Proof. The radial symmetry of \(u(t)\) for \(t>0\) is obvious from the rotational invariance of Eq. 2 . As for the monotonicity of the profile, we provide a formal proof, which can be made rigorous by an approximation argument as usual. Introducing the radially symmetric variable \(r=|x|\) and setting \(u(r)=u(|x|)=u(x)\) for simplicity, we derive the equation solved by \(\partial_r u\). To this end, we start from Eq. 2 written in radially symmetric variables as \[\label{eq146rad} \partial_t u(t,r) = \partial_r^2 u^m(t,r) + \frac{N-1}{r} \partial_r u^m(t,r) - r^{\sigma} u^m(t,r), \quad (t,r)\in (0,\infty)^2,\tag{15}\] and differentiate 15 with respect to \(r\) to find, by direct calculation, that \(w:=\partial_r u\) solves \[\label{eq146deriv} \begin{align} \partial_t w &=mu^{m-1} \partial_r^2 w + 3m(m-1)u^{m-2}w\partial_r w + m(m-1)(m-2)u^{m-3}w^3\\ & \quad -\frac{m(N-1)}{r^2}u^{m-1}w + \frac{m(N-1)(m-1)}{r}u^{m-2}w^2\\ & \quad + \frac{m(N-1)}{r}u^{m-1} \partial_r w - mr^{\sigma}u^{m-1}w - \sigma r^{\sigma-1}u^m. \end{align}\tag{16}\] This is a degenerate parabolic equation allowing for a comparison principle on the positivity set of \(u\). Since \(w\equiv0\) is a strict supersolution to 16 , we infer by comparing on the positivity set of \(u\) that \(w(t)\geq0\) for any \(t\geq0\), provided \(w_0=\partial_r u_0\leq0\), which completes the proof. ◻

Proof. We first recall that the homogeneity regularizing effect derived in the classical paper [21] implies that the non-negative weak solution \(u\) to Eq. 1 satisfies \[\label{bcest} \partial_t u(t,x) \geq - \frac{u(t,x)}{(m-1)t}, \quad (t,x)\in(0,\infty)\times\mathbb{R}^N,\tag{17}\] in the sense of distributions, from which we deduce that the map \(t\mapsto t^{1/(m-1)} u(t,x)\) is a non-decreasing function of time for all \(x\in\mathbb{R}^N\), whence ?? .

We next infer from ?? that, for \(t>0\) and \(\zeta\in C_c^1(\mathbb{R}^N)\), \[\begin{align} \int_{\mathbb{R}^N} \zeta(x) u(t,x)\;dx & = \int_{\mathbb{R}^N} \zeta(x) u_0(x)\;dx \\ & \quad - \int_0^t \int_{\mathbb{R}^N} \left[ \nabla \zeta(x)\cdot \nabla u^m(s,x) + \zeta(x) |x|^\sigma u^m(s,x) \right]\;dxds. \end{align}\] Since \(u^m\in L^2\big((0,t),H^1_{\text{loc}}(\mathbb{R}^N)\big)\) and \((s,x)\mapsto |x|^\sigma u^m(s,x)\in L^2\big((0,t),L^2_{\text{loc}}(\mathbb{R}^N)\big)\), we may let \(t\to 0\) in the above identity and deduce from the Lebesgue dominated convergence theorem that \[\lim_{t\to 0} \int_{\mathbb{R}^N} \zeta(x) u(t,x)\;dx = \int_{\mathbb{R}^N} \zeta(x) u_0(x)\;dx.\] A density argument then implies that, for any \(R>0\), \[\lim_{t\to 0} \int_{B(0,R)} u(t,x)\;dx = \int_{B(0,R)} u_0(x)\;dx. \label{x10}\tag{18}\] In particular, since \(u_0\not\equiv 0\), there is \(R_0>0\) large enough such that \[\int_{B(0,R_0)} u_0(x)\;dx > 0\] and we infer from 18 that there is \(T_0>0\) such that \[\int_{B(0,R_0)} u(t,x) \ge \frac{1}{2} \int_{B(0,R_0)} u_0(x)\;dx, \quad t\in (0,T_0).\] In other words, \(u(t)\not\equiv 0\) for \(t\in (0,T_0)\), a property which extends to all positive times due to ?? . ◻

Proof. Let \(t>0\) and \(x_0\in\mathbb{R}^N\). We infer from Eq. 2 and 17 that \[\Delta u^m(t,x) = \partial_t u(t,x) + |x|^\sigma u^m(t,x) \ge - \frac{u(t,x)}{(m-1)t} \ge - \frac{\|u(t)\|_\infty}{(m-1)t}, \quad x\in\mathbb{R}^N,\] from which we deduce that \[\Delta \left[ u^m(t,x) + \frac{\|u(t)\|_\infty}{(m-1)t} \frac{|x-x_0|^2}{2N} \right] \ge 0, \quad x\in \mathbb{R}^N.\] Therefore, \(x\mapsto u^m(t,x) + \|u(t)\|_\infty |x-x_0|^2/(2N(m-1)t)\) is a subharmonic function in \(\mathbb{R}^N\) and the mean-value theorem entails that, for \(R>0\), \[u^m(t,x_0) \le \frac{1}{\varpi_N R^N} \int_{B(x_0,R)} \left[ u^m(t,x) + \frac{\|u(t)\|_\infty}{(m-1)t} \frac{|x-x_0|^2}{2N} \right]\;dx,\] where \(\varpi_N\) denotes the volume of the unit ball in \(\mathbb{R}^N\), see [23]. We then infer from Hölder’s inequality that \[\begin{align} u^m(t,x_0) & \le \big(\varpi_N R^N\big)^{-m/(m+1)} \left( \int_{B(0,R)} u^{m+1}(t,x)\;dx \right)^{m/(m+1)} \\ & \quad + \frac{\|u(t)\|_\infty}{2(m-1)R^N t} \int_0^R r^{N+1}\;dr \\ & \le \big(\varpi_N R^N\big)^{-m/(m+1)} \|u(t)\|_{m+1}^m + \frac{\|u(t)\|_\infty}{(m-1)t} R^2. \end{align}\] We optimize with respect to \(R\) by choosing \[R=R(t) = \left[ \frac{t\|u(t)\|_{m+1}^m}{\|u(t)\|_\infty} \right]^{(m+1)/[mN+2(m+1)]}\] in the above inequality to obtain \[u^m(t,x_0) \le \left[ C_1^m \frac{\|u(t)\|_{m+1}^{2m(m+1)} \|u(t)\|_\infty^{Nm}}{t^{Nm}} \right]^{1/[Nm+2(m+1)]},\] with \[C_1 := \left( \frac{1}{\varpi_N^{m/(m+1)}} + \frac{1}{m-1} \right)^{[Nm+2(m+1)]/m}.\] The above inequality being valid for all \(x_0\in\mathbb{R}^N\), we further obtain \[\|u(t)\|_\infty^m \le \left[ C_1^m \frac{\|u(t)\|_{m+1}^{2m(m+1)} \|u(t)\|_\infty^{Nm}}{t^{Nm}} \right]^{1/[Nm+2(m+1)]},\] whence \[\|u(t)\|_\infty^{N(m-1)+2(m+1)} \le C_1 \|u(t)\|_{m+1}^{2(m+1)} t^{-N},\] as claimed. ◻

Proof. We may apply Theorem 6 with \[(q_1,q_2,q_3,\gamma_1,\gamma_2,\gamma_3)=\left(s,2,2,0,0,\frac{\sigma}{2}\right).\] Indeed, with this choice of parameters, the condition 21 is obviously satisfied, while 22 reads \[\begin{align} \frac{1}{s}&=a\left(\frac{1}{2}-\frac{1}{N}\right)+(1-a)\left(\frac{1}{2}+\frac{\sigma}{2N}\right)\\ &=\frac{N+\sigma}{2N}-\frac{a(\sigma+2)}{2N}, \end{align}\] which readily gives \[a=\frac{\sigma s+N(s-2)}{(\sigma+2)s}.\] In order to fulfill the condition 20 , we are left to check that \(a\in[0,1]\), as the other positivity conditions are obvious. On the one hand, we observe that \[1-a=\frac{2N-(N-2)s}{s(\sigma+2)}>0,\] since \(s<2^*\). On the other hand, taking into account that \(\sigma>\sigma_0\), we deduce from ?? that \[\begin{align} \sigma s + N(s-2) & > \frac{N(m-1)s}{m+1}+N(s-2) = \frac{N}{m+1}[2ms-2(m+1)] \\ & = \frac{2mN}{m+1}\left(s-\frac{m+1}{m}\right) \geq 0, \end{align}\] proving thus that \(a>0\). Thus, the condition 20 is satisfied and then the choice of the exponents \(\gamma_1=\gamma_2=0\) obviously ensures the condition 23 . We are thus in a position to complete the proof by applying ?? and observing that \[\||x|^{\gamma_3}z\|_{q_3}=\left(\int_{\mathbb{R}^N}|x|^{\sigma}|z(x)|^2\,dx\right)^{1/2}=N_{\sigma}(z),\] which leads to ?? . ◻

