1 Introduction
It is well known that inequalities are important tools in classical analysis [2‐6, 13, 14, 26‐29, 31‐39, 41‐43, 45]. One application of inequalities is to study the properties of partial differential equations. Li and his coauthors [15‐23] studied the global existence and uniqueness, limit behavior, uniform stability, and blow-up of solutions for partial differential equations by using various inequalities. Liu [11, 24, 25] showed the stability and convergence results of evolution equations and Du [8, 9] studied obstacle problems by using various inequalities.
In recent decades, there have been many results on the extension and refinement of inequality [7, 10, 12, 30, 40, 44]. Qin [30] summarized a large number of inequalities and applications, but Hardy’s inequality was not included. The authors [7, 40] generalized the summation form Hardy inequality, Zhang [44] extended Hardy inequalities using Littlewood–Paley theory and nonlinear estimates method in Besov spaces, and the results improved and extended the well-known results in [1].
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The first edition of classic textbook [10] does not contain Hardy’s inequality, we see that the very significant Hardy’s inequality
holds if \(u\in H^{1}(B(o,r))\), \(n\geq 3\), and \(r>0\) in the second edition of [10]. The proof of Hardy’s inequality given in [10, 12] is very ingenious, but it is not easy to master for the reader. Therefore, we refine the proof of Hardy’s inequality for readers to grasp the essence of the proof and extend Hardy’s inequality in Sobolev space from two aspects. That is to say, we extend the integral estimation function from \(\frac{u}{|x|}\) to \(\frac{u}{|x|^{ \sigma }}\) with suitable \(\sigma >0\) and extend the space dimension from \(n \geq 3\) to \(n \geq 2\). Hardy’s inequality in [10, 12] is the special case of our results.
$$ \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2}}\,dx \leq C \int _{B(o,r)} \biggl( \vert Du \vert ^{2}+ \frac{u ^{2}}{r^{2}} \biggr)\,dx $$
Let \(B(o,r)\) be a closed ball in \(\mathbf{R}^{n}\) with center o and radius \(r>0\), \(x=(x_{1},x_{2},\ldots , x_{n})\) be a vector in \(B(o,r)\), \(\nu =(\nu _{1},\nu _{2},\ldots ,\nu _{n})= (\frac{x_{1}}{r},\frac{x _{2}}{r},\ldots ,\frac{x_{n}}{r} )\) be the unit outward normal to \(\partial B(o,r)\). \(W^{k,p}(\varOmega )\) and \(H^{1}(\varOmega )\) denote the Sobolev spaces. We write
$$\begin{aligned} Du=(u_{x_{1}},u_{x_{2}},\ldots , u_{x_{n}}), \quad \vert x \vert = \bigl(x_{1}^{2}+x_{2}^{2}+ \cdots +x_{n}^{2} \bigr)^{\frac{1}{2}}. \end{aligned}$$
2 Main results
Now, we present the global approximation theorem and Hardy’s inequality in Sobolev space.
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Lemma 2.1
([10], Global approximation theorem)
Assume that
Ω
is bounded and
∂Ω
is
\(C^{1}\). Let
\(u\in W^{k,p}(\varOmega )\)
for some
\(1\leq p<\infty \). Then there exist functions
\(u_{m}\in C^{\infty }(\overline{\varOmega })\)
such that
$$\begin{aligned} u_{m}\rightarrow u \quad \mathit{in } W^{k,p}(\varOmega ). \end{aligned}$$
Lemma 2.2
Assume
\(n\geq 3\)
and
\(r>0\). Let
\(u\in H^{1}(B(o,r))\). Then
\(\frac{u}{|x|} \in L^{2}(B(o,r))\)
with the estimate
$$\begin{aligned} \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2}}\,dx \leq C \int _{B(o,r)} \biggl( \vert Du \vert ^{2}+ \frac{u ^{2}}{r^{2}} \biggr)\,dx. \end{aligned}$$
(2.1)
For readers to grasp the essence of the proof, we give the refined proof below.
