}{p_1} {\left\Vert v_1\right\Vert}_{L^{p_1}(\mathbb{R}^2)}^{p_1}- \frac{\tau^{p_2}}{p_2} {\left\Vert v_2\right\Vert}_{L^{p_2}(\mathbb{R}^2)}^{p_2}. \end{align}\] Computing the derivative with respect to \(\tau\) and using that \(V\in N_{\omega}\), we get \[\frac{d}{d\tau} S_{\omega}(\tau V)=\tau\left[(1-\tau^{p_1-2}){\left\Vert v_1\right\Vert}_{p_1}^{p_1} + (1-\tau^{p_2-2}){\left\Vert v_2\right\Vert}_{p_2}^{p_2}\right],\] which is greater than equal to zero if and only if \(0<\tau\leqslant 1\). Hence \(S_{\omega}(\tau V)\leqslant S_{\omega}(V)<S_{\omega}(U)\) for every \(\tau>0\) which, combined with \(||\tau V||_{2}^2={\left\Vert U\right\Vert}_{2}^2=\mu\), entails that \(F(\tau V)<F(U)\). Since this contradicts the fact that \(U\) is a ground state of \(F\) at mass \(\mu\), it follows that \(U\) is a minimizer of the action at frequency \(\omega\).

Finally, 32 follows from \[\begin{align} -\omega &= \frac{Q(U) - {\left\Vert u_1\right\Vert}^{p_1}_{p_1} - {\left\Vert u_2\right\Vert}^{p_2}_{p_2}}{\mu}=\frac{2{F(U)} - \frac{p_1-2}{p_1}{\left\Vert u_1\right\Vert}_{p_1}^{p_1} - \frac{p_2-2}{p_2}{\left\Vert u_2\right\Vert}_{p_2}^{p_2}}{\mu}\\ &<2\frac{\mathcal{F}(\mu)}{\mu}< \inf_{U\in \Lambda_\mu}\frac{Q(U)}{\mu}=- \omega^\star. \end{align}\] ◻

Proof of Theorem 2. Let us focus first on the proof of 7 and 8 , when \(\beta=0\) and \(\beta>0\) respectively. In particular, if \(\beta=0\), then \[\mathcal{F}(\mu)=\inf_{U\in\Lambda_\mu} \left\{F_1(u) + F_2(v)\right\}=\inf_{t\in[0,\mu]} \left\{\mathcal{F}_{1}(t) + \mathcal{F}_{2}(\mu-t)\right\},\] hence by applying 23 we deduce 7 and the fact that every ground state is either of the form \(U=(u_1,0)\) or of the form \(U=(0,u_2)\), depending on which is the minimum between \(\mathcal{F}_1(\mu)\) and \(\mathcal{F}_2(\mu)\): this entails in particular \((i)\), \((ii)\) and \((iii)\) and the properties of the non-zero component.

If instead \(\beta>0\), then \[\mathcal{F}(\mu)=\inf_{U\in\Lambda_\mu} \left\{F_1(u) + F_2(v)-\beta|q_1||q_2|\}\right\}\leqslant\inf_{t\in[0,\mu]} \left\{\mathcal{F}_{1}(t) + \mathcal{F}_{2}(\mu-t)\right\},\] thus by 23 we get that \[\mathcal{F}(\mu)\leqslant\min\{\mathcal{F}_{1}(\mu), \mathcal{F}_{2}(\mu)\}.\]

In order to prove 8 , it is sufficient to exclude that \(\mathcal{F}(\mu)=\min\{\mathcal{F}_1(\mu), \mathcal{F}_2(\mu)\}\). If we suppose by contradiction that \(\mathcal{F}(\mu)=\min\{\mathcal{F}_1(\mu), \mathcal{F}_2(\mu)\}\), then there exist also ground states either of the form \(U=(u_1,0)\) or of the form \(U=(0,u_2)\). But this contradicts Proposition 8, hence 8 holds.

Fix now \(\beta>0\) and let \(U=(u_1,u_2)\) be a ground state of \(F\) at mass \(\mu\): its existence is ensured by Theorem 1. By proposition 8, we deduce that the associated charges \(q_i\) satisfy \(q_i>0\) for \(i=1,2\), up to multiplication by a phase factor. Moreover, by Lemma 2 \(U\) is also a minimizer of the action at frequency \(\omega\), with \(\omega>\omega^\star\).