Proof. We infer from ?? and Young’s inequality that \[\|z\|_s^2\leq\kappa_s^2\|\nabla z\|_2^{2a}N_{\sigma}^{2(1-a)}(z)\leq\kappa_s^2 \left[a\|\nabla z\|_2^2+(1-a)N_{\sigma}^2(z)\right],\] and the inequality ?? follows from the fact that \(a\in(0,1)\). ◻

Proof. Let \(z\in\mathcal{X}\). Then \(|z|^{m-1}z\in \dot{H}^1(\mathbb{R}^N)\cap L^2_{\sigma}(\mathbb{R}^N)\) and we infer from ?? with \(|z|^{m-1}z\) instead of \(z\) and \(s=(m+1)/m\) that \[\label{interm1} \begin{align} \frac{1}{4} \left( \|\nabla (|z|^{m-1}z)\|_{2}^2 + N_{\sigma}(|z|^m)^{2} \right) & \ge \frac{\kappa_{(m+1)/m}^2}{4} \||z|^m\|_{(m+1)/m}^2 \\ & = \frac{\kappa_{(m+1)/m}^2}{4} \|z\|_{m+1}^{2m}. \end{align}\tag{25}\] Since \(m+1<2m\), Young’s inequality entails that there is a positive constant \(\mathcal{I}_0\) depending only on \(N\), \(m\) and \(\sigma\) such that \[\label{interm2} \frac{m}{m+1} \|z\|_{m+1}^{m+1} \le \frac{\kappa_{(m+1)/m}^2}{4} \|z\|_{m+1}^{2m} + \mathcal{I}_0.\tag{26}\] Combining 25 and 26 completes the proof. ◻

Proof. We only provide a formal proof of ?? , classical approximation arguments being needed for a rigorous justification. We multiply Eq. 11 by \(\partial_sv^m\) and integrate over \(\mathbb{R}^N\) to obtain \[\begin{align} 0\leq\int_{\mathbb{R}^N} \partial_s v \partial_s v^m\,dx&=\int_{\mathbb{R}^N} \left[\Delta v^m \partial_s v^m - |x|^{\sigma} v^m \partial_s v^m + v \partial_s v^m\right]\,dx\\ & = \int_{\mathbb{R}^N} \left[-\frac{1}{2} \partial_s|\nabla v^m|^2 - \frac{1}{2} |x|^{\sigma} \partial_s v^{2m} + \frac{m}{m+1} \partial_s v^{m+1}\right]\,dx\\ &= - \frac{d}{ds} \left( \frac{\|\nabla v^m\|_2^2}{2} + \frac{N_{\sigma}^2(v^m)}{2} - \frac{m}{m+1} \|v\|_{m+1}^{m+1}\right)\\& = - \frac{d}{ds} \mathcal{I}(v), \end{align}\] from which ?? follows after integrating with respect to \(s\), the above identity becoming an inequality in the approximation process. Now, for \(s\ge 0\), we combine Lemma 4 and ?? to find that \[\mathcal{I}(u_0) \ge \mathcal{I}(v(s)) \ge \frac{1}{4}\big(\|\nabla v^m(s)\|_2^2 + N_{\sigma}^2(v^m(s))\big) - \mathcal{I}_0,\] which proves that \(v^m(s)\) belongs to \(\dot{H}^1(\mathbb{R}^N)\cap L_\sigma^2(\mathbb{R}^N)\) with \[\|\nabla v^m(s)\|_2^2 + N_{\sigma}^2(v^m(s)) \le 4 \left[ \mathcal{I}(u_0) + \mathcal{I}_0 \right], \quad s\ge 0. \label{x13}\tag{27}\] We then infer from Corollary 2 (with \(s=(m+1)/m\)) and 27 that, for \(s\ge 0\), \(v^m(s)\) belongs to \(L^{(m+1)/m}(\mathbb{R}^N)\) with \[\begin{align} \|v(s)\|_{m+1}^{2m} & = \|v^m(s)\|_{(m+1)/m}^2 \le \kappa_{(m+1)/m}^2 \left[ \|\nabla v^m(s)\|_2^2 + N_\sigma^2(v^m(s)) \right] \nonumber \\ & \le 4 \kappa_{(m+1)/m}^2 \left[ \mathcal{I}(u_0) + \mathcal{I}_0 \right]. \label{x14} \end{align}\tag{28}\] Therefore, \(v(s)\in\mathcal{X}\) for all \(s\ge 0\) by 27 and 28 and \(v(s)\in L^\infty(\mathbb{R}^N)\) by 9 and Proposition 4. Finally, it follows from ?? and 28 that, for \(s\ge \ln{m}/(m-1)\), \[\begin{align} \|v(s)\|_\infty & \le C_1 (m-1)^N \left( \frac{e^{(m-1)s}}{e^{(m-1)s}-1} \right)^N \|v(s)\|_{m+1}^{2(m+1)} \\ & \le 4^{(m+1)/m} C_1 m^N \kappa_{(m+1)/m}^{2(m+1)/m} \left[ \mathcal{I}(u_0) + \mathcal{I}_0 \right]^{(m+1)/m} \end{align}\] which, together with 27 and 28 , entails ?? and complete the proof. ◻

Proof. We multiply Eq. 29 by \(V^{p-1}\) and integrate over \(\mathbb{R}^N\) to obtain \[\int_{\mathbb{R}^N}\nabla V^{p-1}(x)\cdot\nabla V(x)\,dx+\int_{\mathbb{R}^N}|x|^{\sigma}V^p(x)\,dx=\int_{\mathbb{R}^N}V^{p-1+1/m}(x)\,dx,\] which is equivalent, after easy manipulations, to \[\label{interm5} \frac{4(p-1)}{p^2}\|\nabla V^{p/2}\|_2^2+N_{\sigma}^2(V^{p/2})=\int_{\mathbb{R}^N}V^{p-1+1/m}(x)\,dx.\tag{30}\] Since \(p\geq2\), we also have \[\label{interm6} \frac{4(p-1)}{p^2}\|\nabla V^{p/2}\|_2^2+N_{\sigma}^2(V^{p/2})\geq\frac{2}{p}\left[\|\nabla V^{p/2}\|_2^2+N_{\sigma}^2(V^{p/2})\right],\tag{31}\] and we deduce by combining 30 , 31 and the inequality ?? applied to \(z=V^{p/2}\) that \[\|V^{p/2}\|_s^2\leq Cp\int_{\mathbb{R}^N}V^{p-1+1/m}(x)\,dx,\] which is equivalent to ?? . ◻

Proof. Pick \(p\geq2\) and \(s\in(2,2^*)\). We set \[p_1 := \frac{m+1}{m}, \quad p_{j+1} := \frac{s}{2}\left( p_j+\frac{m-1}{m} \right), \quad j\ge 1,\] and infer from the inequality ?? with \(p=p_j+(m-1)/m\) that, for \(j\ge 1\), \[\left(\int_{\mathbb{R}^N}V^{p_{j+1}}(x)\,dx\right)^{2/s}\leq C\left(p_j+\frac{m-1}{m}\right)\int_{\mathbb{R}^N}V^{p_j}(x)\,dx;\] that is, \[\label{interm7} \|V\|_{p_{j+1}}^{p_{j+1}}\leq C\left(p_j+\frac{m-1}{m}\right)^{s/2}\|V\|_{p_j}^{sp_j/2}.\tag{32}\] Set next \(\gamma_j:=\|V\|_{p_j}^{p_j}\) for \(j\ge 1\). In this notation, the estimate 32 becomes \[\gamma_{j+1}\leq C\left(\frac{2}{s}p_{j+1}\right)^{s/2}\gamma_j^{s/2} \le C p_{j+1}^{s/2}\gamma_j^{s/2}, \quad j\ge 1.\] Since \(\|V\|_{p_1}\) is finite, we are in a position to apply Lemma 6 (with \(a=s/2>1\), \(C_0=C\) and \(C_1=1+\|V\|_{p_1}\)) and conclude that the sequence \(\big\{\gamma_j^{1/p_j}\big\}_{j\geq1}\) is bounded. Since \[\lim\limits_{j\to\infty}\gamma_j^{1/p_j}=\lim\limits_{j\to\infty}\|V\|_{p_j}=\|V\|_{\infty},\] we have just shown that \(V\in L^{\infty}(\mathbb{R}^N)\).