Proof
By the global approximation theorem Lemma 2.1, we may assume \(u\in C^{\infty }(B(o,r))\). Noting that \(D (\frac{1}{|x|^{ \rho }} )=-\rho \frac{x}{|x|^{\rho +2}}\) for any \(\rho >0\) and integrating by parts, we have
Therefore
$$\begin{aligned} \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2}}\,dx &= -\frac{1}{\rho } \int _{B(o,r)}u ^{2}D \biggl(\frac{1}{ \vert x \vert ^{\rho }} \biggr) \cdot \frac{x}{ \vert x \vert ^{2-\rho }}\,dx \\ &= -\frac{1}{\rho } \int _{B(o,r)}u^{2}\sum_{i=1}^{n} \biggl(\frac{1}{ \vert x \vert ^{ \rho }} \biggr)_{x_{i}} \frac{x_{i}}{ \vert x \vert ^{2-\rho }}\,dx \\ &= -\frac{1}{\rho } \int _{\partial B(o,r)}\sum_{i=1}^{n}u^{2} \nu _{i} \cdot \frac{x_{i}}{ \vert x \vert ^{2}}\,dS \\ &\quad {}+\frac{1}{\rho } \int _{B(o,r)}\sum_{i=1}^{n} \frac{1}{ \vert x \vert ^{\rho }} \biggl(u^{2}\frac{x_{i}}{ \vert x \vert ^{2-\rho }} \biggr)_{x_{i}} \,dx \\ &= -\frac{1}{\rho r} \int _{\partial B(o,r)}u^{2}\,dS \\ &\quad {}+\frac{1}{\rho } \int _{B(o,r)} \biggl[2uDu\cdot \frac{x}{ \vert x \vert ^{2}}+(n+ \rho -2) \frac{u^{2}}{ \vert x \vert ^{2}} \biggr]\,dx. \end{aligned}$$
(2.2)
$$\begin{aligned} (n-2) \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2}}\,dx =-2 \int _{B(o,r)}uDu\cdot \frac{x}{ \vert x \vert ^{2}}\,dx+\frac{1}{r} \int _{\partial B(o,r)}u^{2}\,dS. \end{aligned}$$
(2.3)
For any \(\varepsilon >0\), using the Cauchy inequality and Schwarz inequality, we obtain
Fixing \(\varepsilon >0\) such that \(n-2-2\varepsilon >0\), we conclude
According to the divergence theorem, we have
Using the Cauchy inequality and Schwarz inequality, we get
Combining (2.5) and (2.6), we obtain the trace inequality
Employing this inequality (2.7) in (2.4) finishes the proof of (2.1). □
$$\begin{aligned} -2 \int _{B(o,r)}uDu\cdot \frac{x}{ \vert x \vert ^{2}}\,dx &= -2 \int _{B(o,r)} \frac{u}{ \vert x \vert }Du\cdot \frac{x}{ \vert x \vert }\,dx \\ &\leq 2 \int _{B(o,r)}\frac{ \vert u \vert }{ \vert x \vert } \vert Du \vert \biggl\vert \frac{x}{ \vert x \vert } \biggr\vert \,dx \\ &= 2 \int _{B(o,r)}\frac{ \vert u \vert }{ \vert x \vert } \vert Du \vert \,dx \\ &\leq 2 \varepsilon \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2}}\,dx+\frac{1}{2 \varepsilon } \int _{B(o,r)} \vert Du \vert ^{2}\,dx. \end{aligned}$$
$$\begin{aligned} \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2}}\,dx \leq C \int _{B(o,r)} \vert Du \vert ^{2}\,dx+ \frac{C}{r} \int _{\partial B(o,r)}u^{2}\,dS. \end{aligned}$$
(2.4)
$$\begin{aligned} \int _{B(o,r)}\operatorname{div}\bigl(xu^{2}\bigr)\,dx &= \int _{\partial B(o,r)}xu^{2} \cdot \nu \,dS= \int _{\partial B(o,r)}u^{2}x\cdot \frac{x}{r}\,dS \\ &=r \int _{\partial B(o,r)}u^{2}\,dS. \end{aligned}$$
(2.5)
$$\begin{aligned} \int _{B(o,r)}\operatorname{div}\bigl(xu^{2}\bigr)\,dx = & \int _{B(o,r)} \bigl[u^{2} \operatorname{div}(x)+D \bigl(u^{2}\bigr)\cdot x \bigr]\,dx \\ = & \int _{B(o,r)} \bigl(nu^{2}+2uDu\cdot x \bigr)\,dx \\ \leq & \int _{B(o,r)} \bigl(nu^{2}+u^{2}+ \vert x \vert ^{2} \vert Du \vert ^{2} \bigr)\,dx \\ \leq & \int _{B(o,r)} \bigl[(n+1)u^{2}+r^{2} \vert Du \vert ^{2} \bigr]\,dx. \end{aligned}$$
(2.6)
$$\begin{aligned} \frac{1}{r} \int _{\partial B(o,r)}u^{2}\,dS \leq C \int _{B(o,r)} \biggl( \vert Du \vert ^{2}+ \frac{u ^{2}}{r^{2}} \biggr)\,dx. \end{aligned}$$
(2.7)
Under the circumstance, we extend the space dimension n and parameter σ in \(\frac{u}{|x|^{\sigma }}\) of Hardy’s inequality. Now we show our main results.
