Remarks on inequalities of Hardy-Sobolev Type

Abstract

We obtain the sharp constants of some Hardy-Sobolev-type inequalities proved by Balinsky et al. (Banach J Math Anal 2(2):94-106).

2000 Mathematics Subject Classification: Primary 26D10; 46E35.

1. Introduction

Hardy inequality in ℝⁿreads, for all $f\in C_0^{\infty }\left({ℝ}^n\right)$ and n ≥ 3,

$$\underset{{ℝ}^n}{\int }|\nabla f{|}^2\mathsf{\text{d}}x\ge \frac{{\left(n-2\right)}^2}{4}\underset{{ℝ}^n}{\int }\frac{f^2}{|x{|}^2}\mathsf{\text{d}}x.$$

(1.1)

The Sobolev inequality states that, for all $f\in C_0^{\infty }\left({ℝ}^n\right)$ and n ≥ 3,

$$\underset{{ℝ}^n}{\int }|\nabla f{|}^2\mathsf{\text{d}}x\ge S_n{\left(\underset{{ℝ}^n}{\int }|f{|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}},$$

(1.2)

where $2^{*}=\frac{2n}{n-2}$ and $S_n=\pi n\left(n-2\right){\left(\Gamma \left(\frac{n}{2}\right)∕\Gamma \left(n\right)\right)}^{\frac{2}{n}}$ is the best constant (cf. [1, 2]). A result of Stubbe [3] states that for $0\le \delta <\frac{{\left(n-2\right)}^2}{4}$,

$$\underset{{ℝ}^n}{\int }|\nabla f{|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }\frac{f^2}{|x{|}^2}\mathsf{\text{d}}x\ge \frac{{\left(\frac{{\left(n-2\right)}^2}{4}-\delta \right)}^{\frac{n-1}{n}}}{{\left(\frac{{\left(n-2\right)}^2}{4}\right)}^{\frac{n-1}{n}}}{S}_n{\left(\underset{{ℝ}^n}{\int }|f{|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}$$

(1.3)

and the constant in (1.3) is sharp. Recently, Balinsky et al. [4] prove analogous inequalities for the operator $\mathcal{L}:=x\cdot \nabla$. One of the results states that, for 0 ≤ δ < n²/4 and $f\in C_0^{\infty }\left({ℝ}^n\right)$,

$$\underset{{ℝ}^n}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }{f}^2\mathsf{\text{d}}x\ge C{\left(\frac{n^2}{4}-\delta \right)}^{\frac{n-1}{n}}{S}_n{\left(\underset{{ℝ}^n}{\int }|rF{|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}},$$

(1.4)

where F(r) is the integral mean of f over the unit sphere $S^{n-1}$, i.e.,

$$F\left(r\right)=\frac{1}{\left|S^{n-1}\right|}\underset{S^{n-1}}{\int }f\left(r\omega \right)\mathsf{\text{d}}\omega ,$$

and $\left|S^{n-1}\right|={\int }_{S^{n-1}}d\omega =\frac{2{\pi }^{n∕2}}{\Gamma \left(n∕2\right)}$. Here, we use the polar coordinates x = rω. The aim of this note is to look for the sharp constant of inequality (1.4). To this end, we have:

Theorem 1.1. Let$f\in C_0^{\infty }\left({ℝ}^n\right)$and n ≥ 3. There holds, for$0\le \delta <\frac{n^2}{4}$,

$$\underset{{ℝ}^n}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }{f}^2\mathsf{\text{d}}x\ge \frac{{\left(\frac{n^2}{4}-\delta \right)}^{\frac{n-1}{n}}}{{\left(\frac{{\left(n-2\right)}^2}{4}\right)}^{\frac{n-1}{n}}}{S}_n{\left(\underset{{ℝ}^n}{\int }|rF\left(r\right){|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}$$

(1.5)

and the constant in (1.5) is sharp.

When δ = n²/4, we have the following Theorem, which generalize the results of [4], Corollary 4.6.

