Skip to main content

Hardy inequalities with Aharonov-Bohm type magnetic field on the Heisenberg group

Abstract

We introduce an Aharonov-Bohm type magnetic field on three-dimensional Heisenberg group and show this quadratic form satisfy an improved Hardy inequality with weights.

Introduction

The classical Hardy inequality states that, for \(N\geq3\) and for all \(u\in C^{\infty}_{0}(\mathbb{R}^{N})\),

$$ \int_{\mathbb{R}^{N}}|\nabla u|^{2}\, dx\geq \biggl(\frac{N-2}{2} \biggr)^{2}\int_{\mathbb{R}^{N}} \frac{u^{2}}{|x|^{2}}\, dx $$
(1.1)

and \((\frac{N-2}{2} )^{2}\) is the best constant in (1.1). If \(N=2\), the classical Hardy inequality fails. However, for some magnetic forms in dimension two, the Hardy inequality becomes possible. In fact, if β a is the Aharonov-Bohm magnetic field

$$\beta\mathbf{a}=\beta \biggl(-\frac{x_{2}}{x^{2}_{1}+x^{2}_{2}},\frac {x_{1}}{x^{2}_{1}+x^{2}_{2}} \biggr), $$

then for all \(u\in C^{\infty}_{0}(\mathbb{R}^{2}\setminus\{0\})\) (cf. [1]),

$$ \int_{\mathbb{R}^{2}}\bigl\vert (\nabla+i\beta\mathbf{a}) u\bigr\vert ^{2}\, dx\geq \min_{k\in\mathbb{Z}}|k- \beta|^{2}\int_{\mathbb{R}^{2}}\frac{u^{2}}{|x|^{2}}\, dx. $$
(1.2)

Recently, a version of the Aharonov-Bohm magnetic field for a Grushin subelliptic operator has been introduced by Aermark and Laptev [2]. Furthermore, such quadratic form also satisfies an improved Hardy inequality. In the same paper, they asked the following question: does there exist a similar result for the Heisenberg quadratic form?

Recall that the three-dimension Heisenberg group \(\mathbb {H}_{1}=(\mathbb{R}^{2}\times\mathbb{R},\circ)\) is a step-two nilpotent group whose group structure is given by

$$(x,y,t)\circ\bigl(x',y',t'\bigr)= \biggl(x+x',y+y',t+t'-\frac{1}{2} \bigl(xy'-yx'\bigr) \biggr). $$

The vector fields

$$X_{=}\frac{\partial}{\partial x}+\frac{y}{2}\frac{\partial}{\partial t},\qquad Y= \frac{\partial}{\partial y}-\frac{x}{2}\frac{\partial}{\partial t} $$

are left invariant and generate the Lie algebra of \(\mathbb{H}_{1}\). The Kohn sub-Laplacian on \(\mathbb{H}_{1}\) is

$$\Delta_{\mathbb{H}}=X^{2}+Y^{2}=\frac{\partial^{2}}{\partial x^{2}}+ \frac{\partial^{2}}{\partial y^{2}}+ \frac{x^{2}+y^{2}}{4}\frac{\partial^{2}}{\partial t^{2}}+\frac{1}{4} \biggl(y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y} \biggr)\frac{\partial}{\partial t} $$

and the subgradient is the vector given by \(\nabla_{\mathbb{H}}=(X,Y)\). For simplicity, we let \(z=x+yi\). Then \(|z|=\sqrt{x^{2}+y^{2}}\). Denote

$$\rho:=\rho(z,t)=\bigl(|z|^{4}+16t^{2}\bigr)^{\frac{1}{4}}. $$

Similar as in [1, 2], we define an Aharonov-Bohm type magnetic field \(\mathcal{A}\) on \(\mathbb{H}_{1}\):

$$ \mathcal{A}=(\mathcal{A}_{1},\mathcal{A}_{2})= \biggl(-\frac{Y\rho}{\rho },\frac{X\rho}{\rho} \biggr). $$
(1.3)

