# Quantitative Magnetic Isoperimetric Inequality

Rohan Ghanta, Lukas Junge and Léo Morin  Institut for Matematik, Aarhus
Universitet, Ny Munkegade 8000 Aarhus C, Danmark<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
(Date: May 9, 2023)

###### Abstract.

In 1996 Erdős showed that among planar domains of fixed area, the smallest
principal eigenvalue of the Dirichlet Laplacian with a constant magnetic field
is uniquely achieved on the disk. We establish a quantitative version of this
inequality, with an explicit remainder term depending on the field strength
that measures how much the domain deviates from the disk.

###  1\. Introduction

To solve a problem in Probability and Mathematical Physics [11],[12], Erdős
developed the magnetic isoperimetric inequality [10]. It generalizes the
Faber-Krahn inequality to the magnetic Laplacian. Starting with Pólya and
Szegő [19], Faber-Krahn-type results have been established by proving
rearrangement inequalities. The inclusion of a magnetic field, however, makes
it notoriously difficult to implement the standard symmetrization methods.
Erdős met the challenge head on: he managed to prove a magnetic rearrangement
inequality, which is reminiscent of the celebrated Pólya-Szegő inequality but
with an interesting caveat. Such symmetry results with a magnetic field
are–alas!–very few and far between [1],[5].

Still another compelling feature is that rearrangements alone are not
sufficient for arguing the magnetic isoperimetric inequality. This stands in
sharp contrast to the classical Faber-Krahn setting. To complete the proof
Erdős introduced a new inequality, tailored specifically for a magnetic
Schrödinger operator on a disk and for which there exists no analog in the
absence of a magnetic field.

We improve Erdős’ result. He showed that if a planar domain is not a disk,
then the principal eigenvalue of the Dirichlet magnetic Laplacian is strictly
larger on that domain than on the disk of same area. We take the next step and
establish stability: if the principal eigenvalue of the magnetic Laplacian is
just slightly larger on a planar domain than on the disk of same area, then
that domain is only slightly different from the disk. Faint perturbations of
the smallest principal eigenvalue will not induce a dramatic change in the
underlying geometry–and this dynamic is very sensitive to the field strength.
We prove our stability estimate with a remainder term that quantifies the
difference between the domain and the disk.

Quantitative Faber-Krahn-type inequalities have been developed almost
exclusively around the classical theory of rearrangements. Fueled in large
part by the seminal work of Fusco et al. [13], the last decade has given rise
to an entire industry now devoted to the stability of a remarkable range of
geometric and functional inequalities. Our paper provides the first stability
result with a magnetic field. And here, the well-established rearrangement
framework is no longer sufficient.

###  2\. Statement of Problem and Main Result

Let $\Omega\subset\mathbb{R}^{2}$ be a bounded, connected open set with a
smooth boundary. The principal eigenvalue of the Dirichlet magnetic Laplacian
on the planar domain $\Omega$ is

(2.1) $\lambda(B,\Omega):=\inf_{f\in
H_{0}^{1}(\Omega)}\frac{\int_{\Omega}|(-i\nabla-\alpha)f|^{2}dx}{\int_{\Omega}|f|^{2}dx},$

where $\alpha=\frac{B}{2}\left(-x_{2},x_{1}\right)$ is a magnetic vector
potential generating a homogeneous magnetic field of strength $B\geq 0$, i.e.
$\text{rot(}\alpha)=B$. We denote by $D_{R}$ a disk of radius $R$, centered at
the origin, with the same area as $\Omega$, i.e.
$\left|\Omega\right|=\left|D_{R}\right|=\pi R^{2}$.

In 1996 Erdős [10] proved the magnetic isoperimetric inequality

(2.2) $\lambda(B,\Omega)\geq\lambda(B,D_{R}),$

with equality if and only if $\Omega$ is a disk. In the absence of a magnetic
field, i.e. $B=0$, his result reduces to the usual Faber-Krahn inequality.

In this paper, we want to add to the right-hand side of (2.2) a remainder term
that measures how much the planar domain $\Omega$ deviates from being a disk.
This would make it possible to understand the shape of $\Omega$ now in terms
of how close it is to achieving equality in (2.2). Cf. [7] & references
therein.

We measure the difference between $\Omega$ and the disk in the usual way in
terms of the interior deficiency and the Fraenkel asymmetry of the domain.

######  Definition.

The interior deficiency (asymmetry) of a set is defined as

$\mathcal{A}_{I}\left(\Omega\right):=\frac{R-\rho_{-}(\Omega)}{R},$

where $\rho_{-}(\Omega)$ denotes the radius of the largest ball contained in
$\Omega$, and $R$ as above is the radius of $D_{R}$.

######  Definition.

The Fraenkel asymmetry of a set is defined as

$\mathcal{A}_{F}\left(\Omega\right):=\inf_{x_{0}\in\mathbb{R}^{2}}\frac{|\Omega\Delta(x_{0}+D_{R})|}{2|\Omega|}.$

Both asymmetries are bounded by one and vanish if and only if the set is a
disk.

Our main result is a quantitative version of the magnetic isoperimetric
inequality.

######  Theorem 2.1.

Let $\mathcal{A}\left(\Omega\right)$ denote either the interior asymmetry or
the Fraenkel asymmetry. In the case of the interior asymmetry we also assume
$\Omega$ is simply connected. Then there is a universal constant $c>0$,
independent of $\Omega$ and $B$, such that

(2.3)
$\lambda(B,\Omega)\geq\lambda(B,D_{R})(1+ce^{-\frac{5}{6}BR^{2}}\mathcal{A}(\Omega)^{\frac{10}{3}})\,.$

Moreover, if $0\leq BR^{2}\leq\frac{1}{\pi}$, then

(2.4) $\lambda(B,\Omega)\geq\lambda(B,D_{R})(1+c\mathcal{A}(\Omega)^{3})\,.$

######  Remark 2.2.

The quantity $\mathcal{A}\left(\Omega\right)$ is scale invariant. Furthermore
$\lambda$ scales like $t^{2}\lambda(B,t\Omega)=\lambda(t^{2}B,\Omega)$ for
$t>0$, so the factor $BR^{2}$ appearing in our constant is the natural
parameter for this problem.

In the absence of a magnetic field, i.e. $B=0$, the estimate in (2.4) reduces
to Hansen and Nadirashvili’s quantitative Faber-Krahn inequality with the
asymmetry cubed [15],[3]. More recently, Brasco et al. [8] proved it with the
square power: this is the sharp form, because the exponent cannot be any
smaller [4],[18]. Our magnetic version in (2.3) should likewise instead have
the square of the asymmetry and, in principle, one could adapt Brasco et al.’s
argument to achieve this. Their state-of-the-art methods, however, are
nonconstructive and will not yield an explicit constant. This would make it
impossible to understand the pertinent role of the magnetic field strength $B$
in the stability of Erdős’ inequality.

Our methods, on the other hand, yield an explicit constant with a natural
dependence on the field strength. Physical intuition suggests that as
$B\rightarrow\infty$ the principal eigenfunctions start to localize on a
length scale proportional to $1/\sqrt{B}$, away from the boundary, and
therefore $\lambda(B,\cdot)$ becomes less sensitive to the shape of the
domain: it can but faintly distinguish between even very dissimilar shapes,
and the little sensitivity that remains comes from the fact that these
eigenfunctions can still feel about near the boundary with their exponentially
small tails. Now $\Omega$ can look rather different from $D_{R}$ and yet
$\lambda(B,\Omega)\approx\lambda(B,D_{R})$: a strong magnetic field
compromises stability. We manage to capture this picture in (2.3) with our
constant which vanishes, exponentially, as $B\rightarrow\infty$.