Owing to the just established boundedness of \(V\), the function \(V^{1/m}\) belongs to \(L_{\mathrm{loc}}^p(\mathbb{R}^N)\) for all \(p\in (1,\infty)\). Elliptic regularity applied to 29 then implies that \(V\in W_{\mathrm{loc}}^{2,p}(\mathbb{R}^N)\) for all \(p\in (1,\infty)\), see [23]. Sobolev embeddings guarantee that \(V\in C^{1,\alpha}(\mathbb{R}^N)\) for all \(\alpha\in (0,1)\), from which we deduce that \(V^{1/m}\in C^{\alpha}(\mathbb{R}^N)\) for all \(\alpha\in (0,1/m)\). We then infer from [23] that \(V\in C^{2,\alpha}(\mathbb{R}^N)\) for all \(\alpha\in (0,1/m)\), which completes the proof of the regularity estimates on \(V\).

Finally, assume for contradiction that there is \(x_0\in\mathbb{R}^N\) such that \[V(x_0)=0=\min_{\mathbb{R}^N} V,\] the last equality being a consequence of the non-negativity of \(V\). Let \(R>0\). Since \[\Delta V(x) - |x|^\sigma V(x) \le 0 \;\;\text{ and }\;\; - R^\sigma \le - |x|^\sigma \le 0, \quad x\in B(x_0,R),\] it follows from the strong comparison principle that \(V\equiv 0\) in \(B(x_0,R)\), see [23] and its consequences. As \(R>0\) is arbitrary, we conclude that \(V\) vanishes identically in \(\mathbb{R}^N\), and a contradiction. Therefore, \(V>0\) in \(\mathbb{R}^N\) and the proof is complete. ◻

Proof. Let \(\varepsilon\in (0,1)\) and \(V_\varepsilon(x) = V(x) - \varepsilon |x|^2\) for \(x\in\mathbb{R}^N\). For \(a>1\) and \(R_2>R_1>1\), we define the open set \[\begin{align} \Omega_{a,R_1,R_2} & := \left\{ x\in B(0,R_2)\setminus \bar{B}(0,R_1)\;:\;|x|^\sigma V^{(m-1)/m}(x)>a \right\} \\ & = \left\{ x\in B(0,R_2)\setminus \bar{B}(0,R_1)\;:\;|x|^\sigma V(x)>a V^{1/m}(x) \right\} \end{align}\] and assume that \[\begin{align} R_1^{\sigma+2} & \ge R_1^{\sigma+2}(a) := \frac{a}{a-1} \max\left\{ 2N, \frac{\sigma m}{m-1} \left( 1+ \frac{\sigma m}{m-1} \right) \right\}, \\ R_2^{2} & \ge R_2^2(\varepsilon) := \frac{\|V\|_\infty}{\varepsilon}. \end{align} \label{ph01}\tag{33}\] On the one hand, we deduce from 29 and 33 that, for \(x\in \Omega_{a,R_1,R_2}\), \[\begin{align} -\Delta V_\varepsilon(x) + \frac{a-1}{a} |x|^\sigma V_\varepsilon(x) & = V^{1/m}(x) - \frac{|x|^\sigma}{a} V(x) + \varepsilon \left[ 2N - \frac{a-1}{a} |x|^{\sigma+2} \right] \\ & \le \varepsilon \left[ 2N - \frac{a-1}{a} R_1^{\sigma+2} \right] \le 0. \end{align}\] Thus, \[-\Delta V_\varepsilon(x) + \frac{a-1}{a} |x|^\sigma V_\varepsilon(x) \le 0, \quad x\in \Omega_{a,R_1,R_2}. \label{ph02}\tag{34}\] On the other hand, setting \(\theta=\sigma m/(m-1)\) and \(S_A(x) := A |x|^{-\theta}\) for \(x\in\mathbb{R}^N\setminus\{0\}\) with \[A\ge A_{a,R_1} := \max\left\{ a^{m/(m-1)}, \|V\|_\infty R_1^\theta \right\}, \label{ph04}\tag{35}\] we observe that, for \(x\in \Omega_{a,R_1,R_2}\), \[\begin{align} -\Delta S_A(x) + \frac{a-1}{a} |x|^\sigma S_A(x) & = - A\theta ( 1 + \theta) |x|^{-\theta-2} + A(N-1)\theta |x|^{-\theta-2} \\ & \qquad + \frac{A(a-1)}{a} |x|^{\sigma-\theta} \\ & \ge A \left[ \frac{a-1}{a} |x|^{\sigma+2} - \theta(1+\theta)\right] |x|^{-\theta-2} \\ & \ge A \left[ \frac{a-1}{a} R_1^{\sigma+2} - \theta(1+\theta)\right] |x|^{-\theta-2}, \end{align}\] which implies that, thanks to the choice 33 of \(R_1\), \[-\Delta S_A(x) + \frac{a-1}{a} |x|^\sigma S_A(x) \ge 0, \quad x\in \Omega_{a,R_1,R_2}. \label{ph03}\tag{36}\] Consider now \(x\in\partial\Omega_{a,R_1,R_2}\). Either \(|x|\in\{R_1,R_2\}\) and \[\begin{align} S_A(x) \ge 0 \ge V_\varepsilon(x), \quad |x|=R_2, \\ S_A(x) = A {R_1}^{-\theta} \ge \|V\|_\infty \ge V_\varepsilon(x), \quad |x|=R_1, \end{align}\] owing to 33 and the choice 35 of \(A\). Or \(|x|^\sigma V^{(m-1)/m}(x)=a\), which also reads \(V(x)=a^{m/(m-1)} |x|^{-\theta}\) and we infer from 35 that \(S_A(x) \ge V(x) \ge V_\varepsilon(x)\). We have thus shown that \(S_A(x)\ge V_\varepsilon(x)\) for \(x\in\partial\Omega_{a,R_1,R_2}\), which we combine with 34 , 36 and the comparison principle to conclude that \[V_\varepsilon(x) \le S_A(x) = A |x|^{-\theta}, \qquad x\in\Omega_{a,R_1,R_2}. \label{ph05}\tag{37}\] Now, fix \(x\in\mathbb{R}^N\). Either \(|x|^\sigma V^{(m-1)/m}(x) \le 2\), so that \(V(x)\le 2^{m/(m-1)} |x|^{-\theta}\). Or \(|x|^\sigma V^{(m-1)/m}(x) > 2\) and we distinguish two cases: if \(|x|\le R_1(2)\), then \(V(x) \le \|V\|_\infty \le \|V\|_\infty R_1^\theta(2) |x|^{-\theta}\). Otherwise, \(|x|>R_1(2)\) and \[x\in \Omega_{2,R_1(2),R_2(\varepsilon)} \;\;\text{ for any }\;\; \varepsilon\in \big( 0, \min\{1,\|V\|_\infty |x|^{-2}\}\big).\] It then follows from 35 and 37 that \[V(x) - \varepsilon |x|^2 \le A_{2,R_1(2)} |x|^{-\theta}, \quad \varepsilon\in \big( 0, \min\{1,\|V\|_\infty |x|^{-2}\}\big).\] Letting \(\varepsilon\to 0\) in the above inequality gives \(V(x) \le A_{2,R_1(2)} |x|^{-\theta}\). Gathering the outcome of the above analysis completes the proof with \(C= \max\left\{2^{m/(m-1)}, \|V\|_\infty R_1^\theta(2),A_{2,R_1(2)} \right\}\). ◻

Proof. Let \(R>0\). We estimate the \(L^{(m+1)/m}\)-norm in the left-hand side of ?? by splitting the integral into a sum of integrals on \(B(0,R)\) and its complement and applying suitable Hölder inequalities for each of the two terms, as follows: \[\begin{align} \|&z\|_{(m+1)/m}^{(m+1)/m}=\int_{B(0,R)}|z|^{(m+1)/m}(x)\,dx\\ & \quad +\int_{\mathbb{R}^N\setminus B(0,R)}|x|^{\sigma(m+1)/(2m)} |z|^{(m+1)/m}(x)|x|^{-\sigma(m+1)/(2m)}\,dx\\ &\leq\left(\int_{B(0,R)} z^2(x)\,dx\right)^{(m+1)/(2m)}|B(0,R)|^{(m-1)/(2m)}\\ & \quad +\left(\int_{\mathbb{R}^N\setminus B(0,R)}|x|^{\sigma} z^2(x)\,dx\right)^{(m+1)/(2m)}\left(\int_{\mathbb{R}^N\setminus B(0,R)}|x|^{-\sigma(m+1)/(m-1)}\,dx\right)^{(m-1)/(2m)}\\ &\leq\big(\varpi_N R^N\big)^{(m-1)/(2m)} \|z\|_2^{(m+1)/(2m)}\\ & \quad + \left[ \frac{(m-1)N\varpi_N}{\sigma(m+1) - N(m-1)}\right]^{(m-1)/(2m)} \left(R^{N-\sigma(m+1)/(m-1)}\right)^{(m-1)/(2m)}N_{\sigma}^{(m+1)/(2m)}(z), \end{align}\] where \(\varpi_N\) denotes the volume of the unit ball in \(\mathbb{R}^N\) and we have taken into account that \(\sigma>\sigma_0\), which entails the positivity of \(\sigma(m+1)-N(m-1)\) and the finiteness of the integral over \(\mathbb{R}^N\setminus B(0,R)\). Setting \[C_3 := \max\left\{\varpi_N^{(m-1)/(2m)} , \left[\frac{(m-1)N\varpi_N}{\sigma(m+1)-N(m-1)}\right]^{(m-1)/2m}\right\}>0\] completes the proof. ◻