To our surprise, for such magnetic field (1.3), we cannot deal with the Hardy inequality

$$ \int_{\mathbb{H}_{1}}|\nabla_{\mathbb{H}} u|^{2}\, dz\, dt\geq\int_{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho^{2}}| \nabla _{\mathbb{H}} \rho|^{2}\, dz\, dt,\quad u\in C^{\infty}_{0}(\mathbb{H}_{1}), $$
(1.4)

but the Hardy inequality with weight

$$ \int_{\mathbb{H}_{1}}\frac{|\nabla_{\mathbb{H}} u|^{2}}{|\nabla_{\mathbb{H}} \rho|^{2}}\, dz\, dt\geq\int _{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho^{2}}\, dz\, dt,\quad u\in C^{\infty}_{0}(\mathbb{H}_{1}). $$

For the reasons, see Remark 2.2. For more Hardy inequalities on Heisenberg groups, we refer to [310].

The main result is the following theorem.

Theorem 1.1

We have, for \(u\in C^{\infty}_{0}(\mathbb{H}_{1})\),

$$ \int_{\mathbb{H}_{1}}\frac{|(\nabla_{\mathbb{H}}+i\beta\mathcal{A}) u|^{2}}{|\nabla_{\mathbb{H}}\rho|^{2}}\, dz\, dt\geq \Bigl(1+\min_{k\in \mathbb{Z}}|k-\beta|^{2} \Bigr)\int _{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho ^{2}}\, dz\, dt. $$
(1.5)

Proof of Theorem 1.1

Before the proof of Theorem 1.1, we need a polar coordinate associated with ρ on \(\mathbb{H}_{1}\). We describe it as follows. For each real number \(\lambda>0\), there is a dilation naturally associated with the group structure which is usually denoted \(\delta_{\lambda}(x,y,t)=(\lambda x, \lambda y,\lambda^{2}t)\). The Jacobian determinant of \(\delta_{\lambda}\) is \(\lambda^{Q}\), where \(Q=4\) is the homogeneous dimension of \(\mathbb{H}_{1}\). For simplicity, we use the notation \(\lambda (x,y,t)=\delta_{\lambda}(x,y,t)\). Given any \(\xi=(x,y,t)\in \mathbb{H}_{1}\), set \(x^{\ast}=\frac{x}{\rho}\), \(y^{\ast}=\frac{y}{\rho}\), \(t^{\ast}=\frac{t}{\rho^{2}}\), and \(\xi^{\ast}=(x^{\ast},y^{\ast},t^{\ast})\) if \(\rho(\xi)\neq0\). The polar coordinate on \(\mathbb{H}_{1}\) associated with ρ is the following (cf. [11], Proposition (1.15)):

$$\int_{\mathbb{H}_{1}}f(\xi)\, dz\, dt=\int^{\infty}_{0} \int_{\Sigma}f\bigl(\lambda \xi^{\ast}\bigr) \lambda^{3}\, d\sigma\, d\lambda,\quad f\in L^{1}( \mathbb{H}_{1}), $$

where \(\Sigma=\{(x,y,t)\in\mathbb{H}_{1}: \rho(x,y,t)=1\}\) is the unit sphere associated with ρ. Moreover, there is a parametrization of this polar coordinate (cf. [12], Theorem 5.12):

$$ \left \{ \begin{array}{l} x=\rho\sqrt{\cos\alpha}\cos\theta; \\ y=\rho\sqrt{\cos\alpha}\sin\theta; \\ t=\frac{1}{4}\rho^{2}\sin\alpha, \end{array} \right . $$
(2.1)

where \(\alpha\in[-\pi/2,\pi/2)\), \(\theta\in[0,2\pi)\), and \(0\leq\rho <\infty\). Using this parametrization, we can rewrite the polar coordinate as follows:

$$ \int_{\mathbb{H}_{1}}f(\xi)\, dz\, dt=\frac{1}{4} \int^{\infty}_{0}\int^{\pi /2}_{-\pi/2} \int^{2\pi}_{0}f(\xi)\rho^{3}\, d\rho\, d \alpha\, d\theta,\quad f\in L^{1}(\mathbb{H}_{1}). $$
(2.2)