Theorem 1.2. If f is supported in the annulus A _R := {x ∈ ℝⁿ: R^-1 < |x| < R}, then

$$\underset{A_R}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{A_R}{\int }{f}^2\mathsf{\text{d}}x\ge {\left[2\left(n-2\right)lnR\right]}^{-\frac{2\left(n-1\right)}{n}}{S}_n{\left(\underset{A_R}{\int }|rF\left(r\right){|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}.$$

2. The proofs

We first recall the Bliss lemma [5]:

Lemma 2.1. For s ≥ 0, q > p > 1 and r = q/p - 1,

$${\left(\underset{0}{\overset{\infty }{\int }}{\left|\underset{0}{\overset{s}{\int }}g\left(t\right)\mathsf{\text{d}}t\right|}^q{s}^{r-q}\mathsf{\text{d}}s\right)}^{p∕q}\le C_{p,q}\underset{0}{\overset{\infty }{\int }}|g\left(t\right){|}^p\mathsf{\text{d}}t,$$

where

$$C_{p,q}={\left(q-r-1\right)}^{-p∕q}{\left(\frac{r\Gamma \left(q∕r\right)}{\Gamma \left(1∕r\right)\Gamma \left(\left(q-1\right)∕r\right)}\right)}^{rp∕q}$$

is the sharp constant. Equality is attained for functions of the form

$$g\left(t\right)=c_1{\left(c_2{s}^r+1\right)}^{-\frac{r+1}{r}},c_1>0,c_2>0.$$

Using the Bliss lemma, we can prove the Theorem 1.1 for the radial function f, i.e., $f\left(x\right)=\tilde{f}\left(\left|x\right|\right)$ for some $\tilde{f}\in C_0^{\infty }\left(\left[0,\infty \right)\right)$.

Lemma 2.2. Let$f\left(\left|x\right|\right)\in C_0^{\infty }\left({ℝ}^n\right)$and n ≥ 3. There holds, for$0\le \delta <\frac{n^2}{4}$,

$$\underset{{ℝ}^n}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }{f}^2\mathsf{\text{d}}x\ge \frac{{\left(\frac{n^2}{4}-\delta \right)}^{\frac{n-1}{n}}}{{\left(\frac{{\left(n-2\right)}^2}{4}\right)}^{\frac{n-1}{n}}}{S}_n{\left(\underset{{ℝ}^n}{\int }|rF\left(r\right){|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}$$

(2.1)

and the constant in (2.1) is sharp.

Proof. We note if f is radial, then F(r) = f(r) and $\mathcal{L}f=r{f}^{\prime }\left(r\right)$. Therefore, inequality (2.1) is equivalent to

$$\begin{array}{c}\underset{0}{\overset{\infty }{\int }}|f^{\prime }\left(r\right){|}^2{r}^{n+1}\mathsf{\text{d}}r-\delta \underset{0}{\overset{\infty }{\int }}|f\left(r\right){|}^2{r}^{n-1}\mathsf{\text{d}}r\\ \ge \frac{{\left(\frac{n^2}{4}-\delta \right)}^{\frac{n-1}{n}}}{{\left(\frac{{\left(n-2\right)}^2}{4}\right)}^{\frac{n-1}{n}}}{S}_n\cdot |S^{n-1}{|}^{-\frac{2}{n}}{\left(\underset{0}{\overset{\infty }{\int }}|f\left(r\right){|}^{2^{*}}{r}^{2^{*}+n-1}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}.\end{array}$$

(2.2)

Let 0 ≤ β < n/2 and set g(r) = r^βf(r). Through integration by parts, we have that

$$\underset{0}{\overset{\infty }{\int }}|g^{\prime }\left(r\right){|}^2{r}^{n+1-2\beta }\mathsf{\text{d}}r=\underset{0}{\overset{\infty }{\int }}|f^{\prime }\left(r\right){|}^2{r}^{n+1}\mathsf{\text{d}}r-\beta \left(n-\beta \right)\underset{0}{\overset{\infty }{\int }}|f\left(r\right){|}^2{r}^{n-1}\mathsf{\text{d}}r.$$

