Introduction
We are interested in linear variational problems on domains Ω ⊂ R d \Omega\subset\mathbb{R}^{d} Ω ⊂ R d Δ 2 \Delta^{2} Δ 2 ∣ x ∣ − 4 |x|^{-4} ∣ x ∣ − 4 [9 ] .
To avoid trivial situations, we will always assume that either 0 ∈ Ω 0 \in\Omega 0 ∈ Ω 0 ∈ ∂ Ω 0 \in\partial\Omega 0 ∈ ∂ Ω
Our work was initially inspired by the celebrated result of Brezis and Vázquez [5 Theorem 4.1] , where it is proved that for 0 ∈ Ω 0 \in\Omega 0 ∈ Ω d ≥ 3 d \ge3 d ≥ 3
∫ Ω ∣ ∇ u ∣ 2 d x − ( d − 2 2 ) 2 ∫ Ω ∣ u ∣ 2 ∣ x ∣ 2 d x ≥ ( ∣ B 1 ∣ ∣ Ω ∣ ) 2 d λ ( − Δ , D 2 ) ∫ Ω ∣ u ∣ 2 d x , u ∈ H 0 1 ( Ω ) . (1) \int_{\Omega} |\nabla u|^{2}\,dx-\left(\frac{d-2}{2}\right)^{2}\int_{\Omega}\frac{|u|^{2}}{|x|^{2}}\,dx \ge\left(\frac{|B_{1}|}{|\Omega|}\right)^{\frac{2}{d}}\lambda(-\Delta,\mathbb{D}^{2})\int_{\Omega}|u|^{2}\,dx,\qquad u\in H_{0}^{1}(\Omega). \tag*{(1)} ∫ Ω ∣∇ u ∣ 2 d x − ( 2 d − 2 ) 2 ∫ Ω ∣ x ∣ 2 ∣ u ∣ 2 d x ≥ ( ∣Ω∣ ∣ B 1 ∣ ) d 2 λ ( − Δ , D 2 ) ∫ Ω ∣ u ∣ 2 d x , u ∈ H 0 1 ( Ω ) . (1) Here ∣ Ω ∣ |\Omega| ∣Ω∣ ∣ B 1 ∣ |B_{1}| ∣ B 1 ∣ Ω \Omega Ω B 1 ⊂ R d B_{1}\subset\mathbb{R}^{d} B 1 ⊂ R d λ ( − Δ , D 2 ) \lambda(-\Delta,\mathbb{D}^{2}) λ ( − Δ , D 2 ) D 2 ⊂ R 2 \mathbb{D}^{2} \subset\mathbb{R}^{2} D 2 ⊂ R 2
λ ( − Δ , D 2 ) = inf v ∈ H 0 1 ( D 2 ) v ≠ 0 ∫ D 2 ∣ ∇ v ∣ 2 d x ∫ D 2 ∣ v ∣ 2 d x . (2) \lambda(-\Delta,\mathbb{D}^{2})=\inf_{\substack{v\in H_{0}^{1}(\mathbb{D}^{2})\\v\ne0}}\frac{\int_{\mathbb{D}^{2}}|\nabla v|^{2}\,dx}{\int_{\mathbb{D}^{2}}|v|^{2}\,dx}. \tag*{(2)} λ ( − Δ , D 2 ) = v ∈ H 0 1 ( D 2 ) v = 0 inf ∫ D 2 ∣ v ∣ 2 d x ∫ D 2 ∣∇ v ∣ 2 d x . (2) We also refer to [12 , 10 ] , where the case 0 ∈ ∂ Ω 0\in\partial\Omega 0 ∈ ∂ Ω [11 ] for a related problem.
Let us describe the problem we are interested in and our main results (see § 2 for the notation and main definitions). Throughout the paper, we assume that at least one of the following conditions is satisfied:
(A1) d ≥ 5 d\ge5 d ≥ 5 0 ∈ ∂ Ω 0\in\partial\Omega 0 ∈ ∂ Ω Ω \Omega Ω 0 0 0
This allows us to prove in Lemma 3.2 , that the Rellich constant of Ω \Omega Ω
μ 0 ( Ω ) = inf u ∈ H 0 2 ( Ω ) u ≠ 0 ∫ Ω ∣ Δ u ∣ 2 d x ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x , (3) \mu_{0}(\Omega)=\inf_{\substack{u\in H_{0}^{2}(\Omega)\\u\ne0}}\frac{\int_{\Omega}|\Delta u|^{2}\,dx}{\int_{\Omega}|x|^{-4}|u|^{2}\,dx}, \tag*{(3)} μ 0 ( Ω ) = u ∈ H 0 2 ( Ω ) u = 0 inf ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x ∫ Ω ∣Δ u ∣ 2 d x , (3) is positive.
If d ≥ 5 d\ge5 d ≥ 5 ∣ x ∣ − 4 |x|^{-4} ∣ x ∣ − 4 R d \mathbb{R}^{d} R d [15 ] yields
μ 0 ( R d ) = inf u ∈ C c ∞ ( R d ) u ≠ 0 ∫ R d ∣ Δ u ∣ 2 d x ∫ R d ∣ x ∣ − 4 ∣ u ∣ 2 d x = γ d 2 , where γ d = d ( d − 4 ) 4 . (4) \mu_{0}(\mathbb{R}^{d})=\inf_{\substack{u\in C_{c}^{\infty}(\mathbb{R}^{d})\\u\ne0}}\frac{\int_{\mathbb{R}^{d}}|\Delta u|^{2}\,dx}{\int_{\mathbb{R}^{d}}|x|^{-4}|u|^{2}\,dx}=\gamma_{d}^{2}, \qquad\text{where}\qquad\gamma_{d}=\frac{d(d-4)}{4}. \tag*{(4)} μ 0 ( R d ) = u ∈ C c ∞ ( R d ) u = 0 inf ∫ R d ∣ x ∣ − 4 ∣ u ∣ 2 d x ∫ R d ∣Δ u ∣ 2 d x = γ d 2 , where γ d = 4 d ( d − 4 ) . (4) It follows that μ 0 ( Ω ) ≥ γ d 2 > 0 \mu_{0}(\Omega)\ge\gamma_{d}^{2}>0 μ 0 ( Ω ) ≥ γ d 2 > 0 0 ∈ Ω 0\in\Omega 0 ∈ Ω μ 0 ( Ω ) = μ 0 ( R d ) = γ d 2 \mu_{0}(\Omega)=\mu_{0}(\mathbb{R}^{d})=\gamma_{d}^{2} μ 0 ( Ω ) = μ 0 ( R d ) = γ d 2 μ 0 ( Ω ) \mu_{0}(\Omega) μ 0 ( Ω ) H 0 2 ( Ω ) H_{0}^{2}(\Omega) H 0 2 ( Ω ) (4) .
If (A2) is satisfied, then no restriction on the dimension d d d μ 0 ( Ω ) > 0 \mu_{0}(\Omega)>0 μ 0 ( Ω ) > 0 μ 0 ( Ω ) \mu_{0}(\Omega) μ 0 ( Ω ) § 7 .
We then introduce, following [4 , 12 ] , the minimization problem
μ λ ( Ω ) = inf u ∈ H 0 2 ( Ω ) u ≠ 0 J λ ( u ) , J λ ( u ) : = ∫ Ω ∣ Δ u ∣ 2 d x − λ ∫ Ω ∣ u ∣ 2 d x ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x . (5) \mu_{\lambda}(\Omega)=\inf_{\substack{u\in H_{0}^{2}(\Omega)\\u\ne0}}J_{\lambda}(u), \qquad J_{\lambda}(u):=\frac{\int_{\Omega}|\Delta u|^{2}\,dx-\lambda\int_{\Omega}|u|^{2}\,dx}{\int_{\Omega}|x|^{-4}|u|^{2}\,dx}. \tag*{(5)} μ λ ( Ω ) = u ∈ H 0 2 ( Ω ) u = 0 inf J λ ( u ) , J λ ( u ) := ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x ∫ Ω ∣Δ u ∣ 2 d x − λ ∫ Ω ∣ u ∣ 2 d x . (5) If λ \lambda λ Δ 2 \Delta^{2} Δ 2 H 0 2 ( Ω ) H_{0}^{2}(\Omega) H 0 2 ( Ω ) μ λ ( Ω ) = 0 \mu_{\lambda}(\Omega)=0 μ λ ( Ω ) = 0 μ λ ( Ω ) \mu_{\lambda}(\Omega) μ λ ( Ω )
λ Ω ∗ = inf { λ ∈ R ∣ μ λ ( Ω ) is achieved } ∈ [ − ∞ , ∞ ) . \lambda_{\Omega}^{*}=\inf\{\lambda\in\mathbb{R}\mid\mu_{\lambda}(\Omega)\ \text{is achieved}\}\in[-\infty,\infty). λ Ω ∗ = inf { λ ∈ R ∣ μ λ ( Ω ) is achieved } ∈ [ − ∞ , ∞ ) . In our first result we describe the main features of the function λ ↦ μ λ ( Ω ) \lambda\mapsto\mu_\lambda(\Omega) λ ↦ μ λ ( Ω ) (5) . The variational problem defining μ λ ( Ω ) \mu_\lambda(\Omega) μ λ ( Ω ) R d \mathbb{R}^d R d μ λ ( Ω ) \mu_\lambda(\Omega) μ λ ( Ω )
μ Ω ∞ = lim λ → λ Ω ∗ μ λ ( Ω ) . \mu_\Omega^\infty= \lim_{\lambda\to\lambda_\Omega^*} \mu_\lambda(\Omega). μ Ω ∞ = λ → λ Ω ∗ lim μ λ ( Ω ) . Indeed, in our first result we show, in particular, that μ λ ( Ω ) ≤ μ Ω ∞ \mu_\lambda(\Omega) \leq\mu_\Omega^\infty μ λ ( Ω ) ≤ μ Ω ∞ λ ∈ R \lambda\in\mathbb{R} λ ∈ R μ λ ( Ω ) < μ Ω ∞ \mu_\lambda(\Omega) < \mu_\Omega^\infty μ λ ( Ω ) < μ Ω ∞ μ λ ( Ω ) \mu_\lambda(\Omega) μ λ ( Ω )
Theorem 1.1 Let Ω ⊂ R d \Omega\subset\mathbb{R}^d Ω ⊂ R d d ≥ 1 d \geq1 d ≥ 1 λ Ω ∗ < λ ( Δ 2 , Ω ) \lambda_\Omega^* < \lambda(\Delta^2,\Omega) λ Ω ∗ < λ ( Δ 2 , Ω ) μ Ω ∞ > 0 \mu_\Omega^\infty> 0 μ Ω ∞ > 0
i) If λ > λ Ω ∗ \lambda> \lambda_\Omega^* λ > λ Ω ∗ μ λ ( Ω ) < μ Ω ∞ \mu_\lambda(\Omega) < \mu_\Omega^\infty μ λ ( Ω ) < μ Ω ∞ μ λ ( Ω ) \mu_\lambda(\Omega) μ λ ( Ω )
ii) The function λ ↦ μ λ ( Ω ) \lambda\mapsto\mu_\lambda(\Omega) λ ↦ μ λ ( Ω ) ( λ Ω ∗ , ∞ ) (\lambda_\Omega^*,\infty) ( λ Ω ∗ , ∞ )
iii) If, in addition, λ Ω ∗ > − ∞ \lambda_\Omega^* > -\infty λ Ω ∗ > − ∞ μ λ ( Ω ) = μ Ω ∞ \mu_\lambda(\Omega) = \mu_\Omega^\infty μ λ ( Ω ) = μ Ω ∞ λ ≤ λ Ω ∗ \lambda\leq\lambda_\Omega^* λ ≤ λ Ω ∗
In Appendix A we show that there exist domains Ω \Omega Ω μ Ω ∞ = ∞ \mu_\Omega^\infty= \infty μ Ω ∞ = ∞ λ Ω ∗ = − ∞ \lambda_\Omega^* = -\infty λ Ω ∗ = − ∞
Further information can be obtained either in case of an interior singularity or when Ω \Omega Ω 0 ∈ ∂ Ω 0 \in\partial\Omega 0 ∈ ∂ Ω
If 0 ∈ Ω 0 \in\Omega 0 ∈ Ω d ≥ 5 d \geq5 d ≥ 5 λ Ω ∗ \lambda_\Omega^* λ Ω ∗ [5 ] . Our bound involves the radius
R = sup x ∈ Ω ∣ x ∣ , R = \sup_{x \in\Omega} |x|, R = x ∈ Ω sup ∣ x ∣ , the Poincaré constant on the unit disk already appeared in (1) , and the principal eigenvalue of the biharmonic operator Δ 2 \Delta^2 Δ 2
λ ( Δ 2 , B 4 ) = inf v ∈ H 0 2 ( B 4 ) v ≠ 0 ∫ B 4 ∣ Δ v ∣ 2 d x ∫ B 4 ∣ v ∣ 2 d x . (6) \lambda(\Delta^2,B^4) = \inf_{\substack{v \in H_0^2(B^4)\\v\ne0}} \frac{\int_{B^4} |\Delta v|^2\,dx}{\int_{B^4} |v|^2\,dx}. \tag*{(6)} λ ( Δ 2 , B 4 ) = v ∈ H 0 2 ( B 4 ) v = 0 inf ∫ B 4 ∣ v ∣ 2 d x ∫ B 4 ∣Δ v ∣ 2 d x . (6) Theorem 1.2 Let Ω \Omega Ω R d \mathbb{R}^d R d d ≥ 5 d \geq5 d ≥ 5 0 ∈ Ω 0 \in\Omega 0 ∈ Ω μ Ω ∞ = μ 0 ( R d ) \mu_\Omega^\infty= \mu_0(\mathbb{R}^d) μ Ω ∞ = μ 0 ( R d )
λ Ω ∗ ≥ λ ( Δ 2 , B 4 ) R 4 + 2 d ( d − 4 ) λ ( − Δ , D 2 ) R 4 . (7) \lambda_\Omega^* \geq\frac{\lambda(\Delta^2,B^4)}{R^4} + 2d(d-4)\frac{\lambda(-\Delta,D^2)}{R^4}. \tag*{(7)} λ Ω ∗ ≥ R 4 λ ( Δ 2 , B 4 ) + 2 d ( d − 4 ) R 4 λ ( − Δ , D 2 ) . (7) Therefore, for any u ∈ H 0 2 ( Ω ) u \in H^2_0(\Omega) u ∈ H 0 2 ( Ω )
∫ Ω ∣ Δ u ∣ 2 d x − ( d ( d − 4 ) 4 ) 2 ∫ Ω ∣ u ∣ 2 ∣ x ∣ 4 d x ≥ [ λ ( Δ 2 , B 4 ) R 4 + 2 d ( d − 4 ) λ ( − Δ , D 2 ) R 2 ] ∫ Ω ∣ u ∣ 2 d x . (8) \int_{\Omega} |\Delta u|^2\,dx-\left(\frac{d(d-4)}{4}\right)^2\int_{\Omega}\frac{|u|^2}{|x|^4}\,dx\geq\left[\frac{\lambda(\Delta^2,\mathbb{B}^4)}{R^4}+2d(d-4)\frac{\lambda(-\Delta,\mathbb{D}^2)}{R^2}\right]\int_{\Omega}|u|^2\,dx. \tag*{(8)} ∫ Ω ∣Δ u ∣ 2 d x − ( 4 d ( d − 4 ) ) 2 ∫ Ω ∣ x ∣ 4 ∣ u ∣ 2 d x ≥ [ R 4 λ ( Δ 2 , B 4 ) + 2 d ( d − 4 ) R 2 λ ( − Δ , D 2 ) ] ∫ Ω ∣ u ∣ 2 d x . (8) We conjecture that the infimum μ λ ( Ω ) \mu_\lambda(\Omega) μ λ ( Ω ) λ > λ Ω ∗ \lambda>\lambda^*_\Omega λ > λ Ω ∗ λ = λ Ω ∗ \lambda=\lambda^*_\Omega λ = λ Ω ∗ μ λ ( Ω ) \mu_\lambda(\Omega) μ λ ( Ω )
Inequality (1.8) is complementary to the inequality of Gazzola–Grunau–Mitidieri [13 ] , namely,
∫ Ω ∣ Δ u ∣ 2 d x − ( d ( d − 4 ) 4 ) 2 ∫ Ω ∣ u ∣ 2 ∣ x ∣ 4 d x ≥ λ ( Δ 2 , B 4 ) R 4 ∫ Ω ∣ u ∣ 2 d x + d ( d − 4 ) λ ( − Δ , D 2 ) 2 R 2 ∫ Ω ∣ u ∣ 2 ∣ x ∣ 2 d x . \int_{\Omega}|\Delta u|^2\,dx-\left(\frac{d(d-4)}{4}\right)^2\int_{\Omega}\frac{|u|^2}{|x|^4}\,dx\geq\frac{\lambda(\Delta^2,\mathbb{B}^4)}{R^4}\int_{\Omega}|u|^2\,dx+\frac{d(d-4)\lambda(-\Delta,\mathbb{D}^2)}{2R^2}\int_{\Omega}\frac{|u|^2}{|x|^2}\,dx. ∫ Ω ∣Δ u ∣ 2 d x − ( 4 d ( d − 4 ) ) 2 ∫ Ω ∣ x ∣ 4 ∣ u ∣ 2 d x ≥ R 4 λ ( Δ 2 , B 4 ) ∫ Ω ∣ u ∣ 2 d x + 2 R 2 d ( d − 4 ) λ ( − Δ , D 2 ) ∫ Ω ∣ x ∣ 2 ∣ u ∣ 2 d x . although the two estimates are not directly comparable.
In case of a boundary singularity, the critical threshold μ Ω ∞ \mu^\infty_\Omega μ Ω ∞ ∂ Ω \partial\Omega ∂ Ω μ Ω ∞ \mu^\infty_\Omega μ Ω ∞
μ 0 ( R + d ) = inf u ∈ C c ∞ ( R + d ) u ≠ 0 ∫ R + d ∣ Δ u ∣ 2 d x ∫ R + d ∣ x ∣ − 4 ∣ u ∣ 2 d x . \mu_0(\mathbb{R}^d_+)=\inf_{\substack{u\in C^\infty_c(\mathbb{R}^d_+)\\u\ne0}}\frac{\int_{\mathbb{R}^d_+}|\Delta u|^2\,dx}{\int_{\mathbb{R}^d_+}|x|^{-4}|u|^2\,dx}. μ 0 ( R + d ) = u ∈ C c ∞ ( R + d ) u = 0 inf ∫ R + d ∣ x ∣ − 4 ∣ u ∣ 2 d x ∫ R + d ∣Δ u ∣ 2 d x . By the results in [8 ] , we have that μ 0 ( R + d ) \mu_0(\mathbb{R}^d_+) μ 0 ( R + d ) d ≥ 2 d\geq2 d ≥ 2
For the sake of clarity, we state our results separately for the cases d ≥ 2 d\geq2 d ≥ 2 d = 1 d=1 d = 1
Theorem 1.3 Let Ω \Omega Ω R d \mathbb{R}^d R d d ≥ 2 d\geq2 d ≥ 2 0 ∈ ∂ Ω 0\in\partial\Omega 0 ∈ ∂ Ω ∂ Ω \partial\Omega ∂ Ω
μ Ω ∞ = μ 0 ( R + d ) . (9) \mu^\infty_\Omega=\mu_0(\mathbb{R}^d_+). \tag*{(9)} μ Ω ∞ = μ 0 ( R + d ) . (9) If in addition Ω \Omega Ω
λ Ω ∗ ≥ λ ( Δ 2 , B 4 ) R 4 + 2 ( d 2 − 4 ) λ ( − Δ , D 2 ) R 4 . (10) \lambda^*_\Omega\geq\frac{\lambda(\Delta^2,\mathbb{B}^4)}{R^4}+2(d^2-4)\frac{\lambda(-\Delta,\mathbb{D}^2)}{R^4}. \tag*{(10)} λ Ω ∗ ≥ R 4 λ ( Δ 2 , B 4 ) + 2 ( d 2 − 4 ) R 4 λ ( − Δ , D 2 ) . (10) In particular, μ 0 ( Ω ) = μ 0 ( R + d ) \mu_0(\Omega)=\mu_0(\mathbb{R}^d_+) μ 0 ( Ω ) = μ 0 ( R + d ) Ω \Omega Ω u ∈ H 0 2 ( Ω ) u\in H^2_0(\Omega) u ∈ H 0 2 ( Ω )
∫ Ω ∣ Δ u ∣ 2 d x − μ 0 ( R + d ) ∫ Ω ∣ u ∣ 2 ∣ x ∣ 4 d x ≥ ( λ ( Δ 2 , B 4 ) R 4 + 2 ( d 2 − 4 ) λ ( − Δ , D 2 ) R 4 ) ∫ Ω ∣ u ∣ 2 d x . \int_{\Omega}|\Delta u|^2\,dx-\mu_0(\mathbb{R}^d_+)\int_{\Omega}\frac{|u|^2}{|x|^4}\,dx\geq\left(\frac{\lambda(\Delta^2,\mathbb{B}^4)}{R^4}+2(d^2-4)\frac{\lambda(-\Delta,\mathbb{D}^2)}{R^4}\right)\int_{\Omega}|u|^2\,dx. ∫ Ω ∣Δ u ∣ 2 d x − μ 0 ( R + d ) ∫ Ω ∣ x ∣ 4 ∣ u ∣ 2 d x ≥ ( R 4 λ ( Δ 2 , B 4 ) + 2 ( d 2 − 4 ) R 4 λ ( − Δ , D 2 ) ) ∫ Ω ∣ u ∣ 2 d x . By analogy with [10 ] , we conjecture that λ Ω ∗ \lambda_{\Omega}^{*} λ Ω ∗ Ω \Omega Ω μ λ ( Ω ) \mu_{\lambda}(\Omega) μ λ ( Ω ) λ = λ Ω ∗ \lambda=\lambda_{\Omega}^{*} λ = λ Ω ∗ [12 ] for the corresponding lower-order problem. For both questions, the arguments in [10 ] , [12 ] do not apply in the present setting, due to the failure of the maximum principle, and to severe additional technical difficulties. Finally, we notice that the improved Rellich inequality we obtain in case Ω ⊂ R + d \Omega\subset\mathbb{R}_{+}^{d} Ω ⊂ R + d [12 Theorem 5.2] .