Let us now consider the decomposition of \(u_i\) corresponding to the Lagrange multiplier \(\omega\), namely \(u_i=\phi_{i,\omega}+q_i\mathcal{G}_\omega\). Since the proof of the positivity and the radiality of \(\phi_{i,\omega}\) and \(u_i\) is quite long, we divide it into three steps.
Step 1. Positivity of \(\phi_{i,\omega}\) and \(u_i\). We first prove that \(\phi_{1,\omega}>0\). Let us start defining the two sets \[\Omega := \{x\in\mathbb{R}^2\setminus\{0\}:\phi_{1,\omega}(x)\neq 0\}\] and \[\widehat{\Omega}:=\{x\in\Omega : \alpha(x)\neq 0\},\] where \(\alpha:\Omega\to [0,2\pi)\) is such that \(\phi_{1,\omega}(x)=e^{i\alpha(x)}|\phi_{1,\omega}(x)|\). To prove that \(\phi_{1,\omega}>0\), we are reduced to prove that \(\Omega=\mathbb{R}^2\) and \(|\widehat{\Omega}|=0\).

We start proving that \(|\widehat{\Omega}|=0\). Assume by contradiction that \(\abs{\widehat{\Omega}}>0\) and introduce the function \[U_\tau:=(u_{1,\tau},u_2),\quad \text{with}\quad u_{1,\tau}:=\tau\abs{\phi_{1,\omega}}+q_1\mathcal{G}_\omega.\] Let us observe that, since \(\|\nabla\abs{\phi_{1,\omega}}\|_2^2\leqslant\|\nabla\phi_{1,\omega}\|_2^2\), then for every \(\tau\in (-1,1)\) \[\begin{align} Q_{\omega}(U_{\tau})&=\tau^2\left({\left\Vert\nabla \abs{\phi_{1,\omega}}\right\Vert}_{L^2(\mathbb{R}^2)}^2 + \omega {\left\Vert\phi_{1,\omega}\right\Vert}_2^2\right) + \abs{q_1}^2(\sigma_1 + \theta_{\omega})+Q_{\sigma_2}(u_2) + \omega {\left\Vert u_2\right\Vert}_2^2-\beta\, q_1 \,q_2\\ &<{\left\Vert\nabla \abs{\phi_{1,\omega}}\right\Vert}_{L^2(\mathbb{R}^2)}^2 + \omega {\left\Vert\phi_{1,\omega}\right\Vert}_2^2 + \abs{q_1}^2(\sigma_1 + \theta_{\omega})+Q_{\sigma_2}(u_2) + \omega {\left\Vert u_2\right\Vert}_2^2-\beta\, q_1 \,q_2\\ &=Q_{\omega}(U_{1})\leqslant Q_\omega(U). \end{align}\] which entails in particular that \[\label{contr:lem-pos} A_\omega(U_\tau)<A_\omega(U_1)\leqslant A_\omega(U)\quad\forall\,\tau\in(-1,1).\tag{33}\]