To prove his Faber-Krahn-type inequality in (2.2), Erdős started out in the
usual way by establishing a rearrangement inequality. See Lemma 3.1. While
there are certainly nontrivial magnetic aspects to the argument, Erdős
essentially mimicked the standard proof [20] of the analogous Pólya-Szegő
inequality using the coarea formula and the isoperimetric inequality. But in
imposing the Pólya-Szegő scheme on his problem, he was forced to change the
magnetic field on the disk. The vector potential on the right-hand side of
(3.1) is no longer the same: and thus his magnetic rearrangement inequality
cannot readily imply (2.2) in the same way that the Pólya-Szegő inequality
yields Faber-Krahn.

To deal with this mis-match between the magnetic fields on $\Omega$ and
$D_{R}$, Erdős developed the comparison lemma on the disk. See Remark 4.2. It
compares the ground-state energies of the operator on the right-hand side of
(3.1) corresponding to different magnetic fields. This in turn allowed him to
recover the original magnetic field on $D_{R}$ and finish proving (2.2). His
comparison lemma is built on the variational principle and has nothing to do
with rearrangements. And unlike his rearrangement inequality, it has no analog
in the absence of a magnetic field.

To prove our stability estimate in Theorem 2.1, we also start out in the usual
way by establishing a quantitative version of Erdős’ rearrangement inequality.
See Proposition 3.2. This is nothing new: in the absence of a magnetic field,
i.e. $B=0$, it just reduces to the quantitative version of the Pólya-Szegő
inequality that was used in proving stability of Faber-Krahn [7]. Here we
mimic Erdős’ proof but instead apply the quantitative isoperimetric inequality
on the level sets.

######  Theorem 2.3.

Let $U\subset\mathbb{R}^{2}$ be a bounded set with smooth boundary, and let
$\mathcal{P}(U)$ denote the perimeter of $U$. Let $\mathcal{A}(U)$ denote
either the interior asymmetry or the Fraenkel asymmetry. In the case of the
interior asymmetry we also assume $U$ is simply connected. Then there is a
universal constant $c>0$ such that

$\mathcal{P}(U)\geq
2\sqrt{\pi}\left|U\right|^{\frac{1}{2}}\left(1+c\mathcal{A}(U)^{2}\right).$

This was first proved by Bonnesen in 1924 for simply connected planar sets
using the interior asymmetry [6],[18]. In 2008 Fusco et al. proved a more
general version using the Fraenkel asymmetry [13]. Theorem 2.3 forms the
backbone of the first part of the paper.

In Lemma 4.1 we establish a quantitative version of Erdős’ comparison lemma.
Now this is really a new estimate, which stands completely outside of the
rearrangement framework–and it only enters the scene when $B$ is large.

In Corollary 4.3 we present two very different lower bounds on the quantity
$\lambda(B,\Omega)-\lambda(B,D_{R})$, both involving the asymmetry of the
level sets of the principal eigenfunction corresponding to
$\lambda(B,\Omega)$. The first bound, (4.7), is based on our quantitative
version of the rearrangement inequality. The second bound, (4.8), is based on
our quantitative version of the comparison lemma.

As usual, the main difficulty lies in going from the asymmetry of these level
sets in Corollary 4.3 to the asymmetry of the whole domain. We deal with this
in the second part of the paper. When $B$ is small, we operate entirely within
the rearrangement framework just as in the classical Faber-Krahn setting. Here
our argument is a direct perturbation of Hansen and Nadirashvili’s proof of
their quantitative Faber-Krahn inequality [15]. We only use the first bound,
given in (4.7), of Corollary 4.3 which is based on the quantitative version of
the rearrangement inequality. This is enough to prove the estimate in (2.4) of
Theorem 2.1.

But as $B$ increases, our weak-field adaptation of Hansen and Nadirashvili’s
technique breaks down: with a strong magnetic field, the rearrangement
framework alone is no longer sufficient for establishing stability. Here we
make full use of both the quantitative version of the rearrangement inequality
and now our quantitative version of the comparison lemma. A distinctive
feature of our argument is the necessary interplay between the traditional
bound in (4.7)–rooted firmly within the paradigmatic framework of
rearrangement inequalities–and our magnetic bound in (4.8), which is unique to
our problem and irreducible to any other estimate used in establishing
stability of a Faber-Krahn-type inequality.

##  Part I The Magnetic Isoperimetric Inequality

Here we re-prove Erdős’ magnetic isoperimetric inequality but with a remainder
term involving the asymmetry of the level sets of the principal eigenfunction
corresponding to $\lambda\left(B,\Omega\right)$. This is given as Corollary
4.3. The quantitative isoperimetric inequality plays an essential role.

###  3\. The Magnetic Rearrangement Inequality

Standard elliptic theory tells us that the principal eigenfunction
corresponding to $\lambda(B,\Omega)$ is a complex-valued analytic function.
The first ingredient in Erdős’ proof is a rearrangement inequality. He proved
the following.

######  Lemma 3.1.

Let $f$, $\|f\|_{2}=1$ be a complex-valued analytic function on $\Omega$ that
vanishes on the boundary, and let $\left|f\right|^{*}$ denote the symmetric
decreasing rearrangement of $\left|f\right|$. Then there exists a vector
potential
$\tilde{\alpha}(x)=\frac{a\left(\left|x\right|\right)}{\left|x\right|}\left(-x_{2},x_{1}\right)$,
where $a\left(\left|x\right|\right)$ is a function satisfying $0\leq
a\left(\left|x\right|\right)\leq\frac{B\left|x\right|}{2}$, such that

(3.1)
$\int_{\Omega}\left|\left(-i\nabla-\alpha\right)f\right|^{2}dx\geq\int_{D_{R}}\left|\left(-i\nabla-\tilde{\alpha}\right)\left|f\right|^{*}\right|^{2}dx+B-\int_{D_{R}}\text{rot}\left(\tilde{\alpha}\right)\left|f\right|^{*2}dx.$

This is analogous to the celebrated Pólya-Szegő inequality but with some
caveats:

  1. (1)

The magnetic field on the disk is no longer the same. Our vector potential
$\alpha=\frac{B}{2}\left(-x_{2},x_{1}\right)$ corresponds to a homogeneous
field of strength $B$. Now $\tilde{\alpha}$ corresponds to a radially
symmetric but inhomogeneous field.

  2. (2)

The potential $\tilde{\alpha}$ depends on $f$, because Erdős constructed
$a\left(\left|x\right|\right)$ from the level sets of $\left|f\right|$;

  3. (3)

in particular, if $a\left(\left|x\right|\right)=\frac{B\left|x\right|}{2}$,
then the level set $\left\\{\left|f\right|>\left|f\right|^{*}(x)\right\\}$ is
a disk.

Lemma 3.1 yields a lower bound on $\lambda(B,\Omega)$. Had the vector
potential remained unchanged, (3.1) would have readily implied
$\lambda(B,\Omega)\geq\lambda(B,D_{R})$.

In this section we prove a quantitative version of his rearrangement
inequality, and we write the right-hand side more conveniently in terms of
polar coordinates.