The lowest dimensional case d = 1 d=1 d = 1
μ λ ( 0 , R ) = inf u ∈ H 0 2 ( 0 , R ) u ≠ 0 ∫ 0 R ∣ u ′ ′ ∣ 2 d x − λ ∫ 0 R ∣ u ∣ 2 d x ∫ 0 R x − 4 ∣ u ∣ 2 d x . (11) \mu_{\lambda}(0,R)=\inf_{\substack{u\in H_{0}^{2}(0,R)\\u\ne0}}\frac{\int_{0}^{R}|u^{\prime\prime}|^{2}\,dx-\lambda\int_{0}^{R}|u|^{2}\,dx}{\int_{0}^{R}x^{-4}|u|^{2}\,dx}. \tag*{(11)} μ λ ( 0 , R ) = u ∈ H 0 2 ( 0 , R ) u = 0 inf ∫ 0 R x − 4 ∣ u ∣ 2 d x ∫ 0 R ∣ u ′′ ∣ 2 d x − λ ∫ 0 R ∣ u ∣ 2 d x . (11) In our last result we show in particular that μ ( 0 , R ) ∞ = μ 0 ( R + ) \mu_{(0,R)}^{\infty}=\mu_{0}(\mathbb{R}_{+}) μ ( 0 , R ) ∞ = μ 0 ( R + ) μ 0 ( R + ) \mu_{0}(\mathbb{R}_{+}) μ 0 ( R + ) (19) , one readily obtains
μ 0 ( R + ) : = inf u ∈ C c ∞ ( R + ) u ≠ 0 ∫ 0 ∞ ∣ u ′ ′ ∣ 2 d x ∫ 0 ∞ x − 4 ∣ u ∣ 2 d x = 9 16 . (12) \mu_{0}(\mathbb{R}_{+}):=\inf_{\substack{u\in C_{c}^{\infty}(\mathbb{R}_{+})\\u\ne0}}\frac{\int_{0}^{\infty}|u^{\prime\prime}|^{2}\,dx}{\int_{0}^{\infty}x^{-4}|u|^{2}\,dx}=\frac{9}{16}. \tag*{(12)} μ 0 ( R + ) := u ∈ C c ∞ ( R + ) u = 0 inf ∫ 0 ∞ x − 4 ∣ u ∣ 2 d x ∫ 0 ∞ ∣ u ′′ ∣ 2 d x = 16 9 . (12) Our result for (11) involves the eigenvalues in (2) , (6) , and the principal eigenvalue of the Laplace operator − Δ -\Delta − Δ H 0 1 ( B 4 ) H_{0}^{1}(\mathbb{B}^{4}) H 0 1 ( B 4 ) λ ( − Δ , B 4 ) \lambda(-\Delta,\mathbb{B}^{4}) λ ( − Δ , B 4 )
Theorem 1.4 Consider the domain ( 0 , R ) ⊂ R (0,R)\subset\mathbb{R} ( 0 , R ) ⊂ R μ ( 0 , R ) ∞ = 9 / 16 \mu_{(0,R)}^{\infty}=9/16 μ ( 0 , R ) ∞ = 9/16
λ ( 0 , R ) ∗ ≥ 5 8 R 4 λ ( Δ 2 , B 4 ) + 3 4 R 4 λ ( − Δ , B 4 ) λ ( − Δ , D 2 ) . (13) \lambda_{(0,R)}^{*}\ge\frac{5}{8R^{4}}\lambda(\Delta^{2},\mathbb{B}^{4})+\frac{3}{4R^{4}}\lambda(-\Delta,\mathbb{B}^{4})\lambda(-\Delta,\mathbb{D}^{2}). \tag*{(13)} λ ( 0 , R ) ∗ ≥ 8 R 4 5 λ ( Δ 2 , B 4 ) + 4 R 4 3 λ ( − Δ , B 4 ) λ ( − Δ , D 2 ) . (13) Therefore, for any u ∈ H 0 2 ( 0 , R ) u\in H_{0}^{2}(0,R) u ∈ H 0 2 ( 0 , R )
∫ 0 R ∣ u ′ ′ ∣ 2 d x − 9 16 ∫ 0 R ∣ u ∣ 2 x 4 d x ≥ ( 5 8 R 4 λ ( Δ 2 , B 4 ) + 3 8 R 4 λ ( − Δ , B 4 ) λ ( − Δ , D 2 ) ) ∫ 0 R ∣ u ∣ 2 d x . \int_{0}^{R}|u^{\prime\prime}|^{2}\,dx-\frac{9}{16}\int_{0}^{R}\frac{|u|^{2}}{x^{4}}\,dx\ge\left(\frac{5}{8R^{4}}\lambda(\Delta^{2},\mathbb{B}^{4})+\frac{3}{8R^{4}}\lambda(-\Delta,\mathbb{B}^{4})\lambda(-\Delta,\mathbb{D}^{2})\right)\int_{0}^{R}|u|^{2}\,dx. ∫ 0 R ∣ u ′′ ∣ 2 d x − 16 9 ∫ 0 R x 4 ∣ u ∣ 2 d x ≥ ( 8 R 4 5 λ ( Δ 2 , B 4 ) + 8 R 4 3 λ ( − Δ , B 4 ) λ ( − Δ , D 2 ) ) ∫ 0 R ∣ u ∣ 2 d x . For the convenience of the reader, we provide here the approximate values of the constants appearing in our improved Rellich inequalities. It is well known that λ ( − Δ , D 2 ) = j 0 , 1 2 \lambda(-\Delta,\mathbb{D}^{2})=j_{0,1}^{2} λ ( − Δ , D 2 ) = j 0 , 1 2 λ ( − Δ , B 4 ) = j 1 , 1 2 \lambda(-\Delta,\mathbb{B}^{4})=j_{1,1}^{2} λ ( − Δ , B 4 ) = j 1 , 1 2 j ν , 1 j_{\nu,1} j ν , 1 J ν J_{\nu} J ν λ ( Δ 2 , B 4 ) \lambda(\Delta^{2},\mathbb{B}^{4}) λ ( Δ 2 , B 4 ) [16 ] . We have
λ ( − Δ , D 2 ) ≈ 5.78 , λ ( − Δ , B 4 ) ≈ 14.68 , λ ( Δ 2 , B 4 ) ≈ 452. \lambda(-\Delta,\mathbb{D}^{2})\approx5.78,\qquad\lambda(-\Delta,\mathbb{B}^{4})\approx14.68,\qquad\lambda(\Delta^{2},\mathbb{B}^{4})\approx452. λ ( − Δ , D 2 ) ≈ 5.78 , λ ( − Δ , B 4 ) ≈ 14.68 , λ ( Δ 2 , B 4 ) ≈ 452. The paper is organized as follows.
After the preliminary § 2 , in § 3 we start to study the infimum μ λ ( Ω ) \mu_\lambda(\Omega) μ λ ( Ω ) Lemma 3.3 , which is a crucial step in our arguments. Finally, we prove Theorem 1.1 .
In § 5 we deal with the critical parameter λ Ω ∗ \lambda_\Omega^* λ Ω ∗ Ω \Omega Ω conic domain , namely, it is the intersection of a cone with a ball about the origin. In order to prove Lemma 4.1 , which is the main step in the proof of our improved Rellich inequalities, we recall a few facts about the Emden-Fowler transform and study some auxiliary variational problems. The section concludes with the proofs of Theorem 1.2 and of Theorem 1.4 .
§ 6 contains the proof of Theorem 1.3 .
Appendix § 7 contains two examples. In Appendix § 8 we prove two inequalities for radial functions in H 0 2 ( B 4 ) H^2_0(\mathbb{B}^4) H 0 2 ( B 4 ) λ Ω ∗ \lambda_\Omega^* λ Ω ∗
Notation and preliminaries
We denote by B R ⊂ R d B_R \subset\mathbb{R}^d B R ⊂ R d R > 0 R > 0 R > 0 d = 2 , 4 d = 2, 4 d = 2 , 4 D 2 \mathbb{D}^2 D 2 B 4 \mathbb{B}^4 B 4 B 1 B_1 B 1
Ω \Omega Ω R d \mathbb{R}^d R d 0 ∈ ∂ Ω 0 \in\partial\Omega 0 ∈ ∂ Ω Ω \Omega Ω 0 0 0 δ > 0 \delta> 0 δ > 0 Ω ∩ B δ \Omega\cap B_\delta Ω ∩ B δ e ∈ S d − 1 e \in S^{d-1} e ∈ S d − 1
Ω ∩ B δ ⊂ { r σ ∈ R d ∣ r > 0 , σ ∈ S d − 1 ∖ { e } } . \Omega\cap B_\delta\subset\{r\sigma\in\mathbb{R}^d \mid r > 0,\ \sigma\in S^{d-1} \setminus\{e\}\}. Ω ∩ B δ ⊂ { r σ ∈ R d ∣ r > 0 , σ ∈ S d − 1 ∖ { e }} . We say that ∂ Ω \partial\Omega ∂ Ω differentiable at 0 if there exists δ 0 > 0 \delta_0 > 0 δ 0 > 0 F : B δ 0 → R F:B_{\delta_0} \to\mathbb{R} F : B δ 0 → R 0 0 0
∣ ∇ F ( 0 ) ∣ = 1 , Ω ∩ B δ 0 = { x ∈ B δ 0 : F ( x ) < 0 } , ∂ Ω ∩ B δ 0 = { x ∈ B δ 0 : F ( x ) = 0 } . (14) |\nabla F(0)| = 1,\quad\Omega\cap B_{\delta_0} = \{x \in B_{\delta_0}: F(x) < 0\},\quad\partial\Omega\cap B_{\delta_0} = \{x \in B_{\delta_0}: F(x) = 0\}. \tag*{(14)} ∣∇ F ( 0 ) ∣ = 1 , Ω ∩ B δ 0 = { x ∈ B δ 0 : F ( x ) < 0 } , ∂ Ω ∩ B δ 0 = { x ∈ B δ 0 : F ( x ) = 0 } . (14) The norm in L 2 ( Ω ) L^2(\Omega) L 2 ( Ω ) ∥ ⋅ ∥ 2 \|\cdot\|_2 ∥ ⋅ ∥ 2 L 2 L^2 L 2 H 0 1 ( Ω ) H^1_0(\Omega) H 0 1 ( Ω ) H 0 2 ( Ω ) H^2_0(\Omega) H 0 2 ( Ω ) H 0 1 ( Ω ) H^1_0(\Omega) H 0 1 ( Ω ) H 0 2 ( Ω ) H^2_0(\Omega) H 0 2 ( Ω ) 0 0 0 Ω \Omega Ω H 1 ( R d ) H^1(\mathbb{R}^d) H 1 ( R d ) H 2 ( R d ) H^2(\mathbb{R}^d) H 2 ( R d )
Since the ( m , 2 ) (m,2) ( m , 2 ) { 0 } \{0\} { 0 } 2 m ≤ d 2m \le d 2 m ≤ d [1 Theorem 9.9.1] , the next lemma hold.
Lemma 2.1 If d ≥ 2 d \ge2 d ≥ 2 C c ∞ ( B 1 ∖ { 0 } ) C^\infty_c(B_1 \setminus\{0\}) C c ∞ ( B 1 ∖ { 0 }) H 0 1 ( B 1 ) H^1_0(B_1) H 0 1 ( B 1 ) d ≥ 4 d \ge4 d ≥ 4 C c ∞ ( B 1 ∖ { 0 } ) C^\infty_c(B_1 \setminus\{0\}) C c ∞ ( B 1 ∖ { 0 }) H 0 2 ( B 1 ) H^2_0(B_1) H 0 2 ( B 1 )
if d ≥ 2 d \ge2 d ≥ 2 ∇ σ \nabla_\sigma ∇ σ Δ σ \Delta_\sigma Δ σ d σ d\sigma d σ S d − 1 S^{d-1} S d − 1 ω ⊆ S d − 1 \omega\subseteq S^{d-1} ω ⊆ S d − 1 H 0 1 ( ω ) , H 0 2 ( ω ) H^1_0(\omega), H^2_0(\omega) H 0 1 ( ω ) , H 0 2 ( ω )
E ( ω ) = inf φ ∈ H 0 1 ( ω ) φ ≠ 0 ∫ ω ∣ ∇ σ φ ∣ 2 d σ ∫ ω ∣ φ ∣ 2 d σ . (15) E(\omega) = \inf_{\substack{\varphi\in H^1_0(\omega) \\ \varphi\ne0}} \frac{\int_\omega|\nabla_\sigma\varphi|^2\,d\sigma}{\int_\omega|\varphi|^2\,d\sigma}. \tag*{(15)} E ( ω ) = φ ∈ H 0 1 ( ω ) φ = 0 inf ∫ ω ∣ φ ∣ 2 d σ ∫ ω ∣ ∇ σ φ ∣ 2 d σ . (15) The infimum E ( ω ) E(\omega) E ( ω ) E ( S d − 1 ) = 0 E(S^{d-1}) = 0 E ( S d − 1 ) = 0 H 0 1 ( S d − 1 ) = H 1 ( S d − 1 ) H^1_0(S^{d-1}) = H^1(S^{d-1}) H 0 1 ( S d − 1 ) = H 1 ( S d − 1 ) S + d − 1 S^{d-1}_+ S + d − 1 E ( S + d − 1 ) = d − 1 E(S^{d-1}_+) = d-1 E ( S + d − 1 ) = d − 1
m γ d ( ω ) = inf φ ∈ H 0 2 ( ω ) φ ≠ 0 ∫ ω ∣ Δ σ φ − γ d φ ∣ 2 d σ ∫ ω ∣ φ ∣ 2 d σ , (16) m_{\gamma_d}(\omega) = \inf_{\substack{\varphi\in H^2_0(\omega) \\ \varphi\ne0}} \frac{\int_\omega|\Delta_\sigma\varphi- \gamma_d\varphi|^2\,d\sigma}{\int_\omega|\varphi|^2\,d\sigma}, \tag*{(16)} m γ d ( ω ) = φ ∈ H 0 2 ( ω ) φ = 0 inf ∫ ω ∣ φ ∣ 2 d σ ∫ ω ∣ Δ σ φ − γ d φ ∣ 2 d σ , (16) where γ d = d ( d − 4 ) / 4 \gamma_d = d(d-4)/4 γ d = d ( d − 4 ) /4 m γ d ( ω ) m_{\gamma_d}(\omega) m γ d ( ω ) d ≥ 5 d \ge5 d ≥ 5
m γ d ( ω ) ≥ m γ d ( S d − 1 ) = γ d 2 > 0. m_{\gamma_d}(\omega) \ge m_{\gamma_d}(S^{d-1}) = \gamma_d^2 > 0. m γ d ( ω ) ≥ m γ d ( S d − 1 ) = γ d 2 > 0. Further, we have m γ d ( S 3 ) = 0 m_{\gamma_d}(S^3) = 0 m γ d ( S 3 ) = 0 m γ d ( S 2 ) = 9 / 16 m_{\gamma_d}(S^2) = 9/16 m γ d ( S 2 ) = 9/16 m γ d ( S 1 ) = 0 m_{\gamma_d}(S^1) = 0 m γ d ( S 1 ) = 0 [8 ] .
Let e ∈ S d − 1 e \in S^{d-1} e ∈ S d − 1 d ≥ 5 d \ge5 d ≥ 5 H 0 2 ( S d − 1 ∖ { e } ) = H 2 ( S d − 1 ) H^2_0(S^{d-1} \setminus\{e\}) = H^2(S^{d-1}) H 0 2 ( S d − 1 ∖ { e }) = H 2 ( S d − 1 ) m γ d ( S d − 1 ∖ { e } ) = m γ d ( S d − 1 ) = γ d 2 m_{\gamma_d}(S^{d-1} \setminus\{e\}) = m_{\gamma_d}(S^{d-1}) = \gamma_d^2 m γ d ( S d − 1 ∖ { e }) = m γ d ( S d − 1 ) = γ d 2 d ∈ { 2 , 3 , 4 } d \in\{2,3,4\} d ∈ { 2 , 3 , 4 } H 0 2 ( S d − 1 ∖ { e } ) ⊊ H 2 ( S d − 1 ) H^2_0(S^{d-1} \setminus\{e\}) \subsetneq H^2(S^{d-1}) H 0 2 ( S d − 1 ∖ { e }) ⊊ H 2 ( S d − 1 )
m γ d ( S d − 1 ∖ { e } ) > m γ d ( S d − 1 ) ≥ 0. m_{\gamma_d}(S^{d-1} \setminus\{e\}) > m_{\gamma_d}(S^{d-1}) \ge0. m γ d ( S d − 1 ∖ { e }) > m γ d ( S d − 1 ) ≥ 0. Thus in any dimension, m γ d ( ω ) > 0 m_{\gamma_d}(\omega) > 0 m γ d ( ω ) > 0 ω \omega ω S d − 1 S^{d-1} S d − 1
μ 0 ( C ) : = inf u ∈ C c ∞ ( C ) u ≠ 0 ∫ C ∣ Δ u ∣ 2 d x ∫ C ∣ x ∣ − 4 ∣ u ∣ 2 d x . \mu_0(\mathcal{C}) := \inf_{\substack{u \in C^\infty_c(\mathcal{C}) \\ u \ne0}} \frac{\int_{\mathcal{C}} |\Delta u|^2\,dx}{\int_{\mathcal{C}} |x|^{-4}|u|^2\,dx}. μ 0 ( C ) := u ∈ C c ∞ ( C ) u = 0 inf ∫ C ∣ x ∣ − 4 ∣ u ∣ 2 d x ∫ C ∣Δ u ∣ 2 d x . If d ≥ 2 d \ge2 d ≥ 2
ω C = C ∩ S d − 1 , (17) \omega_{\mathcal C} = \mathcal C \cap S^{d-1}, \tag*{(17)} ω C = C ∩ S d − 1 , (17) and we say that C \mathcal C C proper cone if C ⊊ R d ∖ { 0 } \mathcal C \subsetneq\mathbb{R}^{d} \setminus\{0\} C ⊊ R d ∖ { 0 }
μ 0 ( C ) = m γ d ( ω C ) > 0. (18) \mu_0(\mathcal C) = m_{\gamma_d}(\omega_{\mathcal C}) > 0. \tag*{(18)} μ 0 ( C ) = m γ d ( ω C ) > 0. (18) The equality in (2.5) can be proved by adapting the proof of [8 Theorem 3.1] . In particular, denoting by R + d \mathbb{R}^{d}_{+} R + d R d \mathbb{R}^{d} R d μ 0 ( R + d ) > 0 \mu_0(\mathbb{R}^{d}_{+}) > 0 μ 0 ( R + d ) > 0 R \mathbb{R} R R ± = { ± x > 0 } \mathbb{R}_{\pm} = \{\pm x > 0\} R ± = { ± x > 0 } α ∈ R \alpha\in\mathbb{R} α ∈ R
∫ 0 ∞ x α ∣ f ′ ∣ 2 d x ≥ ( α − 1 2 ) 2 ∫ 0 ∞ x α − 2 ∣ f ∣ 2 d x , f ∈ C c ∞ ( R + ) , (19) \int_{0}^{\infty} x^{\alpha}|f'|^{2}\,dx \ge\left(\frac{\alpha-1}{2}\right)^{2}\int_{0}^{\infty} x^{\alpha-2}|f|^{2}\,dx,\qquad f \in C_{c}^{\infty}(\mathbb{R}_{+}), \tag*{(19)} ∫ 0 ∞ x α ∣ f ′ ∣ 2 d x ≥ ( 2 α − 1 ) 2 ∫ 0 ∞ x α − 2 ∣ f ∣ 2 d x , f ∈ C c ∞ ( R + ) , (19) holds with a sharp constant in the right hand side.
A first look on μ λ ( Ω ) \mu_{\lambda}(\Omega) μ λ ( Ω )
We start by studying, for any domain Ω ⊂ R d \Omega\subset\mathbb{R}^{d} Ω ⊂ R d Rellich constant μ 0 ( Ω ) \mu_0(\Omega) μ 0 ( Ω ) Hardy-Rellich constant
μ 0 H R ( Ω ) : = inf u ∈ H 0 2 ( Ω ) u ≠ 0 ∫ Ω ∣ Δ u ∣ 2 d x ∫ Ω ∣ x ∣ − 2 ∣ ∇ u ∣ 2 d x , \mu_{0}^{\mathrm{HR}}(\Omega) := \inf_{\substack{u \in H_{0}^{2}(\Omega)\\u\ne0}} \frac{\int_{\Omega}|\Delta u|^{2}\,dx}{\int_{\Omega}|x|^{-2}|\nabla u|^{2}\,dx}, μ 0 HR ( Ω ) := u ∈ H 0 2 ( Ω ) u = 0 inf ∫ Ω ∣ x ∣ − 2 ∣∇ u ∣ 2 d x ∫ Ω ∣Δ u ∣ 2 d x , compare with [2 , 3 , 7 , 14 ] .