Proof of Theorem 1.1

Using the identity

$$\bigl(a^{2}+b^{2}\bigr) \bigl(|z_{1}|^{2}+|z_{2}|^{2} \bigr)=|az_{1}+bz_{2}|^{2}+|az_{2}-bz_{1}|^{2}, \quad a,b\in\mathbb{R}, z_{1},z_{2}\in\mathbb{C}, $$

we have

$$\begin{aligned}& |\nabla_{\mathbb{H}}\rho|^{2}\bigl\vert ( \nabla_{\mathbb{H}}+i\beta\mathcal{A}) u\bigr\vert ^{2} \\& \quad = \bigl(|X \rho|^{2}+|Y\rho|^{2}\bigr) \biggl(\biggl\vert Xu-i\beta \frac{Y\rho}{\rho}\biggr\vert ^{2}+\biggl\vert Yu+i\beta \frac {X\rho}{\rho}\biggr\vert ^{2} \biggr) \\& \quad = |X\rho\cdot Xu+Y\rho\cdot Yu|^{2}+\biggl\vert X\rho\cdot Yu-Y\rho \cdot Xu+i\beta\frac{|\nabla_{\mathbb{H}}\rho|^{2}u}{\rho}\biggr\vert ^{2}. \end{aligned}$$
(2.3)

By (2.1),

$$ \frac{\partial}{\partial\theta}= \frac{\partial x}{\partial\theta }\frac{\partial}{\partial x}+ \frac{\partial y}{\partial\theta} \frac{\partial}{\partial y} =x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x} $$
(2.4)

and

$$\begin{aligned} \frac{\partial}{\partial\alpha} =&\frac{\partial x}{\partial \alpha}\frac{\partial}{\partial x}+ \frac{\partial y}{\partial \alpha}\frac{\partial}{\partial y} +\frac{\partial t}{\partial\alpha}\frac{\partial}{\partial t} \\ =&-\frac{\rho\sin\alpha\cos\theta}{2\sqrt{\cos\alpha}}\frac{\partial }{\partial x}-\frac{\rho\sin\alpha\sin\theta}{2\sqrt{\cos\alpha}}\frac{\partial }{\partial y} + \frac{1}{4}\rho^{2}\cos\alpha\frac{\partial}{\partial t} \\ =&-\frac{2xt}{|z|^{2}}\frac{\partial}{\partial x}-\frac {2yt}{|z|^{2}}\frac{\partial}{\partial y}+ \frac{|z|^{2}}{4}\frac {\partial}{\partial t}. \end{aligned}$$
(2.5)

Therefore, by (2.4) and (2.5)

$$\begin{aligned} \begin{aligned}[b] X\rho\cdot Yu-Y\rho\cdot Xu&=\frac{|z|^{2}x+4yt}{\rho^{3}} \biggl( \frac {\partial u}{\partial y}-\frac{x}{2}\frac{\partial u}{\partial t} \biggr)- \frac{|z|^{2}y-4xt}{\rho^{3}} \biggl(\frac{\partial u}{\partial x}+\frac {y}{2}\frac{\partial u}{\partial t} \biggr) \\ &=\frac{|z|^{2}}{\rho^{3}} \biggl(x\frac{\partial}{\partial y}-y\frac {\partial}{\partial x} \biggr)+ \frac{4xt}{\rho^{3}} \frac{\partial u}{\partial x}+\frac{4yt}{\rho^{3}} \frac{\partial u}{\partial y}- \frac{|z|^{4}}{2\rho^{3}}\frac{\partial u}{\partial t} \\ &=\frac{|z|^{2}}{\rho^{3}}\frac{\partial u}{\partial\theta}+\frac {|z|^{2}}{\rho^{3}} \biggl( \frac{4xt}{|z|^{2}}\frac{\partial}{\partial x}+\frac{4yt}{|z|^{2}}\frac {\partial}{\partial y}- \frac{|z|^{2}}{2}\frac{\partial}{\partial t} \biggr) \\ &=\frac{|z|^{2}}{\rho^{3}} \biggl(\frac{\partial u}{\partial\theta }-2\frac{\partial u}{\partial\alpha} \biggr)= \frac{|\nabla_{\mathbb {H}}\rho|^{2}}{\rho} \biggl(\frac{\partial u}{\partial\theta}-2\frac {\partial u}{\partial\alpha} \biggr). \end{aligned} \end{aligned}$$
(2.6)