(2.3)

Make the change of variables s = r^n-2β,

$$\underset{0}{\overset{\infty }{\int }}|g^{\prime }\left(r\right){|}^2{r}^{n+1-2\beta }\mathsf{\text{d}}r=\left(n-2\beta \right)\underset{0}{\overset{\infty }{\int }}{s}^2{\left|\frac{\partial g}{\partial s}\right|}^2\mathsf{\text{d}}s.$$

(2.4)

On the other hand, set $h\left(s\right)=\frac{\partial g}{\partial s}$ so that $g=-{\int }_s^{+\infty }h\left(t\right)\mathsf{\text{d}}t$, we have

$$\underset{0}{\overset{\infty }{\int }}{s}^2{\left|\frac{\partial g}{\partial s}\right|}^2\mathsf{\text{d}}s=\underset{0}{\overset{\infty }{\int }}{s}^2{h}^2\mathsf{\text{d}}s=\underset{0}{\overset{\infty }{\int }}|w\left(s\right){|}^2\mathsf{\text{d}}s,$$

where w(s) = s^-2h(s^-1). By Bliss lemma,

$$\underset{0}{\overset{\infty }{\int }}|w\left(s\right){|}^2\mathsf{\text{d}}s\ge {\left(\frac{n}{n-2}\right)}^{\frac{n-2}{n}}{\left(\frac{\Gamma \left(n∕2\right)\Gamma \left(1+n∕2\right)}{\Gamma \left(n\right)}\right)}^{\frac{2}{n}}{\left(\underset{0}{\overset{\infty }{\int }}{\left|\underset{0}{\overset{s}{\int }}\left|w\left(t\right)\right|\mathsf{\text{d}}t\right|}^{2^{*}}{s}^{\frac{2-2n}{n-2}}\mathsf{\text{d}}s\right)}^{\frac{2}{2^{*}}},$$

i.e.,

$$\begin{array}{c}\underset{0}{\overset{\infty }{\int }}{s}^2{\left|\frac{\partial g}{\partial s}\right|}^2\mathsf{\text{d}}s=\underset{0}{\overset{\infty }{\int }}{s}^2{h}^2\mathsf{\text{d}}s=\underset{0}{\overset{\infty }{\int }}|w\left(s\right){|}^2\mathsf{\text{d}}s\\ \ge {\left(\frac{n}{n-2}\right)}^{\frac{n-2}{n}}{\left(\frac{\Gamma \left(n∕2\right)\Gamma \left(1+n∕2\right)}{\Gamma \left(n\right)}\right)}^{\frac{2}{n}}{\left(\underset{0}{\overset{\infty }{\int }}{\left|\underset{0}{\overset{s}{\int }}\left|w\left(t\right)\right|\mathsf{\text{d}}t\right|}^{2^{*}}{s}^{\frac{2-2n}{n-2}}\mathsf{\text{d}}s\right)}^{\frac{2}{2^{*}}}\\ ={\left(\frac{n}{n-2}\right)}^{\frac{n-2}{n}}{\left(\frac{\Gamma \left(n∕2\right)\Gamma \left(1+n∕2\right)}{\Gamma \left(n\right)}\right)}^{\frac{2}{n}}{\left(\underset{0}{\overset{\infty }{\int }}{\left|\underset{s}{\overset{+\infty }{\int }}\left|h\left(t\right)\right|\mathsf{\text{d}}t\right|}^{2^{*}}{s}^{\frac{2}{n-2}}\mathsf{\text{d}}s\right)}^{\frac{2}{2^{*}}}\\ \ge {\left(\frac{n}{n-2}\right)}^{\frac{n-2}{n}}{\left(\frac{\Gamma \left(n∕2\right)\Gamma \left(1+n∕2\right)}{\Gamma \left(n\right)}\right)}^{\frac{2}{n}}{\left(\underset{0}{\overset{\infty }{\int }}{\left|g\right|}^{2^{*}}{s}^{\frac{2}{n-2}}\mathsf{\text{d}}s\right)}^{\frac{2}{2^{*}}}.\end{array}$$