Lemma 3.1 If μ 0 ( Ω ) > 0 \mu_0(\Omega)>0 μ 0 ( Ω ) > 0 μ 0 H R ( Ω ) > 0 \mu_{0}^{\mathrm{HR}}(\Omega)>0 μ 0 HR ( Ω ) > 0
Proof. For u ∈ C c ∞ ( Ω ) u \in C_{c}^{\infty}(\Omega) u ∈ C c ∞ ( Ω )
∫ Ω ∣ x ∣ − 2 ∣ ∇ u ∣ 2 d x = − ∫ Ω ∣ x ∣ − 2 u Δ u d x − ( d − 4 ) ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x ≤ ( ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x ) 1 2 ( ∫ Ω ∣ Δ u ∣ 2 d x ) 1 2 + ( 4 − d ) + μ 0 ( Ω ) ∫ Ω ∣ x ∣ − 2 ∣ Δ u ∣ 2 d x ≤ ( 1 μ 0 ( Ω ) + ( 4 − d ) + μ 0 ( Ω ) ) ∫ Ω ∣ Δ u ∣ 2 d x . \begin{aligned}
\int_{\Omega}|x|^{-2}|\nabla u|^{2}\,dx
= -\int_{\Omega}|x|^{-2}u\Delta u\,dx-(d-4)\int_{\Omega}|x|^{-4}|u|^{2}\,dx \\
&\le\left(\int_{\Omega}|x|^{-4}|u|^{2}\,dx\right)^{\frac12}\left(\int_{\Omega}|\Delta u|^{2}\,dx\right)^{\frac12}
+\frac{(4-d)^{+}}{\mu_0(\Omega)}\int_{\Omega}|x|^{-2}|\Delta u|^{2}\,dx \\
&\le\left(\frac{1}{\sqrt{\mu_0(\Omega)}}+\frac{(4-d)^{+}}{\mu_0(\Omega)}\right)\int_{\Omega}|\Delta u|^{2}\,dx.
\end{aligned} ∫ Ω ∣ x ∣ − 2 ∣∇ u ∣ 2 d x = − ∫ Ω ∣ x ∣ − 2 u Δ u d x − ( d − 4 ) ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x ≤ ( ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x ) 2 1 ( ∫ Ω ∣Δ u ∣ 2 d x ) 2 1 + μ 0 ( Ω ) ( 4 − d ) + ∫ Ω ∣ x ∣ − 2 ∣Δ u ∣ 2 d x ≤ ( μ 0 ( Ω ) 1 + μ 0 ( Ω ) ( 4 − d ) + ) ∫ Ω ∣Δ u ∣ 2 d x . Since C c ∞ ( Ω ) C_{c}^{\infty}(\Omega) C c ∞ ( Ω ) H 0 2 ( Ω ) H_{0}^{2}(\Omega) H 0 2 ( Ω )
Lemma 3.2 Let Ω ⊂ R d \Omega\subset\mathbb{R}^{d} Ω ⊂ R d
i) μ 0 ( Ω ) > 0 \mu_{0}(\Omega)>0 μ 0 ( Ω ) > 0
ii) μ λ ( Ω ) \mu_{\lambda}(\Omega) μ λ ( Ω ) λ ∈ R \lambda\in\mathbb{R} λ ∈ R
iii) The function λ ↦ μ λ ( Ω ) \lambda\mapsto\mu_{\lambda}(\Omega) λ ↦ μ λ ( Ω )
Proof. If d ≥ 5 d \ge5 d ≥ 5 μ 0 ( Ω ) ≥ μ 0 ( R d ) = γ d 2 > 0 \mu_{0}(\Omega) \ge\mu_{0}(\mathbb{R}^{d}) = \gamma_{d}^{2} > 0 μ 0 ( Ω ) ≥ μ 0 ( R d ) = γ d 2 > 0
If d = 1 d=1 d = 1 Ω = ( 0 , R ) \Omega=(0,R) Ω = ( 0 , R ) μ 0 ( 0 , R ) = μ 0 ( R + ) = 9 / 16 \mu_{0}(0,R)=\mu_{0}(\mathbb{R}_{+})=9/16 μ 0 ( 0 , R ) = μ 0 ( R + ) = 9/16
To complete the proof of i) assume that d ∈ { 2 , 3 , 4 } d \in\{2,3,4\} d ∈ { 2 , 3 , 4 } e ∈ S d − 1 e \in\mathbb{S}^{d-1} e ∈ S d − 1 δ > 0 \delta>0 δ > 0
Ω ∩ B δ ⊂ C e where C e : = { r σ ∈ R d ∣ r > 0 , σ ∈ S d − 1 ∖ { e } } . \Omega\cap B_{\delta} \subset C^{e} \quad\text{where } C^{e}:=\{r\sigma\in\mathbb{R}^{d}\mid r>0,\ \sigma\in\mathbb{S}^{d-1}\setminus\{e\}\}. Ω ∩ B δ ⊂ C e where C e := { r σ ∈ R d ∣ r > 0 , σ ∈ S d − 1 ∖ { e }} . Hence μ 0 ( Ω ∩ B 2 δ ) ≥ μ 0 ( C e ) \mu_{0}(\Omega\cap B_{2\delta}) \ge\mu_{0}(C^{e}) μ 0 ( Ω ∩ B 2 δ ) ≥ μ 0 ( C e ) μ 0 ( C e ) > 0 \mu_{0}(C^{e})>0 μ 0 ( C e ) > 0
We now take a cut-off function ψ ∈ C 0 ∞ ( B 2 δ ) \psi\in C_{0}^{\infty}(B_{2\delta}) ψ ∈ C 0 ∞ ( B 2 δ ) 0 ≤ ψ ≤ 1 0 \le\psi\le1 0 ≤ ψ ≤ 1 ψ ≡ 1 \psi\equiv1 ψ ≡ 1 B δ B_{\delta} B δ u ∈ C 0 ∞ ( Ω ) u \in C_{0}^{\infty}(\Omega) u ∈ C 0 ∞ ( Ω ) ψ u ∈ C 0 ∞ ( Ω ∩ B 2 δ ) \psi u \in C_{0}^{\infty}(\Omega\cap B_{2\delta}) ψ u ∈ C 0 ∞ ( Ω ∩ B 2 δ ) H 0 2 ( Ω ) ↪ H 0 1 ( Ω ) ↪ L 2 ( Ω ) H_{0}^{2}(\Omega) \hookrightarrow H_{0}^{1}(\Omega) \hookrightarrow L^{2}(\Omega) H 0 2 ( Ω ) ↪ H 0 1 ( Ω ) ↪ L 2 ( Ω )
μ 0 ( Ω ∩ B 2 δ ) ∥ ∣ x ∣ − 2 ( ψ u ) ∥ 2 ≤ ∥ Δ ( ψ u ) ∥ 2 ≤ ∥ Δ u ∥ 2 + 2 ∥ ∇ ψ ⋅ ∇ u ∥ 2 + ∥ Δ ψ ∥ 2 ∥ u ∥ 2 ≤ c ∥ Δ u ∥ 2 . \sqrt{\mu_{0}(\Omega\cap B_{2\delta})}\lVert|x|^{-2}(\psi u)\rVert_{2} \le\lVert\Delta(\psi u)\rVert_{2} \le\lVert\Delta u\rVert_{2}+2\lVert\nabla\psi\cdot\nabla u\rVert_{2}+\lVert\Delta\psi\rVert_{2}\lVert u\rVert_{2}\le c\lVert\Delta u\rVert_{2}. μ 0 ( Ω ∩ B 2 δ ) ∥ ∣ x ∣ − 2 ( ψ u ) ∥ 2 ≤ ∥ Δ ( ψ u ) ∥ 2 ≤ ∥ Δ u ∥ 2 + 2 ∥ ∇ ψ ⋅ ∇ u ∥ 2 + ∥ Δ ψ ∥ 2 ∥ u ∥ 2 ≤ c ∥ Δ u ∥ 2 . where c > 0 c>0 c > 0 Ω \Omega Ω ψ \psi ψ supp ( 1 − ψ ) ⊆ { ∣ x ∣ ≥ δ } \operatorname{supp}(1-\psi)\subseteq\{|x|\ge\delta\} supp ( 1 − ψ ) ⊆ { ∣ x ∣ ≥ δ }
∥ ∣ x ∣ − 2 ( 1 − ψ ) u ∥ 2 ≤ δ − 2 ∥ u ∥ 2 ≤ c ∥ Δ u ∥ 2 , \lVert|x|^{-2}(1-\psi)u\rVert_{2}\le\delta^{-2}\lVert u\rVert_{2}\le c\lVert\Delta u\rVert_{2}, ∥ ∣ x ∣ − 2 ( 1 − ψ ) u ∥ 2 ≤ δ − 2 ∥ u ∥ 2 ≤ c ∥ Δ u ∥ 2 , for another constant c > 0 c>0 c > 0 δ \delta δ Ω \Omega Ω u u u C 0 ∞ ( Ω ) C_{0}^{\infty}(\Omega) C 0 ∞ ( Ω ) ∥ ∣ x ∣ − 2 u ∥ 2 ≤ ∥ ∣ x ∣ − 2 ( ψ u ) ∥ 2 + ∥ ∣ x ∣ − 2 ( 1 − ψ ) u ∥ 2 \lVert|x|^{-2}u\rVert_{2}\le\lVert|x|^{-2}(\psi u)\rVert_{2}+\lVert|x|^{-2}(1-\psi)u\rVert_{2} ∥ ∣ x ∣ − 2 u ∥ 2 ≤ ∥ ∣ x ∣ − 2 ( ψ u ) ∥ 2 + ∥ ∣ x ∣ − 2 ( 1 − ψ ) u ∥ 2 μ 0 ( Ω ) > 0 \mu_{0}(\Omega)>0 μ 0 ( Ω ) > 0
Next, we take R > 0 R>0 R > 0 Ω ⊆ B R \Omega\subseteq B_{R} Ω ⊆ B R λ ∈ R \lambda\in\mathbb{R} λ ∈ R u ∈ H 0 2 ( Ω ) u\in H_{0}^{2}(\Omega) u ∈ H 0 2 ( Ω ) ∥ u ∥ 2 ≤ R − 2 ∥ ∣ x ∣ − 2 u ∥ 2 \lVert u\rVert_{2}\le R^{-2}\lVert|x|^{-2}u\rVert_{2} ∥ u ∥ 2 ≤ R − 2 ∥ ∣ x ∣ − 2 u ∥ 2 J λ ( u ) ≥ μ 0 ( Ω ) − ∣ λ ∣ R 4 J_{\lambda}(u)\ge\mu_{0}(\Omega)-|\lambda|R^{4} J λ ( u ) ≥ μ 0 ( Ω ) − ∣ λ ∣ R 4 u u u μ λ ( Ω ) ≥ μ 0 ( Ω ) − ∣ λ ∣ R 4 > − ∞ \mu_{\lambda}(\Omega)\ge\mu_{0}(\Omega)-|\lambda|R^{4}>-\infty μ λ ( Ω ) ≥ μ 0 ( Ω ) − ∣ λ ∣ R 4 > − ∞
Lastly, notice that for any u ∈ H 0 2 ( Ω ) ∖ { 0 } u\in H_{0}^{2}(\Omega)\setminus\{0\} u ∈ H 0 2 ( Ω ) ∖ { 0 } J λ ( u ) J_{\lambda}(u) J λ ( u ) λ \lambda λ
The following lemma will play a crucial role in proving our main results.
Lemma 3.3 (compactness). Let Ω ⊂ R d \Omega\subset\mathbb{R}^{d} Ω ⊂ R d d ≥ 1 d \ge1 d ≥ 1 bounded domain satisfying (A1) or (A2). Let λ 0 < λ 1 \lambda_{0} < \lambda_{1} λ 0 < λ 1 μ λ 1 ( Ω ) ≤ μ λ 0 ( Ω ) \mu_{\lambda_{1}}(\Omega) \le\mu_{\lambda_{0}}(\Omega) μ λ 1 ( Ω ) ≤ μ λ 0 ( Ω )
i) If μ λ 1 ( Ω ) < μ λ 0 ( Ω ) \mu_{\lambda_{1}}(\Omega) < \mu_{\lambda_{0}}(\Omega) μ λ 1 ( Ω ) < μ λ 0 ( Ω ) μ λ ( Ω ) \mu_{\lambda}(\Omega) μ λ ( Ω ) λ ≥ λ 1 \lambda\ge\lambda_{1} λ ≥ λ 1
ii) If μ λ 0 ( Ω ) \mu_{\lambda_{0}}(\Omega) μ λ 0 ( Ω ) μ λ 1 ( Ω ) < μ λ 0 ( Ω ) \mu_{\lambda_{1}}(\Omega) < \mu_{\lambda_{0}}(\Omega) μ λ 1 ( Ω ) < μ λ 0 ( Ω ) μ λ 1 ( Ω ) \mu_{\lambda_{1}}(\Omega) μ λ 1 ( Ω )
iii) If μ λ 1 ( Ω ) = μ λ 0 ( Ω ) \mu_{\lambda_{1}}(\Omega) = \mu_{\lambda_{0}}(\Omega) μ λ 1 ( Ω ) = μ λ 0 ( Ω ) μ λ 0 ( Ω ) \mu_{\lambda_{0}}(\Omega) μ λ 0 ( Ω )
Proof. We take any λ ≥ λ 1 \lambda\ge\lambda_{1} λ ≥ λ 1 μ λ ( Ω ) ≤ μ λ 1 ( Ω ) < μ λ 0 ( Ω ) \mu_{\lambda}(\Omega) \le\mu_{\lambda_{1}}(\Omega) < \mu_{\lambda_{0}}(\Omega) μ λ ( Ω ) ≤ μ λ 1 ( Ω ) < μ λ 0 ( Ω ) ( u n ) ⊂ H 0 2 ( Ω ) (u_{n}) \subset H_{0}^{2}(\Omega) ( u n ) ⊂ H 0 2 ( Ω ) μ λ ( Ω ) \mu_{\lambda}(\Omega) μ λ ( Ω )
∫ Ω ∣ x ∣ − 4 ∣ u n ∣ 2 d x = 1 ∀ n ≥ 1 , J λ ( u n ) = μ λ ( Ω ) + o ( 1 ) \int_{\Omega} |x|^{-4}|u_{n}|^{2}\,dx = 1 \qquad\forall n \ge1,\qquad J_{\lambda}(u_{n}) = \mu_{\lambda}(\Omega) + o(1) ∫ Ω ∣ x ∣ − 4 ∣ u n ∣ 2 d x = 1 ∀ n ≥ 1 , J λ ( u n ) = μ λ ( Ω ) + o ( 1 ) as n → ∞ n \to\infty n → ∞ R > 0 R > 0 R > 0 B R ⊇ Ω B_{R} \supseteq\Omega B R ⊇ Ω
∫ Ω ∣ Δ u n ∣ 2 d x = J λ ( u n ) + λ ∫ Ω ∣ u n ∣ 2 d x ≤ μ λ ( Ω ) + ∣ λ ∣ R 4 ∫ Ω ∣ x ∣ − 4 ∣ u n ∣ 2 d x + o ( 1 ) , \int_{\Omega} |\Delta u_{n}|^{2}\,dx = J_{\lambda}(u_{n}) + \lambda\int_{\Omega} |u_{n}|^{2}\,dx \le\mu_{\lambda}(\Omega) + |\lambda|R^{4}\int_{\Omega} |x|^{-4}|u_{n}|^{2}\,dx + o(1), ∫ Ω ∣Δ u n ∣ 2 d x = J λ ( u n ) + λ ∫ Ω ∣ u n ∣ 2 d x ≤ μ λ ( Ω ) + ∣ λ ∣ R 4 ∫ Ω ∣ x ∣ − 4 ∣ u n ∣ 2 d x + o ( 1 ) , we see that u n u_{n} u n H 0 2 ( Ω ) H_{0}^{2}(\Omega) H 0 2 ( Ω ) u ∈ H 0 2 ( Ω ) u \in H_{0}^{2}(\Omega) u ∈ H 0 2 ( Ω )
u n → u weakly in H 0 2 ( Ω ) , u n → u in L 2 ( Ω ) and pointwise a.e. , u_{n} \to u \quad\text{weakly in } H_{0}^{2}(\Omega),\qquad u_{n} \to u \quad\text{in } L^{2}(\Omega)\text{ and pointwise a.e.}, u n → u weakly in H 0 2 ( Ω ) , u n → u in L 2 ( Ω ) and pointwise a.e. , ∣ x ∣ − 2 u n → ∣ x ∣ − 2 u weakly in L 2 ( Ω ) . |x|^{-2}u_{n} \to|x|^{-2}u \quad\text{weakly in } L^{2}(\Omega). ∣ x ∣ − 2 u n → ∣ x ∣ − 2 u weakly in L 2 ( Ω ) . Consider the sequence θ n = u n − u \theta_{n}=u_{n}-u θ n = u n − u
Δ θ n → 0 weakly in L 2 ( Ω ) , θ n → 0 in L 2 ( Ω ) , ∣ x ∣ − 2 θ n → 0 weakly in L 2 ( Ω ) , \Delta\theta_{n} \to0 \quad\text{weakly in } L^{2}(\Omega),\qquad\theta_{n} \to0 \quad\text{in } L^{2}(\Omega),\qquad|x|^{-2}\theta_{n} \to0 \quad\text{weakly in } L^{2}(\Omega), Δ θ n → 0 weakly in L 2 ( Ω ) , θ n → 0 in L 2 ( Ω ) , ∣ x ∣ − 2 θ n → 0 weakly in L 2 ( Ω ) , which imply
∫ Ω ∣ Δ u n ∣ 2 d x = ∫ Ω ∣ Δ u ∣ 2 d x + ∫ Ω ∣ Δ θ n ∣ 2 d x + o ( 1 ) , ∫ Ω ∣ u n ∣ 2 d x = ∫ Ω ∣ u ∣ 2 d x + o ( 1 ) \int_{\Omega} |\Delta u_{n}|^{2}\,dx = \int_{\Omega} |\Delta u|^{2}\,dx + \int_{\Omega} |\Delta\theta_{n}|^{2}\,dx + o(1),\qquad\int_{\Omega} |u_{n}|^{2}\,dx = \int_{\Omega} |u|^{2}\,dx + o(1) ∫ Ω ∣Δ u n ∣ 2 d x = ∫ Ω ∣Δ u ∣ 2 d x + ∫ Ω ∣Δ θ n ∣ 2 d x + o ( 1 ) , ∫ Ω ∣ u n ∣ 2 d x = ∫ Ω ∣ u ∣ 2 d x + o ( 1 ) ∫ Ω ∣ Δ θ n ∣ 2 d x = ∫ Ω ∣ Δ θ n ∣ 2 d x − λ 0 ∫ Ω ∣ θ n ∣ 2 d x + o ( 1 ) ≥ μ λ 0 ( Ω ) ∫ Ω ∣ x ∣ − 4 ∣ θ n ∣ 2 d x + o ( 1 ) , \int_{\Omega} |\Delta\theta_{n}|^{2}\,dx = \int_{\Omega} |\Delta\theta_{n}|^{2}\,dx - \lambda_{0}\int_{\Omega} |\theta_{n}|^{2}\,dx + o(1) \ge\mu_{\lambda_{0}}(\Omega)\int_{\Omega} |x|^{-4}|\theta_{n}|^{2}\,dx + o(1), ∫ Ω ∣Δ θ n ∣ 2 d x = ∫ Ω ∣Δ θ n ∣ 2 d x − λ 0 ∫ Ω ∣ θ n ∣ 2 d x + o ( 1 ) ≥ μ λ 0 ( Ω ) ∫ Ω ∣ x ∣ − 4 ∣ θ n ∣ 2 d x + o ( 1 ) , ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x = ∫ Ω ∣ x ∣ − 4 ∣ u n ∣ 2 d x − ∫ Ω ∣ x ∣ − 4 ∣ θ n ∣ 2 d x + o ( 1 ) = 1 − ∫ Ω ∣ x ∣ − 4 ∣ θ n ∣ 2 d x + o ( 1 ) . (20) \int_{\Omega} |x|^{-4}|u|^{2}\,dx = \int_{\Omega} |x|^{-4}|u_{n}|^{2}\,dx - \int_{\Omega} |x|^{-4}|\theta_{n}|^{2}\,dx + o(1) = 1 - \int_{\Omega} |x|^{-4}|\theta_{n}|^{2}\,dx + o(1). \tag*{(20)} ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x = ∫ Ω ∣ x ∣ − 4 ∣ u n ∣ 2 d x − ∫ Ω ∣ x ∣ − 4 ∣ θ n ∣ 2 d x + o ( 1 ) = 1 − ∫ Ω ∣ x ∣ − 4 ∣ θ n ∣ 2 d x + o ( 1 ) . (20) Therefore, we easily get
μ λ ( Ω ) = J λ ( u n ) + o ( 1 ) = ∫ Ω ∣ Δ u ∣ 2 d x − λ ∫ Ω ∣ u ∣ 2 d x + ∫ Ω ∣ Δ θ n ∣ 2 d x + o ( 1 ) ≥ μ λ ( Ω ) ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x + μ λ 0 ( Ω ) ∫ Ω ∣ x ∣ − 4 ∣ θ n ∣ 2 d x + o ( 1 ) = μ λ ( Ω ) + [ μ λ 0 ( Ω ) − μ λ ( Ω ) ] ∫ Ω ∣ x ∣ − 4 ∣ θ n ∣ 2 d x + o ( 1 ) . \begin{aligned}
\mu_{\lambda}(\Omega)=J_{\lambda}(u_n)+o(1)=\int_{\Omega}|\Delta u|^2\,dx-\lambda\int_{\Omega}|u|^2\,dx+\int_{\Omega}|\Delta\theta_n|^2\,dx+o(1) \\
&\geq\mu_{\lambda}(\Omega)\int_{\Omega}|x|^{-4}|u|^2\,dx+\mu_{\lambda_0}(\Omega)\int_{\Omega}|x|^{-4}|\theta_n|^2\,dx+o(1) \\
&=\mu_{\lambda}(\Omega)+\left[\mu_{\lambda_0}(\Omega)-\mu_{\lambda}(\Omega)\right]\int_{\Omega}|x|^{-4}|\theta_n|^2\,dx+o(1).