We claim (and prove in Step 2) that there exists \(\overline{\tau}\in (-1,1)\) such that \[\label{claim:lem-pos} \|u_{1,\overline{\tau}}\|_{p_1}^{p_1}=\|u_1\|_{p_1}^{p_1}.\tag{34}\] In fact, if we prove 34 , then \(U_{\overline{\tau}}\in \Pi_\omega\) and \(A_\omega(U_\tau)<A_\omega(U)\). On the other hand, \(U\) is a minimizer of \(A_\omega\) on \(\Pi_\omega\) by Lemma 4, but this generates a contradiction, ensuring that \(|\widehat{\Omega}|=0\) and \(\phi_{1,\omega}>0\) on \(\Omega\). Therefore, it is left to prove that \(\Omega=\mathbb{R}^2\), i.e. \(\phi_{1,\omega}(x)>0\) for every \(x\in\mathbb{R}^2\). First of all, \(|\Omega|>0\) by Proposition 8. Then, since \(u_1\) solves 47 , \(\phi_{1,\omega}\) satisfies \[(-\Delta+ \omega) \phi_{1,\omega} =\abs{u_1}^{p_1-2}(\phi_{1,\omega} + q_1\mathcal{G}_{\omega}).\] Therefore, as \(\phi_{1,\omega}\) is non-negative, \((-\Delta+ \omega) \phi_{1,\omega}\geqslant 0\) in \(H^{-1}(\mathbb{R}^2)\) and \(\phi_{1,\omega}\not\equiv 0\) by Proposition 8, then by the Strong Maximum Principle (e.g.,[36]) \(\phi_{1,\omega}>0\) on \(\mathbb{R}^2\). To prove that \(\phi_{2,\omega}(x)>0\) for every \(x\in\mathbb{R}^2\), we repeat the same arguments using the functional \(B_{\omega}\) instead of \({A}_{\omega}\). The positivity of \(u_i\) follows from the positivity of \(\phi_{i,\omega}\) and \(q_i\).
Step 2. Proof of the claim 34 . First of all, let us observe that the function \(\tau\mapsto \|u_{1,\tau}\|_{p_1}^{p_1}\) is continuous on \([-1,1]\). Moreover, since \[\abs{u_1(x)}^2=\abs{\phi_{1,\omega}(x)}^2 + q_1^2\mathcal{G}_{\omega}^2(x)+ 2\cos(\alpha) q_1 \abs{\phi_{1,\omega}(x)}\mathcal{G}_{\omega}(x)< \abs{u_{1,1}(x)}^2,\quad \forall x\in\widehat{\Omega},\] and \(\abs{\widehat{\Omega}}>0\), there results that \({\left\Vert u_1\right\Vert}_{p_1}^{p_1}<{\left\Vert u_{1,1}\right\Vert}_{p_1}^{p_1}\). In order to prove 34 , we need to distinguish two cases. Suppose first that there exists \(\widetilde{\Omega}\), with \(\abs{\widetilde{\Omega}}>0\), such that \(\alpha\neq \pi\) on \(\widetilde{\Omega}\). In this case, we have that \[\begin{align} {\left\Vert u_{1,-1}\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2)}&={\left\Vert u_{1,-1}\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2\setminus{\widetilde{\Omega}})}+{\left\Vert u_{1,-1}\right\Vert}^{p_1}_{L^{p_1}(\widetilde{\Omega})}\\ &={\left\Vert u_{1}\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2\setminus{\widetilde{\Omega}})}+\int_{\widetilde{\Omega}} \Bigl \lvert \abs{\phi_{1,\omega}}^2 + q_1^2\mathcal{G}_{\omega}^2 - 2\, q_1 \abs{\phi_{1,\omega}}\mathcal{G}_{\omega} \Bigr \rvert^{\frac{p_1}{2}}\, dx\\ &<{\left\Vert u_{1}\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2\setminus{\widetilde{\Omega}})}+\int_{\widetilde{\Omega}} \Bigl \lvert \abs{\phi_{1,\omega}}^2 + q_1^2\mathcal{G}_{\omega}^2 +2 \cos(\alpha) q_1 \abs{\phi_{1,\omega}}\mathcal{G}_{\omega} \Bigr \rvert^{\frac{p_1}{2}}\, dx\\ &={\left\Vert u_{1}\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2)} \end{align}\] hence, since \({\left\Vert u_{1,-1}\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2)}< {\left\Vert u_1\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2)} <{\left\Vert u_{1,1}\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2)}\), there exists \(\overline{\tau}\in (-1,1)\) such that 34 holds.