Proof. As already mentioned in the Introduction, it is more convenient to work with the functional \(\mathcal{J}\) defined by \[\mathcal{J}(V) = \mathcal{I}(|V|^{(1-m)/m}V) =\frac{1}{2} \left( \|\nabla V\|_2^2 + N_{\sigma}^2(V) \right) - \frac{m}{m+1} \|V\|_{(m+1)/m}^{(m+1)/m}, \quad V\in\mathcal{Y}.\] According to Corollary 2 (with \(s=(m+1)/m\)) and Lemma 4, the functional \(\mathcal{J}\) is bounded from below and coercive on \(\mathcal{Y}\), so that there is a minimizing sequence \(\{V_j\}_{j\ge 1}\) in \(\mathcal{Y}\) satisfying \[\lim_{j\to\infty} \mathcal{J}(V_j) = \inf_{\mathcal{Y}} \mathcal{J} \;\;\text{ and }\;\; \|\nabla V_j\|_2+\|V_j\|_{(m+1)/m}+N_{\sigma}(V_j)\leq C_4, \quad j\ge 1, \label{x15}\tag{38}\] for some \(C_4>0\) depending only on \(N\), \(m\) and \(\sigma\). In particular, \(\{V_j\}_{j\geq1}\) is bounded in \(\dot{H}^1(\mathbb{R}^N)\) and in \(L^2_{\sigma}(\mathbb{R}^N)\) and thus relatively compact in \(L^2(\mathbb{R}^N)\) since \(\sigma>0\). Consequently, there are \(V\in \dot{H}^1(\mathbb{R}^N)\cap L^2_{\sigma}(\mathbb{R}^N)\) and a subsequence of \(\{V_j\}_{j\ge 1}\) (not relabeled) such that \[\lim_{j\to\infty} \|V_j-V\|_2 = 0 \label{x16}\tag{39}\] and \(\{V_j\}_{j\ge 1}\) converges weakly to \(V\) in \(\dot{H}^1(\mathbb{R}^N)\cap L^2_{\sigma}(\mathbb{R}^N)\). This weak convergence, along with the lower semicontinuity of the norms in \(\dot{H}^1(\mathbb{R}^N)\) and \(L_\sigma^2(\mathbb{R}^N)\) and 38 , ensures that \[\|\nabla V\|_2\leq\liminf\limits_{j\to\infty}\|\nabla V_j\|_2, \quad N_\sigma(V) \le \liminf\limits_{j\to\infty} N_\sigma(V_j) \le C_4. \label{x17}\tag{40}\] Finally, we derive the strong convergence in the \(L^{(m+1)/m}\)-norm by employing ?? . Indeed, for \(R>0\), it follows from ?? , 38 and 40 that \[\begin{align} \|V_j-V\|_{(m+1)/m}^{(m+1)/m} & \leq C_3 R^{N(m-1)/(2m)} \|V_j-V\|_{2}^{(m+1)/m} \\ & \quad + C_3 R^{[(m+1)(\sigma_0-\sigma)]/(2m)} N_\sigma^{(m+1)/m}(V_j-V)\\ & \le C_3 R^{N(m-1)/(2m)} \|V_j-V\|_{2}^{(m+1)/m} \\ & \quad + C_3 R^{[(m+1)(\sigma_0-\sigma)]/(2m)} (2C_4)^{(m+1)/m}. \end{align}\] Thanks to the convergence 39 , we may take the limit \(j\to\infty\) in the above inequality to obtain \[\limsup\limits_{j\to\infty}\|V_j-V\|_{(m+1)/m}^{(m+1)/m} \leq C_3 (2C_4)^{(m+1)/m} R^{[(m+1)(\sigma_0-\sigma)]/(2m)}.\] The above inequality being valid for any \(R>0\), we take the limit \(R\to\infty\) and, taking into account the range \(\sigma>\sigma_0\), we conclude that \(\{V_j\}_{j\ge 1}\) converges to \(V\) in \(L^{(m+1)/m}(\mathbb{R}^N)\). Together with 38 and 40 , this convergence implies that \[\inf_{\mathcal{Y}} \mathcal{J} \le \mathcal{J}(V) \le \liminf_{j\to\infty} \mathcal{J}(V_j) = \inf_{\mathcal{Y}} \mathcal{J},\] so that \(V\) is a minimizer for \(\mathcal{J}\) on \(\mathcal{Y}\). In particular, \(V\) is a weak solution to its associated Euler-Lagrange equation, which is Eq. 29 . Moreover, the properties of the symmetric decreasing rearrangement and the Pólya-Szegö inequality, see [27] for instance, directly imply that \[\mathcal{J}(V)=\mathcal{J}(|V|)\geq\mathcal{J}(|V|^*),\] which ensures that there exists a radially symmetric minimizer \(V\in\mathcal{Y}_+\) to \(\mathcal{J}\) on \(\mathcal{Y}\) with non-increasing profile. We finally observe that, for any \(W\in\mathcal{Y}_+\), \(W\not\equiv 0\), and \(\lambda>0\), \[\frac{\mathcal{J}(\lambda W)}{\lambda^2} = \frac{1}{2} \left( \|\nabla W\|_2^2 + N_{\sigma}^2(W) \right) - \frac{m}{m+1} \lambda^{-(m-1)/m} \|W\|_{(m+1)/m}^{(m+1)/m},\] which is negative for \(\lambda\) small enough (depending on \(W\)). Consequently, \(\mathcal{J}\) takes negative values on \(\mathcal{Y}\) and thus \(\mathcal{J}(V)<0\), which implies that \(V\not\equiv 0\). Finally, since \(V\in\mathcal{Y}_+\) is a non-trivial solution to 29 , we infer from Proposition 9 and Lemma 7 that \(V\) satisfies the properties listed in Proposition 10. ◻

Proof. We introduce \[\label{def46W} W(r):=V(r)r^{m\sigma/(m-1)}, \quad r\ge 0,\tag{42}\] and find by direct calculation that \[V'(r)=\left(rW'(r)-\frac{m\sigma}{m-1}W(r)\right)r^{-m\sigma/(m-1)-1}\] and \[V''(r)=\left[r^2W''(r)-\frac{2m\sigma}{m-1}rW'(r)+\frac{m\sigma}{m-1}\left(1+\frac{m\sigma}{m-1}\right)W(r)\right]r^{-m\sigma/(m-1)-2}.\] We thus find that \(W\) solves the following differential equation: \[\label{W46ODE} \begin{align} r^2W''(r)&+\left(N-1-\frac{2m\sigma}{m-1}\right)rW'(r)+\frac{m\sigma}{m-1}\left(2-N+\frac{m\sigma}{m-1}\right)W(r)\\ &-r^{\sigma+2}W(r)+r^{\sigma+2}W^{1/m}(r)=0, \quad r\in (0,\infty). \end{align}\tag{43}\]