\end{aligned} μ λ ( Ω ) = J λ ( u n ) + o ( 1 ) = ∫ Ω ∣Δ u ∣ 2 d x − λ ∫ Ω ∣ u ∣ 2 d x + ∫ Ω ∣Δ θ n ∣ 2 d x + o ( 1 ) ≥ μ λ ( Ω ) ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x + μ λ 0 ( Ω ) ∫ Ω ∣ x ∣ − 4 ∣ θ n ∣ 2 d x + o ( 1 ) = μ λ ( Ω ) + [ μ λ 0 ( Ω ) − μ λ ( Ω ) ] ∫ Ω ∣ x ∣ − 4 ∣ θ n ∣ 2 d x + o ( 1 ) . Since μ λ ( Ω ) < μ λ 0 ( Ω ) \mu_{\lambda}(\Omega)<\mu_{\lambda_0}(\Omega) μ λ ( Ω ) < μ λ 0 ( Ω ) ∣ x ∣ − 2 θ n → 0 |x|^{-2}\theta_n\to0 ∣ x ∣ − 2 θ n → 0 L 2 ( Ω ) L^2(\Omega) L 2 ( Ω ) (20) into account, we can conclude that
∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x = 1 and J λ ( u ) = μ λ ( Ω ) , \int_{\Omega}|x|^{-4}|u|^2\,dx=1\quad\text{and}\quad J_{\lambda}(u)=\mu_{\lambda}(\Omega), ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x = 1 and J λ ( u ) = μ λ ( Ω ) , that is, u u u μ λ ( Ω ) \mu_{\lambda}(\Omega) μ λ ( Ω ) i ) i) i )
Assume that μ λ 0 ( Ω ) \mu_{\lambda_0}(\Omega) μ λ 0 ( Ω ) u ∈ H 0 2 ( Ω ) u\in H_0^2(\Omega) u ∈ H 0 2 ( Ω ) L 2 ( Ω ; ∣ x ∣ − 4 d x ) L^2(\Omega;|x|^{-4}dx) L 2 ( Ω ; ∣ x ∣ − 4 d x )
μ λ 1 ( Ω ) ≤ J λ 1 ( u 0 ) = J λ 0 ( u 0 ) − ( λ 1 − λ 0 ) ∫ Ω ∣ u ∣ 2 d x = μ λ 0 ( Ω ) − ( λ 1 − λ 0 ) ∫ Ω ∣ u ∣ 2 d x < μ λ 0 ( Ω ) , \mu_{\lambda_1}(\Omega)\leq J_{\lambda_1}(u_0)=J_{\lambda_0}(u_0)-(\lambda_1-\lambda_0)\int_{\Omega}|u|^2\,dx=\mu_{\lambda_0}(\Omega)-(\lambda_1-\lambda_0)\int_{\Omega}|u|^2\,dx<\mu_{\lambda_0}(\Omega), μ λ 1 ( Ω ) ≤ J λ 1 ( u 0 ) = J λ 0 ( u 0 ) − ( λ 1 − λ 0 ) ∫ Ω ∣ u ∣ 2 d x = μ λ 0 ( Ω ) − ( λ 1 − λ 0 ) ∫ Ω ∣ u ∣ 2 d x < μ λ 0 ( Ω ) , hence μ λ 1 ( Ω ) < μ λ 0 ( Ω ) \mu_{\lambda_1}(\Omega)<\mu_{\lambda_0}(\Omega) μ λ 1 ( Ω ) < μ λ 0 ( Ω ) i i ) ii) ii )
Since i i i ) iii) iii ) i i ) ii) ii )
Proof of Theorem 1.1
We denote by λ ( Δ 2 , Ω ) \lambda(\Delta^2,\Omega) λ ( Δ 2 , Ω ) Δ 2 \Delta^2 Δ 2 H 0 2 ( Ω ) H_0^2(\Omega) H 0 2 ( Ω )
λ ( Δ 2 , Ω ) = inf v ∈ H 0 2 ( Ω ) v ≠ 0 ∫ Ω ∣ Δ v ∣ 2 d x ∫ Ω ∣ v ∣ 2 d x . \lambda(\Delta^2,\Omega)=\inf_{\substack{v\in H_0^2(\Omega)\\v\ne0}}\frac{\int_{\Omega}|\Delta v|^2\,dx}{\int_{\Omega}|v|^2\,dx}. λ ( Δ 2 , Ω ) = v ∈ H 0 2 ( Ω ) v = 0 inf ∫ Ω ∣ v ∣ 2 d x ∫ Ω ∣Δ v ∣ 2 d x . By Rellich’s theorem, the infimum λ ( Δ 2 , Ω ) \lambda(\Delta^2,\Omega) λ ( Δ 2 , Ω )
Evidently μ λ ( Ω ) = 0 \mu_{\lambda}(\Omega)=0 μ λ ( Ω ) = 0 λ = λ ( Δ 2 , Ω ) \lambda=\lambda(\Delta^2,\Omega) λ = λ ( Δ 2 , Ω ) μ λ ( Ω ) > 0 \mu_{\lambda}(\Omega)>0 μ λ ( Ω ) > 0 λ < λ ( Δ 2 , Ω ) \lambda<\lambda(\Delta^2,\Omega) λ < λ ( Δ 2 , Ω )
Thanks to Lemma 3.2 , this is enough to conclude that μ Ω ∞ > 0 \mu_{\Omega}^{\infty}>0 μ Ω ∞ > 0
To prove i ) i) i ) λ > λ Ω ∗ \lambda>\lambda_{\Omega}^{*} λ > λ Ω ∗ λ 0 ∈ ( λ Ω ∗ , λ ) \lambda_0\in(\lambda_{\Omega}^{*},\lambda) λ 0 ∈ ( λ Ω ∗ , λ ) μ λ 0 ( Ω ) \mu_{\lambda_0}(\Omega) μ λ 0 ( Ω ) μ λ ( Ω ) \mu_{\lambda}(\Omega) μ λ ( Ω ) μ λ ( Ω ) < μ λ 0 ( Ω ) ≤ μ Ω ∞ \mu_{\lambda}(\Omega)<\mu_{\lambda_0}(\Omega)\leq\mu_{\Omega}^{\infty} μ λ ( Ω ) < μ λ 0 ( Ω ) ≤ μ Ω ∞ i i ) ii) ii ) Lemma 3.3 .
Take any λ 0 > λ Ω ∗ \lambda_0>\lambda_{\Omega}^{*} λ 0 > λ Ω ∗ i ) i) i ) Theorem 1.1 is proved, we can infer that μ λ 0 ( Ω ) \mu_{\lambda_0}(\Omega) μ λ 0 ( Ω ) μ λ ( Ω ) < μ λ 0 ( Ω ) \mu_{\lambda}(\Omega)<\mu_{\lambda_0}(\Omega) μ λ ( Ω ) < μ λ 0 ( Ω ) λ > λ 0 \lambda>\lambda_0 λ > λ 0 i i ) ii) ii ) Lemma 3.3 . This shows that the function λ ↦ μ λ ( Ω ) \lambda\mapsto\mu_{\lambda}(\Omega) λ ↦ μ λ ( Ω ) ( λ Ω ∗ , ∞ ) (\lambda_{\Omega}^{*},\infty) ( λ Ω ∗ , ∞ )
Suppose that it is affine on [ λ 0 , λ 1 ] ⊂ ( λ Ω ∗ , ∞ ) [\lambda_0,\lambda_1]\subset(\lambda_\Omega^*,\infty) [ λ 0 , λ 1 ] ⊂ ( λ Ω ∗ , ∞ ) λ = 1 2 ( λ 0 + λ 1 ) \lambda=\frac{1}{2}(\lambda_0+\lambda_1) λ = 2 1 ( λ 0 + λ 1 ) 1 2 ( μ λ 0 ( Ω ) + μ λ 1 ( Ω ) ) = μ λ ( Ω ) \frac{1}{2}(\mu_{\lambda_0}(\Omega)+\mu_{\lambda_1}(\Omega))=\mu_\lambda(\Omega) 2 1 ( μ λ 0 ( Ω ) + μ λ 1 ( Ω )) = μ λ ( Ω ) u ∈ H 0 2 ( Ω ) ∖ { 0 } u\in H^2_0(\Omega)\setminus\{0\} u ∈ H 0 2 ( Ω ) ∖ { 0 } μ λ ( Ω ) \mu_\lambda(\Omega) μ λ ( Ω )
μ λ ( Ω ) = J λ ( u ) = 1 2 ( J λ 0 ( u ) + J λ 1 ( u ) ) ≥ 1 2 ( μ λ 0 ( Ω ) + μ λ 1 ( Ω ) ) = μ λ ( Ω ) . \mu_\lambda(\Omega)=J_\lambda(u)=\frac{1}{2}\left(J_{\lambda_0}(u)+J_{\lambda_1}(u)\right)\ge\frac{1}{2}\left(\mu_{\lambda_0}(\Omega)+\mu_{\lambda_1}(\Omega)\right)=\mu_\lambda(\Omega). μ λ ( Ω ) = J λ ( u ) = 2 1 ( J λ 0 ( u ) + J λ 1 ( u ) ) ≥ 2 1 ( μ λ 0 ( Ω ) + μ λ 1 ( Ω ) ) = μ λ ( Ω ) . We see that u u u μ λ 0 ( Ω ) \mu_{\lambda_0}(\Omega) μ λ 0 ( Ω ) μ λ 1 ( Ω ) \mu_{\lambda_1}(\Omega) μ λ 1 ( Ω ) μ λ ( Ω ) \mu_\lambda(\Omega) μ λ ( Ω ) u u u
Δ 2 u = λ 0 u + μ λ 0 ( Ω ) ∣ x ∣ − 4 u and Δ 2 u = λ 1 u + μ λ 1 ( Ω ) ∣ x ∣ − 4 u in Ω . \Delta^2u=\lambda_0u+\mu_{\lambda_0}(\Omega)|x|^{-4}u\quad\text{and}\quad\Delta^2u=\lambda_1u+\mu_{\lambda_1}(\Omega)|x|^{-4}u\quad\text{in }\Omega. Δ 2 u = λ 0 u + μ λ 0 ( Ω ) ∣ x ∣ − 4 u and Δ 2 u = λ 1 u + μ λ 1 ( Ω ) ∣ x ∣ − 4 u in Ω. Since u ≠ 0 u\ne0 u = 0 A ⊂ Ω A\subset\Omega A ⊂ Ω
λ 1 − λ 0 = ( μ λ 0 ( Ω ) − μ λ 1 ( Ω ) ) ∣ x ∣ − 4 for any x ∈ A . \lambda_1-\lambda_0=\left(\mu_{\lambda_0}(\Omega)-\mu_{\lambda_1}(\Omega)\right)|x|^{-4}\quad\text{for any }x\in A. λ 1 − λ 0 = ( μ λ 0 ( Ω ) − μ λ 1 ( Ω ) ) ∣ x ∣ − 4 for any x ∈ A . This implies that λ 1 = λ 0 \lambda_1=\lambda_0 λ 1 = λ 0 i i ) ii) ii )
If λ Ω ∗ > − ∞ \lambda_\Omega^*>-\infty λ Ω ∗ > − ∞ λ ↦ μ λ ( Ω ) \lambda\mapsto\mu_\lambda(\Omega) λ ↦ μ λ ( Ω ) ( − ∞ , λ Ω ∗ ] (-\infty,\lambda_\Omega^*] ( − ∞ , λ Ω ∗ ] i ) i) i )
Conic domains
Following [12 ] , we say that a conic domain is the intersection of a cone with a ball about the origin. We consider here general cones, although only the conic domains B R B_R B R B R ∩ R + d B_R\cap\mathbb{R}^d_+ B R ∩ R + d
Since μ λ ( B R ∩ C ) = μ R 4 λ ( B 1 ∩ C ) \mu_\lambda(B_R\cap C)=\mu_{R^4\lambda}(B_1\cap C) μ λ ( B R ∩ C ) = μ R 4 λ ( B 1 ∩ C ) R = 1 R=1 R = 1 d = 1 d=1 d = 1 ( 0 , 1 ) (0,1) ( 0 , 1 )
If d ≥ 5 d\ge5 d ≥ 5 C ⊊ R d ∖ { 0 } C\subsetneq\mathbb{R}^d\setminus\{0\} C ⊊ R d ∖ { 0 } μ 0 ( C ) \mu_0(C) μ 0 ( C ) Ω = B 1 ∩ C \Omega=B_1\cap C Ω = B 1 ∩ C
The main result of this section needs the notation introduced in (2.4) and (2.2).
Lemma 4.1 Let C C C R d \mathbb{R}^d R d d ≤ 4 d\le4 d ≤ 4 C C C μ λ ( B 1 ∩ C ) ≤ μ 0 ( C ) \mu_\lambda(B_1\cap C)\le\mu_0(C) μ λ ( B 1 ∩ C ) ≤ μ 0 ( C ) λ ∈ R \lambda\in\mathbb{R} λ ∈ R
λ B 1 ∩ C ∗ = inf { λ ∈ R ∣ μ λ ( B 1 ∩ C ) is achieved } , \lambda^*_{B_1\cap C}=\inf\{\lambda\in\mathbb{R}\mid\mu_\lambda(B_1\cap C)\text{ is achieved}\}, λ B 1 ∩ C ∗ = inf { λ ∈ R ∣ μ λ ( B 1 ∩ C ) is achieved } , we have that μ B 1 ∩ C ∞ = μ 0 ( C ) = μ λ ( B 1 ∩ C ) \mu^\infty_{B_1\cap C}=\mu_0(C)=\mu_\lambda(B_1\cap C) μ B 1 ∩ C ∞ = μ 0 ( C ) = μ λ ( B 1 ∩ C ) λ ≤ λ B 1 ∩ C ∗ \lambda\le\lambda^*_{B_1\cap C} λ ≤ λ B 1 ∩ C ∗
i ) i) i ) d ≥ 2 d\ge2 d ≥ 2 E ( ω C ) + γ d ≥ 0 \mathrm{E}(\omega_C)+\gamma_d\ge0 E ( ω C ) + γ d ≥ 0
λ B 1 ∩ C ∗ ≥ λ ( Δ 2 , B 4 ) + 8 ( E ( ω C ) + γ d ) λ ( − Δ , D 2 ) ; \lambda^*_{B_1\cap C}\ge\lambda(\Delta^2,\mathbb{B}^4)+8\left(\mathrm{E}(\omega_C)+\gamma_d\right)\lambda(-\Delta,\mathbb{D}^2); λ B 1 ∩ C ∗ ≥ λ ( Δ 2 , B 4 ) + 8 ( E ( ω C ) + γ d ) λ ( − Δ , D 2 ) ; ii) If d ≥ 2 d \ge2 d ≥ 2 E ( ω C ) + γ d < 0 \mathrm{E}(\omega_C)+\gamma_d<0 E ( ω C ) + γ d < 0
λ B 1 ∩ C ∗ ≥ E ( ω C ) + γ d + 2 2 λ ( Δ 2 , B 4 ) + ∣ E ( ω C ) + γ d ∣ 2 λ ( − Δ , B 4 ) λ ( − Δ , D 2 ) ; \lambda^*_{B_1\cap C}\ge\frac{\mathrm{E}(\omega_C)+\gamma_d+2}{2}\lambda(\Delta^2,\mathbb{B}^4)+\frac{|\mathrm{E}(\omega_C)+\gamma_d|}{2}\lambda(-\Delta,\mathbb{B}^4)\lambda(-\Delta,\mathbb{D}^2); λ B 1 ∩ C ∗ ≥ 2 E ( ω C ) + γ d + 2 λ ( Δ 2 , B 4 ) + 2 ∣ E ( ω C ) + γ d ∣ λ ( − Δ , B 4 ) λ ( − Δ , D 2 ) ; iii) If d = 1 d=1 d = 1
λ ( 0 , 1 ) ∗ ≥ 5 8 λ ( Δ 2 , B 4 ) + 3 8 λ ( − Δ , B 4 ) λ ( − Δ , D 2 ) . \lambda^*_{(0,1)}\ge\frac{5}{8}\lambda(\Delta^2,\mathbb{B}^4)+\frac{3}{8}\lambda(-\Delta,\mathbb{B}^4)\lambda(-\Delta,\mathbb{D}^2). λ ( 0 , 1 ) ∗ ≥ 8 5 λ ( Δ 2 , B 4 ) + 8 3 λ ( − Δ , B 4 ) λ ( − Δ , D 2 ) . The proof of Lemma 4.1 needs some preliminary results.
The Emden-Fowler transform
The Emden-Fowler transform is an essential tool in [8 ] , on which the present paper is based.