######  Proposition 3.2.

Let $f$, $\|f\|_{2}=1$ be as in the statement of Lemma 3.1, and
$q\left(\left|x\right|\right):=\left|f\right|^{*}(x)$. Then there exists a
bounded function $a\left(\left|x\right|\right)$, depending on $f$ and $B$,
such that111We use here the following convention for the interior asymmetry.
If the open set $U$ is not simply connected, we define $\mathcal{A}_{I}(U)$ to
be the asymmetry of the smallest simply connected set containing $U$. Since
$\Omega$ is simply connected, this will not change the final value of
$\mathcal{A}_{I}(\Omega)$. This convention allows us to use Theorem 2.3 for
the level sets of $\left|f\right|$.

$\int_{\Omega}\left|\left(-i\nabla-\alpha\right)f\right|^{2}dx\geq
B+2\pi\int_{0}^{R}\left(q^{\prime}(r)+a(r)q(r)\right)^{2}\left(1+c\mathcal{A}^{2}\left(\left\\{\left|f\right|>q(r)\right\\}\right)\right)^{2}rdr,$

and

(3.2) $0\leq
a(r)\leq\frac{Br}{2}\left(1+c\mathcal{A}^{2}\left(\left\\{\left|f\right|>q(r)\right\\}\right)\right)^{-2}\leq\frac{Br}{2}\,,$

where $c>0$ is a universal constant independent of $B$ and $\Omega$.

In the absence of the asymmetry term, the expression on the right-hand side
indeed coincides with that of (3.1). See Proof of Lemma A.2 in the appendix.

####  3.1. The Proof of Proposition 3.2

Erdős proved his rearrangement inequality within the standard Pólya-Szegő
scheme [20] using the coarea formula and the isoperimetric inequality, which
we replace with its quantitative version.

To use the coarea formula, first we need a real-valued function. By modifying
the magnetic vector potential, we can work with $\left|f\right|$ instead.

######  Lemma 3.3.

Let $f$ be as in the statement of Lemma 3.1, and
$\Omega_{0}:=\Omega\setminus\\{f=0\\}$. Let
$\theta:\Omega_{0}\mapsto\left[0,2\pi\right)$ be such that
$f=\left|f\right|e^{i\theta}$. Since $\Omega_{0}$ has full measure,
$w:=\alpha-\nabla\theta$ is defined almost everywhere and
$\text{rot}\left(w\right)=B$. Then, with
$w^{\perp}:=\left(-w_{2},w_{1}\right)$,

$\int_{\Omega}\left|\left(-i\nabla-\alpha\right)f\right|^{2}dx=B+\int_{\Omega}\left|\nabla\left|f\right|+w^{\perp}\left|f\right|\right|^{2}dx.$

###### Proof.

Since $\left|f\right|\in H_{0}^{1}\left(\Omega\right)$ and $w$ is real-valued,

$\int_{\Omega}\left|\left(-i\nabla-\alpha\right)f\right|^{2}dx=\int_{\Omega}\left|\left(-i\nabla-w\right)\left|f\right|\right|^{2}dx=\int_{\Omega}\left(\left|\nabla\left|f\right|\right|^{2}+\left|w^{\perp}\right|^{2}\left|f\right|^{2}\right)dx.$

Note $w$ is smooth a.e. By completing the square and integrating by parts,

$\displaystyle\int_{\Omega}\left|\left(-i\nabla-\alpha\right)f\right|^{2}dx$
$\displaystyle=\int_{\Omega}\left(\left|\nabla\left|f\right|+w^{\perp}\left|f\right|\right|^{2}-2\left|f\right|w^{\perp}\cdot\nabla\left|f\right|\right)dx$
$\displaystyle=\int_{\Omega}\left(\left|\nabla\left|f\right|+w^{\perp}\left|f\right|\right|^{2}+\left|f\right|^{2}\text{div}(w^{\perp})\right)dx.$

Since $\text{rot}\left(w\right)=B$, the lemma follows. ∎

Then we use the coarea formula and arrive at an expression involving an
integral over the level sets of $\left|f\right|$.

######  Lemma 3.4.

Let $f,w^{\perp}$ be as in the statement of Lemma 3.3. Then,

(3.3)
$\int_{\Omega}\left|\nabla\left|f\right|+w^{\perp}\left|f\right|\right|^{2}dx\geq\int_{0}^{\infty}dz\
(1-B\Phi(z)z)^{2}\int_{\\{\left|f\right|=z\\}}|\nabla\left|f\right||,$

with

(3.4)
$\Phi(z):=\frac{\left|\left\\{\left|f\right|>z\right\\}\right|}{\int_{\left\\{\left|f\right|=z\right\\}}\left|\nabla\left|f\right|\right|}.$

If there is no magnetic field, i.e. $B=0$, and $f$ is a positive function,
then the relation in (3.3) reduces to the usual coarea formula used in the
proof of the Pólya-Szegő inequality [20].

###### Proof of Lemma 3.4.

There exists $w^{\prime}$ orthogonal to $\nabla\left|f\right|$ and
$\varphi:\Omega\mapsto\mathbb{R}$ such that
$w^{\perp}=-\varphi\nabla\left|f\right|+w^{\prime}$. By the Pythagorean
theorem,

$\displaystyle\int_{\Omega}\left|\nabla\left|f\right|+w^{\perp}\left|f\right|\right|^{2}dx$
$\displaystyle=\int_{\Omega}\left(\left|\left(1-\varphi\left|f\right|\right)\nabla\left|f\right|\right|^{2}+\left|w^{\prime}\left|f\right|\right|^{2}\right)dx$
$\displaystyle\geq\int_{\Omega}\left|\left(1-\varphi\left|f\right|\right)\nabla\left|f\right|\right|^{2}dx.$

Now we are in a position to use the coarea formula:

$\displaystyle\int_{\Omega}\left|\left(1-\varphi\left|f\right|\right)\nabla\left|f\right|\right|^{2}dx$
$\displaystyle=\int_{0}^{\infty}dz\int_{\left\\{\left|f\right|=z\right\\}}\left(1-\varphi
z\right)^{2}\left|\nabla\left|f\right|\right|$
$\displaystyle\geq\int_{0}^{\infty}dz\,\frac{\left(\int_{\\{\left|f\right|=z\\}}(1-\varphi
z)|\nabla\left|f\right||\right)^{2}}{\int_{\\{\left|f\right|=z\\}}|\nabla\left|f\right||}.$

We use Stokes’ theorem on the level sets. For almost all $z>0$, the level set
$\\{\left|f\right|=z\\}$ is regular by Sard’s theorem. Thus

$B|\\{\left|f\right|>z\\}|=\int_{\\{\left|f\right|>z\\}}\text{rot}\left(w\right)=\int_{\\{\left|f\right|=z\\}}w\cdot\tau,$

where
$\tau=\frac{(\nabla\left|f\right|)^{\perp}}{\left|(\nabla\left|f\right|)^{\perp}\right|}$.
Since $w\cdot\tau=\varphi\left|\nabla\left|f\right|\right|$, we conclude

$\int_{\Omega}\left|\nabla\left|f\right|+w^{\perp}\left|f\right|\right|^{2}dx\geq\int_{0}^{\infty}dz\,\frac{\left(\int_{\\{\left|f\right|=z\\}}|\nabla\left|f\right||-Bz|\\{\left|f\right|>z\\}|\right)^{2}}{\int_{\\{\left|f\right|=z\\}}|\nabla\left|f\right||}.$

The lemma follows from the definition of $\Phi$ in (3.4). ∎

With the coarea-type estimate in (3.3), Erdős applied the isoperimetric
inequality on the level sets of $\left|f\right|$ to prove his rearrangement
inequality; and when $B=0$, his argument reduces to the standard proof of the
Pólya-Szegő inequality [20]. Below we instead apply the quantitative
isoperimetric inequality on these level sets.