Step 1. Monotonicity for \(r\) large. In a first step, we prove that there is \(R_0>0\) sufficiently large such that, either \(W'>0\) on \((R_0,\infty)\), or \(W'<0\) on \((R_0,\infty)\). In order to prove this claim, pick \(R_0\) large enough (to be chosen later) and assume that \(W\) has a point of maximum at \(r_1>R_0\). Then \(r_1W'(r_1)=0\) and \(r_1^2W''(r_1)<0\) and it follows from 43 that \[\begin{align} 0&\leq\left(\frac{m\sigma}{m-1}\left(2-N+\frac{m\sigma}{m-1}\right)-r_1^{\sigma+2}\right)W(r_1)+r_1^{\sigma+2}W^{1/m}(r_1)\\ &=r_0^{\sigma+2}W^{1/m}(r_1)\left[1-\left(1-\frac{C_5}{r_1^{\sigma+2}}\right)W^{(m-1)/m}(r_1)\right], \end{align}\] where \[C_5 := \frac{m\sigma}{m-1}\left[2-N+\frac{m\sigma}{m-1}\right].\] The previous estimate implies that, if \(r_1>R_0>C_5^{1/(\sigma+2)}\) is a point of maximum for \(W\), then \[\label{interm46max} W^{(m-1)/m}(r_1) \leq\frac{r_1^{\sigma+2}}{r_1^{\sigma+2}-C_5}.\tag{44}\] Similarly, if \(W\) has a point of minimum at \(r_0>R_0>C_5^{1/(\sigma+2)}\), then \[\label{interm46min} W^{(m-1)/m}(r_0) \geq \frac{r_0^{\sigma+2}}{r_0^{\sigma+2}-C_5}.\tag{45}\] Assume now for contradiction that \(W\) has a point of minimum at \(r_0>R_0>C_5^{1/(\sigma+2)}\) and then a point of maximum at \(r_1>r_0\). Combining 44 and 45 gives \[W(r_1)^{(m-1)/m}\leq \frac{r_1^{\sigma+2}}{r_1^{\sigma+2}-C_5} \leq\frac{r_0^{\sigma+2}}{r_0^{\sigma+2}-C_5}\leq W(r_0)^{(m-1)/m},\] which implies that \(W\) is constant on \((r_0,r_1)\) and thus (by standard theory of ordinary differential equations) \(W\) is constant for \(r\in(0,\infty)\). This implies that, either \(W\equiv1\), which is a contradiction since \(W(0)=0\) by construction, or \(W\equiv 0\), which contradicts \(V\not\equiv 0\). Therefore, \(W(r)\) cannot have an oscillatory behavior as \(r\to\infty\) and we have thus established that there is \(R_0>0\) such that, either \(W'>0\) on \((R_0,\infty)\), or \(W'<0\) on \((R_0,\infty)\).

Step 2. Behavior at infinity. We infer from the previous step that there exists \[L:=\lim\limits_{r\to\infty}W(r)\in [0,\infty].\] The goal of this step is to prove that \(L=1\). To this end, we want to rule out first the possibility that either \(L=0\) or \(L=\infty\). The latter is actually discarded by Proposition 10, so that we turn to the proof of the former by a contradiction argument, as indicated below.

Assume for contradiction that \(L=0\). This fact, together with Step 1, entails that \(W'<0\) on \((R_0,\infty)\) for some \(R_0>0\). We deduce from Eq. 43 that \[\begin{align} r^2W''(r)&=r^{\sigma+2}W^{1/m}(r) \left[ W^{(m-1)/m}(r) - 1 - C_5 r^{-(\sigma+2)} W^{(m-1)/m}(r) \right]\\ & \quad + \left(\frac{2m\sigma}{m-1}-N+1\right)rW'(r). \end{align}\] Since \(\sigma>\sigma_0\), we find that \[\frac{2m\sigma}{m-1}-N+1>\frac{2mN(m-1)}{(m-1)(m+1)}-N+1=\frac{N(m-1)}{m+1}+1>0,\] and it readily follows from the previous equality and the assumption \(L=0\) that there exist \(\delta>0\) and \(R_1>R_0>0\) such that \[r^2W''(r)\leq-\delta, \quad r\in(R_1,\infty).\] This is equivalent to \[\label{interm8} W''(r)\leq-\frac{\delta}{r^2}, \quad r\in(R_1,\infty),\tag{46}\] and, by integrating twice 46 over \((R_1,\infty)\), we easily deduce that \[W(r) \leq W(R_1) + \left( W'(R_1) -\frac{\delta}{R_1} \right)(r-R_1) + \delta\ln\left(\frac{r}{R_1}\right), \quad r>R_1.\] Recalling that \(W'(R_1)\le 0\), we readily infer from the above inequality that \(W(r)<0\) for \(r>R_1\) sufficiently large, in contradiction to the positivity of \(W\).

We have thus shown that \(L\in(0,\infty)\). We now argue exactly as in [28] and, since \(|W'|\in L^1((R_0,\infty))\), we infer that there exists a sequence \(\{r_k\}_{k\geq1}\) such that \(r_k\to\infty\) and \(r_kW'(r_k)\to 0\) as \(k\to\infty\). Applying [28] to the function \(r\mapsto rW'(r)\) if \(W\) is non-decreasing on \((R_0,\infty)\) or to \(r\mapsto -rW'(r)\) otherwise, we deduce that there is a sequence \(\{\rho_k\}_{k\geq1}\) such that \(\rho_k\to\infty\), \(\rho_k W'(\rho_k)\to 0\) and \(\rho_k(\rho_k W''(\rho_k)+W'(\rho_k))\to 0\) as \(k\to\infty\). The latter limit also ensures that \(\rho_k^2W''(\rho_k)\to 0\) as \(k\to\infty\). We next obtain by evaluating 43 at \(r=\rho_k\) and passing to the limit as \(k\to\infty\) that \(L^{1/m}=L\) and, since \(L\in(0,\infty)\), it follows that \(L=1\). The claimed behavior ?? follows then from 42 . ◻

Proof. The existence and regularity statements in Proposition 11 follow from Proposition 10, while the precise behavior ?? for large values of \(|x|\) is provided by Lemma 9. We are thus left with proving the uniqueness statement. To this end, we employ an optimal barrier argument stemming from [29], which is completely similar to the one used in [8], [20]. We give a sketch of the proof for the sake of completeness. Assume for contradiction that there exist two solutions \(V_1\) and \(V_2\) to 29 with the properties described in Proposition 11. Owing to their radial symmetry, \(V_1\) and \(V_2\) are actually two solutions to 41 . We may also assume that \(V_2(0)<V_1(0)\) (otherwise, since \(V_1'(0)=V_2'(0)=0\) by radial symmetry, we have \(V_1\equiv V_2\)). For \(\lambda>0\), we set \[W_{\lambda}(r)=\lambda^{-2m/(m-1)}V_2(\lambda r), \quad r\in[0,\infty),\] and deduce by direct calculation that \(W_{\lambda}\) is a solution to \[\label{eq46resc} W_{\lambda}''(r)+\frac{N-1}{r}W_{\lambda}'(r)+W_{\lambda}^{1/m}(r)-\lambda^{\sigma+2} r^\sigma W_{\lambda}(r)=0, \quad r\in (0,\infty).\tag{47}\] Moreover, we infer from ?? (which is satisfied by \(V_2\)) that \[\label{interm10} \begin{align} \lim\limits_{r\to\infty}r^{m\sigma/(m-1)}W_{\lambda}(r)&=\lim\limits_{r\to\infty}(\lambda r)^{m\sigma/(m-1)}\lambda^{-2m/(m-1)-m\sigma/(m-1)}V_2(\lambda r)\\ &=\lambda^{-m(\sigma+2)/(m-1)}. \end{align}\tag{48}\] Picking \(\lambda\in(0,1)\), we obtain from ?? (which is satisfied by \(V_1\)) and 48 that there is \(r_{\lambda}>0\) such that \[\label{interm11} W_{\lambda}(r)>V_1(r), \quad r\in(r_{\lambda},\infty),\tag{49}\] and we take the smallest value of \(r_{\lambda}\) for which 49 holds true. Since, for \(\lambda'<\lambda\), the monotonicity of \(V_2\) implies that \[(\lambda')^{-2m/(m-1)}>\lambda^{-2m/(m-1)}, \quad V_2(\lambda'r)\geq V_2(\lambda r),\] we infer that \(W_{\lambda'}(r)>W_{\lambda}(r)\) for any \(r\in[0,\infty)\) and thus \(r_{\lambda'}<r_{\lambda}\). Moreover, \[\lim\limits_{\lambda\to0}\inf\limits_{[0,r_{1/2}]} W_{\lambda} = \infty,\] so that there is \(\lambda_*\in (0,1/2)\) such that \[\inf_{[0,r_{1/2}]} W_{\lambda} > \max_{[0,r_{1/2}]} V_1, \quad \lambda\in (0,\lambda_*).\] Since \[W_{\lambda}(r) > W_{\lambda_*}(r) > W_{1/2}(r) > V_1(r), \quad (r,\lambda)\in (r_{1/2},\infty)\times (0,\lambda_*),\] we conclude that \(W_{\lambda}>V_1\) in \([0,\infty)\) for \(\lambda>0\) small enough. We can thus introduce the optimal parameter \[\lambda_0:=\sup\{\lambda\in(0,1): W_{\lambda}>V_1 \;\text{ in } \;[0,\infty)\},\] and observe that \(\lambda_0\in(0,1)\), since \(V_2(0)<V_1(0)\). Assume for contradiction that \(r_{\lambda_0}>0\). It then follows by the definitions of \(\lambda_0\) and \(r_{\lambda_0}\) that \[W_{\lambda_0}(r)\geq V_1(r), \quad r\in[0,\infty), \quad W_{\lambda_0}(r_{\lambda_0})=V_1(r_{\lambda_0}),\] which also means that \(r_{\lambda_0}\) is a point of minimum for \(W_{\lambda_0}-V_1\) and thus \[W_{\lambda_0}'(r_{\lambda_0})=V_1'(r_{\lambda_0}), \quad W_{\lambda_0}''(r_{\lambda_0})\geq V_1''(r_{\lambda_0}).\] Evaluating 47 (with the solution \(W_{\lambda_0}\)) and 41 (with the solution \(V_1\)) at \(r=r_{\lambda_0}\), we deduce that \[0\geq(1-\lambda_0^{\sigma+2})r_{\lambda_0}^\sigma V_1(r_{\lambda_0}),\] which contradicts the strict positivity of \(V_1\). Consequently, \(r_{\lambda_0}=0\) and the optimality of \(\lambda_0\) ensures that \[W_{\lambda_0}(0)=\lambda_0^{-2m/(m-1)}V_2(0)=V_1(0).\] Arguing as in [20] and using the expansions of \(V_1\) and \(V_2\) as \(r\to 0\) (which are the same as the ones derived in [20]), we discard this possibility and conclude the proof of the uniqueness. ◻