Let C ⊆ R d C\subseteq\mathbb{R}^d C ⊆ R d d < 4 d<4 d < 4 C C C d ≥ 4 d\ge4 d ≥ 4 C = R d C=\mathbb{R}^d C = R d B 1 ∩ C = B 1 B_1\cap C=B_1 B 1 ∩ C = B 1 d ≥ 4 d\ge4 d ≥ 4 C 0 ∞ ( B 1 ∖ { 0 } ) C^\infty_0(B_1\setminus\{0\}) C 0 ∞ ( B 1 ∖ { 0 }) H 0 2 ( B 1 ) H^2_0(B_1) H 0 2 ( B 1 )
We first consider the case d ≥ 2 d\ge2 d ≥ 2 r > 0 r>0 r > 0 σ ∈ S d − 1 \sigma\in\mathbb{S}^{d-1} σ ∈ S d − 1 R d \mathbb{R}^d R d [8 Lemma 3.6] , one can show that the transform u ↦ T d u u\mapsto\mathcal{T}_d u u ↦ T d u
u ( r σ ) = r 4 − d 2 ( T d u ) ( − log r , σ ) , u(r\sigma)=r^{\frac{4-d}{2}}(\mathcal{T}_d u)(-\log r,\sigma), u ( r σ ) = r 2 4 − d ( T d u ) ( − log r , σ ) , is an invertible isomorphism
T d : H 0 2 ( B 1 ∩ C ) ⟶ H 0 2 ( R + × ω C ) . \mathcal{T}_d:H^2_0(B_1\cap C)\longrightarrow H^2_0(\mathbb{R}_+\times\omega_C). T d : H 0 2 ( B 1 ∩ C ) ⟶ H 0 2 ( R + × ω C ) . Setting v = T d u v=\mathcal{T}_d u v = T d u u ∈ H 0 2 ( B 1 ∩ C ) u\in H^2_0(B_1\cap C) u ∈ H 0 2 ( B 1 ∩ C )
∫ B 1 ∩ C ∣ x ∣ − 4 ∣ u ∣ 2 d x = ∬ R + × ω C ∣ v ∣ 2 d t d σ , ∫ B 1 ∩ C ∣ u ∣ 2 d x = ∬ R + × ω C e − 4 t ∣ v ∣ 2 d t d σ , ∫ B 1 ∩ C ∣ Δ u ∣ 2 d x = ∬ R + × ω C ∣ Δ σ v − γ d v ∣ 2 d t d σ + ∬ R + × ω C [ ∣ v t t ∣ 2 + 2 ∣ ∇ σ v t ∣ 2 + 2 ( γ d + 2 ) ∣ v t ∣ 2 ] d t d σ , (21) \begin{aligned}
\int_{B_1\cap C}|x|^{-4}|u|^2\,dx&=\iint_{\mathbb{R}_+\times\omega_C}|v|^2\,dt\,d\sigma,\qquad
\int_{B_1\cap C}|u|^2\,dx=\iint_{\mathbb{R}_+\times\omega_C}e^{-4t}|v|^2\,dt\,d\sigma,\\
\int_{B_1\cap C}|\Delta u|^2\,dx&=\iint_{\mathbb{R}_+\times\omega_C}|\Delta_\sigma v-\gamma_dv|^2\,dt\,d\sigma\\
&\quad+\iint_{\mathbb{R}_+\times\omega_C}\left[|v_{tt}|^2+2|\nabla_\sigma v_t|^2+2(\gamma_d+2)|v_t|^2\right]\,dt\,d\sigma, \tag*{(21)}
\end{aligned} ∫ B 1 ∩ C ∣ x ∣ − 4 ∣ u ∣ 2 d x ∫ B 1 ∩ C ∣Δ u ∣ 2 d x = ∬ R + × ω C ∣ v ∣ 2 d t d σ , ∫ B 1 ∩ C ∣ u ∣ 2 d x = ∬ R + × ω C e − 4 t ∣ v ∣ 2 d t d σ , = ∬ R + × ω C ∣ Δ σ v − γ d v ∣ 2 d t d σ + ∬ R + × ω C [ ∣ v tt ∣ 2 + 2∣ ∇ σ v t ∣ 2 + 2 ( γ d + 2 ) ∣ v t ∣ 2 ] d t d σ , (21) so that
μ λ ( B 1 ∩ C ) = inf v ∈ H 0 2 ( R + × ω C ) v ≠ 0 [ ∬ R + × ω C ∣ Δ σ v − γ d v ∣ 2 d t d σ ∬ R + × ω C ∣ v ∣ 2 d t d σ + G ( v ) − λ ∬ R + × ω C e − 4 t ∣ v ∣ 2 d t d σ ∬ R + × ω C ∣ v ∣ 2 d t d σ ] . (22) \mu_\lambda(B_1\cap C)=\inf_{\substack{v\in H^2_0(\mathbb{R}_+\times\omega_C)\\v\ne0}}\left[\frac{\iint_{\mathbb{R}_+\times\omega_C}|\Delta_\sigma v-\gamma_dv|^2\,dt\,d\sigma}{\iint_{\mathbb{R}_+\times\omega_C}|v|^2\,dt\,d\sigma}+G(v)-\lambda\frac{\iint_{\mathbb{R}_+\times\omega_C}e^{-4t}|v|^2\,dt\,d\sigma}{\iint_{\mathbb{R}_+\times\omega_C}|v|^2\,dt\,d\sigma}\right]. \tag*{(22)} μ λ ( B 1 ∩ C ) = v ∈ H 0 2 ( R + × ω C ) v = 0 inf [ ∬ R + × ω C ∣ v ∣ 2 d t d σ ∬ R + × ω C ∣ Δ σ v − γ d v ∣ 2 d t d σ + G ( v ) − λ ∬ R + × ω C ∣ v ∣ 2 d t d σ ∬ R + × ω C e − 4 t ∣ v ∣ 2 d t d σ ] . (22) Moreover, the minimization problems in (22) and in (5) , with Ω = B 1 ∩ C \Omega= B_1 \cap\mathcal{C} Ω = B 1 ∩ C
In the lowest dimensional case d = 1 d = 1 d = 1 C = R + \mathcal{C} = \mathbb{R}_+ C = R +
T 1 : H 0 2 ( 0 , 1 ) → H 0 2 ( R + ) , u ( x ) = r 3 2 ( T d u ) ( − log x ) . \mathcal{T}_1 : H_0^2(0,1) \to H_0^2(\mathbb{R}_+), \qquad u(x) = r^{\frac{3}{2}}(\mathcal{T}_d u)(-\log x). T 1 : H 0 2 ( 0 , 1 ) → H 0 2 ( R + ) , u ( x ) = r 2 3 ( T d u ) ( − log x ) . Formula (22) then reduces to the corresponding one-dimensional expression upon suppressing integration and derivatives with respect to the σ \sigma σ γ 1 = − 3 / 4 \gamma_1 = -3/4 γ 1 = − 3/4
μ λ ( 0 , 1 ) = 9 16 + inf v ∈ H 0 2 ( R + ) v ≠ 0 ∫ 0 ∞ [ ∣ v t t ∣ 2 + 5 2 ∣ v t ∣ 2 ] d t − λ ∫ 0 ∞ e − 4 t ∣ v ∣ 2 d t ∫ 0 ∞ ∣ v ∣ 2 d t . (23) \mu_\lambda(0,1) = \frac{9}{16} + \inf_{\substack{v \in H_0^2(\mathbb{R}_+) \\ v \ne0}} \frac{ \int_0^\infty\left[|v_{tt}|^2 + \frac{5}{2}|v_t|^2\right]\,dt - \lambda\int_0^\infty e^{-4t}|v|^2\,dt}{ \int_0^\infty|v|^2\,dt}. \tag*{(23)} μ λ ( 0 , 1 ) = 16 9 + v ∈ H 0 2 ( R + ) v = 0 inf ∫ 0 ∞ ∣ v ∣ 2 d t ∫ 0 ∞ [ ∣ v tt ∣ 2 + 2 5 ∣ v t ∣ 2 ] d t − λ ∫ 0 ∞ e − 4 t ∣ v ∣ 2 d t . (23) Auxiliary best constants
In order to face the cases d ≥ 2 d \ge2 d ≥ 2 d = 1 d = 1 d = 1
Λ # ( β ) = inf f ∈ H 0 2 ( R + ) f ≠ 0 ∫ 0 ∞ [ ∣ f t t ∣ 2 + 2 β ∣ f t ∣ 2 ] d t ∫ 0 ∞ e − 4 t ∣ f ∣ 2 d t , β ≥ 0 , (24) \Lambda_{\#}(\beta) = \inf_{\substack{f \in H_0^2(\mathbb{R}_+) \\ f \ne0}} \frac{ \int_0^\infty\left[|f_{tt}|^2 + 2\beta|f_t|^2\right]\,dt}{ \int_0^\infty e^{-4t}|f|^2\,dt}, \qquad\beta\ge0, \tag*{(24)} Λ # ( β ) = f ∈ H 0 2 ( R + ) f = 0 inf ∫ 0 ∞ e − 4 t ∣ f ∣ 2 d t ∫ 0 ∞ [ ∣ f tt ∣ 2 + 2 β ∣ f t ∣ 2 ] d t , β ≥ 0 , (24) which is related to the infima in (22) , (23) for β = γ d + 2 \beta= \gamma_d + 2 β = γ d + 2 (19) , one easily obtains
Λ # ( β ) ≥ inf f ∈ H 0 2 ( R + ) f ≠ 0 9 16 ∫ 0 ∞ t − 4 ∣ f ∣ 2 d t + 2 β 1 4 ∫ 0 ∞ t − 2 ∣ f ∣ 2 d t ∫ 0 ∞ e − 4 t ∣ f ∣ 2 d t ≥ 9 16 e 4 + 2 β e 2 . \Lambda_{\#}(\beta) \ge\inf_{\substack{f \in H_0^2(\mathbb{R}_+) \\ f \ne0}} \frac{ \frac{9}{16}\int_0^\infty t^{-4}|f|^2\,dt + 2\beta\frac{1}{4}\int_0^\infty t^{-2}|f|^2\,dt}{ \int_0^\infty e^{-4t}|f|^2\,dt} \ge\frac{9}{16}e^4 + 2\beta e^2. Λ # ( β ) ≥ f ∈ H 0 2 ( R + ) f = 0 inf ∫ 0 ∞ e − 4 t ∣ f ∣ 2 d t 16 9 ∫ 0 ∞ t − 4 ∣ f ∣ 2 d t + 2 β 4 1 ∫ 0 ∞ t − 2 ∣ f ∣ 2 d t ≥ 16 9 e 4 + 2 β e 2 . The proof of Lemma 4.1 needs a more refined estimate.
Lemma 4.2 Let β ≥ 0 \beta\ge0 β ≥ 0
Λ # ( β ) ≥ { λ ( Δ 2 , B 4 ) + 8 ( β − 2 ) λ ( − Δ , D 2 ) if β ≥ 2 , β 2 λ ( Δ 2 , B 4 ) + 2 − β 2 λ ( − Δ , B 4 ) λ ( − Δ , D 2 ) if 0 ≤ β < 2. (25) \Lambda_{\#}(\beta) \ge
\begin{cases}
\lambda(\Delta^2,\mathbb{B}^4) + 8(\beta- 2)\lambda(-\Delta,\mathbb{D}^2) & \text{if } \beta\ge2, \\
\frac{\beta}{2}\lambda(\Delta^2,\mathbb{B}^4) + \frac{2-\beta}{2}\lambda(-\Delta,\mathbb{B}^4)\lambda(-\Delta,\mathbb{D}^2) & \text{if } 0 \le\beta< 2.
\end{cases} \tag*{(25)} Λ # ( β ) ≥ { λ ( Δ 2 , B 4 ) + 8 ( β − 2 ) λ ( − Δ , D 2 ) 2 β λ ( Δ 2 , B 4 ) + 2 2 − β λ ( − Δ , B 4 ) λ ( − Δ , D 2 ) if β ≥ 2 , if 0 ≤ β < 2. (25) Proof. We reformulate (24) as a minimization problem over radial functions in H 0 2 ( B 4 ) H_0^2(\mathbb{B}^4) H 0 2 ( B 4 )
( T 4 w ) ( x ) = f ( − log ∣ x ∣ ) , w ∈ H 0 , r a d 2 ( B 4 ) : = { w ∈ H 0 2 ( B 4 ) ∣ w = w ( ∣ x ∣ ) } . (\mathcal{T}_4 w)(x) = f(-\log|x|), \qquad w \in H_{0,\mathrm{rad}}^2(\mathbb{B}^4) := \{w \in H_0^2(\mathbb{B}^4) \mid w = w(|x|)\}. ( T 4 w ) ( x ) = f ( − log ∣ x ∣ ) , w ∈ H 0 , rad 2 ( B 4 ) := { w ∈ H 0 2 ( B 4 ) ∣ w = w ( ∣ x ∣ )} . It turns out that T 4 \mathcal{T}_4 T 4 H 0 , r a d 2 ( B 4 ) → H 0 2 ( R + ) H^2_{0,\mathrm{rad}}(\mathbb{B}^4) \to H^2_0(\mathbb{R}_+) H 0 , rad 2 ( B 4 ) → H 0 2 ( R + )
Let w ∈ H 0 , r a d 2 ( B 4 ) w \in H^2_{0,\mathrm{rad}}(\mathbb{B}^4) w ∈ H 0 , rad 2 ( B 4 ) f = T 4 w f=\mathcal{T}_4w f = T 4 w σ ∈ S 3 \sigma\in\mathbb{S}^3 σ ∈ S 3 γ 4 = 0 \gamma_4=0 γ 4 = 0 C = B 4 C=\mathbb{B}^4 C = B 4 ω C = S 3 \omega_C=\mathbb{S}^3 ω C = S 3
∫ B 4 ∣ Δ w ∣ 2 d x = ∣ S 3 ∣ ∫ 0 ∞ ( ∣ f t t ∣ 2 + 4 ∣ f t ∣ 2 ) d t , ∫ B 4 ∣ w ∣ 2 d x = ∣ S 3 ∣ ∫ 0 ∞ e − 4 t ∣ f ∣ 2 d t . \int_{\mathbb{B}^4}|\Delta w|^2\,dx=|\mathbb{S}^3|\int_0^\infty\left(|f_{tt}|^2+4|f_t|^2\right)\,dt,\qquad\int_{\mathbb{B}^4}|w|^2\,dx=|\mathbb{S}^3|\int_0^\infty e^{-4t}|f|^2\,dt. ∫ B 4 ∣Δ w ∣ 2 d x = ∣ S 3 ∣ ∫ 0 ∞ ( ∣ f tt ∣ 2 + 4∣ f t ∣ 2 ) d t , ∫ B 4 ∣ w ∣ 2 d x = ∣ S 3 ∣ ∫ 0 ∞ e − 4 t ∣ f ∣ 2 d t . Since in addition
∫ B 4 ∣ x ∣ − 2 ∣ ∇ w ∣ 2 d x = ∣ S 3 ∣ ∫ 0 ∞ ∣ f t ∣ 2 d t , \int_{\mathbb{B}^4}|x|^{-2}|\nabla w|^2\,dx=|\mathbb{S}^3|\int_0^\infty|f_t|^2\,dt, ∫ B 4 ∣ x ∣ − 2 ∣∇ w ∣ 2 d x = ∣ S 3 ∣ ∫ 0 ∞ ∣ f t ∣ 2 d t , we infer that
Λ # ( β ) = inf w ∈ H 0 , r a d 2 ( B 4 ) w ≠ 0 ∫ B 4 ∣ Δ w ∣ 2 d x + 2 ( β − 2 ) ∫ B 4 ∣ x ∣ − 2 ∣ ∇ w ∣ 2 d x ∫ B 4 ∣ w ∣ 2 d x . \Lambda_{\#}(\beta)=\inf_{\substack{w\in H^2_{0,\mathrm{rad}}(\mathbb{B}^4)\\w\ne0}}\frac{\int_{\mathbb{B}^4}|\Delta w|^2\,dx+2(\beta-2)\int_{\mathbb{B}^4}|x|^{-2}|\nabla w|^2\,dx}{\int_{\mathbb{B}^4}|w|^2\,dx}. Λ # ( β ) = w ∈ H 0 , rad 2 ( B 4 ) w = 0 inf ∫ B 4 ∣ w ∣ 2 d x ∫ B 4 ∣Δ w ∣ 2 d x + 2 ( β − 2 ) ∫ B 4 ∣ x ∣ − 2 ∣∇ w ∣ 2 d x . If β ≥ 2 \beta\ge2 β ≥ 2
Λ # ( β ) ≥ inf w ∈ H 0 , r a d 2 ( B 4 ) w ≠ 0 ∫ B 4 ∣ Δ w ∣ 2 d x + 8 ( β − 2 ) λ ( − Δ , D 2 ) ∫ B 4 ∣ w ∣ 2 d x ∫ B 4 ∣ w ∣ 2 d x = λ ( Δ 2 , B 4 ) + 8 ( β − 2 ) λ ( − Δ , D 2 ) . \Lambda_{\#}(\beta)\ge\inf_{\substack{w\in H^2_{0,\mathrm{rad}}(\mathbb{B}^4)\\w\ne0}}\frac{\int_{\mathbb{B}^4}|\Delta w|^2\,dx+8(\beta-2)\lambda(-\Delta,\mathbb{D}^2)\int_{\mathbb{B}^4}|w|^2\,dx}{\int_{\mathbb{B}^4}|w|^2\,dx}=\lambda(\Delta^2,\mathbb{B}^4)+8(\beta-2)\lambda(-\Delta,\mathbb{D}^2). Λ # ( β ) ≥ w ∈ H 0 , rad 2 ( B 4 ) w = 0 inf ∫ B 4 ∣ w ∣ 2 d x ∫ B 4 ∣Δ w ∣ 2 d x + 8 ( β − 2 ) λ ( − Δ , D 2 ) ∫ B 4 ∣ w ∣ 2 d x = λ ( Δ 2 , B 4 ) + 8 ( β − 2 ) λ ( − Δ , D 2 ) . If β ∈ [ 0 , 2 ) \beta\in[0,2) β ∈ [ 0 , 2 )
Λ # ( β ) ≥ inf w ∈ H 0 , r a d 2 ( B 4 ) w ≠ 0 β 2 ∫ B 4 ∣ Δ w ∣ 2 d x + 2 − β 2 λ ( − Δ , D 2 ) ∫ B 4 ∣ ∇ w ∣ 2 d x ∫ B 4 ∣ w ∣ 2 d x ≥ β 2 λ ( Δ 2 , B 4 ) + 2 − β 2 λ ( − Δ , D 2 ) λ ( − Δ , B 4 ) . \begin{aligned}
\Lambda_{\#}(\beta)\ge\inf_{\substack{w\in H^2_{0,\mathrm{rad}}(\mathbb{B}^4)\\w\ne0}}\frac{\frac{\beta}{2}\int_{\mathbb{B}^4}|\Delta w|^2\,dx+\frac{2-\beta}{2}\lambda(-\Delta,\mathbb{D}^2)\int_{\mathbb{B}^4}|\nabla w|^2\,dx}{\int_{\mathbb{B}^4}|w|^2\,dx} \\
&\ge\frac{\beta}{2}\lambda(\Delta^2,\mathbb{B}^4)+\frac{2-\beta}{2}\lambda(-\Delta,\mathbb{D}^2)\lambda(-\Delta,\mathbb{B}^4).
\end{aligned} Λ # ( β ) ≥ w ∈ H 0 , rad 2 ( B 4 ) w = 0 inf ∫ B 4 ∣ w ∣ 2 d x 2 β ∫ B 4 ∣Δ w ∣ 2 d x + 2 2 − β λ ( − Δ , D 2 ) ∫ B 4 ∣∇ w ∣ 2 d x ≥ 2 β λ ( Δ 2 , B 4 ) + 2 2 − β λ ( − Δ , D 2 ) λ ( − Δ , B 4 ) . The lemma is completely proved.
Remark 4.3. For completeness, we observe that the infimum Λ # ( β ) \Lambda_{\#}(\beta) Λ # ( β ) H 0 2 ( R + ) H^2_0(\mathbb{R}_+) H 0 2 ( R + ) L 2 ( R + ; e − 4 t d t ) L^2(\mathbb{R}_+;e^{-4t}dt) L 2 ( R + ; e − 4 t d t ) H 0 2 ( B 4 ) ↪ L 2 ( B 4 ) H^2_0(\mathbb{B}^4)\hookrightarrow L^2(\mathbb{B}^4) H 0 2 ( B 4 ) ↪ L 2 ( B 4 ) T d : H 0 , r a d 2 ( B 4 ) → H 0 2 ( R + ) \mathcal{T}_d:H^2_{0,\mathrm{rad}}(\mathbb{B}^4)\to H^2_0(\mathbb{R}_+) T d : H 0 , rad 2 ( B 4 ) → H 0 2 ( R + ) T d − 1 : L 2 ( R + ; e − 4 t d t ) → L 2 ( B 4 ) \mathcal{T}_d^{-1}:L^2(\mathbb{R}_+;e^{-4t}dt)\to L^2(\mathbb{B}^4) T d − 1 : L 2 ( R + ; e − 4 t d t ) → L 2 ( B 4 )
Lemma 4.4 Let ω \omega ω S d − 1 \mathbb{S}^{d-1} S d − 1 d ≥ 2 d\ge2 d ≥ 2
M d ( ω ) : = inf v ∈ H 0 2 ( R + × ω ) v ≠ 0 ∫ 0 ∞ d t ∫ ω [ ∣ v t t ∣ 2 + 2 ∣ ∇ σ v t ∣ 2 + 2 ( γ d + 2 ) ∣ v t ∣ 2 ] d σ ∫ 0 ∞ e − 4 t d t ∫ ω ∣ v ∣ 2 d σ ≥ Λ # ( E ( ω ) + γ d + 2 ) . M_d(\omega):=\inf_{\substack{v\in H^2_0(\mathbb{R}_+\times\omega)\\v\ne0}}\frac{\int_0^\infty dt\int_\omega\left[|v_{tt}|^2+2|\nabla_\sigma v_t|^2+2(\gamma_d+2)|v_t|^2\right]\,d\sigma}{\int_0^\infty e^{-4t}dt\int_\omega|v|^2\,d\sigma}\ge\Lambda_{\#}(E(\omega)+\gamma_d+2). M d ( ω ) := v ∈ H 0 2 ( R + × ω ) v = 0 inf ∫ 0 ∞ e − 4 t d t ∫ ω ∣ v ∣ 2 d σ ∫ 0 ∞ d t ∫ ω [ ∣ v tt ∣ 2 + 2∣ ∇ σ v t ∣ 2 + 2 ( γ d + 2 ) ∣ v t ∣ 2 ] d σ ≥ Λ # ( E ( ω ) + γ d + 2 ) . Proof. Let v ∈ H 0 2 ( R + × ω ) v \in H^2_0(\mathbb{R}_+ \times\omega) v ∈ H 0 2 ( R + × ω ) v t ( ⋅ , t ) ∈ H 0 1 ( ω ) v_t(\cdot,t) \in H^1_0(\omega) v t ( ⋅ , t ) ∈ H 0 1 ( ω ) t ∈ R + t \in\mathbb{R}_+ t ∈ R + v t ( x , ⋅ ) ∈ H 0 1 ( R + ) v_t(x,\cdot) \in H^1_0(\mathbb{R}_+) v t ( x , ⋅ ) ∈ H 0 1 ( R + ) x ∈ ω x \in\omega x ∈ ω E ( ω ) E(\omega) E ( ω ) (16) and by (24) , we can estimate
∫ 0 ∞ d t ∫ ω [ ∣ v t t ∣ 2 + 2 ∣ ∇ σ v t ∣ 2 + 2 ( γ d + 2 ) ∣ v t ∣ 2 ] d σ ≥ ∫ ω d σ ∫ 0 ∞ [ ∣ v t ∣ 2 + 2 ( E ( ω ) + γ d + 2 ) ∣ v ∣ 2 ] d t ≥ Λ # ( E ( ω ) + γ d + 2 ) ∫ 0 ∞ e − 4 t d t ∫ ω ∣ v ∣ 2 d σ . \begin{aligned}
\int_0^\infty\!dt \int_\omega\left[|v_{tt}|^2+2|\nabla_\sigma v_t|^2+2(\gamma_d+2)|v_t|^2\right]d\sigma\ge\int_\omega d\sigma\int_0^\infty\left[|v_t|^2+2\bigl(E(\omega)+\gamma_d+2\bigr)|v|^2\right]dt \\
&\ge\Lambda_\#\bigl(E(\omega)+\gamma_d+2\bigr)\int_0^\infty e^{-4t}\,dt\int_\omega|v|^2d\sigma.