Suppose instead that \(\alpha(x)=\pi\) for a.e. \(x\in \Omega\): this choice of \(\alpha\) can be reformulated also as \(u_1(x)=u_{1,-1}(x)\) for a.e. \(x\in\Omega\) or as \(\phi_{1,\omega}=-|\phi_{1,\omega}|\) for a.e. \(x\in\Omega\). Differently from the previous case, we cannot conclude immediately using the continuity of \(\|u_{1,\tau}\|_{p_1}^{p_1}\) since \(\|u_{1,-1}\|_{p_1}^{p_1}=\|u_{1}\|_{p_1}^{p_1}\). To solve this issue, we observe that for every \(\tau\in [-1,1]\) \[\begin{align} \left| \frac{d}{d\tau}\abs{u_{1,\tau}}^{p_1}\right|&= \left| \frac{d}{d\tau}(\tau^2\abs{\phi_{1,\omega}}^2+ q_1^2 \mathcal{G}_{\omega}^2 + 2\tau q_1 \abs{\phi_{1,\omega}}\mathcal{G}_{\omega})^{\frac{p_1}{2}} \right| \\ &=\left|p_1\left|\phi_{1,\omega}\right|\left|u_{1,\tau}\right|^{p_1-2}u_{1,\tau}\right|=p_1\left|\phi_{1,\omega}\right|\left|u_{1,\tau}\right|^{p_1-1} \leqslant p_1\abs{\phi_{1,\omega}}\abs{u_{1,1}}^{p_1-1}\in L^1(\mathbb{R}^2), \end{align}\] hence by dominated convergence \[\label{eq:d-dtau} \begin{align} \frac{d}{d\tau}{{\left\Vert u_{1,\tau}\right\Vert}^{p_ 1}_{L^{p_1}(\mathbb{R}^2)}}\Bigr|_{\tau=-1^+} = \int_{\mathbb{R}^2} \frac{d}{d\tau}\abs{u_{1,\tau}}^{p_1}\Bigr|_{\tau=-1} \,dx &= \int_{\mathbb{R}^2}\, p_1\abs{\phi_{1,\omega}}\abs{u_{1,-1}}^{p_1-2} u_{1,-1}\,dx\\ &=\int_{\mathbb{R}^2}\, p_1\abs{\phi_{1,\omega}}\abs{u_{1}}^{p_1-2} u_{1}\,dx. \end{align}\tag{35}\]

Since \(U\) is a ground state and solves 47 with \(\lambda_1=\omega\), there results that \[\label{EL-plane} \langle \nabla\chi,\nabla\phi_{1,\omega}\rangle_{L^{2}(\mathbb{R}^2)}+\langle \chi,\omega \phi_{1,\omega}-|v|^{r-2}v\rangle_{L^{2}(\mathbb{R}^2)}=0\quad\forall\, \chi\in H^1(\mathbb{R}^2).\tag{36}\] By choosing \(\chi=|\phi_{1,\omega}|\) in 36 and recalling that \(\phi_{1,\omega}=-|\phi_{1,\omega}|\) on \(\Omega\) (and consequently in \(\mathbb{R}^2\)), it follows that \[\int_{\mathbb{R}^2} \abs{\phi_{1,\omega}(x)}\abs{u_1(x)}^{p_1-2}u_1(x)\, dx =- \int_{\mathbb{R}^2} \abs{\nabla\abs{\phi_{1,\omega}(x)}}^2 \,dx- \omega\int_{\mathbb{R}^2} \abs{\phi_{1,\omega}(x)}^2\, dx,\] thus \[\label{eq:der-tau600} \frac{d}{d\tau}{\left\Vert u_{1,\tau}\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2)} \,\biggr\rvert_{\tau=-1^+}< 0.\tag{37}\] Therefore, by combining 37 and the fact that \({\left\Vert u_{1,-1}\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2)}= {\left\Vert u_1\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2)} <{\left\Vert u_{1,1}\right\Vert}^{p_1}_{L^{p_1}(\mathbb{R}^2)}\), 34 follows.
Step 3. Radiality of \(\phi_{i,\omega}\) and \(u_i\). As done in Step 1, we recall that by lemma 4 \(U\) is also a minimizer of \(A_{\omega}\) in \(\Pi_{\omega}\). In order to prove that \(\phi_{1,\omega}\) and \(u_1\) are radially symmetric and decreasing along the radial direction, it is sufficient to show that \(\phi_{1,\omega}=\phi_{1,\omega}^\star\) a.e. on \(\mathbb{R}^2\), where \(\phi_{1,\omega}^\star\) denotes the symmetric decreasing rearrangement of \(\phi_{1,\omega}\).