Proof. We deduce from Proposition 8 and ?? that \(v(s)\in L^\infty_+(\mathbb{R}^N)\cap \mathcal{X}\) for any \(s\geq0\) and that \(\{v^m(s)\}_{s\geq0}\) is uniformly bounded in \(\dot{H^1}(\mathbb{R}^N)\cap L^{2}_{\sigma}(\mathbb{R}^N)\cap L^{(m+1)/m}(\mathbb{R}^N)\). On the one hand, since \((m+1)/m<2\), it follows that \(\dot{H^1}(\mathbb{R}^N)\cap L^{(m+1)/m}(\mathbb{R}^N)\) embeds continuously in \(H^1(\mathbb{R}^N)\). On the other hand, since \(\sigma>\sigma_0>0\), \(H^1(\mathbb{R}^N)\cap L^2_{\sigma}(\mathbb{R}^N)\) is compactly embedded in \(L^2(\mathbb{R}^N)\). Putting together these embeddings, we infer that the set \(\{v^m(s)\}_{s\geq0}\) is relatively compact in \(L^2(\mathbb{R}^N)\), from which we readily deduce that \(\omega(u_0)\) is non-empty.

Pick \(V\in\omega(u_0)\) and let \(\{s_j\}_{j\geq1}\) be a sequence such that 50 holds true. To show that \(V\) solves 29 , we proceed along the lines of [30] and first observe that Lemma 4 and ?? imply that \[\begin{align} \int_0^{s}\Big\|\partial_s\big(v^{(m+1)/2}\big)(s_*)\Big\|_2^2\,ds_* & \leq \frac{(m+1)^2}{4m} \left[\mathcal{I}(u_0)-\mathcal{I}(v(s))\right] \\ & \le \frac{(m+1)^2}{4m} \left[\mathcal{I}(u_0) + \mathcal{I}_0 \right] \end{align}\] for any \(s\geq 0\), from which we readily deduce that \[\label{interm13} K_1 := \int_0^{\infty}\Big\|\partial_s\big(v^{(m+1)/2}\big)(s)\Big\|_2^2\,ds < \infty.\tag{51}\] Set now \[v_j(s):=v(s+s_j), \quad s\in[0,1], \quad j\geq1.\] On the one hand, the following numerical inequality, valid for any \((X,Y)\in[0,\infty)^2\) since \(m>(m+1)/2\), \[\begin{align} \big|X^m - Y^m\big| & \le \frac{2m}{m+1} \max\left\{ X^{(m+1)/2} , Y^{(m+1)/2} \right\}^{(m-1)/(m+1)} \left| X^{(m+1)/2} - Y^{(m+1)/2} \right| \\ & = \frac{2m}{m+1} \max\left\{X,Y \right\}^{(m-1)/2} \left| X^{(m+1)/2} - Y^{(m+1)/2} \right| \end{align}\] implies that, for any \(s\in[0,1]\), \(j\geq1\) and \(x\in\mathbb{R}^N\), \[\begin{align} \big|v_j^m(s,x) - v_j^m(0,x)\big|&\leq\frac{2m}{m+1} \max\left\{ \|v_j(s)\|_\infty,\|v_j(0)\|_\infty \right\}^{(m-1)/2}\\ &\quad \times\big|v_j^{(m+1)/2}(s,x) - v_j^{(m+1)/2}(0,x)\big| \\ &\leq\frac{2m}{m+1} \max\left\{ \|v(s_j+s)\|_\infty,\|v(s_j)\|_\infty \right\}^{(m-1)/2}\\ &\quad \times\int_{s_j}^{s_j+s} \left| \partial_s \left( v^{(m+1)/2} \right)(s_*,x) \right|\,ds_*. \end{align}\] The uniform boundedness ?? of \(v\) in \(L^{\infty}((\ln{m}/(m-1),\infty)\times\mathbb{R}^N)\), together with the previous estimate and an application of Hölder’s inequality, further gives \[\sup_{s\in [0,1]} \big\| v_j^m(s) - v^m(s_j) \big\|_2 \le K_2\left( \int_{s_j}^{\infty} \left\| \partial_s \left( v^{(m+1)/2} \right)(s_*,x) \right\|_2^2\;ds_* \right)^{1/2},\] with \(K_2 := 2C_2(u_0)^{(m-1)/2}\), where we recall that the constant \(C_2(u_0)\) is introduced in ?? . Therefore, by 50 and 51 , \[\label{conv1} \lim_{j\to\infty}\sup_{s\in [0,1]}\big\|v_j^m(s)-V\big\|_2 = 0.\tag{52}\] Moreover, combining 52 and the numerical inequality \[\label{num46ineq} |X-Y|^m \le|X^m-Y^m|, \quad (X,Y)\in[0,\infty)^2,\tag{53}\] which is valid due to \(m>1\), gives \[\label{conv3} \lim_{j\to\infty} \sup_{s\in [0,1]} \big\|v_j(s) - V^{1/m}\big\|_{2m} = 0.\tag{54}\] On the other hand, the uniform boundedness ?? of \(v^m\) in \(\dot{H}^1(\mathbb{R}^N)\cap L^2_{\sigma}(\mathbb{R}^N)\) ensures that \(\big\{\nabla v_j^m\big\}_{j\geq1}\) and \(\{v_j^m\}_{j\geq1}\) are uniformly bounded in \(L^{\infty}((0,1),L^2(\mathbb{R}^N))\) and in \(L^{\infty}((0,1),L^2_{\sigma}(\mathbb{R}^N))\), respectively. We infer from these bounds and 52 that there exists a subsequence of \(\{v_j\}_{j\geq1}\) (not relabeled for simplicity) such that \[\label{conv2} \begin{align} \nabla v^m(s_j) \rightharpoonup \nabla V & \;\;\text{ in }\;\; L^2(\mathbb{R}^N), \\ \nabla v^m_j \stackrel{*}{\rightharpoonup} \nabla V & \;\;\text{ in }\;\; L^\infty((0,1),L^2(\mathbb{R}^N)), \\ v^m_j \stackrel{*}{\rightharpoonup} V & \;\;\text{ in }\;\; L^\infty((0,1),L_\sigma^2(\mathbb{R}^N)). \end{align}\tag{55}\]

We finally write the weak formulation of Eq. 11 on \((s_j,s_j+1)\) with \(\vartheta\in C_c^{\infty}(\mathbb{R}^N)\) as test function, which yields \[\begin{align} \int_{\mathbb{R}^N} \big[v_j(1,x) - v_j(0,x)] \vartheta(x)\;dx & + \int_0^1 \int_{\mathbb{R}^N} \nabla v_j^m(s,x)\cdot \nabla\vartheta\;dxds \\ & \quad = \int_0^1 \int_{\mathbb{R}^N} \left[v_j(s,x)-|x|^\sigma v_j^m(s,x)\right] \vartheta(x)\;dxds. \end{align}\] We can thus pass to the limit as \(j\to\infty\) in the previous identity and deduce from 52 , 55 and 54 that \[\begin{align} \int_{\mathbb{R}^N}\nabla V(x)\cdot\nabla\vartheta(x)\,dxds =\int_{\mathbb{R}^N}\left[V^{1/m}(x)- |x|^\sigma V(x)\right]\vartheta(x)\,dxds, \end{align}\] whence \(V\) is a weak solution to 29 . In order to check that \(V\not\equiv 0\), we fix \(s_0>0\) and recall the time monotonicity ?? \[\left( \frac{e^{(m-1)s_j} - 1}{e^{(m-1)s_j}} \right)^{1/(m-1)} v(s_j,x) \ge \left( \frac{e^{(m-1)s_0} - 1}{e^{(m-1)s_0}} \right)^{1/(m-1)} v(s_0,x), \quad x\in\mathbb{R}^N,\] which is valid for any \(j\ge 1\) such that \(s_j>s_0\). We then let \(j\to\infty\) in the above inequality and deduce from the convergence 54 that \[V^{1/m}(x) \ge \left( \frac{e^{(m-1)s_0} - 1}{e^{(m-1)s_0}} \right)^{1/(m-1)} v(s_0,x), \quad x\in\mathbb{R}^N,\] which readily implies ?? , as well as \(V\not\equiv 0\), since \(v(s_0)\not\equiv 0\) for all \(s_0>0\) by Proposition 5.