\end{aligned} ∫ 0 ∞ d t ∫ ω [ ∣ v tt ∣ 2 + 2∣ ∇ σ v t ∣ 2 + 2 ( γ d + 2 ) ∣ v t ∣ 2 ] d σ ≥ ∫ ω d σ ∫ 0 ∞ [ ∣ v t ∣ 2 + 2 ( E ( ω ) + γ d + 2 ) ∣ v ∣ 2 ] d t ≥ Λ # ( E ( ω ) + γ d + 2 ) ∫ 0 ∞ e − 4 t d t ∫ ω ∣ v ∣ 2 d σ . which concludes the proof. □
Remark 4.5 For completeness, we observe that M d ( ω ) = Λ # ( E ( ω ) + γ d + 2 ) M_d(\omega)=\Lambda_\#(E(\omega)+\gamma_d+2) M d ( ω ) = Λ # ( E ( ω ) + γ d + 2 ) f ∈ H 0 2 ( R + ) f \in H^2_0(\mathbb{R}_+) f ∈ H 0 2 ( R + ) Λ # ( E ( ω ) + γ d + 2 ) \Lambda_\#(E(\omega)+\gamma_d+2) Λ # ( E ( ω ) + γ d + 2 ) ( φ n ) n (\varphi_n)_n ( φ n ) n C c ∞ ( ω ) C^\infty_c(\omega) C c ∞ ( ω ) φ n → φ 1 \varphi_n \to\varphi_1 φ n → φ 1 H 0 1 ( ω c ) H^1_0(\omega^c) H 0 1 ( ω c ) φ 1 ∈ H 2 ( ω c ) ∩ H 0 1 ( ω c ) \varphi_1 \in H^2(\omega^c)\cap H^1_0(\omega^c) φ 1 ∈ H 2 ( ω c ) ∩ H 0 1 ( ω c ) − Δ σ -\Delta_\sigma − Δ σ H 0 1 ( ω c ) H^1_0(\omega^c) H 0 1 ( ω c ) M d ( ω ) M_d(\omega) M d ( ω ) f φ n ∈ H 0 2 ( R + × ω ) f\varphi_n \in H^2_0(\mathbb{R}_+\times\omega) f φ n ∈ H 0 2 ( R + × ω ) n → ∞ n\to\infty n → ∞ M d ( ω ) ≤ Λ # ( E ( ω ) + γ d + 2 ) M_d(\omega)\le\Lambda_\#(E(\omega)+\gamma_d+2) M d ( ω ) ≤ Λ # ( E ( ω ) + γ d + 2 )
Finally, notice that from the equality M d ( ω ) = Λ # ( E ( ω ) + γ d + 2 ) M_d(\omega)=\Lambda_\#(E(\omega)+\gamma_d+2) M d ( ω ) = Λ # ( E ( ω ) + γ d + 2 ) M d ( ω ) M_d(\omega) M d ( ω )
Proof of Lemma 4.1
We fix any nontrivial function u ∈ C c ∞ ( C ) u\in C^\infty_c(\mathcal{C}) u ∈ C c ∞ ( C ) u h ( x ) = u ( h x ) u_h(x)=u(hx) u h ( x ) = u ( h x ) h > 0 h>0 h > 0 h h h u h ∈ C c ∞ ( B 1 ∩ C ) u_h\in C^\infty_c(B_1\cap\mathcal{C}) u h ∈ C c ∞ ( B 1 ∩ C ) λ ∈ R \lambda\in\mathbb{R} λ ∈ R
μ λ ( B 1 ∩ C ) ≤ ∫ C ∣ Δ u h ∣ 2 d x − λ ∫ C ∣ u h ∣ 2 d x ∫ C ∣ x ∣ − 4 ∣ u h ∣ 2 d x = h 4 ∫ C ∣ Δ u ∣ 2 d x − λ ∫ C ∣ u ∣ 2 d x h 4 ∫ C ∣ x ∣ − 4 ∣ u ∣ 2 d x = ∫ C ∣ Δ u ∣ 2 d x ∫ C ∣ x ∣ − 4 ∣ u ∣ 2 d x + o ( 1 ) , \mu_\lambda(B_1\cap\mathcal{C})\le\frac{\int_\mathcal{C}|\Delta u_h|^2\,dx-\lambda\int_\mathcal{C}|u_h|^2\,dx}{\int_\mathcal{C}|x|^{-4}|u_h|^2\,dx}=\frac{h^4\int_\mathcal{C}|\Delta u|^2\,dx-\lambda\int_\mathcal{C}|u|^2\,dx}{h^4\int_\mathcal{C}|x|^{-4}|u|^2\,dx}=\frac{\int_\mathcal{C}|\Delta u|^2\,dx}{\int_\mathcal{C}|x|^{-4}|u|^2\,dx}+o(1), μ λ ( B 1 ∩ C ) ≤ ∫ C ∣ x ∣ − 4 ∣ u h ∣ 2 d x ∫ C ∣Δ u h ∣ 2 d x − λ ∫ C ∣ u h ∣ 2 d x = h 4 ∫ C ∣ x ∣ − 4 ∣ u ∣ 2 d x h 4 ∫ C ∣Δ u ∣ 2 d x − λ ∫ C ∣ u ∣ 2 d x = ∫ C ∣ x ∣ − 4 ∣ u ∣ 2 d x ∫ C ∣Δ u ∣ 2 d x + o ( 1 ) , as h → ∞ h\to\infty h → ∞ u u u
μ λ ( B 1 ∩ C ) ≤ inf u ∈ C c ∞ ( C ) u ≠ 0 ∫ C ∣ Δ u ∣ 2 d x ∫ C ∣ x ∣ − 4 ∣ u ∣ 2 d x = μ 0 ( C ) . \mu_\lambda(B_1\cap\mathcal{C})\le\inf_{\substack{u\in C^\infty_c(\mathcal{C})\\u\ne0}}\frac{\int_\mathcal{C}|\Delta u|^2\,dx}{\int_\mathcal{C}|x|^{-4}|u|^2\,dx}=\mu_0(\mathcal{C}). μ λ ( B 1 ∩ C ) ≤ u ∈ C c ∞ ( C ) u = 0 inf ∫ C ∣ x ∣ − 4 ∣ u ∣ 2 d x ∫ C ∣Δ u ∣ 2 d x = μ 0 ( C ) . Easily, for any λ ≤ 0 \lambda\le0 λ ≤ 0 μ λ ( B 1 ∩ C ) ≥ μ 0 ( B 1 ∩ C ) ≥ μ 0 ( C ) \mu_\lambda(B_1\cap\mathcal{C})\ge\mu_0(B_1\cap\mathcal{C})\ge\mu_0(\mathcal{C}) μ λ ( B 1 ∩ C ) ≥ μ 0 ( B 1 ∩ C ) ≥ μ 0 ( C ) μ λ ( B 1 ∩ C ) = μ 0 ( C ) \mu_\lambda(B_1\cap\mathcal{C})=\mu_0(\mathcal{C}) μ λ ( B 1 ∩ C ) = μ 0 ( C ) λ ≤ 0 \lambda\le0 λ ≤ 0 λ B 1 ∩ C ∗ \lambda^*_{B_1\cap\mathcal{C}} λ B 1 ∩ C ∗ μ Ω ∞ = μ 0 ( C ) = μ λ ( B 1 ∩ C ) \mu^\infty_\Omega=\mu_0(\mathcal{C})=\mu_\lambda(B_1\cap\mathcal{C}) μ Ω ∞ = μ 0 ( C ) = μ λ ( B 1 ∩ C ) λ ≤ λ B 1 ∩ C ∗ \lambda\le\lambda^*_{B_1\cap\mathcal{C}} λ ≤ λ B 1 ∩ C ∗
We now assume that d ≥ 2 d\ge2 d ≥ 2 i ) i) i ) i i ) ii) ii ) λ > λ B 1 ∩ C ∗ \lambda>\lambda^*_{B_1\cap\mathcal{C}} λ > λ B 1 ∩ C ∗
μ λ ( B 1 ∩ C ) < μ 0 ( C ) (26) \mu_\lambda(B_1\cap\mathcal{C})<\mu_0(\mathcal{C}) \tag*{(26)} μ λ ( B 1 ∩ C ) < μ 0 ( C ) (26) and μ λ ( B 1 ∩ C ) \mu_{\lambda}(B_1 \cap\mathcal{C}) μ λ ( B 1 ∩ C ) u λ ∈ H 0 2 ( B 1 ∩ C ) ∖ { 0 } u^{\lambda} \in H_0^2(B_1 \cap\mathcal{C}) \setminus\{0\} u λ ∈ H 0 2 ( B 1 ∩ C ) ∖ { 0 } μ λ ( B 1 ∩ C ) \mu_{\lambda}(B_1 \cap\mathcal{C}) μ λ ( B 1 ∩ C )
v λ : = T d u λ , v^{\lambda} := \mathcal{T}_d u^{\lambda}, v λ := T d u λ , where T d \mathcal{T}_d T d v λ ∈ H 0 2 ( R + × ω C ) v^{\lambda} \in H_0^2(\mathbb{R}_+ \times\omega_{\mathcal{C}}) v λ ∈ H 0 2 ( R + × ω C ) (22) . We can assume that v λ v^{\lambda} v λ L 2 ( R + × ω C ) L^2(\mathbb{R}_+ \times\omega_{\mathcal{C}}) L 2 ( R + × ω C )
μ λ ( B 1 ∩ C ) = ∬ R + × ω C ∣ Δ σ v λ − γ d v λ ∣ 2 d t d σ + ∬ R + × ω C [ ∣ v t t λ ∣ 2 + 2 ∣ ∇ σ v t λ ∣ 2 + 2 ( γ d + 2 ) ∣ v t λ ∣ 2 ] d t d σ − λ ∬ R + × ω C e − 4 t ∣ v λ ∣ 2 d t d σ . \begin{aligned}
\mu_{\lambda}(B_1 \cap\mathcal{C})={}&\iint_{\mathbb{R}_+ \times\omega_{\mathcal{C}}} \left|\Delta_{\sigma}v^{\lambda}-\gamma_d v^{\lambda}\right|^2\,dt\,d\sigma\\
&+\iint_{\mathbb{R}_+ \times\omega_{\mathcal{C}}}\left[|v_{tt}^{\lambda}|^2+2|\nabla_{\sigma}v_t^{\lambda}|^2+2(\gamma_d+2)|v_t^{\lambda}|^2\right]\,dt\,d\sigma-\lambda\iint_{\mathbb{R}_+ \times\omega_{\mathcal{C}}}e^{-4t}|v^{\lambda}|^2\,dt\,d\sigma.
\end{aligned} μ λ ( B 1 ∩ C ) = ∬ R + × ω C Δ σ v λ − γ d v λ 2 d t d σ + ∬ R + × ω C [ ∣ v tt λ ∣ 2 + 2∣ ∇ σ v t λ ∣ 2 + 2 ( γ d + 2 ) ∣ v t λ ∣ 2 ] d t d σ − λ ∬ R + × ω C e − 4 t ∣ v λ ∣ 2 d t d σ . By the definition of m γ d ( ω C ) m_{\gamma_d}(\omega_{\mathcal{C}}) m γ d ( ω C ) (16) , thanks to (18) , and using Lemma 4.4, we can estimate
∬ R + × ω C ∣ Δ σ v λ − γ d v λ ∣ 2 d t d σ ≥ m γ d ( ω C ) = μ 0 ( C ) , \iint_{\mathbb{R}_+ \times\omega_{\mathcal{C}}}\left|\Delta_{\sigma}v^{\lambda}-\gamma_d v^{\lambda}\right|^2\,dt\,d\sigma\ge m_{\gamma_d}(\omega_{\mathcal{C}})=\mu_0(\mathcal{C}), ∬ R + × ω C Δ σ v λ − γ d v λ 2 d t d σ ≥ m γ d ( ω C ) = μ 0 ( C ) , ∬ R + × ω C [ ∣ v t t λ ∣ 2 + 2 ∣ ∇ σ v t λ ∣ 2 + 2 ( γ d + 2 ) ∣ v t λ ∣ 2 ] d t d σ ≥ Λ # ( E ( ω C ) + γ d + 2 ) ∬ R + × ω C e − 4 t ∣ v λ ∣ 2 d t d σ . \iint_{\mathbb{R}_+ \times\omega_{\mathcal{C}}}\left[|v_{tt}^{\lambda}|^2+2|\nabla_{\sigma}v_t^{\lambda}|^2+2(\gamma_d+2)|v_t^{\lambda}|^2\right]\,dt\,d\sigma\ge\Lambda_{\#}\bigl(E(\omega_{\mathcal{C}})+\gamma_d+2\bigr)\iint_{\mathbb{R}_+ \times\omega_{\mathcal{C}}}e^{-4t}|v^{\lambda}|^2\,dt\,d\sigma. ∬ R + × ω C [ ∣ v tt λ ∣ 2 + 2∣ ∇ σ v t λ ∣ 2 + 2 ( γ d + 2 ) ∣ v t λ ∣ 2 ] d t d σ ≥ Λ # ( E ( ω C ) + γ d + 2 ) ∬ R + × ω C e − 4 t ∣ v λ ∣ 2 d t d σ . Taking also (26) into account, we infer that
0 > μ λ ( B 1 ∩ C ) − μ 0 ( C ) ≥ [ Λ # ( E ( ω C ) + γ d + 2 ) − λ ] ∬ R + × ω C e − 4 t ∣ v λ ∣ 2 d t d σ . 0>\mu_{\lambda}(B_1\cap\mathcal{C})-\mu_0(\mathcal{C})\ge\left[\Lambda_{\#}\bigl(E(\omega_{\mathcal{C}})+\gamma_d+2\bigr)-\lambda\right]\iint_{\mathbb{R}_+\times\omega_{\mathcal{C}}}e^{-4t}|v^{\lambda}|^2\,dt\,d\sigma. 0 > μ λ ( B 1 ∩ C ) − μ 0 ( C ) ≥ [ Λ # ( E ( ω C ) + γ d + 2 ) − λ ] ∬ R + × ω C e − 4 t ∣ v λ ∣ 2 d t d σ . Thus λ > Λ # ( E ( ω C ) + γ d + 2 ) \lambda>\Lambda_{\#}\bigl(E(\omega_{\mathcal{C}})+\gamma_d+2\bigr) λ > Λ # ( E ( ω C ) + γ d + 2 ) λ > λ B 1 ∩ C ∗ \lambda>\lambda^*_{B_1\cap\mathcal{C}} λ > λ B 1 ∩ C ∗ λ B 1 ∩ C ∗ ≥ Λ # ( E ( ω C ) + γ d + 2 ) \lambda^*_{B_1\cap\mathcal{C}}\ge\Lambda_{\#}\bigl(E(\omega_{\mathcal{C}})+\gamma_d+2\bigr) λ B 1 ∩ C ∗ ≥ Λ # ( E ( ω C ) + γ d + 2 ) β = E ( ω ) + γ d + 2 \beta=E(\omega)+\gamma_d+2 β = E ( ω ) + γ d + 2
If d = 1 d=1 d = 1 ( 0 , 1 ) (0,1) ( 0 , 1 ) (23) and the second inequality in (25) , with β = 5 / 4 \beta=5/4 β = 5/4 □ \square □
Proof of Theorem 1.2. Since μ λ ( Ω ) = μ R 4 λ ( R − 1 Ω ) \mu_{\lambda}(\Omega)=\mu_{R^4\lambda}(R^{-1}\Omega) μ λ ( Ω ) = μ R 4 λ ( R − 1 Ω ) Ω ⊂ B 1 \Omega\subset B_1 Ω ⊂ B 1
Let δ > 0 \delta>0 δ > 0 B δ ⊂ Ω B_{\delta}\subset\Omega B δ ⊂ Ω μ λ ( B 1 ) ≤ μ λ ( Ω ) ≤ μ λ ( B δ ) = μ δ − 4 λ ( B 1 ) \mu_{\lambda}(B_1)\le\mu_{\lambda}(\Omega)\le\mu_{\lambda}(B_{\delta})=\mu_{\delta^{-4}\lambda}(B_1) μ λ ( B 1 ) ≤ μ λ ( Ω ) ≤ μ λ ( B δ ) = μ δ − 4 λ ( B 1 ) λ ∈ R \lambda\in\mathbb{R} λ ∈ R μ λ ( Ω ) = μ 0 ( R d ) \mu_{\lambda}(\Omega)=\mu_0(\mathbb{R}^d) μ λ ( Ω ) = μ 0 ( R d ) λ ≤ λ B 1 ∗ = δ − 4 λ B δ ∗ \lambda\le\lambda^*_{B_1}=\delta^{-4}\lambda^*_{B_{\delta}} λ ≤ λ B 1 ∗ = δ − 4 λ B δ ∗ μ Ω ∞ = μ 0 ( R d ) \mu^{\infty}_{\Omega}=\mu_0(\mathbb{R}^d) μ Ω ∞ = μ 0 ( R d ) λ Ω ∗ ≥ λ B 1 ∗ \lambda^*_{\Omega}\ge\lambda^*_{B_1} λ Ω ∗ ≥ λ B 1 ∗
To conclude the proof, notice that λ B 1 ∗ ≥ λ ( Δ 2 , B 4 ) + 2 ( d − 4 ) λ ( − Δ , D 2 ) \lambda^*_{B_1}\ge\lambda(\Delta^2,\mathbb{B}^4)+2(d-4)\lambda(-\Delta,\mathbb{D}^2) λ B 1 ∗ ≥ λ ( Δ 2 , B 4 ) + 2 ( d − 4 ) λ ( − Δ , D 2 ) C = R d \mathcal{C}=\mathbb{R}^d C = R d γ d = d ( d − 4 ) / 4 > 0 \gamma_d=d(d-4)/4>0 γ d = d ( d − 4 ) /4 > 0 □ \square □
Proof of Theorem 1.4 Since μ λ ( 0 , R ) = μ R 4 λ ( 0 , 1 ) \mu_\lambda(0,R)=\mu_{R^4\lambda}(0,1) μ λ ( 0 , R ) = μ R 4 λ ( 0 , 1 ) R = 1 R=1 R = 1 μ 0 ( 0 , 1 ) = μ 0 ( R + ) = 9 / 16 \mu_0(0,1)=\mu_0(\mathbb{R}_{+})=9/16 μ 0 ( 0 , 1 ) = μ 0 ( R + ) = 9/16 μ 0 , R ∞ = 9 / 16 \mu^\infty_{0,R}=9/16 μ 0 , R ∞ = 9/16 i v ) iv) i v ) i i i ) iii) iii )
Proof of Theorem 1.3
For any ε > 0 \varepsilon>0 ε > 0 C ε + C_\varepsilon^{+} C ε + C ε − C_\varepsilon^{-} C ε − δ > 0 \delta>0 δ > 0
0 < μ 0 ( R + d ) − μ 0 ( C ε + ) < ε and Ω ∩ B δ ⊂ C ε + , 0 < μ 0 ( C ε − ) − μ 0 ( R + d ) < ε and B δ ∩ C ε − ⊂ Ω . (27) \begin{aligned}
0<\mu_0(\mathbb{R}_{+}^{d})-\mu_0(C_\varepsilon^{+})<\varepsilon&\quad\text{and}\quad\Omega\cap B_\delta\subset C_\varepsilon^{+},\\
0<\mu_0(C_\varepsilon^{-})-\mu_0(\mathbb{R}_{+}^{d})<\varepsilon&\quad\text{and}\quad B_\delta\cap C_\varepsilon^{-}\subset\Omega. \tag*{(27)}
\end{aligned} 0 < μ 0 ( R + d ) − μ 0 ( C ε + ) < ε 0 < μ 0 ( C ε − ) − μ 0 ( R + d ) < ε and Ω ∩ B δ ⊂ C ε + , and B δ ∩ C ε − ⊂ Ω. (27) This elementary fact can be proved as follows. Take δ 0 > 0 \delta_0>0 δ 0 > 0 F : B δ 0 → R F:B_{\delta_0}\to\mathbb{R} F : B δ 0 → R 0 0 0
lim ρ ↓ 0 μ 0 ( C ρ ) = μ 0 ( R + d ) , where C ρ = { x ∈ R d : x ⋅ ∇ F ( 0 ) < ρ ∣ x ∣ } ⊃ R + d . \lim_{\rho\downarrow0}\mu_0(C_\rho)=\mu_0(\mathbb{R}_{+}^{d}),\qquad\text{where}\qquad C_\rho=\{x\in\mathbb{R}^{d}:x\cdot\nabla F(0)<\rho\lvert x\rvert\}\supset\mathbb{R}_{+}^{d}. ρ ↓ 0 lim μ 0 ( C ρ ) = μ 0 ( R + d ) , where C ρ = { x ∈ R d : x ⋅ ∇ F ( 0 ) < ρ ∣ x ∣} ⊃ R + d . Hence μ 0 ( C ρ ε ) > μ 0 ( R + d ) − ε \mu_0(C_{\rho_\varepsilon})>\mu_0(\mathbb{R}_{+}^{d})-\varepsilon μ 0 ( C ρ ε ) > μ 0 ( R + d ) − ε ρ ε > 0 \rho_\varepsilon>0 ρ ε > 0 C ε + : = C ρ ε C_\varepsilon^{+}:=C_{\rho_\varepsilon} C ε + := C ρ ε 0 > F ( x ) = x ⋅ ∇ F ( 0 ) + o ( ∣ x ∣ ) 0>F(x)=x\cdot\nabla F(0)+o(\lvert x\rvert) 0 > F ( x ) = x ⋅ ∇ F ( 0 ) + o (∣ x ∣) x ∈ Ω x\in\Omega x ∈ Ω x → 0 x\to0 x → 0 δ ≤ δ 0 \delta\leq\delta_0 δ ≤ δ 0 x ∈ Ω ∩ B δ x\in\Omega\cap B_\delta x ∈ Ω ∩ B δ x ∈ C ε + x\in C_\varepsilon^{+} x ∈ C ε + C ε + C_\varepsilon^{+} C ε + C ε − C_\varepsilon^{-} C ε −
We are in position to prove (1.9). In fact, it is equivalent to establish that
lim λ → − ∞ μ λ ( Ω ) = μ 0 ( R + d ) , (28) \lim_{\lambda\to-\infty}\mu_\lambda(\Omega)=\mu_0(\mathbb{R}_{+}^{d}), \tag*{(28)} λ → − ∞ lim μ λ ( Ω ) = μ 0 ( R + d ) , (28) use i i i ) iii) iii ) λ Ω ∗ \lambda_\Omega^* λ Ω ∗
Take any ε > 0 \varepsilon>0 ε > 0 C ε ± C_\varepsilon^{\pm} C ε ± δ > 0 \delta>0 δ > 0 μ λ \mu_\lambda μ λ λ ≤ 0 \lambda\leq0 λ ≤ 0 μ 0 ( C ε + ) = μ 0 ( B δ ∩ C ε + ) ≤ μ 0 ( Ω ∩ B δ ) \mu_0(C_\varepsilon^{+})=\mu_0(B_\delta\cap C_\varepsilon^{+})\leq\mu_0(\Omega\cap B_\delta) μ 0 ( C ε + ) = μ 0 ( B δ ∩ C ε + ) ≤ μ 0 ( Ω ∩ B δ ) μ λ ( Ω ) ≤ μ λ ( B δ ∩ C ε − ) = μ 0 ( C ε − ) \mu_\lambda(\Omega)\leq\mu_\lambda(B_\delta\cap C_\varepsilon^{-})=\mu_0(C_\varepsilon^{-}) μ λ ( Ω ) ≤ μ λ ( B δ ∩ C ε − ) = μ 0 ( C ε − ) λ ≤ 0 \lambda\leq0 λ ≤ 0
μ 0 ( R + d ) − ε < μ 0 ( Ω ∩ B δ ) (29) \mu_0(\mathbb{R}_{+}^{d})-\varepsilon<\mu_0(\Omega\cap B_\delta) \tag*{(29)} μ 0 ( R + d ) − ε < μ 0 ( Ω ∩ B δ ) (29) μ λ ( Ω ) < μ 0 ( R + d ) + ε for any λ ≤ 0. (30) \mu_\lambda(\Omega)<\mu_0(\mathbb{R}_{+}^{d})+\varepsilon\quad\text{for any }\lambda\leq0. \tag*{(30)} μ λ ( Ω ) < μ 0 ( R + d ) + ε for any λ ≤ 0. (30) Now we take a function η ∈ C ∞ ( R ) \eta\in C^\infty(\mathbb{R}) η ∈ C ∞ ( R ) 0 ≤ η ≤ 1 0\leq\eta\leq1 0 ≤ η ≤ 1 η ≡ 0 \eta\equiv0 η ≡ 0 ( − ∞ , 1 ) (-\infty,1) ( − ∞ , 1 ) η ≡ 1 \eta\equiv1 η ≡ 1 ( 2 , ∞ ) (2,\infty) ( 2 , ∞ )
ψ h ( x ) = η ( − log ∣ x ∣ h ) = { 0 if ∣ x ∣ ≥ e − h , 1 if ∣ x ∣ ≤ e − 2 h . (31) \psi_h(x)=\eta\left(\frac{-\log\lvert x\rvert}{h}\right)=
\begin{cases}
0 & \text{if }\lvert x\rvert\geq e^{-h},\\
1 & \text{if }\lvert x\rvert\leq e^{-2h}.