###### Proof of Proposition 3.2.

From Lemma 3.3, Lemma 3.4 and Hölder’s inequality

$\int_{\Omega}\left|\left(-i\nabla-\alpha\right)f\right|^{2}dx\geq
B+\int_{0}^{\infty}dz\
(1-B\Phi(z)z)^{2}\frac{\left|\left\\{\left|f\right|=z\right\\}\right|^{2}}{\int_{\left\\{\left|f\right|=z\right\\}}\left|\nabla\left|f\right|\right|^{-1}}\,.$

By Sard’s theorem, the denominator is non-vanishing for almost all $z>0$. And
since $q$ is the rearrangement of $\left|f\right|$,

(3.5) $q(r)=F^{-1}\left(\pi r^{2}\right)\ \text{where}\
F(z):=\left|\left\\{\left|f\right|>z\right\\}\right|.$

By the coarea formula, again for almost all $z>0$

(3.6)
$F(z)=\int_{z}^{\infty}d\xi\int_{\left\\{\left|f\right|=\xi\right\\}}\left|\nabla\left|f\right|\right|^{-1}\
\text{and}\
F^{\prime}(z)=-\int_{\left\\{\left|f\right|=z\right\\}}\left|\nabla\left|f\right|\right|^{-1}.$

Then,

(3.7) $\int_{\Omega}\left|\left(-i\nabla-\alpha\right)f\right|^{2}dx\geq
B-\int_{0}^{\infty}\left(1-B\Phi(z)z\right)^{2}\left|\left\\{\left|f\right|=z\right\\}\right|^{2}F^{\prime}(z)^{-1}\,dz.$

Now we do a change of variable $z=q(r)$ and apply the isoperimetric
inequality, Theorem 2.3, on the level sets:
$\left|\left\\{\left|f\right|=q(r)\right\\}\right|\geq 2\pi
r\left(1+c\mathcal{A}^{2}\left(\left\\{\left|f\right|>q(r)\right\\}\right)\right)$.
We write $\mathcal{A}^{2}$ for short. Then,

$\int_{\Omega}\left|\left(-i\nabla-\alpha\right)f\right|^{2}dx\geq
B+\int_{0}^{R}\left(1-B\Phi\left(q(r)\right)q(r)\right)^{2}\frac{\left(2\pi
r\right)^{2}q^{\prime}(r)}{F^{\prime}(q(r))}\left(1+c\mathcal{A}^{2}\right)^{2}dr.$

Since $q^{\prime}(r)=2\pi rF^{\prime}(q(r))^{-1}$,

$\int_{\Omega}\left|\left(-i\nabla-\alpha\right)f\right|^{2}dx\geq
B+2\pi\int_{0}^{R}\left[q^{\prime}(r)-\frac{2\pi
rB\Phi(q(r))}{F^{\prime}(q(r))}q(r)\right]^{2}(1+c\mathcal{A}^{2})^{2}rdr.$

Writing $a(r):=-2\pi rBF^{\prime}(q(r))^{-1}\Phi(q(r))$, we deduce our
rearrangement inequality.

It remains to prove the upper bound in (3.2). By Hölder’s inequality

$-F^{\prime}(q(r))=\int_{\\{\left|f\right|=q(r)\\}}\left|\nabla\left|f\right|\right|^{-1}\geq|\left\\{\left|f\right|=q(r)\right\\}|^{2}\left(\int_{\\{\left|f\right|=q(r)\\}}|\nabla\left|f\right||\right)^{-1},$

and by the isoperimetric inequality, Theorem 2.3,

$a(r)\leq 2\pi
rB\frac{\left|\left\\{\left|f\right|>q(r)\right\\}\right|}{\left|\left\\{\left|f\right|=q(r)\right\\}\right|^{2}}\leq\frac{Br}{2}\left(1+c\mathcal{A}^{2}\left(\left\\{\left|f\right|>q(r)\right\\}\right)\right)^{-2}.$

This concludes the proof of Proposition 3.2. ∎

###  4\. The Comparison Lemma

The second ingredient in Erdős’ proof is a comparison lemma, which makes it
possible to recover from the right-hand side of (3.1) the original potential
$\alpha$ on the disk. In this section we prove a quantitative version of his
comparison lemma.

For a potential
$\tilde{\alpha}=\frac{a\left(\left|x\right|\right)}{\left|x\right|}\left(-x_{2},x_{1}\right)$,
with $a\in L^{\infty}\left(\left(0,R\right)\right)$, we consider the ground-
state energy of the operator
$\left(-i\nabla-\tilde{\alpha}\right)^{2}-\text{rot}\left(\tilde{\alpha}\right)$
restricted to radial functions on the disk, again written more conveniently in
terms of polar coordinates

(4.1) $\mathfrak{e}\left(a(r)\right):=\inf_{q\in
H_{0}^{1,\text{rad}}\left(D_{R}\right)}\frac{2\pi\int_{0}^{R}\left(q^{\prime}(r)+a(r)q(r)\right)^{2}rdr}{2\pi\int_{0}^{R}q(r)^{2}rdr},$

where $H_{0}^{1,\text{rad}}(D_{R}):=\left\\{q:[0,R]\rightarrow\mathbb{R}\
\text{such that}\ x\mapsto q(|x|)\ \text{belongs to}\
H_{0}^{1}(D_{R})\right\\}.$

The function $a(r)=\frac{Br}{2}$ corresponds to the original potential
$\alpha=\frac{B}{2}\left(-x_{2},x_{1}\right)$, and since
$\text{rot}\left(\alpha\right)=B$,

(4.2) $B+\mathfrak{e}\left(Br/2\right)=\inf_{q\in
H_{0}^{1,\text{rad}}\left(D_{R}\right)}\frac{\int_{D_{R}}\left|\left(-i\nabla-\alpha\right)q(\left|x\right|)\right|^{2}dx}{\int_{D_{R}}q\left(\left|x\right|\right)^{2}dx}\geq\lambda(B,D_{R}).$

We compare the ground-state energies for different potentials on the disk.

######  Lemma 4.1.

Let $q_{a}$ be a normalized minimizer for the energy
$\mathfrak{e}\left(a(r)\right)$ in (4.1). Let

(4.3) $u_{a}(r):=\exp\left(-2\int_{0}^{r}a(s)\,ds\right)\ \ \text{and\ }\
p_{a}(r):=q_{a}(r)u_{a}(r)^{-\frac{1}{2}}.$

Then for $a,\tilde{a}\in L^{\infty}\left(\left(0,R\right)\right)$,

(4.4)
$\mathfrak{e}\left(a(r)\right)\geq\mathfrak{e}\left(\tilde{a}(r)\right)+\frac{2\int_{0}^{R}\left(\tilde{a}-a\right)p_{a}\left|p_{a}^{\prime}\right|u_{\tilde{a}}rdr}{\int_{0}^{R}p_{a}^{2}u_{\tilde{a}}rdr}.$

######  Remark 4.2.

Our bound in (4.4) implies Erdős’ comparison lemma: if $a\leq\tilde{a}$, then
$\mathfrak{e}\left(a(r)\right)\geq\mathfrak{e}\left(\tilde{a}\left(r\right)\right)$.
See Lemma 3.1 in [10].