Finally, if \(u_0\) is radially symmetric with non-increasing profile, then it follows from Lemma 1 that \(v(s)\) shares the same properties for \(s\geq0\) and thus \(V\) is radially symmetric with non-increasing profile, completing the proof. ◻

Proof of Theorems 1 and 2. Several steps are needed to be able to handle arbitrary initial conditions \(u_0\) in \(L^\infty_+(\mathbb{R}^N)\cap \mathcal{X}\).

Step 1. Assume first that \(u_0\in L^\infty_+(\mathbb{R}^N)\cap \mathcal{X}\) is a radially symmetric initial condition with non-increasing profile. It follows from Proposition 12 that any element \(V\in\omega(u_0)\) is a radially symmetric function with non-increasing profile in \(L^\infty_+(\mathbb{R}^N)\cap \mathcal{X}\), as well as a non-trivial weak solution to 29 . Proposition 11 then entails that \(\omega(u_0)=\{V_*\}\) is a singleton and a classical compactness argument leads us to the convergence \[\lim_{s\to\infty} \|v^m(s) - V_*\|_2 = 0. \label{conv4}\tag{56}\]

Step 2. Assume now that \(u_0\in L^\infty_+(\mathbb{R}^N)\cap \mathcal{X}\) is such that there are two radially symmetric initial conditions with non-increasing profile \(u_{0,b}\in L^\infty_+(\mathbb{R}^N)\cap \mathcal{X}\) and \(u_{0,a}\in L^\infty_+(\mathbb{R}^N)\cap \mathcal{X}\) satisfying \(u_{0,b} \le u_0 \le u_{0,a}\) in \(\mathbb{R}^N\). Denoting the corresponding solutions to 10 by \(v_{b}\) and \(v_{a}\), respectively, the comparison principle recalled in Proposition 4 implies that \[v_{b}(s,x) \le v(s,x) \le v_{a}(s,x), \quad (s,x)\in (0,\infty)\times\mathbb{R}^N,\] so that \[\begin{align} \|v^m(s)-V_*\|_2^2 & = \big\|\big(v^m(s)-V_*\big)_+\big\|_2^2 + \big\|\big(V_*-v^m(s)\big)_+\big\|_2^2 \\ & \le \big\|\big(v_{a}^m(s)-V_*\big)_+\big\|_2^2 + \big\|\big(V_*-v_{b}^m(s)\big)_+\big\|_2^2 \\ & \le \|v_{a}^m(s)-V_*\|_2^2 + \|v_{b}^m(s)-V_*\|_2^2. \end{align}\] Since \(v_{b}\) and \(v_{a}\) satisfy 56 by Step 1, an immediate consequence of the above inequality is that \(\{v^m(s)\}_{s>0}\) converges to \(V_*\) in \(L^2(\mathbb{R})\); that is, \(\omega(u_0)=\{V_*\}\).

Step 3. We now complete the proof of Theorem 1. Let \(V\in\mathcal{Y}_+\) be a non-trivial stationary solution to 29 . Then \(V\in L^\infty(\mathbb{R}^N)\) by Proposition 9, so that \(V^{1/m}\) is a non-trivial function in \(L^\infty_+(\mathbb{R}^N)\cap \mathcal{X}\) and a stationary solution to 11 . Moreover, the positivity of \(V\) established in Proposition 9 guarantees that there is a radially symmetric initial condition with non-increasing profile \(u_{0,b}\in L^\infty_+(\mathbb{R}^N)\cap \mathcal{X}\) such that \(u_{0,b}\le V^{1/m}\) in \(\mathbb{R}^N\). Also, owing to Proposition 9 and Lemma 7, there is \(C>0\) depending on \(N\), \(m\), \(\sigma\) and \(\|V\|_\infty\) such that \(V^{1/m}(x) \le C u_{0,a}(x)\) with \(u_{0,a}(x) := (1+|x|)^{-\sigma/(m-1)}\) for \(x\in\mathbb{R}^N\). Taking into account that \(\sigma>\sigma_0\), we realize that \(u_{0,a}\) belongs to \(L^\infty_+(\mathbb{R}^N)\cap \mathcal{X}\) and is non-negative and radially symmetric with non-increasing profile. We are then in a position to apply Step 2 to conclude that \(V=V_*\), thereby completing the proof of Theorem 1 with \(\varphi=V_*^{1/m}\).

Step 4. Consider now an arbitrary initial condition \(u_0\in L^\infty_+(\mathbb{R}^N)\cap \mathcal{X}\) and let \(v\) be the corresponding solution to 10 . A direct consequence of Proposition 12 and Step 3 is that \(\omega(u_0)=\{V_*\}\) and the same classical compactness argument used in Step 1 then implies that 56 holds true for \(v\) as well. Furthermore, by ?? , \[v(s,x) \le \left( \frac{e^{(m-1)s}}{e^{(m-1)s} - 1} \right)^{1/(m-1)} V_*^{1/m}(x), \quad (s,x)\in (0,\infty)\times \mathbb{R}^N. \label{fg03}\tag{57}\] We then infer from 53 , 57 and Hölder’s inequality that, for \(s>0\) and \(R\ge 1\), \[\begin{align} & \big\|v(s)-V_*^{1/m}\big\|_{m+1}^{m+1} \le \|v^m(s)-V_*\|_{(m+1)/m}^{(m+1)/m} \\ & \quad \le \int_{B(0,R)} |v^m(s)-V_*|^{(m+1)/m}\;dx \\ & \quad\quad + \int_{\mathbb{R}^N\setminus B(0,R)} |v^m(s)+V_*|^{(m+1)/m}\;dx \\ & \quad\le |B(0,R)|^{(m-1)/(2m)} \left( \int_{B(0,R)} |v^m(s)-V_*|^{2}\;dx \right)^{(m+1)/(2m)} \\ & \quad\quad + 2^{1/m} \int_{\mathbb{R}^N\setminus B(0,R)} \left( v^{m+1}(s) + V_*^{(m+1)/m}\right)\;dx \\ & \quad\le \big(\varpi_N R^N \big)^{(m-1)/(2m)} \big\|v^m(s)-V_*\big\|_2^{(m+1)/m} \\ & \quad\quad + 2^{1/m} \left( 1 + \left[ \frac{e^{(m-1)s}}{e^{(m-1)s}-1} \right]^{(m+1)/(m-1)} \right)\int_{\mathbb{R}^N\setminus B(0,R)} V_*^{(m+1)/m}\;dx. \end{align}\] According to 56 , we may take the limit \(s\to\infty\) in the above inequality and find \[\limsup_{s\to\infty} \big\|v(s)-V_*^{1/m}\big\|_{m+1}^{m+1} \le 2^{(m+1)/m} \int_{\mathbb{R}^N\setminus B(0,R)} V_*^{(m+1)/m}\;dx.\] Recalling that \(V_*\in L^{(m+1)/m}(\mathbb{R}^N)\), we may take the limit \(R\to\infty\) in the above inequality and end up with \[\lim_{s\to\infty} \big\|v(s)-V_*^{1/m}\big\|_{m+1}^{m+1} = 0. \label{fg04}\tag{58}\] Combining 58 and the boundedness ?? of \(v\) in \(L^\infty((\ln{m}/(m-1),\infty)\times\mathbb{R}^N)\) gives the convergence ?? (with \(\varphi:=V_*^{1/m}\)) for all \(r\in [m+1,\infty)\).

