1 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 [3–10].

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)

2 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).