###### Proof.

We write

(4.5) $\mathfrak{e}\left(a(r)\right)=\inf_{p\in
H_{0}^{1,\text{rad}}\left(D_{R}\right)}\frac{\int_{0}^{R}\left(p^{\prime}\right)^{2}u_{a}rdr}{\int_{0}^{R}p^{2}u_{a}rdr}=\frac{\int_{0}^{R}\left(p_{a}^{\prime}\right)^{2}u_{a}rdr}{\int_{0}^{R}p_{a}^{2}u_{a}rdr}.$

Since $p_{a}$ is the minimizer in (4.5), it solves the Euler-Lagrange equation

(4.6)
$-p_{a}^{\prime\prime}u_{a}r-p_{a}^{\prime}u_{a}^{\prime}r-p_{a}^{\prime}u_{a}=\mathfrak{e}\left(a(r)\right)p_{a}u_{a}r.$

Now we consider $\mathfrak{e}\left(\tilde{a}(r)\right)$. It follows from the
variational principle and (4.6) that

$\displaystyle\mathfrak{e}\left(\tilde{a}(r)\right)$
$\displaystyle\leq\frac{\int_{0}^{R}(p_{a}^{\prime})^{2}u_{\tilde{a}}rdr}{\int_{0}^{R}p_{a}^{2}u_{\tilde{a}}rdr}=\frac{\int_{0}^{R}(-p_{a}^{\prime\prime}u_{a}r-p_{a}^{\prime}u_{a}^{\prime}r-p_{a}^{\prime}u_{a})\frac{u_{\tilde{a}}}{u_{a}}p_{a}-p_{a}^{\prime}p_{a}u_{a}r(\frac{u_{\tilde{a}}}{u_{a}})^{\prime}dr}{\int_{0}^{R}p_{a}^{2}u_{\tilde{a}}rdr}$
$\displaystyle=\mathfrak{e}\left(a(r)\right)+\frac{2\int_{0}^{R}p_{a}^{\prime}p_{a}(\tilde{a}-a)u_{\tilde{a}}rdr}{\int_{0}^{R}p_{a}^{2}u_{\tilde{a}}rdr}.$

Note that $p_{a}^{\prime}<0$ by Hopf’s Lemma. ∎

Proposition 3.2, Lemma 4.1 and the observation in (4.2) allow us to conclude
with the following corollary.

######  Corollary 4.3.

Now let $f$ be a principal eigenfunction corresponding to $\lambda(B,\Omega)$
and $q\left(\left|x\right|\right):=\left|f\right|^{*}\left(x\right)$. Let
$a(r)$ be as in Proposition 3.2 above, and let $q_{a}$ be a normalized
minimizer for the energy $\mathfrak{e}\left(a(r)\right)$ in (4.1). Then there
is a universal constant $c>0$, independent of $B$ and $\Omega$, such that

(4.7)
$\lambda(B,\Omega)\geq\lambda(B,D_{R})+c\int_{0}^{R}\left(q^{\prime}(r)+a(r)q(r)\right)^{2}\mathcal{A}^{2}\left(\left\\{\left|f\right|>q(r)\right\\}\right)rdr,$

and

(4.8)
$\lambda(B,\Omega)\geq\lambda(B,D_{R})+cB\frac{\int_{0}^{R}p_{a}\left|p_{a}^{\prime}\right|e^{-\frac{Br^{2}}{2}}\mathcal{A}^{2}\left(\left\\{\left|f\right|>q(r)\right\\}\right)r^{2}dr}{\int_{0}^{R}p_{a}^{2}e^{-\frac{Br^{2}}{2}}rdr},$

where $p_{a}$ is as given in Lemma 4.1 above.

Corollary 4.3 implies $\lambda(B,\Omega)\geq\lambda(B,D_{R})$. Furthermore if
$\lambda(B,\Omega)=\lambda(B,D_{R})$, then either (4.7) or (4.8) can be used
to deduce that almost all of the level sets of $\left|f\right|$ are disks; and
since $f$ is an analytic function, this implies $\Omega$ is a disk.

The first bound, given in (4.7), is established with our quantitative version
of the rearrangement inequality and with Erdős’ comparison lemma. In the
absence of a magnetic field, i.e. $B=0$, this bound reduces to the usual
estimate used in all the proofs of the quantitative Faber-Krahn inequality,
e.g., [3], [14] and [15].

Our second bound, given in (4.8), is established with Erdős’ rearrangement
inequality, our quantitative version of the comparison lemma and our estimate
in (3.2), which follows from the quantitative isoperimetric inequality. This
bound, on the other hand, has no such analog in the absence of a magnetic
field.

##  Part II The Quantitative Version

Here we prove Theorem 2.1 from Corollary 4.3 by extracting the asymmetry of
the whole domain from the asymmetry of the level sets in (4.7) and (4.8). Let

(4.9)
$\left|\left\\{q\left(\left|x\right|\right)>s\right\\}\right|=\left|\Omega\right|\left(1-\frac{1}{2}\mathcal{A}\left(\Omega\right)\right).$

Following Hansen and Nadirashvili [15] we split the proof into two cases,
depending on whether $s$ is small or large. Lemma B.1 in the appendix will be
useful.

###  5\. The First Case: $s\lesssim e^{-BR^{2}}\mathcal{A}\left(\Omega\right)$

We assume

(5.1)
$s\leq\frac{1}{8}\left|\Omega\right|^{-\frac{1}{2}}e^{-\frac{BR^{2}}{4}}\mathcal{A}\left(\Omega\right).$

We use the representation in (4.5), which allows us to adapt the usual
strategy for dealing with the Dirichlet Laplacian; and when $B=0$, the
argument reduces to Hansen and Nadirashvili’s proof of their quantitative
Faber-Krahn inequality [15].

We write $E(B,\Omega):=\lambda(B,\Omega)-B$. Let $p:=qu_{a}^{-\frac{1}{2}}$
with $q,a$ as in Corollary 4.3 and $u_{a}$ as in (4.3), and let
$\tilde{p}(r):=p(r)-se^{\int_{0}^{q^{-1}(s)}a(\tau)d\tau}$. Since
$\tilde{p}^{\prime}=p^{\prime}$, it follows from the rearrangement inequality
that

$E(B,\Omega)\geq
2\pi\int_{0}^{R}\left(q^{\prime}+aq\right)^{2}rdr=2\pi\int_{0}^{R}\left(\tilde{p}^{\prime}\right)^{2}u_{a}rdr\geq
2\pi\int_{0}^{q^{-1}(s)}\left(\tilde{p}^{\prime}\right)^{2}u_{a}rdr.$

Since $\tilde{p}$ vanishes at $q^{-1}(s)$, it is admissible in the variational
problem in (4.5) but on the disk $\left\\{q>s\right\\}$, and