(2.5)

Recall that s = r^n-2βand g(r) = r^βf(r),

$$\underset{0}{\overset{\infty }{\int }}{g}^{2^{*}}{s}^{\frac{2}{n-2}}\mathsf{\text{d}}s=\left(n-2\beta \right)\underset{0}{\overset{\infty }{\int }}{\left(r^{1-\beta }g\right)}^{2^{*}}{r}^{n-1}\mathsf{\text{d}}r=\left(n-2\beta \right)\underset{0}{\overset{\infty }{\int }}{\left(rf\right)}^{2^{*}}{r}^{n-1}\mathsf{\text{d}}r.$$

(2.6)

Therefore, by (2.3), (2.4), (2.5) and (2.6),

$$\begin{array}{c}\underset{0}{\overset{\infty }{\int }}|f^{\prime }\left(r\right){|}^2{r}^{n+1}\mathsf{\text{d}}r-\beta \left(n-\beta \right)\underset{0}{\overset{\infty }{\int }}|f\left(r\right){|}^2{r}^{n-1}\mathsf{\text{d}}r\\ =\left(n-2\beta \right)\underset{0}{\overset{\infty }{\int }}{s}^2{\left|\frac{\partial g}{\partial s}\right|}^2\mathsf{\text{d}}s\\ \ge {\left(n-2\beta \right)}^{1+\frac{2}{2^{*}}}{\left(\frac{n}{n-2}\right)}^{\frac{n-2}{n}}{\left(\frac{\Gamma \left(n∕2\right)\Gamma \left(1+n∕2\right)}{\Gamma \left(n\right)}\right)}^{\frac{2}{n}}{\left(\underset{0}{\overset{\infty }{\int }}{\left(rf\right)}^{2^{*}}{r}^{n-1}\mathsf{\text{d}}r\right)}^{\frac{2}{2^{*}}}\\ ={\left(n-2\beta \right)}^{\frac{2n-2}{n}}{\left(\frac{n}{n-2}\right)}^{\frac{n-2}{n}}{\left(\frac{\Gamma \left(n∕2\right)\Gamma \left(1+n∕2\right)}{\Gamma \left(n\right)}\right)}^{\frac{2}{n}}{\left(\underset{0}{\overset{\infty }{\int }}{\left(rf\right)}^{2^{*}}{r}^{n-1}\mathsf{\text{d}}r\right)}^{\frac{2}{2^{*}}}.\end{array}$$

Since $\left|S^{n-1}\right|={\int }_{S^{n-1}}\mathsf{\text{d}}\omega =\frac{2{\pi }^{n∕2}}{\Gamma \left(n∕2\right)}$ and $S_n=\pi n\left(n-2\right){\left(\Gamma \left(\frac{n}{2}\right)∕\Gamma \left(n\right)\right)}^{\frac{2}{n}}$, we have

$$\begin{array}{c}\underset{{ℝ}^n}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\beta \left(n-\beta \right)\underset{{ℝ}^n}{\int }{f}^2\mathsf{\text{d}}x\\ =\left|S^{n-1}\right|\underset{0}{\overset{\infty }{\int }}|f^{\prime }\left(r\right){|}^2{r}^{n+1}\mathsf{\text{d}}r-\beta \left(n-\beta \right)|S^{n-1}\left|\underset{0}{\overset{\infty }{\int }}\right|f\left(r\right){|}^2{r}^{n-1}\mathsf{\text{d}}r\\ \ge \left|S^{n-1}\right|\cdot {\left(n-2\beta \right)}^{\frac{2n-2}{n}}{\left(\frac{n}{n-2}\right)}^{\frac{n-2}{n}}{\left(\frac{\Gamma \left(n∕2\right)\Gamma \left(1+n∕2\right)}{\Gamma \left(n\right)}\right)}^{\frac{2}{n}}{\left(\underset{0}{\overset{\infty }{\int }}{\left(rf\right)}^{2^{*}}{r}^{n-1}\mathsf{\text{d}}r\right)}^{\frac{2}{2^{*}}}\\ =|S^{n-1}{|}^{\frac{2}{n}}{\left(n-2\beta \right)}^{\frac{2n-2}{n}}{\left(\frac{n}{n-2}\right)}^{\frac{n-2}{n}}{\left(\frac{\Gamma \left(n∕2\right)\Gamma \left(1+n∕2\right)}{\Gamma \left(n\right)}\right)}^{\frac{2}{n}}{\left(\underset{{ℝ}^n}{\int }|rf\left(r\right){|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}\\ ={\left(\frac{n-2\beta }{n-2}\right)}^{\frac{2n-2}{n}}{S}_n{\left(\underset{{ℝ}^n}{\int }|rf\left(r\right){|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}.\end{array}$$