To get the last inequality above, we use the fact \(|\nabla_{\mathbb {H}}\rho|=\frac{|z|}{\rho}\). Combining (2.3) and (2.6) yields

$$\begin{aligned} \int_{\mathbb{H}_{1}}\frac{|(\nabla_{\mathbb{H}}+i\beta\mathcal{A}) u|^{2}}{|\nabla_{\mathbb{H}}\rho|^{2}}\, dz\, dt =& \int _{\mathbb{H}_{1}}\frac{|X\rho\cdot Xu+Y\rho\cdot Yu|^{2}}{|\nabla _{\mathbb{H}}\rho|^{4}}\, dz\, dt \\ &{}+\int_{\mathbb{H}_{1}} \frac{|\frac{\partial u}{\partial\theta}-2\frac{\partial u}{\partial\alpha}+i\beta u|^{2}}{\rho^{2}}\, dz\, dt \\ =&(I)+(\mathit{II}), \end{aligned}$$
(2.7)

where

$$(I):= \int_{\mathbb{H}_{1}}\frac{|X\rho\cdot Xu+Y\rho\cdot Yu|^{2}}{|\nabla_{\mathbb{H}}\rho|^{4}}\, dz\, dt $$

and

$$\begin{aligned} (\mathit{II})&:= \int_{\mathbb{H}_{1}}\frac{|\frac{\partial u}{\partial \theta}-2\frac{\partial u}{\partial\alpha}+i\beta u|^{2}}{\rho^{2}}\, dz\, dt \\ &= \frac{1}{4}\int^{\infty}_{0}\int ^{\pi/2}_{-\pi/2}\int^{2\pi}_{0} \biggl\vert \frac{\partial u}{\partial\theta}-2\frac{\partial u}{\partial\alpha}+i\beta u\biggr\vert ^{2}\rho \, d\rho\, d\alpha \, d\theta. \end{aligned}$$

If we represent u by the Fourier series

$$u(\rho,\alpha,\theta)=\sum^{+\infty}_{n=-\infty}u_{n}( \rho,\alpha )e^{in\theta}/\sqrt{2\pi}, $$

then

$$ (\mathit{II})=\sum^{+\infty}_{n=-\infty} \frac{1}{4}\int^{\infty}_{0}\int ^{\pi /2}_{-\pi/2}\biggl\vert 2\frac{\partial u_{n} (\rho,\alpha)}{\partial \alpha}-i( \beta+n)u_{n}(\rho,\alpha)\biggr\vert ^{2}\rho\, d\rho \, d \alpha. $$
(2.8)