$\frac{E(B,\Omega)}{2\pi\int_{0}^{q^{-1}(s)}\tilde{p}^{2}u_{a}rdr}\geq\inf_{p\in
H_{0}^{1,\text{rad}}\left(\left\\{q>s\right\\}\right)}\frac{\int_{0}^{q^{-1}(s)}\left(p^{\prime}\right)^{2}u_{a}rdr}{\int_{0}^{q^{-1}(s)}p^{2}u_{a}rdr}\geq
E\left(B,\left\\{q>s\right\\}\right),$

where the last inequality follows from the comparison lemma and the
observation in (4.2). Using the scaling property in Remark 2.2 we further
estimate

(5.2)
$\frac{E(B,\Omega)}{2\pi\int_{0}^{q^{-1}(s)}\tilde{p}^{2}u_{a}rdr}\geq\frac{\left|\Omega\right|}{\left|\left\\{q>s\right\\}\right|}E\left(B\frac{\left|\left\\{q>s\right\\}\right|}{\left|\Omega\right|},D_{R}\right)\geq\frac{\left|\Omega\right|}{\left|\left\\{q>s\right\\}\right|}E(B,D_{R}),$

where the last inequality follows from Lemma A.2 in the appendix and again the
comparison lemma. Finally, we estimate the denominator

$2\pi\int_{0}^{q^{-1}(s)}\tilde{p}^{2}u_{a}rdr=1-2\pi\int_{q^{-1}(s)}^{R}q^{2}rdr\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \
+2\pi\int_{0}^{q^{-1}(s)}\left((se^{\int_{0}^{q^{-1}(s)}a(\tau)d\tau})^{2}-2pse^{\int_{0}^{q^{-1}(s)}a(\tau)d\tau}\right)u_{a}rdr$
$\ \ \ \geq
1-s^{2}|\\{q<s\\}|+s^{2}|\\{q>s\\}|-2se^{\frac{BR^{2}}{4}}|\Omega|^{\frac{1}{2}}$
$\geq 1-2se^{\frac{BR^{2}}{4}}|\Omega|^{\frac{1}{2}}.\ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

At the penultimate inequality we used that
$e^{\int_{0}^{q^{-1}(s)}a(\tau)d\tau}\leq e^{\frac{BR^{2}}{4}}$ and that

$2\pi\int_{0}^{q^{-1(s)}}pu_{a}rdr\leq 2\pi\int_{0}^{R}qrdr\leq
2\pi|\Omega|^{\frac{1}{2}}\int_{0}^{R}q^{2}rdr=|\Omega|^{\frac{1}{2}}.$

Combining the above estimate with (5.2), we have

$E(B,\Omega)\geq
E(B,D_{R})\frac{|\Omega|(1-2se^{\frac{BR^{2}}{4}}|\Omega|^{\frac{1}{2}})}{|\\{q>s\\}|},$

and the choice of $s$ in (4.9) and our assumption in (5.1) give us

$E(B,\Omega)\geq
E(B,D_{R})\frac{1-\frac{1}{4}\mathcal{A}(\Omega)}{1-\frac{1}{2}\mathcal{A}(\Omega)}\geq
E(B,D_{R})\Big{(}1+\frac{1}{4}\mathcal{A}(\Omega)\Big{)}.$

Then using Lemma A.3 in the appendix we find

$\lambda(B,\Omega)\geq\lambda(B,D_{R})\left(1+c\min(1,(BR^{2})^{-1}e^{-\frac{3}{4}BR^{2}})\mathcal{A}(\Omega)\right),$

which yields the desired estimates in (2.3) and (2.4). This concludes the
proof of Theorem 2.1 in the first case.

###  6\. The Second Case: $s\gtrsim e^{-BR^{2}}\mathcal{A}\left(\Omega\right)$

We assume

(6.1)
$s\geq\frac{1}{8}\left|\Omega\right|^{-\frac{1}{2}}e^{-\frac{BR^{2}}{4}}\mathcal{A}\left(\Omega\right).$

Now we have to treat weak and strong magnetic fields separately. When $B$ is
small, we only use the first bound, given in (4.7), of Corollary 4.3. As $B$
increases, it becomes necessary to also make use of our second bound in (4.8).

####  6.1. Weak Magnetic Fields

We consider $0\leq BR^{2}\leq\frac{1}{\pi}$ and prove the stability estimate
in (2.4); and when $B=0$, the argument reduces to Hansen and Nadirashvili’s
proof of their quantitative Faber-Krahn inequality [15].

We work on the annulus $\left\\{q\left(\left|x\right|\right)\leq s\right\\}$,
whose area is proportional to the asymmetry of the domain. From the first
bound, given in (4.7), of Corollary 4.3, the choice of $s$ in (4.9), and Lemma
B.1 we have

$\lambda(B,\Omega)-\lambda(B,D_{R})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
$\geq
c\int_{q^{-1}(s)}^{R}\left(q^{\prime}(r)+a(r)q(r)\right)^{2}\mathcal{A}^{2}\left(\left\\{\left|f\right|>q(r)\right\\}\right)rdr\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ $ $\geq
cR^{2}\mathcal{A}^{2}\left(\Omega\right)\int_{q^{-1}(s)}^{R}\left(q^{\prime}(r)+a(r)q(r)\right)^{2}r^{-1}dr\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ $\ \ \geq
cR^{2}\mathcal{A}^{2}\left(\Omega\right)\left(\sqrt{\int_{q^{-1}(s)}^{R}q^{\prime}(r)^{2}\,r^{-1}dr}-\sqrt{\int_{q^{-1}(s)}^{R}\left(\frac{B}{2}q\right)^{2}rdr}\right)^{2}$
$\geq
cR^{2}\mathcal{A}^{2}\left(\Omega\right)\left(\frac{s}{\sqrt{\left|\left\\{q\left(\left|x\right|\right)\leq
s\right\\}\right|}}-\frac{B}{2}s\sqrt{\left|\left\\{q\left(\left|x\right|\right)\leq
s\right\\}\right|}\right)^{2}\ \ \ \ \ \ \ $ $\geq
cR^{-2}\mathcal{A}^{3}\left(\Omega\right)\left(2-B\left|\Omega\right|\right)^{2}\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ $ $\geq cR^{-2}\mathcal{A}^{3}\left(\Omega\right),\ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ $

since $B\leq\frac{1}{\left|\Omega\right|}=\frac{1}{\pi R^{2}}$. At the
penultimate inequality we also used the assumption in (6.1). Using Lemma A.3,
we conclude
$\lambda(B,\Omega)\geq\lambda(B,D_{R})(1+c\mathcal{A}(\Omega)^{3})$.