Let $\beta =\frac{n-\sqrt{n^2-4\delta }}{2}$ when 0 ≤ δ < n²/4. Then, 0 ≤ β < n/2 and δ = β (n - β). Therefore,

$$\underset{{ℝ}^n}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }{f}^2\mathsf{\text{d}}x\ge {\left(\frac{n^2-4\delta }{{\left(n-2\right)}^2}\right)}^{\frac{n-1}{n}}{S}_n{\left(\underset{{ℝ}^n}{\int }|rf\left(r\right){|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}.$$

Inequality (2.1) follows.

Now we can prove Theorem 1.1.

Proof of Theorem 1.1. Decomposing f into spherical harmonics, we get (see e.g. [6])

$$f=\sum _{k=0}^{\infty }{f}_k:=\sum _{k=0}^{\infty }{g}_k\left(r\right){\varphi }_k\left(\sigma \right),$$

where ϕ _k (σ) are the orthonormal eigenfunctions of the Laplace-Beltrami operator with responding eigenvalues

$$c_k=k\left(N+k-2\right),k\ge 0.$$

The functions g _k (r) belong to $C_0^{\infty }\left({ℝ}^n\right)$, satisfying g _k (r) = O(r^k) and $g_k^{\prime }\left(r\right)=O\left(r^{k-1}\right)$ as r → 0. By orthogonality,

$$F\left(r\right)=\frac{1}{\left|S^{n-1}\right|}\underset{S^{n-1}}{\int }f\left(r\omega \right)\mathsf{\text{d}}\omega =g_0\left(r\right).$$

On the other hand,

$$\mathcal{L}f\left(x\right)=\sum _{k=0}^{\infty }r\frac{\partial \left(g_k\left(r\right){\varphi }_k\right)}{\partial r}=\sum _{k=0}^{\infty }r{g}_k^{\prime }\left(r\right){\varphi }_k\left(\sigma \right).$$

Here, we use the radial derivative $\frac{\partial }{\partial r}=\frac{x\cdot \nabla }{\left|x\right|}=\frac{\mathcal{L}}{\left|x\right|}$. Therefore,

$$\begin{array}{c}\underset{{ℝ}^n}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }{f}^2\mathsf{\text{d}}x=\sum _{k=0}^{\infty }\left(\underset{{ℝ}^n}{\int }{r}^2|{g^{\prime }}_k\left(r\right){|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }{g}_k^2\mathsf{\text{d}}x\right)\\ \ge \underset{{ℝ}^n}{\int }{r}^2|{g^{\prime }}_0\left(r\right){|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }{g}_0^2\mathsf{\text{d}}x=\underset{{ℝ}^n}{\int }{r}^2|F^{\prime }\left(r\right){|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }F{\left(r\right)}^2\mathsf{\text{d}}x\end{array}$$

since

$$\underset{{ℝ}^n}{\int }|\mathcal{L}u{|}^2\mathsf{\text{d}}x\ge \frac{n^2}{4}\underset{{ℝ}^n}{\int }{u}^2\mathsf{\text{d}}x$$

holds for all $u\in C_0^{\infty }\left({ℝ}^n\right)$ and $\mathcal{L}u=r{u}^{\prime }\left(r\right)$ if u is radial. By Lemma 2.2,

$$\underset{{ℝ}^n}{\int }{r}^2|F^{\prime }\left(r\right){|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }F{\left(r\right)}^2\mathsf{\text{d}}x\ge \frac{{\left(\frac{n^2}{4}-\delta \right)}^{\frac{n-1}{n}}}{{\left(\frac{{\left(n-2\right)}^2}{4}\right)}^{\frac{n-1}{n}}}{S}_n{\left(\underset{{ℝ}^n}{\int }|rF\left(r\right){|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}$$