Similarly, representing \(u_{n}(\rho,\alpha)\) by the Fourier series

$$u_{n}(\rho,\alpha)=\sum^{+\infty}_{k=-\infty}u_{n,k}( \rho)e^{i2k\alpha }/\sqrt{\pi}, $$

we have

$$\begin{aligned} (\mathit{II}) =&\sum^{+\infty}_{n=-\infty} \sum^{+\infty}_{k=-\infty}\frac{|4k-n-\beta |^{2}}{4}\int ^{\infty}_{0}\bigl\vert u_{n,k}(\rho)\bigr\vert ^{2}\rho\, d\rho \\ \geq&\min_{k\in \mathbb{Z}}|k-\beta|^{2}\cdot\frac{1}{4} \sum^{+\infty}_{n=-\infty}\sum ^{+\infty}_{k=-\infty}\int^{\infty}_{0} \bigl\vert u_{n,k}(\rho)\bigr\vert ^{2}\rho \, d\rho \\ =&\min_{k\in \mathbb{Z}}|k-\beta|^{2}\cdot\frac{1}{4} \int^{\infty}_{0}\int^{\pi /2}_{-\pi/2} \int^{2\pi}_{0}\bigl\vert u(\rho,\alpha,\theta) \bigr\vert ^{2}\rho \, d\rho\, d\alpha\, d\theta \\ =&\min_{k\in \mathbb{Z}}|k-\beta|^{2}\int_{\mathbb{H}_{1}} \frac{|u|^{2}}{\rho^{2}}\, dz\, dt. \end{aligned}$$
(2.9)

To finish the proof, it is enough to show

$$(I)=\int_{\mathbb{H}_{1}}\frac{|X\rho\cdot Xu+Y\rho\cdot Yu|^{2}}{|\nabla_{\mathbb{H}}\rho|^{4}}\, dx\, dy\, dt \geq \int _{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho^{2}}\, dx\, dy\, dt. $$

This will be done in Lemma 2.1. The proof of Theorem 1.1 is therefore completed. □

Before we turn to the proof of Lemma 2.1, we need the horizontal polar coordinates on \(\mathbb{H}_{1}\) which have been introduced by Korányi and Reimann [13] (see also [14], pp.110-112). Set

$$\gamma_{\xi}(\rho)= \biggl(sze^{4i\frac{t}{|z|^{2}}\log \rho},\frac{1}{4} \rho^{2}t \biggr),\quad \xi=(z,t)\in\Sigma. $$

The horizontal polar coordinate on \(\mathbb{H}_{1}\) is

$$\int_{\mathbb{H}_{1}}f(z,t)\, dz\, dt=\int^{\infty}_{0} \int_{\Sigma}f\bigl(\gamma _{\xi}(\rho)\bigr) \rho^{3}\, ds\, d\sigma,\quad f\in L^{1}( \mathbb{H}_{1}). $$

Furthermore, we can also give a parametrization of this polar coordinate through setting ([14], pp.111-112)

$$ \Phi: \left \{ \begin{array}{l} x=\rho\sqrt{\cos\alpha}\cos(\theta+4\tan\alpha\log\rho); \\ y=\rho\sqrt{\cos\alpha}\sin(\theta+4\tan\alpha\log\rho); \\ t=\rho^{2}\sin\alpha. \end{array} \right . $$

The Jacobian determinant of Φ is \(\rho^{3}\) so that

$$\begin{aligned} \int_{\mathbb{H}_{1}}f(z,t)\, dz\, dt =&\int ^{\infty}_{0}\int_{\Sigma}f\bigl(\gamma _{\xi}(\rho)\bigr)\rho^{3}\, d\rho \, d\xi \\ =&\frac{1}{4}\int^{\infty}_{0}\int ^{\pi/2}_{-\pi/2}\int^{2\pi }_{0}f \bigl(\gamma_{\xi}(\rho)\bigr)\rho^{3}\, d\rho \, d\alpha \, d \theta. \end{aligned}$$
(2.10)