####  6.2. Strong Magnetic Fields

We consider $BR^{2}>\frac{1}{\pi}$ and prove our stability estimate in (2.3);
instead of integrating as above on $\left\\{q\left(\left|x\right|\right)\leq
s\right\\}$, we choose to work closer to the boundary on a smaller annulus
whose area is now proportional to the spectral deficit of the domain

$\mathcal{D}(B,\Omega):=\frac{\lambda(B,\Omega)}{\lambda(B,D_{R})}-1.$

We treat two cases, depending on whether $q$ is large or small near the
boundary:
$q(R(1-\mathcal{D}\left(B,\Omega\right)^{\alpha}))>R^{-1}\mathcal{D}(B,\Omega)^{\beta}$
and $q(R(1-\mathcal{D}\left(B,\Omega\right)^{\alpha}))\leq
R^{-1}\mathcal{D}(B,\Omega)^{\beta}$, where $\alpha=\frac{1}{5}$ and
$\beta=\frac{3}{10}$ are chosen to optimize our result. For proving our
estimate in (2.3), we can assume that the spectral deficit is very small

(6.2)
$\mathcal{D}(B,\Omega)^{\alpha}<\min\left\\{\frac{1}{2BR^{2}},\frac{1}{2}\right\\}.$

#####  6.2.1.  Suppose
$q(R(1-\mathcal{D}\left(B,\Omega\right)^{\alpha}))>R^{-1}\mathcal{D}(B,\Omega)^{\beta}$.

Then by continuity of $q$,

(6.3)
$q(R(1-\mathcal{D}\left(B,\Omega\right)^{\tilde{\alpha}}))=R^{-1}\mathcal{D}(B,\Omega)^{\beta}\
\text{for some}\ \tilde{\alpha}>\alpha.$

If $q(R(1-\mathcal{D}(B,\Omega)^{\tilde{\alpha}}))\geq s$, our assumption in
(6.1) readily yields

$cR^{-1}e^{-\frac{BR^{2}}{4}}\mathcal{A}\left(\Omega\right)\leq s\leq
q(R(1-\mathcal{D}\left(B,\Omega\right)^{\tilde{\alpha}}))=R^{-1}\mathcal{D}(B,\Omega)^{\beta},$

and therefore

(6.4) $\mathcal{D}(B,\Omega)\geq
ce^{-\frac{BR^{2}}{4\beta}}\mathcal{A}\left(\Omega\right)^{\frac{1}{\beta}}.$

If $q(R(1-\mathcal{D}(B,\Omega)^{\tilde{\alpha}}))<s$, then the weak-field
argument from Section 6.1 applies mutatis mutandis. From the first bound,
given in (4.7), of Corollary 4.3, the relation in (6.3), and Lemma B.1 we have

$\displaystyle\lambda(B,D_{R})\mathcal{D}\left(B,\Omega\right)$
$\displaystyle\geq
c\int_{R\left(1-\mathcal{D}(B,\Omega)^{\tilde{\alpha}}\right)}^{R}\left(q^{\prime}(r)+a(r)q(r)\right)^{2}\mathcal{A}^{2}\left(\left\\{\left|f\right|>q(r)\right\\}\right)rdr$
$\displaystyle\geq
cR^{2}\mathcal{A}(\Omega)^{2}\left(\frac{q\left(R\left(1-\mathcal{D}(B,\Omega)^{\tilde{\alpha}}\right)\right)}{\sqrt{2R^{2}\mathcal{D}(B,\Omega)^{\tilde{\alpha}}}}-\frac{B}{2}q\left(R\left(1-\mathcal{D}(B,\Omega)^{\tilde{\alpha}}\right)\right)\sqrt{2R^{2}\mathcal{D}(B,\Omega)^{\tilde{\alpha}}}\right)^{2}$
$\displaystyle=cR^{-2}\mathcal{A}(\Omega)^{2}\mathcal{D}\left(B,\Omega\right)^{2\beta-\tilde{\alpha}}\left(1-BR^{2}\mathcal{D}(B,\Omega)^{\tilde{\alpha}}\right)^{2}.$

However, $\tilde{\alpha}$ depends on $B$ and $\Omega$. Fortunately since
$\tilde{\alpha}>\alpha$ and $\mathcal{D}\left(B,\Omega\right)<1$, we have
$\mathcal{D}\left(B,\Omega\right)^{\tilde{\alpha}}<\mathcal{D}\left(B,\Omega\right)^{\alpha}$;
this allows to replace $\mathcal{D}\left(B,\Omega\right)^{\tilde{\alpha}}$ in
the above with $\mathcal{D}\left(B,\Omega\right)^{\alpha}$. Furthermore, the
bound in (6.2) offsets the large $BR^{2}$ in the parenthetical expression,
which thereby remains positive. Using Lemma A.3,

$\mathcal{D}\left(B,\Omega\right)\geq
c\frac{\mathcal{A}(\Omega)^{2}}{R^{2}\lambda(B,D_{R})}\mathcal{D}\left(B,\Omega\right)^{2\beta-\alpha}\geq
c\frac{\mathcal{A}(\Omega)^{2}}{1+BR^{2}}\mathcal{D}\left(B,\Omega\right)^{2\beta-\alpha},$

and therefore

(6.5) $\mathcal{D}\left(B,\Omega\right)^{1-2\beta+\alpha}\geq
c\frac{\mathcal{A}(\Omega)^{2}}{1+BR^{2}}.$

With our above choice of $\alpha$ and $\beta$, the inequalities in (6.4) and
(6.5) both yield the same desired estimate in (2.3).

Thus far, we have only used the first bound, given in (4.7), of Corollary 4.3
which is based on the quantitative version of the rearrangement inequality.

#####  6.2.2. Suppose $q(R(1-\mathcal{D}\left(B,\Omega\right)^{\alpha}))\leq
R^{-1}\mathcal{D}(B,\Omega)^{\beta}$.

If $q(R(1-\mathcal{D}(B,\Omega)^{\alpha}))\geq s,\ \ $ again our assumption in
(6.1) readily yields

$cR^{-1}e^{-\frac{BR^{2}}{4}}\mathcal{A}\left(\Omega\right)\leq s\leq
q(R(1-\mathcal{D}\left(B,\Omega\right)^{\alpha}))\leq
R^{-1}\mathcal{D}(B,\Omega)^{\beta}$

and therefore, as above,

(6.6) $\mathcal{D}(B,\Omega)\geq
ce^{-\frac{BR^{2}}{4\beta}}\mathcal{A}\left(\Omega\right)^{\frac{1}{\beta}}.$

But when $q(R(1-\mathcal{D}(B,\Omega)^{\alpha}))<s$, the weak-field argument
from Section 6.1 is no longer useful: it requires a lower bound on
$q(R(1-\mathcal{D}(B,\Omega)^{\alpha}))$, as above in Section 6.2.1, to be
effective. That argument, however, is based wholly on the first bound, given
in (4.7), of Corollary 4.3.

Now we instead turn to our second bound, given in (4.8), which is based on our
quantitative version of the comparison lemma. Here there is hope: it is
possible to bound the remainder term in (4.8) from below independently of $q$.

######  Lemma 6.1.

Let $p_{a}$ be as in Corollary 4.3. Then there exists a universal constant
$c>0$, independent of $B$ and $\Omega$, such that for any
$\,0<\varepsilon<\frac{1}{2}$

$\frac{\int_{R\left(1-\varepsilon\right)}^{R}\,p_{a}\left|p_{a}^{\prime}\right|e^{-\frac{Br^{2}}{2}}\mathcal{A}^{2}\left(\left\\{\left|f\right|>q(r)\right\\}\right)r^{2}dr}{\int_{0}^{R}p_{a}^{2}e^{-\frac{Br^{2}}{2}}rdr}\geq
ce^{-\frac{BR^{2}}{2}}\mathcal{M}_{\varepsilon}\varepsilon^{2},$

where
$\mathcal{M}_{\varepsilon}:=\inf\left\\{\mathcal{A}^{2}\left(\left\\{\left|f\right|>q(r)\right\\}\right):\,R\left(1-\varepsilon\right)<r<R\right\\}$.