Therefore,

$$\begin{array}{c}\underset{{ℝ}^n}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }{f}^2\mathsf{\text{d}}x\\ \ge \underset{{ℝ}^n}{\int }{r}^2|F^{\prime }\left(r\right){|}^2\mathsf{\text{d}}x-\delta \underset{{ℝ}^n}{\int }F{\left(r\right)}^2\mathsf{\text{d}}x\\ \ge \frac{{\left(\frac{n^2}{4}-\delta \right)}^{\frac{n-1}{n}}}{{\left(\frac{{\left(n-2\right)}^2}{4}\right)}^{\frac{n-1}{n}}}{S}_n{\left(\underset{{ℝ}^n}{\int }|rF\left(r\right){|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}.\end{array}$$

The proof of Theorem 1.1 is completed.

Proof of Theorem 1.2. We denote by B _R ⊂ ℝ^Nthe unit ball centered at zero.

Step 1. Assume f is radial and $f\in C_0^{\infty }\left(B_R\right)$. Then,

$$\begin{array}{c}\underset{B_R}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{B_R}{\int }{f}^2\mathsf{\text{d}}x=\underset{B_R}{\int }|r{f}^{\prime }\left(r\right){|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{B_R}{\int }{f}^2\left(r\right)\mathsf{\text{d}}x\\ =\underset{B_R}{\int }|{\left(rf\left(r\right)\right)}^{\prime }{|}^2\mathsf{\text{d}}x-\frac{{\left(n-2\right)}^2}{4}\underset{B_R}{\int }\frac{{\left(rf\right)}^2}{|x{|}^2}\mathsf{\text{d}}x.\end{array}$$

Therefore, by Theorem B in [7],

$$\begin{array}{c}\underset{B_R}{\int }|{\left(rf\left(r\right)\right)}^{\prime }{|}^2\mathsf{\text{d}}x-\frac{{\left(n-2\right)}^2}{4}\underset{B_R}{\int }\frac{{\left(rf\right)}^2}{|x{|}^2}\mathsf{\text{d}}x\\ \ge {\left(n-2\right)}^{-\frac{2\left(n-1\right)}{n}}{S}_n{\left(\underset{B_R}{\int }{X}_1^{\frac{2\left(n-1\right)}{n-2}}\left(a,\frac{\left|x\right|}{R}\right)|rf{|}^{\frac{2n}{n-2}}\mathsf{\text{d}}x\right)}^{\frac{n-2}{n}},\end{array}$$

where

$$X_1\left(a,s\right):={\left(a-lns\right)}^{-1},a>0,0<s\le 1.$$

Thus,

$$\begin{array}{c}\underset{B_R}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{B_R}{\int }{f}^2\mathsf{\text{d}}x=\underset{B_R}{\int }|r{f}^{\prime }\left(r\right){|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{B_R}{\int }{f}^2\left(r\right)\mathsf{\text{d}}x\\ \ge {\left(n-2\right)}^{-\frac{2\left(n-1\right)}{n}}{S}_n{\left(\underset{B_R}{\int }{X}_1^{\frac{2\left(n-1\right)}{n-2}}\left(a,\frac{\left|x\right|}{R}\right)|rf{|}^{\frac{2n}{n-2}}\mathsf{\text{d}}x\right)}^{\frac{n-2}{n}}.\end{array}$$