Theorem 2.1
Assume
\(n\geq 2\)
and
\(r>0\), \(u\in H^{1}(B(o,r))\). Then, for
\(\sigma <\frac{n}{2}\), we have
\(\frac{u}{|x|^{\sigma }}\in L^{2}(B(o,r))\)
with the estimate as follows:
If
\(\sigma \leq 1\)
and
\(\sigma < \frac{n}{2}\), we have
$$\begin{aligned} \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2\sigma }}\,dx \leq C \int _{B(o,r)} \biggl(\frac{ \vert Du \vert ^{2}}{r ^{2(\sigma -1)}} +\frac{u^{2}}{r^{2\sigma }} \biggr)\,dx. \end{aligned}$$
If
\(\sigma >1\)
and
\(\sigma < \frac{n}{2}\), we have
$$\begin{aligned} \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2\sigma }}\,dx\leq C \int _{B(o,r)} \biggl(\frac{ \vert Du \vert ^{2}}{ \vert x \vert ^{2( \sigma -1)}} +\frac{u^{2}}{r^{2\sigma }} \biggr)\,dx. \end{aligned}$$
Remark 2.2
If \(n=2\), then \(\sigma <1\). \(B(o,r)\) denotes a closed circular region with center o and radius \(r>0\), \(\partial B(o,r)\) denotes a circle, and \(\int _{B(o,r)}\cdots dS\) denotes curvilinear integration.
3 Proofs of the main results
In this section we show the proofs of the main results Theorem 2.1.
Proof
For any \(\rho >0\), since
which implies
$$\begin{aligned} D \biggl(\frac{1}{ \vert x \vert ^{\rho }} \biggr)=-\rho \frac{x}{ \vert x \vert ^{\rho +2}}, \end{aligned}$$
$$\begin{aligned} \frac{1}{{ \vert x \vert }^{2\sigma }} &= \biggl[-\rho \frac{x}{{ \vert x \vert }^{\rho +2}} \biggr] \cdot \biggl[ \biggl(-\frac{1}{ \rho } \biggr)\frac{x}{{ \vert x \vert }^{2\sigma -\rho }} \biggr] \\ &=-\frac{1}{\rho }D \biggl(\frac{1}{{ \vert x \vert }^{\rho }} \biggr)\cdot \frac{x}{ { \vert x \vert }^{2\sigma -\rho }} . \end{aligned}$$
(3.1)
By the global approximation theorem, we may assume \(u\in C^{\infty }(B(o,r))\). Noting that (3.1) holds, we obtain
Hence
For any \(\varepsilon >0\), using the Cauchy inequality and Schwarz inequality, we obtain
According to the divergence theorem, we have
and using the Cauchy inequality and Schwarz inequality, we get
which implies
By substituting (3.4) and (3.5) into (3.3), fixing ε such that \(n-2\sigma -2\varepsilon >0\), we conclude
$$\begin{aligned} \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2\sigma }}\,dx &= -\frac{1}{\rho } \int _{B(o,r)}u^{2}D \biggl(\frac{1}{ \vert x \vert ^{\rho }} \biggr) \cdot \frac{x}{ \vert x \vert ^{2 \sigma -\rho }}\,dx \\ &= -\frac{1}{\rho } \int _{B(o,r)}\sum_{i=1}^{n} \biggl(\frac{1}{ \vert x \vert ^{ \rho }} \biggr)_{x_{i}} \biggl(u^{2} \frac{x_{i}}{ \vert x \vert ^{2\sigma -\rho }} \biggr)\,dx \\ &= -\frac{1}{\rho } \int _{\partial B(o,r)}\sum_{i=1}^{n}u^{2} \nu _{i} \cdot \frac{x_{i}}{ \vert x \vert ^{2\sigma }}\,dS \\ &\quad {}+\frac{1}{\rho } \int _{B(o,r)}\sum_{i=1}^{n} \frac{1}{ \vert x \vert ^{\rho }} \biggl(u^{2}\frac{x_{i}}{ \vert x \vert ^{2\sigma -\rho }} \biggr)_{x_{i}} \,dx \\ &= -\frac{1}{\rho r^{2\sigma -1}} \int _{\partial B(o,r)}u^{2}\,dS \\ &\quad {}+\frac{1}{\rho } \int _{B(o,r)} \biggl[2uDu\cdot \frac{x}{ \vert x \vert ^{2\sigma }}+(n+\rho -2\sigma )\frac{u^{2}}{ \vert x \vert ^{2\sigma }} \biggr]\,dx. \end{aligned}$$
(3.2)
$$\begin{aligned} (n-2\sigma ) \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2\sigma }}\,dx &= -2 \int _{B(o,r)}uDu \cdot \frac{x}{ \vert x \vert ^{2\sigma }}\,dx \\ &\quad {}+\frac{1}{r^{2\sigma -1}} \int _{\partial B(o,r)}u^{2}\,dS. \end{aligned}$$
(3.3)