Using this parametrization, we have (see [14], p.112)

$$\begin{aligned} \frac{d}{d\rho}f\bigl(\gamma_{\xi}(\rho)\bigr) =& \frac{1}{\rho|z|^{2}} \bigl(\bigl(x|z|^{2}-4yt\bigr)Xf+ \bigl(y|z|^{2}+4xt\bigr)Yf \bigr) \\ =&\frac{1}{4}\frac{\langle\nabla_{\mathbb{H}}\rho^{4},\nabla_{\mathbb{H}} f\rangle}{\rho|z|^{2}}=\frac{\rho^{2}}{|z|^{2}}\langle \nabla_{\mathbb {H}}\rho,\nabla_{\mathbb{H}} f\rangle \\ =&\frac{\langle\nabla_{\mathbb{H}}\rho,\nabla_{\mathbb{H}} f\rangle}{|\nabla_{\mathbb{H}}\rho|^{2}}. \end{aligned}$$
(2.11)

Lemma 2.1

We have, for \(u\in C^{\infty}_{0}(\mathbb{H}_{1})\),

$$\int_{\mathbb{H}_{1}}\frac{|X\rho\cdot Xu+Y\rho\cdot Yu|^{2}}{|\nabla_{\mathbb{H}}\rho|^{4}}\, dz\, dt \geq \int _{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho^{2}}\, dz\, dt. $$

Proof

Notice that

$$\begin{aligned} \begin{aligned}[b] &\int^{\infty}_{0}\biggl\vert \frac{d}{d\rho}f\bigl(\gamma_{\xi}(\rho)\bigr) \biggr\vert ^{2}\rho^{3}\, d\rho- \int^{\infty}_{0} \bigl\vert f\bigl(\gamma_{\xi}(\rho)\bigr)\bigr\vert ^{2} \rho\, d\rho \\ &\quad =\int^{\infty}_{0}\biggl\vert \frac{d(\rho f(\gamma_{\xi}(\rho)))}{d\rho}\biggr\vert ^{2}\rho \, d\rho\geq0. \end{aligned} \end{aligned}$$
(2.12)

Integrating over \(-\pi/2\leq\alpha\leq\pi/2\) and \(0\leq\theta\leq 2\pi\) and using (2.10) yields

$$\int_{\mathbb{H}_{1}}\biggl\vert \frac{d}{d\rho}f\bigl( \gamma_{\xi}(\rho)\bigr) \biggr\vert ^{2}\, dz\, dt\geq \int _{\mathbb{H}_{1}}\frac{|f(\gamma_{\xi}(\rho))|^{2}}{\rho^{2}}\, dz\, dt. $$

Combining the inequality above and (2.11) gives

$$\int_{\mathbb{H}_{1}}\frac{|X\rho\cdot Xu+Y\rho\cdot Yu|^{2}}{|\nabla_{\mathbb{H}}\rho|^{4}}\, dz\, dt \geq \int _{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho^{2}}\, dz\, dt. $$

This completes the proof of Lemma 2.1. □

Remark 2.2

If one considers the Hardy inequality (1.4) with the Aharonov-Bohm type magnetic field \(\mathcal{A}\) on \(\mathbb{H}_{1}\), then, following the proof above, one needs to show

$$ \int^{\pi/2}_{-\pi/2}\biggl\vert \frac{\partial u}{\partial\alpha}-i\beta u\biggr\vert ^{2}\cos\alpha \, d\alpha\geq \min_{k\in\mathbb{Z}}|2k-\beta|^{2} \int^{\pi/2}_{-\pi/2} \vert u\vert ^{2}\cos\alpha\, d\alpha. $$
(2.13)

However, to the best of our knowledge, it is not known whether inequality (2.13) is valid. The reason is that, in this case, \(\{e^{2ki\pi}/\sqrt{\pi } \}\) is not an orthonormal basis because there exists a weight cosα in (2.13).

References

  1. 1.

    Laptev, A, Weidl, T: Hardy inequalities for magnetic Dirichlet forms. In: Mathematical Results in Quantum Mechanics (Prague, 1998). Oper. Theory Adv. Appl., vol. 108, pp. 299-305. Birkhäuser, Basel (1999)

    Google Scholar 

  2. 2.