For the sake of completeness, let us recall that, given a nonnegative function \(f\in L^p(\mathbb{R}^d)\), the symmetric decreasing rearrangement \(f^*\) is defined as \[f^\star(x):=\int_0^{+\infty}\mathbb{1}_{B_{r(t)}}(x)\,dt,\quad r(t):=\sqrt{\frac{|\{x\,:\,f(x)>t\}|}{\pi}}.\] Then the following relations hold for \(\phi_{1,\omega}^\star\) (see for example [26]): \[\label{eq:prop-rearr} \|\nabla\phi_{1,\omega}^\star\|_2^2\leqslant\|\nabla\phi_{1,\omega}\|_2^2,\quad \quad\|\phi_{1,\omega}^\star\|_2^2=\|\phi_{1,\omega}\|_2^2,\tag{38}\] \[\label{eq:prop-rearr2} \forall\, p>1\quad \forall\,g \in L^p(\mathbb{R}^2)\quad\|\phi_{1,\omega}^\star+g^\star\|_p^p\geqslant\|\phi_{1,\omega}+g\|_p^p.\tag{39}\] Moreover, if \(g\) is radially symmetric and decreasing, then the equality in 39 holds if and only if \(\phi_{1,\omega}^\star=\phi_{1,\omega}\) a.e. on \(\mathbb{R}^2\) (see again [26]).

Assume now by contradiction that \(\phi_{1,\omega}\neq\phi^\star_{1,\omega}\) on a set of non-zero Lebesgue measure, and define the function \(U^\star:=(\phi_{1,\omega}^\star+q_1\mathcal{G}_\omega,u_2)\). By 38 and the fact that the equality case in 39 is realized if and only if \(\phi_{1,\omega}^\star=\phi_{1,\omega}\) a.e. on \(\mathbb{R}^2\), we get that \(A_{\omega}(U^\star)\leqslant A_{\omega}(U)\) and \(\|\phi_{1,\omega}^\star+q_1\mathcal{G}_\omega\|_{p_1}^{p_1}>\|\phi_{1,\omega}+q_1\mathcal{G}_\omega\|_{p_1}^{p_1}\). In particular, there exists \(\tau\in(0,1)\) such that, denoted with \(U^\star_\tau=(\tau\phi_{1,\omega}^\star+q_1\mathcal{G}_\omega, u_2)\), there results that \(\|\tau\phi_{1,\omega}^\star+q_1\mathcal{G}_\omega\|_{p_1}^{p_1}=\|\phi_{1,\omega}+q_1\mathcal{G}_\omega\|_{p_1}^{p_1}\), i.e. \(U^\star_\tau\in \Pi_\omega\). Moreover, it holds that \[\begin{align} \frac{p_1-2}{2p_1}Q_{\omega}(U^\star_\tau) - \frac{p_1-2}{2p_1}Q_{\omega}(U^\star) &=-\frac{p_1-2}{2p_1} (1-\tau^2)\left( {\left\Vert\nabla\phi^\star_{1,\omega}\right\Vert}_2^2 + \, \omega {\left\Vert\phi^\star_{1,\omega}\right\Vert}_2^2 \right) <0, \end{align}\] thus \(A_\omega(U^\star_\tau)<A_\omega(U^\star)\leqslant A_\omega(U)\), which contradicts the fact that \(U\) is a minimizer of the action at frequency \(\omega\), hence \(\phi_{1,\omega}\) and \(u_1\) are radially symmetric and decreasing. The same properties can be proved for \(\phi_{2,\omega}\) and \(u_2\) (using the functional \(B_\omega\) instead of \(A_\omega\)), hence the proof is concluded.
 ◻