Step 2. Assume f is not radial and $f\in C_0^{\infty }\left(B_R\right)$. We extend f as zero outside B _R . So $f\in C_0^{\infty }\left({ℝ}^n\right)$. Decomposing f into spherical harmonics, we have

$$f=\sum _{k=0}^{\infty }{f}_k:=\sum _{k=0}^{\infty }{g}_k\left(r\right){\varphi }_k\left(\sigma \right),$$

where ϕ _k (σ) are the orthonormal eigenfunctions of the Laplace-Beltrami operator with responding eigenvalues

$$c_k=k\left(N+k-2\right),k\ge 0.$$

The functions f _k (r) belong to $C_0^{\infty }\left(B_R\right)$. By the proof of Theorem 1.1 and Step 1,

$$\begin{array}{c}\underset{{ℝ}^n}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{{ℝ}^n}{\int }{f}^2\mathsf{\text{d}}x=\sum _{k=0}^{\infty }\left(\underset{{ℝ}^n}{\int }{r}^2|g_k^{\prime }\left(r\right){|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{{ℝ}^n}{\int }{g}_k^2\mathsf{\text{d}}x\right)\\ \ge \underset{{ℝ}^n}{\int }{r}^2|g_0^{\prime }\left(r\right){|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{{ℝ}^n}{\int }{g}_0^2\mathsf{\text{d}}x=\underset{{ℝ}^n}{\int }{r}^2|F^{\prime }\left(r\right){|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{{ℝ}^n}{\int }F{\left(r\right)}^2\mathsf{\text{d}}x\\ \ge {\left(n-2\right)}^{-\frac{2\left(n-1\right)}{n}}{S}_n{\left(\underset{B_R}{\int }{X}_1^{\frac{2\left(n-1\right)}{n-2}}\left(a,\frac{\left|x\right|}{R}\right)|rF{|}^{\frac{2n}{n-2}}\mathsf{\text{d}}x\right)}^{\frac{n-2}{n}}.\end{array}$$

Step 3. By Step 1 and Step 2, the following inequality holds for $f\in C_0^{\infty }\left(B_R\right)$

$$\underset{{ℝ}^n}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{{ℝ}^n}{\int }{f}^2\mathsf{\text{d}}x\ge {\left(n-2\right)}^{-\frac{2\left(n-1\right)}{n}}{S}_n{\left(\underset{B_R}{\int }{X}_1^{\frac{2\left(n-1\right)}{n-2}}\left(a,\frac{\left|x\right|}{R}\right)|rF{|}^{\frac{2n}{n-2}}\mathsf{\text{d}}x\right)}^{\frac{n-2}{n}}.$$

We note if R^-1 < |x| < R, then

$$X_1^{\frac{2\left(N-1\right)}{N-2}}\left(a,\frac{\left|x\right|}{D}\right)={\left(\frac{1}{a-ln\frac{\left|x\right|}{R}}\right)}^{\frac{2\left(N-1\right)}{N-2}}\ge {\left(\frac{1}{a+2lnR}\right)}^{\frac{2\left(N-1\right)}{N-2}}.$$

Therefore, If f is supported in the annulus A _R := {x ∈ ℝⁿ: R^-1 < |x| < R}, then

$$\underset{A_R}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{A_R}{\int }{f}^2\mathsf{\text{d}}x\ge {\left[\left(n-2\right)\left(2lnR+a\right)\right]}^{-\frac{2\left(n-1\right)}{n}}{S}_n{\left(\underset{A_R}{\int }|rF\left(r\right){|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}.$$

Letting a → 0, we have

$$\underset{A_R}{\int }|\mathcal{L}f{|}^2\mathsf{\text{d}}x-\frac{n^2}{4}\underset{A_R}{\int }{f}^2\mathsf{\text{d}}x\ge {\left[2\left(n-2\right)lnR\right]}^{-\frac{2\left(n-1\right)}{n}}{S}_n{\left(\underset{A_R}{\int }|rF\left(r\right){|}^{2^{*}}\mathsf{\text{d}}x\right)}^{\frac{2}{2^{*}}}.$$

The proof of Theorem 2 is completed.