$$\begin{aligned} -2 \int _{B(o,r)}uDu\cdot \frac{x}{ \vert x \vert ^{2\sigma }}\,dx &= -2 \int _{B(o,r)}\frac{u}{ \vert x \vert ^{ \sigma }}Du\cdot \frac{x}{ \vert x \vert ^{\sigma }}\,dx \\ &\leq 2 \int _{B(o,r)}\frac{ \vert u \vert }{ \vert x \vert ^{\sigma }} \vert Du \vert \biggl\vert \frac{x}{ \vert x \vert ^{ \sigma }} \biggr\vert \,dx \\ &= 2 \int _{B(o,r)}\frac{ \vert u \vert }{ \vert x \vert ^{\sigma }} \vert Du \vert \frac{1}{ \vert x \vert ^{\sigma -1}}\,dx \\ &\leq 2\varepsilon \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2\sigma }}\,dx \\ &\quad {}+\frac{1}{2\varepsilon } \int _{B(o,r)} \vert Du \vert ^{2}\frac{1}{ \vert x \vert ^{2(\sigma -1)}} \,dx. \end{aligned}$$
(3.4)
$$\begin{aligned} \int _{B(o,r)}\operatorname{div}\bigl(xu^{2}\bigr)\,dx=r \int _{\partial B(o,r)}u^{2}\,dS, \end{aligned}$$
$$\begin{aligned} \int _{B(o,r)}\operatorname{div}\bigl(xu^{2}\bigr)\,dx &= \int _{B(o,r)} \bigl(nu^{2}+2uDu \cdot x \bigr)\,dx \\ &\leq \int _{B(o,r)} \bigl(nu^{2}+u^{2}+ \vert Du \vert ^{2} \vert x \vert ^{2} \bigr)\,dx \\ &\leq \int _{B(o,r)} \bigl[(n+1)u^{2}+r^{2} \vert Du \vert ^{2} \bigr]\,dx, \end{aligned}$$
$$\begin{aligned} \frac{1}{r^{2\sigma -1}} \int _{\partial B(o,r)}u^{2}\,dS \leq \frac{n+1}{r ^{2\sigma }} \int _{B(o,r)}u^{2}\,dx+\frac{1}{r^{2(\sigma -1)}} \int _{B(o,r)} \vert Du \vert ^{2}\,dx. \end{aligned}$$
(3.5)
$$\begin{aligned} \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2\sigma }}\,dx \leq C \int _{B(o,r)} \biggl[\frac{ \vert Du \vert ^{2}}{ \vert x \vert ^{2( \sigma -1)}} +\frac{ \vert Du \vert ^{2}}{r^{2(\sigma -1)}}+ \frac{u^{2}}{r^{2 \sigma }} \biggr]\,dx. \end{aligned}$$
(3.6)
if \(\sigma \leq 1\), noting that
we obtain
$$\begin{aligned} \frac{ \vert Du \vert ^{2}}{ \vert x \vert ^{2(\sigma -1)}}\leq \frac{ \vert Du \vert ^{2}}{r^{2(\sigma -1)}}, \quad x\in B(o,r), \end{aligned}$$
$$\begin{aligned} \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2\sigma }}\,dx \leq C \int _{B(o,r)} \biggl(\frac{ \vert Du \vert ^{2}}{r ^{2(\sigma -1)}}+\frac{u^{2}}{r^{2\sigma }} \biggr) \,dx. \end{aligned}$$
(3.7)
if \(\sigma >1\), noting that
we obtain
$$\begin{aligned} \frac{ \vert Du \vert ^{2}}{ \vert x \vert ^{2(\sigma -1)}}\geq \frac{ \vert Du \vert ^{2}}{r^{2(\sigma -1)}}, \quad x\in B(o,r), \end{aligned}$$
$$\begin{aligned} \int _{B(o,r)}\frac{u^{2}}{ \vert x \vert ^{2\sigma }}\,dx \leq C \int _{B(o,r)} \biggl(\frac{ \vert Du \vert ^{2}}{ \vert x \vert ^{2( \sigma -1)}} +\frac{u^{2}}{r^{2\sigma }} \biggr)\,dx. \end{aligned}$$
(3.8)
The proof of Theorem 2.1 is completed. □
4 Conclusions
In this paper, we refine the proof of Hardy’s inequality for readers to grasp the essence of the proof and extend Hardy’s inequality in Sobolev space from two aspects. That is to say, we extend the integral estimation function from \(\frac{u}{|x|}\) to \(\frac{u}{|x|^{\sigma }}\) with suitable \(\sigma >0\) and extend the space dimension from \(n \geq 3\) to \(n \geq 2\). Hardy’s inequality in [10, 12] is the special case of our results.
Acknowledgements
The authors would like to thank the referees for their valuable comments and suggestion.
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Competing interests
The authors declare that they have no competing interests.
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