\end{cases} \tag*{(31)} ψ h ( x ) = η ( h − log ∣ x ∣ ) = { 0 1 if ∣ x ∣ ≥ e − h , if ∣ x ∣ ≤ e − 2 h . (31) Then ψ h ∈ C c ∞ ( R d ) \psi_h \in C^\infty_c(\mathbb{R}^d) ψ h ∈ C c ∞ ( R d ) supp ∇ ψ h ⊂ { e − 2 h ≤ ∣ x ∣ ≤ e − h } \operatorname{supp}\nabla\psi_h \subset\{e^{-2h} \le|x| \le e^{-h}\} supp ∇ ψ h ⊂ { e − 2 h ≤ ∣ x ∣ ≤ e − h } c > 0 c > 0 c > 0 η \eta η
∣ ∇ ψ h ( x ) ∣ ≤ c h − 1 ∣ x ∣ − 1 , ∣ Δ ψ h ( x ) ∣ ≤ c h − 1 ∣ x ∣ − 2 if e − 2 h < ∣ x ∣ < e − h . (32) |\nabla\psi_h(x)| \le ch^{-1}|x|^{-1}, \qquad|\Delta\psi_h(x)| \le ch^{-1}|x|^{-2} \qquad\text{if } e^{-2h} < |x| < e^{-h}. \tag*{(32)} ∣∇ ψ h ( x ) ∣ ≤ c h − 1 ∣ x ∣ − 1 , ∣Δ ψ h ( x ) ∣ ≤ c h − 1 ∣ x ∣ − 2 if e − 2 h < ∣ x ∣ < e − h . (32) Further, for h > ∣ log δ ∣ h > |\log\delta| h > ∣ log δ ∣ u ∈ H 0 2 ( Ω ) u \in H^2_0(\Omega) u ∈ H 0 2 ( Ω ) ψ h u ∈ H 0 2 ( Ω ∩ B δ ) \psi_hu \in H^2_0(\Omega\cap B_\delta) ψ h u ∈ H 0 2 ( Ω ∩ B δ ) (29) gives
( μ 0 ( R + d ) − ε ) ∫ Ω ∣ x ∣ − 4 ∣ ψ h u ∣ 2 d x ≤ ∫ Ω ∣ Δ ( ψ h u ) ∣ 2 d x . (33) \left(\mu_0(\mathbb{R}^d_+) - \varepsilon\right)\int_\Omega|x|^{-4}|\psi_hu|^2\,dx \le\int_\Omega|\Delta(\psi_hu)|^2\,dx. \tag*{(33)} ( μ 0 ( R + d ) − ε ) ∫ Ω ∣ x ∣ − 4 ∣ ψ h u ∣ 2 d x ≤ ∫ Ω ∣Δ ( ψ h u ) ∣ 2 d x . (33) We write u = ψ h u + ( 1 − ψ h ) u u=\psi_hu+(1-\psi_h)u u = ψ h u + ( 1 − ψ h ) u
∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x = ∫ Ω ∣ x ∣ − 4 ∣ ψ h u ∣ 2 d x + 2 ∫ Ω ∣ x ∣ − 4 ψ h ( 1 − ψ h ) ∣ u ∣ 2 d x + ∫ Ω ∣ x ∣ − 4 ( 1 − ψ h ) 2 ∣ u ∣ 2 d x ≤ ∫ Ω ∣ x ∣ − 4 ∣ ψ h u ∣ 2 d x + 3 e 8 h ∫ Ω ∣ u ∣ 2 d x . \begin{aligned}
\int_\Omega|x|^{-4}|u|^2\,dx = \int_\Omega|x|^{-4}|\psi_hu|^2\,dx + 2\int_\Omega|x|^{-4}\psi_h(1-\psi_h)|u|^2\,dx + \int_\Omega|x|^{-4}(1-\psi_h)^2|u|^2\,dx \\
&\le\int_\Omega|x|^{-4}|\psi_hu|^2\,dx + 3e^{8h}\int_\Omega|u|^2\,dx.
\end{aligned} ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x = ∫ Ω ∣ x ∣ − 4 ∣ ψ h u ∣ 2 d x + 2 ∫ Ω ∣ x ∣ − 4 ψ h ( 1 − ψ h ) ∣ u ∣ 2 d x + ∫ Ω ∣ x ∣ − 4 ( 1 − ψ h ) 2 ∣ u ∣ 2 d x ≤ ∫ Ω ∣ x ∣ − 4 ∣ ψ h u ∣ 2 d x + 3 e 8 h ∫ Ω ∣ u ∣ 2 d x . Thanks to (32) and since μ 0 H R ( Ω ) > 0 \mu^{\mathrm{HR}}_0(\Omega)>0 μ 0 HR ( Ω ) > 0 μ 0 ( Ω ) > 0 \mu_0(\Omega)>0 μ 0 ( Ω ) > 0
∫ Ω ∣ ∇ ψ h ⋅ ∇ u ∣ 2 d x ≤ c h − 2 ∫ Ω ∣ x ∣ − 2 ∣ ∇ u ∣ 2 d x ≤ c h − 2 ∫ Ω ∣ Δ u ∣ 2 d x , \int_\Omega|\nabla\psi_h\cdot\nabla u|^2\,dx \le ch^{-2}\int_\Omega|x|^{-2}|\nabla u|^2\,dx \le ch^{-2}\int_\Omega|\Delta u|^2\,dx, ∫ Ω ∣∇ ψ h ⋅ ∇ u ∣ 2 d x ≤ c h − 2 ∫ Ω ∣ x ∣ − 2 ∣∇ u ∣ 2 d x ≤ c h − 2 ∫ Ω ∣Δ u ∣ 2 d x , ∫ Ω ∣ u Δ ψ h ∣ 2 d x ≤ c h − 2 ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x ≤ c h − 2 ∫ Ω ∣ Δ u ∣ 2 d x , \int_\Omega|u\Delta\psi_h|^2\,dx \le ch^{-2}\int_\Omega|x|^{-4}|u|^2\,dx \le ch^{-2}\int_\Omega|\Delta u|^2\,dx, ∫ Ω ∣ u Δ ψ h ∣ 2 d x ≤ c h − 2 ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x ≤ c h − 2 ∫ Ω ∣Δ u ∣ 2 d x , where the constants c c c η \eta η Ω \Omega Ω u u u
∫ Ω ∣ Δ ( ψ h u ) ∣ 2 d x ≤ ( ∥ Δ u ∥ 2 + 2 ∥ ∇ ψ h ⋅ ∇ u ∥ 2 + ∥ u Δ ψ h ∥ 2 ) 2 ≤ ( 1 + c h − 1 ) 2 ∫ Ω ∣ Δ u ∣ 2 d x . \int_\Omega|\Delta(\psi_hu)|^2\,dx \le\left(\|\Delta u\|_2 + 2\|\nabla\psi_h\cdot\nabla u\|_2 + \|u\Delta\psi_h\|_2\right)^2 \le(1+ch^{-1})^2\int_\Omega|\Delta u|^2\,dx. ∫ Ω ∣Δ ( ψ h u ) ∣ 2 d x ≤ ( ∥Δ u ∥ 2 + 2∥∇ ψ h ⋅ ∇ u ∥ 2 + ∥ u Δ ψ h ∥ 2 ) 2 ≤ ( 1 + c h − 1 ) 2 ∫ Ω ∣Δ u ∣ 2 d x . Collecting the above estimates, from (33) we infer that
( μ 0 ( R + d ) − ε ) ( ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x − 3 e 8 h ∫ Ω ∣ u ∣ 2 d x ) ≤ ( 1 + c h − 1 ) 2 ∫ Ω ∣ Δ u ∣ 2 d x , \left(\mu_0(\mathbb{R}^d_+) - \varepsilon\right)\left(\int_\Omega|x|^{-4}|u|^2\,dx - 3e^{8h}\int_\Omega|u|^2\,dx\right) \le(1+ch^{-1})^2\int_\Omega|\Delta u|^2\,dx, ( μ 0 ( R + d ) − ε ) ( ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x − 3 e 8 h ∫ Ω ∣ u ∣ 2 d x ) ≤ ( 1 + c h − 1 ) 2 ∫ Ω ∣Δ u ∣ 2 d x , which gives
μ 0 ( R + d ) − ε ( 1 + c h − 1 ) 2 ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x ≤ ∫ Ω ∣ Δ u ∣ 2 d x − λ h ∫ Ω ∣ u ∣ 2 d x , where λ h = − 3 ( μ 0 ( R + d ) − ε ) e 8 h ( 1 + c h − 1 ) 2 . \frac{\mu_0(\mathbb{R}^d_+) - \varepsilon}{(1+ch^{-1})^2}\int_\Omega|x|^{-4}|u|^2\,dx \le\int_\Omega|\Delta u|^2\,dx - \lambda_h\int_\Omega|u|^2\,dx, \qquad\text{where } \lambda_h=-\frac{3(\mu_0(\mathbb{R}^d_+)-\varepsilon)e^{8h}}{(1+ch^{-1})^2}. ( 1 + c h − 1 ) 2 μ 0 ( R + d ) − ε ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x ≤ ∫ Ω ∣Δ u ∣ 2 d x − λ h ∫ Ω ∣ u ∣ 2 d x , where λ h = − ( 1 + c h − 1 ) 2 3 ( μ 0 ( R + d ) − ε ) e 8 h . Notice that λ h → − ∞ \lambda_h\to-\infty λ h → − ∞ h → ∞ h\to\infty h → ∞ u ∈ H 0 2 ( Ω ) u\in H^2_0(\Omega) u ∈ H 0 2 ( Ω )
μ λ h ( Ω ) ≥ μ 0 ( R + d ) − ε ( 1 + c h − 1 ) 2 = μ 0 ( R + d ) − ε + o ( 1 ) as h → ∞ . \mu_{\lambda_h}(\Omega) \ge\frac{\mu_0(\mathbb{R}^d_+)-\varepsilon}{(1+ch^{-1})^2}=\mu_0(\mathbb{R}^d_+)-\varepsilon+o(1) \quad\text{as } h\to\infty. μ λ h ( Ω ) ≥ ( 1 + c h − 1 ) 2 μ 0 ( R + d ) − ε = μ 0 ( R + d ) − ε + o ( 1 ) as h → ∞. Therefore, taking also the monotonicity of λ ↦ μ λ ( Ω ) \lambda\mapsto\mu_\lambda(\Omega) λ ↦ μ λ ( Ω )
μ 0 ( R + d ) − ε ≤ lim h → ∞ μ λ h ( Ω ) = lim λ → − ∞ μ λ ( Ω ) ≤ μ 0 ( R + d ) + ε . \mu_0(\mathbb{R}^d_+) - \varepsilon\le\lim_{h \to\infty} \mu_{\lambda_h}(\Omega) = \lim_{\lambda\to-\infty} \mu_\lambda(\Omega) \le\mu_0(\mathbb{R}^d_+) + \varepsilon. μ 0 ( R + d ) − ε ≤ h → ∞ lim μ λ h ( Ω ) = λ → − ∞ lim μ λ ( Ω ) ≤ μ 0 ( R + d ) + ε . Since ε > 0 \varepsilon> 0 ε > 0
It only remains to prove the estimate in (1.10) in case Ω ⊂ R + d \Omega\subset\mathbb{R}^d_+ Ω ⊂ R + d Ω ⊂ B 1 ∩ R + d \Omega\subset B_1 \cap\mathbb{R}^d_+ Ω ⊂ B 1 ∩ R + d μ Ω ∞ = μ B 1 ∩ R + d ∞ = μ 0 ( R + d ) \mu^\infty_\Omega= \mu^\infty_{B_1 \cap\mathbb{R}^d_+} = \mu_0(\mathbb{R}^d_+) μ Ω ∞ = μ B 1 ∩ R + d ∞ = μ 0 ( R + d ) μ λ ( B 1 ∩ R + d ) ≤ μ λ ( Ω ) \mu_\lambda(B_1 \cap\mathbb{R}^d_+) \le\mu_\lambda(\Omega) μ λ ( B 1 ∩ R + d ) ≤ μ λ ( Ω ) λ ∈ R \lambda\in\mathbb{R} λ ∈ R λ B 1 ∩ R + d ∗ ≤ λ Ω ∗ \lambda^*_{B_1 \cap\mathbb{R}^d_+} \le\lambda^*_\Omega λ B 1 ∩ R + d ∗ ≤ λ Ω ∗ λ B 1 ∩ R + d ∗ \lambda^*_{B_1 \cap\mathbb{R}^d_+} λ B 1 ∩ R + d ∗
Now, we take C = R + d C = \mathbb{R}^d_+ C = R + d ω C = S + d − 1 \omega_C = \mathbb{S}^{d-1}_+ ω C = S + d − 1
E ( S + d − 1 ) + γ d = ( d − 1 ) + d ( d − 4 ) 4 = d 2 − 4 4 ≥ 0 , \mathcal{E}(\mathbb{S}^{d-1}_+) + \gamma_d = (d - 1) + \frac{d(d - 4)}{4} = \frac{d^2 - 4}{4} \ge0, E ( S + d − 1 ) + γ d = ( d − 1 ) + 4 d ( d − 4 ) = 4 d 2 − 4 ≥ 0 , then (1.10) for Ω = B 1 ∩ R + d \Omega= B_1 \cap\mathbb{R}^d_+ Ω = B 1 ∩ R + d
Examples
Example A.1 Let δ ∈ ( 0 , 1 ) \delta\in(0,1) δ ∈ ( 0 , 1 ) ν ∈ S d − 1 \nu\in\mathbb{S}^{d-1} ν ∈ S d − 1 C δ = { x ∈ R N ∣ x ⋅ ν > − δ ∣ x ∣ } C_\delta= \{x \in\mathbb{R}^N \mid x \cdot\nu> -\delta|x|\} C δ = { x ∈ R N ∣ x ⋅ ν > − δ ∣ x ∣ } R δ > 1 R_\delta> 1 R δ > 1 Ω ⊂ R d \Omega\subset\mathbb{R}^d Ω ⊂ R d
Ω ⊃ C δ ∩ { 1 < ∣ x ∣ < R δ } , \Omega\supset C_\delta\cap\{1 < |x| < R_\delta\}, Ω ⊃ C δ ∩ { 1 < ∣ x ∣ < R δ } , it holds that λ Ω ∗ < 0 \lambda^*_\Omega< 0 λ Ω ∗ < 0 μ 0 ( Ω ) \mu_0(\Omega) μ 0 ( Ω )
This is an adaptation of the example in [12 Proposition 6.3] . Notice that the curvature of Ω \Omega Ω 0 ∈ ∂ Ω 0 \in\partial\Omega 0 ∈ ∂ Ω
Since the cone C δ C_\delta C δ μ 0 ( C δ ) \mu_0(C_\delta) μ 0 ( C δ ) μ 0 ( R + d ) \mu_0(\mathbb{R}^d_+) μ 0 ( R + d ) u δ ∈ C c ∞ ( C δ ) u_\delta\in C^\infty_c(C_\delta) u δ ∈ C c ∞ ( C δ )
∫ C δ ∣ x ∣ − 2 ∣ u δ ∣ 2 d x = 1 , ∫ C δ ∣ Δ u δ ∣ 2 d x < μ 0 ( R + d ) . \int_{C_\delta} |x|^{-2}|u_\delta|^2 \, dx = 1, \qquad\int_{C_\delta} |\Delta u_\delta|^2 \, dx < \mu_0(\mathbb{R}^d_+). ∫ C δ ∣ x ∣ − 2 ∣ u δ ∣ 2 d x = 1 , ∫ C δ ∣Δ u δ ∣ 2 d x < μ 0 ( R + d ) . Up to rescaling, we can assume that the support of u δ u_\delta u δ 1 1 1 R δ > 1 R_\delta> 1 R δ > 1
If the domain Ω \Omega Ω C δ ∩ { 1 < ∣ x ∣ < R δ } C_\delta\cap\{1 < |x| < R_\delta\} C δ ∩ { 1 < ∣ x ∣ < R δ } u δ u_\delta u δ Ω \Omega Ω μ 0 ( Ω ) \mu_0(\Omega) μ 0 ( Ω ) u δ u_\delta u δ
μ 0 ( Ω ) ≤ ∫ C δ ∣ Δ u δ ∣ 2 d x < μ 0 ( R + d ) = μ Ω ∞ . \mu_0(\Omega) \le\int_{C_\delta} |\Delta u_\delta|^2 \, dx < \mu_0(\mathbb{R}^d_+) = \mu^\infty_\Omega. μ 0 ( Ω ) ≤ ∫ C δ ∣Δ u δ ∣ 2 d x < μ 0 ( R + d ) = μ Ω ∞ . Example A.2 Set
Ω = { x = ( x ′ , x d ) ∈ R d − 1 × R ∣ 0 < x d < ∣ x ′ ∣ 2 } ∩ B 1 ⊂ R + d . \Omega= \left\{x=(x',x_d)\in\mathbb{R}^{d-1}\times\mathbb{R}\mid0<x_d<|x'|^2\right\}\cap B_1\subset\mathbb{R}_+^d. Ω = { x = ( x ′ , x d ) ∈ R d − 1 × R ∣ 0 < x d < ∣ x ′ ∣ 2 } ∩ B 1 ⊂ R + d . Then μ Ω ∞ = ∞ \mu_\Omega^\infty=\infty μ Ω ∞ = ∞ μ λ ( Ω ) \mu_\lambda(\Omega) μ λ ( Ω ) λ ∈ R \lambda\in\mathbb{R} λ ∈ R Theorem 1.1 , and λ Ω ∗ = − ∞ \lambda_\Omega^*=-\infty λ Ω ∗ = − ∞
In particular, the Rellich constant μ 0 ( Ω ) \mu_0(\Omega) μ 0 ( Ω )
We start by noticing that Ω \Omega Ω Theorem 1.1 applies. To prove that μ Ω ∞ = ∞ \mu_\Omega^\infty=\infty μ Ω ∞ = ∞ θ > 0 \theta>0 θ > 0
C θ = { x = ( x ′ , x d ) ∈ R d − 1 × R ∣ 0 < x d < θ ∣ x ∣ } , ω θ = C θ ∩ S d − 1 = { σ ∈ S d − 1 ∣ 0 < σ ⋅ e < θ } , C_\theta=\left\{x=(x',x_d)\in\mathbb{R}^{d-1}\times\mathbb{R}\mid0<x_d<\theta|x|\right\},\qquad\omega_\theta=C_\theta\cap\mathbb{S}^{d-1}=\left\{\sigma\in\mathbb{S}^{d-1}\mid0<\sigma\cdot e<\theta\right\}, C θ = { x = ( x ′ , x d ) ∈ R d − 1 × R ∣ 0 < x d < θ ∣ x ∣ } , ω θ = C θ ∩ S d − 1 = { σ ∈ S d − 1 ∣ 0 < σ ⋅ e < θ } , where e = ( 0 , … , 0 , 1 ) e=(0,\ldots,0,1) e = ( 0 , … , 0 , 1 )