    Aermark, L, Laptev, A: Hardy’s inequality for the Grushin operator with a magnetic field of Aharanov-Bohm type. Algebra Anal. 23(2), 1-8 (2011). English transl.: St. Petersbg. Math. J. 23(2), 203-208 (2012)

    MathSciNet  Google Scholar 

  3. 3.

    D’Ambrosio, L, Hardy, S: Inequalities on the Heisenberg group. Differ. Uravn. 40(4), 509-521 (2004). English transl.: Differ. Equ. 40(4), 552-564 (2004)

    MathSciNet  Google Scholar 

  4. 4.

    Garofallo, N, Lanconelli, E: Frequency functions on Heisenberg group, the uncertainty principle and unique continuation. Ann. Inst. Fourier (Grenoble) 40, 313-356 (1990)

    Article  MathSciNet  Google Scholar 

  5. 5.

    Lian, B, Yang, Q, Yang, F: Some weighted Hardy-type inequalities on anisotropic Heisenberg groups. J. Inequal. Appl. 2011, Article ID 924840 (2011)

    Article  MathSciNet  Google Scholar 

  6. 6.

    Luan, J, Yang, Q: A Hardy type inequality in the half-space on \(\mathbb{R}^{n}\) and Heisenberg group. J. Math. Anal. Appl. 347, 645-651 (2008)

    Article  MATH  MathSciNet  Google Scholar 

  7. 7.

    Niu, P, Zhang, H, Wang, Y: Hardy type and Rellich type inequalities on the Heisenberg group. Proc. Am. Math. Soc. 129, 3623-3630 (2001)

    Article  MATH  MathSciNet  Google Scholar 

  8. 8.

    Yang, Q: Best constants in the Hardy-Rellich type inequalities on the Heisenberg group. J. Math. Anal. Appl. 342, 423-431 (2008)

    Article  MATH  MathSciNet  Google Scholar 

  9. 9.

    Yang, Q: Hardy type inequalities related to Carnot-Carathéodory distance on the Heisenberg group. Proc. Am. Math. Soc. 141(1), 351-362 (2013)

    Article  MATH  Google Scholar 

  10. 10.

    Yang, Q, Lian, B: On the best constant of weighted Poincaré inequalities. J. Math. Anal. Appl. 377, 207-215 (2011)

    Article  MATH  MathSciNet  Google Scholar 

  11. 11.

    Folland, GB, Stein, EM: Hardy Spaces on Homogeneous Groups. Princeton University Press, Princeton (1982)

    Google Scholar 

  12. 12.

    Balogh, ZM, Tyson, JT: Polar coordinates in the Carnot groups. Math. Z. 241, 697-730 (2002)

    Article  MATH  MathSciNet  Google Scholar 

  13. 13.

    Korányi, A, Reimann, HM: Horizontal normal vectors and conformal capacity of spherical rings in the Heisenberg group. Bull. Sci. Math. 111, 3-21 (1987)

    MATH  MathSciNet  Google Scholar 

  14. 14.

    Capogna, L, Danielli, D, Pauls, SD, Tyson, JT: An Introduction to the Heisenberg Group and the Sub-Riemannian Isoperimetric Problem. Progr. Math. Birkhäuser Boston, Boston (2007)

    Google Scholar 

Download references

Acknowledgements

This research is supported by the key project of Hubei provincial education department in P.R. China (No. D20142701) and the National Natural Science Foundation of China (No. 11201346).

Author information

Affiliations

Authors

Corresponding author

Correspondence to Yingxiong Xiao.

Additional information

Competing interests

The author declares to have no competing interests.

Rights and permissions

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Xiao, Y. Hardy inequalities with Aharonov-Bohm type magnetic field on the Heisenberg group. J Inequal Appl 2015, 95 (2015). https://doi.org/10.1186/s13660-015-0617-4

Download citation

MSC

  • 22E25
  • 35H20

Keywords

  • Hardy inequality
  • Heisenberg group
  • Aharonov-Bohm magnetic field