Proof of the Theorem 4. Let us start with the proof of point (i). Let \(U=(u_1,u_2)\) be a ground state of 4 at mass \(\mu\), \(q_i\) and \(\mu_i\) be the charge and the mass associated with \(u_i\) for \(i=1,2\). By Theorem 2, we can assume that \(q_i\) and \(u_i\) are positive for \(i=1,2\). Suppose now by contradiction that \(q_1\leqslant q_2\) and consider the function \(\widetilde{U}=(u_2,u_1)\in \Lambda_\mu\), which satisfies \[\begin{align} \label{eq:q-contr} F(\widetilde{U})&=F_1(u_2) + F_2(u_1) -\beta\, q_1\,q_2 \\ &=F_1(u_1)+\frac{\sigma_1}{2}\left(q_2^2-q_1^2\right)+F_2(u_2)+\frac{\sigma_2}{2}\left(q_1^2-q_2^2\right)-\beta q_1 q_2\\ &=F_1(u_1) + F_2(u_2)-\beta q_1\,q_2 + \frac{(\sigma_{2}-\sigma_{1})}{2} (q_1^2-q_2^2)\\ &=F(U)+\frac{(\sigma_{2}-\sigma_{1})}{2} (q_1^2-q_2^2). \end{align}\tag{40}\] In particular, if \(q_1<q_2\), then by 40 we get that \(F(\widetilde{U})<F(U)\), which contradicts the fact that \(U\) is a ground state. If instead \(q_1=q_2=:q\), then \(F(U)=F(\widetilde{U})\), hence \(\widetilde{U}\) is a ground state too and solves 47 . In particular, since both \(U\) and \(\widetilde{U}\) solve 47 , there results that \[\begin{cases} \phi_{1,\omega}(0)=(\sigma_1+\theta_\omega-\beta)q\\ \phi_{2,\omega}(0)=(\sigma_2+\theta_\omega-\beta)q, \end{cases} \quad\text{and}\quad \begin{cases} \phi_{2,\omega}(0)=(\sigma_1+\theta_\omega-\beta)q\\ \phi_{1,\omega}(0)=(\sigma_2+\theta_\omega-\beta)q, \end{cases}\] hence \(\sigma_1=\sigma_2\), which is again a contradiction and concludes the proof of point (1).
Let us focus now on the proof of point (ii). Fix \(\beta>0\), \(\sigma_1\in \mathbb{R}\), denote with \(p:=p_1=p_2\) and let \(\sigma_2\) vary in the interval \((\sigma_1,+\infty)\): let us denote with \(\delta:=\sigma_2-\sigma_1\in (0,+\infty)\). We observe that, by Theorem 1, there exists at least a ground state of 4 at mass \(\mu\), that we denote by \(U_\delta=(u_{1,\delta}, u_{2,\delta})\). We consider for \(i=1,2\) the decomposition \(u_{i,\delta}=\phi_{i,\delta}+q_{i,\delta}\mathcal{G}_\lambda\), corresponding to \(\lambda=4e^{4\pi(2\beta-\sigma_1)-2\gamma}\) or, equivalently, to \(\sigma_1+\theta_\lambda=2\beta\), and we call \(m_\delta\) the mass of \(u_{1,\delta}\). By arguing as in 18 and using that \(F(U_{\delta})<0\), we get that \[\sum_{i=1,2}\left\{\frac{1}{4}\mathcal{N}_{\sigma_i}(u_{i,\delta})^2-\frac{C_{p}\mu}{p}\mathcal{N}_{\sigma_i}(u_{i,\delta})^{p-2}\right\}<\frac{\lambda}{2}\mu,\] thus \(\mathcal{N}_{\sigma_i}(u_{i,\delta})\) are equibounded in \(\delta>0\) for \(i=1,2\). In particular, this entails that \((\phi_{i,\delta})_{\delta>0}\) is equibounded in \(H^1(\mathbb{R}^2)\) and both \((u_{i,\delta})_{\delta>0}\) and \((\phi_{i,\delta})_{\delta>0}\) are equibounded in \(L^r(\mathbb{R}^2)\) for every \(r\geqslant 2\) for \(i=1,2\). Moreover, \((q_{1,\delta})_{\delta>0}\) is equibounded and \[(\sigma_2+\theta_\lambda)q_{2,\delta}^2=(2\beta+\delta)q_{2,\delta}^2\leqslant C,\] from which we deduce that \(q_{2,\delta}\to 0\) as \(\delta\to+\infty\). Since by Lagrange theorem it holds that for every \(r\geqslant 2\) \[\left|\|u_{2,\delta}\|_{r}^{r}-\|\phi_{2,\delta}\|_{r}^{r}\right|\leqslant r\sup\left\{\|u_{2,\delta}\|_{r}^{r-1}, \|\phi_{2,\delta}\|_{r}^{r-1}\right\}|q_{2,\delta}|\|\mathcal{G}_{\lambda}\|_{r}\to 0,\quad \delta\to+\infty,\] it follows that \[\begin{align} 0>F_{2}(u_{2,\delta})-E_{p}(\phi_{2,\delta})=&\frac{\lambda}{2}\left({\left\Vert\phi_{2,\delta}\right\Vert}_2^2 - {\left\Vert u_{2,\delta}\right\Vert}_2^2\right) + \frac{q_{2,\delta}^2}{2}(\sigma_1+ \delta + \theta_{\lambda})\\ &-\frac{1}{p}\left({\left\Vert{u_{2,\delta}}\right\Vert}_{p}^{p} - \|\phi_{2,\delta}\|_{p}^{p}\right) = \frac{\delta}{2} q_{2,\delta}^{2}+o(1), \end{align}\] hence \(\delta q_{2,\delta}^{2}\to 0\) as \(\delta\to+\infty\). Therefore, \[\begin{align} \label{eq:Fmu62F1Ep} \mathcal{F}(\mu)=F(U_{\delta})=F_{1}(u_{1,\delta}) + F_{2}(u_{2,\delta}) + o(1) &=F_{1}(u_{1,\delta}) + E_{p}(\phi_{2,\delta}) + o(1)\\ &\geqslant\mathcal{F}_{1}(m_{\delta}) + \mathcal{E}_{p}(\mu-m_{\delta}) + o(1),\quad\delta\to+\infty. \end{align}\tag{41}\] On the other hand, since \(\mathcal{F}_{\sigma,p}\) is an increasing function of \(\sigma\) by [1] and 7 8 hold, we get that \[\begin{align} \mathcal{F}(\mu)=&F(U_{\delta}) \leqslant\min \{\mathcal{F}_{1}(\mu),\mathcal{F}_{2}(\mu)\}= \mathcal{F}_1(\mu), \end{align}\] thus combining with 41 , there results that \[\label{eq:contr-thm3} \mathcal{F}_1(\mu)\geqslant\mathcal{F}_1(m_\delta)+\mathcal{E}_p(\mu-m_\delta)+o(1), \quad \delta\to+\infty.\tag{42}\]