μ 0 ( Ω ∩ B θ ) ≥ μ 0 ( C θ ) , \mu_0(\Omega\cap B_\theta)\geq\mu_0(C_\theta), μ 0 ( Ω ∩ B θ ) ≥ μ 0 ( C θ ) , because Ω ∩ B θ ⊂ C θ \Omega\cap B_\theta\subset C_\theta Ω ∩ B θ ⊂ C θ
lim θ → 0 μ 0 ( C θ ) = ∞ . (34) \lim_{\theta\to0}\mu_0(C_\theta)=\infty. \tag*{(34)} θ → 0 lim μ 0 ( C θ ) = ∞. (34) We prove (A.1) by exploiting the equality (18) , which gives
μ 0 ( C θ ) = inf φ ∈ C c ∞ ( ω θ ) φ ≠ 0 ∫ ω θ ∣ Δ σ φ − γ d φ ∣ 2 d σ ∫ ω θ ∣ φ ∣ 2 d σ = γ d 2 + inf φ ∈ C c ∞ ( ω θ ) φ ≠ 0 ∫ ω θ ∣ Δ σ φ ∣ 2 + 2 γ d ∫ ω θ ∣ ∇ σ φ ∣ 2 d σ ∫ ω θ ∣ φ ∣ 2 d σ . \mu_0(C_\theta)=\inf_{\substack{\varphi\in C_c^\infty(\omega_\theta)\\\varphi\ne0}}\frac{\int_{\omega_\theta}|\Delta_\sigma\varphi-\gamma_d\varphi|^2\,d\sigma}{\int_{\omega_\theta}|\varphi|^2\,d\sigma}=\gamma_d^2+\inf_{\substack{\varphi\in C_c^\infty(\omega_\theta)\\\varphi\ne0}}\frac{\int_{\omega_\theta}|\Delta_\sigma\varphi|^2+2\gamma_d\int_{\omega_\theta}|\nabla_\sigma\varphi|^2\,d\sigma}{\int_{\omega_\theta}|\varphi|^2\,d\sigma}. μ 0 ( C θ ) = φ ∈ C c ∞ ( ω θ ) φ = 0 inf ∫ ω θ ∣ φ ∣ 2 d σ ∫ ω θ ∣ Δ σ φ − γ d φ ∣ 2 d σ = γ d 2 + φ ∈ C c ∞ ( ω θ ) φ = 0 inf ∫ ω θ ∣ φ ∣ 2 d σ ∫ ω θ ∣ Δ σ φ ∣ 2 + 2 γ d ∫ ω θ ∣ ∇ σ φ ∣ 2 d σ . Let E ( ω θ ) E(\omega_\theta) E ( ω θ ) − Δ σ -\Delta_\sigma − Δ σ H 0 1 ( ω θ ) H_0^1(\omega_\theta) H 0 1 ( ω θ ) (15) . It holds that
lim θ → 0 E ( ω θ ) = ∞ . \lim_{\theta\to0}E(\omega_\theta)=\infty. θ → 0 lim E ( ω θ ) = ∞. (Indeed, E ( ω θ ) = O ( θ − 2 ) E(\omega_\theta)=O(\theta^{-2}) E ( ω θ ) = O ( θ − 2 ) [6 Theorem 3.3] ). Since for any φ ∈ C c ∞ ( ω θ ) \varphi\in C_c^\infty(\omega_\theta) φ ∈ C c ∞ ( ω θ )
∫ ω θ ∣ ∇ σ φ ∣ 2 d σ = − ∫ ω θ φ Δ σ φ d σ ≤ 1 E ( ω θ ) ( ∫ ω θ ∣ ∇ σ φ ∣ 2 d σ ) 1 2 ( ∫ ω θ ∣ Δ σ φ ∣ 2 d σ ) 1 2 , \int_{\omega_\theta}|\nabla_\sigma\varphi|^2\,d\sigma=-\int_{\omega_\theta}\varphi\Delta_\sigma\varphi\,d\sigma\leq\frac{1}{\sqrt{E(\omega_\theta)}}\left(\int_{\omega_\theta}|\nabla_\sigma\varphi|^2\,d\sigma\right)^{\frac12}\left(\int_{\omega_\theta}|\Delta_\sigma\varphi|^2\,d\sigma\right)^{\frac12}, ∫ ω θ ∣ ∇ σ φ ∣ 2 d σ = − ∫ ω θ φ Δ σ φ d σ ≤ E ( ω θ ) 1 ( ∫ ω θ ∣ ∇ σ φ ∣ 2 d σ ) 2 1 ( ∫ ω θ ∣ Δ σ φ ∣ 2 d σ ) 2 1 , it follows that for θ > 0 \theta>0 θ > 0 d = 2 , 3 d=2,3 d = 2 , 3
μ 0 ( C θ ) ≥ γ d 2 + ( E ( ω θ ) + 2 γ d ) inf φ ∈ C c ∞ ( ω θ ) φ ≠ 0 ∫ ω θ ∣ ∇ σ φ ∣ 2 d σ ∫ ω θ ∣ φ ∣ 2 d σ ≥ ( E ( ω θ ) + 2 γ d ) E ( ω θ ) , \mu_0(C_\theta)\geq\gamma_d^2+\left(E(\omega_\theta)+2\gamma_d\right)\inf_{\substack{\varphi\in C_c^\infty(\omega_\theta)\\\varphi\ne0}}\frac{\int_{\omega_\theta}|\nabla_\sigma\varphi|^2\,d\sigma}{\int_{\omega_\theta}|\varphi|^2\,d\sigma}\geq\left(E(\omega_\theta)+2\gamma_d\right)E(\omega_\theta), μ 0 ( C θ ) ≥ γ d 2 + ( E ( ω θ ) + 2 γ d ) φ ∈ C c ∞ ( ω θ ) φ = 0 inf ∫ ω θ ∣ φ ∣ 2 d σ ∫ ω θ ∣ ∇ σ φ ∣ 2 d σ ≥ ( E ( ω θ ) + 2 γ d ) E ( ω θ ) , and (A.1) follows.
To conclude the proof of the equality μ Ω ∞ = ∞ \mu^\infty_{\Omega}=\infty μ Ω ∞ = ∞ (28) . We add few details for the convenience of the reader.
For any integer h > ∣ log θ ∣ h>|\log\theta| h > ∣ log θ ∣ ψ h ∈ C c ∞ ( B θ ) \psi_h\in C^\infty_c(B_\theta) ψ h ∈ C c ∞ ( B θ ) (31) . Take any nontrivial function u ∈ H 0 2 ( Ω ) u\in H^2_0(\Omega) u ∈ H 0 2 ( Ω ) μ 0 ( Ω ∩ B θ ) \mu_0(\Omega\cap B_\theta) μ 0 ( Ω ∩ B θ ) ψ h u ∈ H 0 2 ( Ω ∩ B θ ) \psi_hu\in H^2_0(\Omega\cap B_\theta) ψ h u ∈ H 0 2 ( Ω ∩ B θ ) μ 0 ( Ω ∩ B θ ) ≥ μ 0 ( C θ ) \mu_0(\Omega\cap B_\theta)\ge\mu_0(C_\theta) μ 0 ( Ω ∩ B θ ) ≥ μ 0 ( C θ )
μ 0 ( C θ ) ∫ Ω ∣ x ∣ − 4 ∣ ψ h u ∣ 2 d x ≤ ∫ Ω ∣ Δ ( ψ h u ) ∣ 2 d x . \mu_0(C_\theta)\int_{\Omega}|x|^{-4}|\psi_hu|^2\,dx\le\int_{\Omega}|\Delta(\psi_hu)|^2\,dx. μ 0 ( C θ ) ∫ Ω ∣ x ∣ − 4 ∣ ψ h u ∣ 2 d x ≤ ∫ Ω ∣Δ ( ψ h u ) ∣ 2 d x . The estimates in the proof of Theorem 1.3 give
μ 0 ( C θ ) ( ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x − 3 e 8 h ∫ Ω ∣ u ∣ 2 d x ) ≤ ( 1 + c h − 1 ) 2 ∫ Ω ∣ Δ u ∣ 2 d x . \mu_0(C_\theta)\left(\int_{\Omega}|x|^{-4}|u|^2\,dx-3e^{8h}\int_{\Omega}|u|^2\,dx\right)\le(1+ch^{-1})^2\int_{\Omega}|\Delta u|^2\,dx. μ 0 ( C θ ) ( ∫ Ω ∣ x ∣ − 4 ∣ u ∣ 2 d x − 3 e 8 h ∫ Ω ∣ u ∣ 2 d x ) ≤ ( 1 + c h − 1 ) 2 ∫ Ω ∣Δ u ∣ 2 d x . Since u u u H 0 2 ( Ω ) H^2_0(\Omega) H 0 2 ( Ω )
μ λ h ( Ω ) ≥ μ 0 ( C θ ) ( 1 + c h − 1 ) 2 , where λ h = − 3 μ 0 ( C θ ) e 8 h ( 1 + c h − 1 ) 2 . \mu_{\lambda_h}(\Omega)\ge\frac{\mu_0(C_\theta)}{(1+ch^{-1})^2},\qquad\text{where}\qquad\lambda_h=-\frac{3\mu_0(C_\theta)e^{8h}}{(1+ch^{-1})^2}. μ λ h ( Ω ) ≥ ( 1 + c h − 1 ) 2 μ 0 ( C θ ) , where λ h = − ( 1 + c h − 1 ) 2 3 μ 0 ( C θ ) e 8 h . Therefore
μ Ω ∞ = lim λ → − ∞ μ λ ( Ω ) = lim h → ∞ μ λ h ( Ω ) ≥ μ 0 ( C θ ) . \mu^\infty_{\Omega}=\lim_{\lambda\to-\infty}\mu_\lambda(\Omega)=\lim_{h\to\infty}\mu_{\lambda_h}(\Omega)\ge\mu_0(C_\theta). μ Ω ∞ = λ → − ∞ lim μ λ ( Ω ) = h → ∞ lim μ λ h ( Ω ) ≥ μ 0 ( C θ ) . We proved that μ Ω ∞ ≥ μ 0 ( C θ ) \mu^\infty_{\Omega}\ge\mu_0(C_\theta) μ Ω ∞ ≥ μ 0 ( C θ ) θ > 0 \theta>0 θ > 0 μ 0 ∞ = ∞ \mu^\infty_0=\infty μ 0 ∞ = ∞
Two inequalities
Let w ∈ H 0 , r a d 2 ( B 4 ) w\in H^2_{0,\mathrm{rad}}(\mathbb{B}^4) w ∈ H 0 , rad 2 ( B 4 )
∫ B 4 ∣ x ∣ − 2 ∣ ∇ w ∣ 2 d x ≥ 4 λ ( − Δ , D 2 ) ∫ B 4 ∣ w ∣ 2 d x . (35) \int_{\mathbb{B}^4}|x|^{-2}|\nabla w|^2\,dx\ge4\lambda(-\Delta,\mathbb{D}^2)\int_{\mathbb{B}^4}|w|^2\,dx. \tag*{(35)} ∫ B 4 ∣ x ∣ − 2 ∣∇ w ∣ 2 d x ≥ 4 λ ( − Δ , D 2 ) ∫ B 4 ∣ w ∣ 2 d x . (35) ∫ B 4 ∣ Δ w ∣ 2 d x − 4 ∫ B 4 ∣ x ∣ − 2 ∣ ∇ w ∣ 2 d x ≥ λ ( − Δ , D 2 ) ∫ B 4 ∣ ∇ w ∣ 2 d x . (36) \int_{\mathbb{B}^4}|\Delta w|^2\,dx-4\int_{\mathbb{B}^4}|x|^{-2}|\nabla w|^2\,dx\ge\lambda(-\Delta,\mathbb{D}^2)\int_{\mathbb{B}^4}|\nabla w|^2\,dx. \tag*{(36)} ∫ B 4 ∣Δ w ∣ 2 d x − 4 ∫ B 4 ∣ x ∣ − 2 ∣∇ w ∣ 2 d x ≥ λ ( − Δ , D 2 ) ∫ B 4 ∣∇ w ∣ 2 d x . (36) Thanks to Lemma 2.1 we can assume that w ∈ C c ∞ ( B 4 ∖ { 0 } ) w\in C^\infty_c(\mathbb{B}^4\setminus\{0\}) w ∈ C c ∞ ( B 4 ∖ { 0 }) r r r ρ \rho ρ R 4 \mathbb{R}^4 R 4 R 2 \mathbb{R}^2 R 2 w ∈ C ∞ ( B 4 ∖ { 0 } ) w\in C^\infty(\mathbb{B}^4\setminus\{0\}) w ∈ C ∞ ( B 4 ∖ { 0 }) w ∈ C c ∞ ( 0 , 1 ) w\in C^\infty_c(0,1) w ∈ C c ∞ ( 0 , 1 ) w r = ∣ x ∣ − 1 x ⋅ ∇ w ( x ) w_r=|x|^{-1}x\cdot\nabla w(x) w r = ∣ x ∣ − 1 x ⋅ ∇ w ( x ) w w w x ∈ B 4 x\in\mathbb{B}^4 x ∈ B 4
g ( ρ ) = w ( ρ ) . (37) g(\rho)=w(\sqrt{\rho}). \tag*{(37)} g ( ρ ) = w ( ρ ) . (37) We regard g ∈ C c ∞ ( 0 , 1 ) g \in C^\infty_c(0,1) g ∈ C c ∞ ( 0 , 1 ) g ∈ C c ∞ ( D 2 ∖ { 0 } ) g \in C^\infty_c(\mathbb{D}^2 \setminus\{0\}) g ∈ C c ∞ ( D 2 ∖ { 0 }) g ρ = ∣ z ∣ − 1 z ⋅ ∇ g ( z ) g_\rho=|z|^{-1}z\cdot\nabla g(z) g ρ = ∣ z ∣ − 1 z ⋅ ∇ g ( z ) g g g z ∈ D 2 z\in\mathbb{D}^2 z ∈ D 2
∫ B 4 ∣ x ∣ − 2 ∣ ∇ w ∣ 2 d x = ∣ S 3 ∣ ∫ 0 1 r ∣ w r ∣ 2 d r = 2 ∣ S 3 ∣ ∫ 0 1 ρ ∣ g ρ ∣ 2 d ρ = ∣ S 3 ∣ π ∫ D 2 ∣ ∇ g ∣ 2 d z ≥ ∣ S 3 ∣ π λ ( − Δ , D 2 ) ∫ D 2 ∣ g ∣ 2 d z = 4 λ ( − Δ , D 2 ) ∫ B 4 ∣ w ∣ 2 d x . \begin{aligned}
\int_{\mathbb{B}^4}|x|^{-2}|\nabla w|^2\,dx=|\mathbb{S}^3|\int_0^1r|w_r|^2\,dr=2|\mathbb{S}^3|\int_0^1\rho|g_\rho|^2\,d\rho\\
&=\frac{|\mathbb{S}^3|}{\pi}\int_{\mathbb{D}^2}|\nabla g|^2\,dz\ge\frac{|\mathbb{S}^3|}{\pi}\lambda(-\Delta,\mathbb{D}^2)\int_{\mathbb{D}^2}|g|^2\,dz=4\lambda(-\Delta,\mathbb{D}^2)\int_{\mathbb{B}^4}|w|^2\,dx.
\end{aligned} ∫ B 4 ∣ x ∣ − 2 ∣∇ w ∣ 2 d x = ∣ S 3 ∣ ∫ 0 1 r ∣ w r ∣ 2 d r = 2∣ S 3 ∣ ∫ 0 1 ρ ∣ g ρ ∣ 2 d ρ = π ∣ S 3 ∣ ∫ D 2 ∣∇ g ∣ 2 d z ≥ π ∣ S 3 ∣ λ ( − Δ , D 2 ) ∫ D 2 ∣ g ∣ 2 d z = 4 λ ( − Δ , D 2 ) ∫ B 4 ∣ w ∣ 2 d x . Next, we compute
1 ∣ S 3 ∣ ∫ B 4 ∣ Δ w ∣ 2 d x = ∫ 0 1 r 3 ∣ w r r + 3 r − 1 w r ∣ 2 d r = ∫ 0 1 r 3 ∣ w r r ∣ 2 d r + 9 ∫ 0 1 r ∣ w r ∣ 2 d r + 3 ∫ 0 1 r 2 ( ∣ w r ∣ 2 ) r d r = ∫ 0 1 r 3 ∣ w r r ∣ 2 d r + 3 ∫ 0 1 r ∣ w r ∣ 2 d r = 1 ∣ S 3 ∣ [ ∫ B 4 ∣ ∇ w r ∣ 2 d x + 3 ∫ B 4 ∣ x ∣ − 2 ∣ w r ∣ 2 d x ] . \begin{aligned}
\frac{1}{|\mathbb{S}^3|}\int_{\mathbb{B}^4}|\Delta w|^2\,dx=\int_0^1r^3|w_{rr}+3r^{-1}w_r|^2\,dr=\int_0^1r^3|w_{rr}|^2\,dr+9\int_0^1r|w_r|^2\,dr+3\int_0^1r^2(|w_r|^2)_r\,dr \\
&=\int_0^1r^3|w_{rr}|^2\,dr+3\int_0^1r|w_r|^2\,dr=\frac{1}{|\mathbb{S}^3|}\left[\int_{\mathbb{B}^4}|\nabla w_r|^2\,dx+3\int_{\mathbb{B}^4}|x|^{-2}|w_r|^2\,dx\right].
\end{aligned} ∣ S 3 ∣ 1 ∫ B 4 ∣Δ w ∣ 2 d x = ∫ 0 1 r 3 ∣ w r r + 3 r − 1 w r ∣ 2 d r = ∫ 0 1 r 3 ∣ w r r ∣ 2 d r + 9 ∫ 0 1 r ∣ w r ∣ 2 d r + 3 ∫ 0 1 r 2 ( ∣ w r ∣ 2 ) r d r = ∫ 0 1 r 3 ∣ w r r ∣ 2 d r + 3 ∫ 0 1 r ∣ w r ∣ 2 d r = ∣ S 3 ∣ 1 [ ∫ B 4 ∣∇ w r ∣ 2 d x + 3 ∫ B 4 ∣ x ∣ − 2 ∣ w r ∣ 2 d x ] . Since ∣ ∇ w ∣ = ∣ w r ∣ |\nabla w|=|w_r| ∣∇ w ∣ = ∣ w r ∣
∫ B 4 ∣ Δ w ∣ 2 d x − 4 ∫ B 4 ∣ x ∣ − 2 ∣ ∇ w ∣ 2 d x = ∫ B 4 ∣ ∇ w r ∣ 2 d x − ∫ B 4 ∣ x ∣ − 2 ∣ w r ∣ 2 d x . \int_{\mathbb{B}^4}|\Delta w|^2\,dx-4\int_{\mathbb{B}^4}|x|^{-2}|\nabla w|^2\,dx=\int_{\mathbb{B}^4}|\nabla w_r|^2\,dx-\int_{\mathbb{B}^4}|x|^{-2}|w_r|^2\,dx. ∫ B 4 ∣Δ w ∣ 2 d x − 4 ∫ B 4 ∣ x ∣ − 2 ∣∇ w ∣ 2 d x = ∫ B 4 ∣∇ w r ∣ 2 d x − ∫ B 4 ∣ x ∣ − 2 ∣ w r ∣ 2 d x . Inequality (B.2) follows by applying the improved Hardy inequality (1) with d = 4 d=4 d = 4 Ω = B 4 \Omega=\mathbb{B}^4 Ω = B 4 w r ∈ H 0 1 ( B 4 ) w_r\in H^1_0(\mathbb{B}^4) w r ∈ H 0 1 ( B 4 ) u u u
Acknowledgments. The second author thanks the Department of Mathematics of the University of Torino for the hospitality during her visit in June 2026.