Step 5. We are left with the convergence in \(L^\infty(\mathbb{R}^N)\). Let \(\{s_j\}_{j\ge 1}\) be a sequence of positive real numbers such thart \(s_j\to\infty\) as \(j\to\infty\). Without loss of generality, we may assume that \(s_j\ge 1+\ln{m}/(m-1)\) for \(j\ge 1\). Setting \(v_j(s) := v(s+s_j)\) for \(s\in [-1,1]\) and \(j\ge 1\), we argue as in the proof of Proposition 12 with the help of ?? to show that \[\label{fg05} \lim_{j\to\infty} \sup_{s\in [-1,1]} \big\|v_j(s) - V_*^{1/m}\big\|_{2m} = 0,\tag{59}\] taking into account that \(\omega(u_0)=\{V_*^{1/m}\}\) as shown in Step 4. Next, let \(\varepsilon\in (0,1)\) and observe that, owing to 11 and ?? , \(v_j\) is a weak solution to \[\partial_s v_j - \Delta\Phi(v_j) = S_j, \quad (s,x)\in (-1,1)\times B(0,2/\varepsilon),\] with \[\Phi(z) := \max\left\{ z , 2 C_2(u_0) \right\}^m, \quad z\ge 0,\] and \[S_j(s,x) := -|x|^\sigma v_j^m(s,x) + v_j(s,x), \quad (s,x)\in (-1,1)\times B(0,2/\varepsilon).\] Since \[|S_j(s,x)| \le \left( \frac{2}{\varepsilon} \right)^\sigma C_2(u_0)^m + C_2(u_0), \quad (s,x)\in (-1,1)\times B(0,2/\varepsilon),\] the assumptions (\(A_1\))-(\(A_6\)) in [31] are satisfied and we infer from [31] that there are \(\alpha_\varepsilon\in (0,1)\) and \(\gamma_\varepsilon>1\) such that \[|v_j(s,x) - v_j(s_0,x_0)| \le \gamma_\varepsilon \left( |x-x_0|^{\alpha_\varepsilon} + |s-s_0|^{\alpha_\varepsilon/2} \right)\] for \((s,s_0,x,x_0) \in [-1+\varepsilon,1-\varepsilon]^2\times \bar{B}(0,1/\varepsilon)^2\). Consequently, \[\{v_j\}_{j\ge 1} \;\text{ is bounded in }\; C^{\alpha_\varepsilon/2,\alpha_\varepsilon}\big([-1+\varepsilon,1-\varepsilon]\times \bar{B}(0,1/\varepsilon)\big),\] and we deduce from the Arzelà-Ascoli theorem and the convergence 59 by a diagonal process that there is a subsequence \(\{v_{j_k}\}_{k\ge 1}\) of \(\{v_j\}_{j\ge 1}\) such that \[v_{j_k} \longrightarrow V_*^{1/m} \;\;\text{ in }\;\; C\big([-1/2,1/2]\times \bar{B}(0,R)\big) \label{fg06}\tag{60}\] for any \(R>1\).

We now take advantage of the upper bound ?? , along with the tail control ?? , to complete the proof. Indeed, pick \(R>1\), \(k\ge 1\) and \(x\in\mathbb{R}^N\). Then, either \(x\in B(0,R)\) and \[\begin{align} \big|v(s_{j_k},x) - V_*^{1/m}(x)\big| & = \big|v_{j_k}(0,x) - V_*^{1/m}(x)\big| \\ & \le \left\| v_{j_k} - V_*^{1/m} \right\|_{C\big([-1/2,1/2]\times \bar{B}(0,R)\big)}. \end{align}\] Or, \(x\in\mathbb{R}^N\setminus B(0,R)\) and, according to ?? with \(s_0=s_{j_k}\) and Proposition 10, \[\begin{align} \big|v(s_{j_k},x) - V_*^{1/m}(x)\big| & \le \big|v(s_{j_k},x)\big| + \big|V_*^{1/m}(x)\big| \\ & \le V_*^{1/m}(x) + \left( \frac{e^{(m-1)s_{j_k}}}{e^{(m-1)s_{j_k} - 1}} \right)^{1/(m-1)} V_*^{1/m}(s,x) \\ & \le C \left[ 1 + \left( \frac{e^{(m-1)s_{j_k}}}{e^{(m-1)s_{j_k} - 1}} \right)^{1/(m-1)} \right] R^{-\sigma/(m-1)}. \end{align}\] Therefore, \[\begin{align} \big\| v(s_{j_k}) -V_*^{1/m}\big\|_\infty & \le \left\| v_{j_k} - V_*^{1/m} \right\|_{C\big([-1/2,1/2]\times \bar{B}(0,R)\big)} \\ & \quad + C \left[ 1 + \left( \frac{e^{(m-1)s_{j_k}}}{e^{(m-1)s_{j_k} - 1}} \right)^{1/(m-1)} \right] R^{-\sigma/(m-1)}, \end{align}\] and we may take the limit \(k\to\infty\) in view of 60 to obtain \[\limsup_{k\to\infty} \big\| v(s_{j_k}) -V_*^{1/m}\big\|_\infty \le 2C R^{-\sigma/(m-1)}.\] Letting \(R\to\infty\) gives \[\lim_{k\to\infty} \big\| v(s_{j_k}) -V_*^{1/m}\big\|_\infty = 0.\] We have thus proved that, given an arbitrary sequence \(\{s_j\}_{j\ge 1}\) satisfying \(s_j\to\infty\) as \(j\to\infty\), there is a subsequence \(\{s_{j_k}\}_{k\ge 1}\) with \(s_{j_k}\to\infty\) as \(k\to\infty\) such that \(\{v(s_{j_k})\}_{k\ge 1}\) converges in \(L^\infty(\mathbb{R}^N)\) and its limit is \(V_*^{1/m}\). Consequently, \(\{v(s_j)\}_{j\ge 1}\) is relatively compact in \(L^\infty(\mathbb{R}^N)\) with a unique cluster point \(V_*^{1/m}\), whence \(\{v(s_j)\}_{j\ge 1}\) converges to \(V_*^{1/m}\) in \(L^\infty(\mathbb{R}^N)\), and the proof of ?? is complete, recalling that \(\varphi=V_*^{1/m}\). ◻

Proof of Proposition 3. Assume that \(\sigma\in [0,\sigma_0]\) and consider a solution \[V\in \dot{H}^{1}(\mathbb{R}^N)\cap L^2_{\sigma}(\mathbb{R}^N)\cap L_{+}^{(m+1)/m}(\mathbb{R}^N)\] to 14 . We then multiply 14 by \(x\cdot\nabla V\) and integrate by parts. By proceeding exactly as in [32], we derive the following identities \[\begin{align} &-\int_{\mathbb{R}^N}(x\cdot\nabla V)\Delta V\,dx==-\frac{N-2}{2}\|\nabla V\|_2^2,\\ &\int_{\mathbb{R}^N}(x\cdot\nabla V)|x|^{\sigma}V\,dx=-\frac{N+\sigma}{2}\mathcal{N}_{\sigma}^2(V),\\ &\int_{\mathbb{R}^N}(x\cdot\nabla V)V^{1/m}\,dx=-\frac{mN}{m+1}\int_{\mathbb{R}^N}V^{(m+1)/m}\,dx. \end{align}\] Gathering the previous identities, we infer from Eq. 14 that \[\label{poh1} -\frac{N-2}{2}\|\nabla V\|_{2}^2 - \frac{N+\sigma}{2} N_{\sigma}^2(V)=-\frac{mN}{m+1}\|V\|_{(m+1)/m}^{(m+1)/m}.\tag{61}\] Multiplying Eq. 14 by \(V\) and integrating over \(\mathbb{R}^N\), we find \[\label{poh2} \|\nabla V\|_{2}^2+N_{\sigma}^2(V)=\|V\|_{(m+1)/m}^{(m+1)/m}.\tag{62}\] We next multiply 62 by \((N+\sigma)/2\) and add the resulting equality to 61 in order to eliminate the terms containing \(N_{\sigma}^2(V)\). We thus obtain \[\label{poh3} \frac{\sigma+2}{2}\|\nabla V\|_2^{2}+\frac{N(m-1)-\sigma(m+1)}{2(m+1)}\|V\|_{(m+1)/m}^{(m+1)/m}=0.\tag{63}\] Since \(\sigma\in[0,\sigma_0]\), we observe that the coefficient of the first term in 63 is strictly positive, while the coefficient of the second term in 63 is non-negative. This readily implies that \(\|\nabla V\|_{2}=0\), hence \(V\) is constant for a.a. \(x\in\mathbb{R}^N\). We then deduce from the integrability properties of \(V\) that \(V\equiv0\). Letting \(\varphi:=V^{1/m}\) completes the proof. ◻