We claim that \(m_\delta\to\mu\) and \(\mathcal{F}(\mu)\to \mathcal{F}_1(\mu)\). Indeed, if we suppose by contradiction that there exists \(\delta_n\to+\infty\) as \(n\to+\infty\) and \(\varepsilon>0\) such that \(m_{\delta_n}\leqslant\mu-\varepsilon\) for every \(n\in\mathbb{N}\), then by concavity of both \(\mathcal{F}_1\) and \(\mathcal{E}_p\) and since \(\mathcal{F}_1(\mu)<\mathcal{E}_p(\mu)\) for every \(\mu>0\), we have that \[\begin{align} \lim_{n\to+\infty}[\mathcal{F}_{1}(m_{\delta_n}) + \mathcal{E}_p(\mu-m_{\delta_n}) +o(1)] &\geqslant\inf_{m\in[0,\mu-\varepsilon]} \{\mathcal{F}_{1}(m) + \mathcal{E}_p(\mu-m) \} \\ &= \min \left\{ \mathcal{E}_p(\mu), \mathcal{F}_{1}(\mu-\varepsilon) + \mathcal{E}_p(\varepsilon)\right\} \\ &> \min \left\{ \mathcal{E}_p(\mu), \mathcal{F}_{1}(\mu)\right\} =\mathcal{F}_{{1}}(\mu), \end{align}\] which is in contradiction with 42 and entails that \(m_\delta\to \mu\) as \(\delta\to+\infty\). Moreover, by 41 and 42 , we deduce that \(\mathcal{F}(\mu)\to \mathcal{F}_1(\mu)\), concluding the proof of point (ii). ◻

Proof of Theorem 5. Given \(\sigma=\sigma_1=\sigma_2\), we know by Theorem 1 that there exists at least a ground state of 4 at mass \(\mu\), that we denote by \(U_\sigma=(u_{1,\sigma}, u_{2,\sigma})\). Fixed \(\lambda=4e^{-2\gamma}\), we represent \(u_{i,\sigma}\) as \(u_{i,\sigma}=\phi_{i,\sigma}+q_{i,\sigma}\mathcal{G}_\lambda\) for \(i=1,2\). By using 16 and the fact that \(F(U_\sigma)<0\), we get that for \(\sigma\) sufficiently large 19 holds, hence \(\mathcal{N}_{\sigma}(u_{i,\sigma})\) is equibounded in \(\sigma>0\). This entails that \(\sigma q_{i,\sigma}^2\) is bounded, hence \(q_{i,\sigma}\to 0\) as \(\sigma \to+\infty\) for \(i=1,2\).

Therefore, as \(\sigma\to+\infty\) \[\begin{align} \label{eq:FmuleqEpi} \mathcal{F}(\mu)=F(U_\sigma)&=\sum_{i=1,2}\left\{\frac{1}{2}\|\nabla\phi_{i,\sigma}\|_2^2+ \frac{1}{2}\lambda\left(\|\phi_{i,\sigma}\|_2^2-\|u_{i,\sigma}\|_2^2\right)-\frac{1}{p_i}\|u_{i,\sigma}\|_{p_i}^{p_i}\right\}+o(1)\\ &=\sum_{i=1,2} E_{p_i}(\phi_{i,\sigma})+ o(1)\geqslant\sum_{i=1,2}\mathcal{E}_{p_i}(\|\phi_{i,\sigma}\|_2^2) +o(1)\\ &=\sum_{i=1,2}\mathcal{E}_{p_i}(\|u_{i,\sigma}\|_2^2) +o(1)\geqslant\inf_{m\in[0,\mu]} \{\mathcal{E}_{p_1}(m) + \mathcal{E}_{p_2}(\mu-m) \}+o(1)\\ &= \min \{ \mathcal{E}_{p_1}(\mu), \mathcal{E}_{p_2}(\mu) \}+o(1). \end{align}\tag{43}\] On the other hand, by Lemma 1 and Step 2 in the proof of Theorem 1, we have that \(F(U_\sigma)=\mathcal{F}(\mu)<\mathcal{F}^\infty(\mu)=\min\{\mathcal{E}_{p_1}(\mu), \mathcal{E}_{p_2}(\mu)\}\), thus together with 43 gives \[\lim_{\sigma\to+\infty}F(U_\sigma)=\min\{\mathcal{E}_{p_1}(\mu), \mathcal{E}_{p_2}(\mu)\}.\]

Let us distinguish now the cases \(\mu<\mu^\star\) and \(\mu>\mu^\star\). If \(\mu<\mu^\star\), then by Proposition 6 there results that \(\min\{\mathcal{E}_{p_1}(\mu), \mathcal{E}_{p_2}(\mu)\}=\mathcal{E}_{p_1}(\mu)\) and \[\label{eq:FsigmatoEp1} \lim_{\sigma\to+\infty}F(U_\sigma)=\mathcal{E}_{p_1}(\mu).\tag{44}\] 44 entails also that \(\|u_{1,\sigma}\|_2^2\to \mu\) as \(\sigma\to+\infty\). Indeed, if we suppose by contradiction that there exists \(\varepsilon>0\) such that \(\|u_{1,\sigma}\|_2^2\leqslant\mu-\varepsilon\) along a subsequence of \(\sigma\to+\infty\), then arguing as in 43 we get that \[\lim_{\sigma\to+\infty} F(U_\sigma)\geqslant\lim_{\sigma\to+\infty}{\sum_{i=1,2}\mathcal{E}_{p_i}(\|u_{i,\sigma}\|_2^2)}\geqslant\min\{\mathcal{E}_{p_1}(\mu-\varepsilon),\mathcal{E}_{p_2}(\mu)\}>\mathcal{E}_{p_1}(\mu),\] which contradicts 44 .

If instead \(\mu>\mu^\star\), then \(\min\{\mathcal{E}_{p_1}(\mu), \mathcal{E}_{p_2}(\mu)\}=\mathcal{E}_{p_2}(\mu)\), hence it is possible to deduce that \(F(U_\sigma)\to \mathcal{E}_{p_2}(\mu)\) and \(\|u_{2,\sigma}\|_2^2\to \mu\) as done for the case \(\mu<\mu^\star\), concluding the proof. ◻