###### Proof.

Since $p_{a}^{\prime}<0$,

$\displaystyle\int_{R\left(1-\varepsilon\right)}^{R}\,p_{a}\left|p_{a}^{\prime}\right|e^{-\frac{Br^{2}}{2}}\mathcal{A}^{2}\left(\left\\{\left|f\right|>q(r)\right\\}\right)r^{2}dr$
$\displaystyle\geq
cM_{\varepsilon}R^{2}e^{-\frac{BR^{2}}{2}}\int_{R(1-\varepsilon)}^{R}-p_{a}(r)p_{a}^{\prime}(r)dr=cM_{\varepsilon}R^{2}e^{-\frac{BR^{2}}{2}}p_{a}(R(1-\varepsilon))^{2}.$

Furthermore,

$p_{a}(R(1-\varepsilon))=\int_{R(1-\varepsilon)}^{R}-p_{a}^{\prime}(r)dr\geq\frac{1}{R}\int_{R(1-\varepsilon)}^{R}-p_{a}^{\prime}(r)rdr\geq\frac{\varepsilon}{R}\int_{0}^{R}-p_{a}^{\prime}(r)rdr,$

where in the last inequality we used that $r\mapsto-p_{a}^{\prime}(r)r$ is
increasing (see (4.6)). The lemma follows from the Sobolev inequality

$\ \ \ \ \ \ \ \ \ \ \ \ \ \int_{0}^{R}-p_{a}^{\prime}(r)rdr\geq
c\left(\int_{0}^{R}p_{a}^{2}(r)rdr\right)^{\frac{1}{2}}\geq
c\left(\int_{0}^{R}p_{a}^{2}(r)\,e^{-\frac{Br^{2}}{2}}rdr\right)^{\frac{1}{2}}.$
∎

Before proceeding with our argument, we remark that Lemma 6.1 would not have
been useful for dealing with the previous situation in Section 6.2.1.

If $q(R(1-\mathcal{D}(B,\Omega)^{\alpha}))<s$, then we use the above lemma
with $\varepsilon=\mathcal{D}\left(B,\Omega\right)^{\alpha}$. From our second
bound, given in (4.8), of Corollary 4.3, Lemma 6.1, and Lemma B.1 we have

$\lambda(B,D_{R})\mathcal{D}\left(B,\Omega\right)\geq
cBe^{-\frac{BR^{2}}{2}}\mathcal{A}\left(\Omega\right)^{2}\mathcal{D}\left(B,\Omega\right)^{2\alpha}.$

Again using Lemma A.3 and now that $BR^{2}>\frac{1}{\pi}$,

$\mathcal{D}\left(B,\Omega\right)\geq
c\frac{e^{-\frac{BR^{2}}{2}}}{1+\left(BR^{2}\right)^{-1}}\mathcal{A}\left(\Omega\right)^{2}\mathcal{D}\left(B,\Omega\right)^{2\alpha}\geq
ce^{-\frac{BR^{2}}{2}}\mathcal{A}\left(\Omega\right)^{2}\mathcal{D}\left(B,\Omega\right)^{2\alpha},$

and therefore

(6.7) $\mathcal{D}\left(B,\Omega\right)^{1-2\alpha}\geq
ce^{-\frac{BR^{2}}{2}}\mathcal{A}\left(\Omega\right)^{2}.$

With our above choice of $\alpha$ and $\beta$, the inequalities in (6.6) and
(6.7) both yield the same desired estimate in (2.3). This concludes the proof
of Theorem 2.1.

### Acknowledgements

We are most grateful to Søren Fournais for encouraging our collaboration. R.G.
first suggested this problem in October 2017 to Michael Loss, whom he thanks
for the initial encouragement. This paper is based on work partially supported
by the Independent Research Fund Denmark via the project grant “Mathematics of
the dilute Bose gas” No. 0135-00166B (L.J. & L.M.).

###  Appendix A The Magnetic Laplacian on the Disk

It follows from Erdős’ rearrangement inequality and comparison lemma, and from
the observation in (4.2) that the principal eigenfunction of the magnetic
Laplacian on the disk is radially symmetric.

######  Theorem A.1.

As above, let $D_{R}$ be a disk of radius $R$ centered at the origin. Then

$\lambda(B,D_{R})=\inf_{q\in
H_{0}^{1,\text{rad}}\left(D_{R}\right)}\frac{\int_{D_{R}}\left|\left(-i\nabla-\alpha\right)q(\left|x\right|)\right|^{2}dx}{\int_{D_{R}}q\left(\left|x\right|\right)^{2}dx},$

where $H_{0}^{1,\text{rad}}(D_{R}):=\left\\{q:[0,R]\rightarrow\mathbb{R}\
\text{such that}\ x\mapsto q(|x|)\ \text{belongs to}\
H_{0}^{1}(D_{R})\right\\}.$

Thus we write $\lambda(B,D_{R})$ more conveniently in terms of polar
coordinates.

######  Lemma A.2.

Let $H_{0}^{1,\text{rad}}(D_{R})$ be as in Theorem A.1. Then

$\lambda(B,D_{R})=B+\inf_{q\in
H_{0}^{1,\text{rad}}(D_{R})}\frac{2\pi\int_{0}^{R}(q^{\prime}(r)+\frac{Br}{2}q(r))^{2}rdr}{2\pi\int_{0}^{R}q(r)^{2}rdr}=:B+\mathfrak{e}\left(Br/2\right).$

###### Proof.

First we consider a broader class of vector potentials
$\tilde{\alpha}(x):=\frac{a\left(\left|x\right|\right)}{\left|x\right|}\left(-x_{2},x_{1}\right)$
on the disk, with $a\left(\left|x\right|\right)$ bounded. These correspond to
radially symmetric but possibly inhomogeneous magnetic fields that show up in
the rearrangement inequality. Written in polar coordinates,
$\tilde{\alpha}(r,\theta)=a(r)\left(-\sin\theta,\cos\theta\right)$ and for
$f\in H_{0}^{1}(D_{R})$

$\int_{D_{R}}\left|\left(-i\nabla-\tilde{\alpha}\right)f\right|^{2}dx=\int_{0}^{R}\int_{0}^{2\pi}\left(|\partial_{r}f|^{2}+|\frac{i}{r}\partial_{\theta}f+af|^{2}\right)rd\theta
dr.$

Thus for any $q\in H_{0}^{1,\text{rad}}(D_{R})$,

$\displaystyle\int_{D_{R}}\left|\left(-i\nabla-\tilde{\alpha}\right)q(\left|x\right|)\right|^{2}dx$
$\displaystyle=2\pi\int_{0}^{R}\left((q^{\prime}(r)^{2}+\left(a(r)q(r)\right)^{2}\right)rdr$
$\displaystyle=2\pi\int_{0}^{R}\left(q^{\prime}(r)+a(r)q(r)\right)^{2}rdr-2\pi\int_{0}^{R}\left(q^{2}\right)^{\prime}a(r)rdr,$

and after integrating by parts

$\displaystyle\int_{D_{R}}\left|\left(-i\nabla-\tilde{\alpha}\right)q(\left|x\right|)\right|^{2}dx$
$\displaystyle=2\pi\int_{0}^{R}\left(q^{\prime}(r)+a(r)q(r)\right)^{2}rdr+2\pi\int_{0}^{R}q^{2}\left(a(r)r\right)^{\prime}dr$
$\displaystyle=2\pi\int_{0}^{R}\left(q^{\prime}(r)+a(r)q(r)\right)^{2}rdr+\int_{D_{R}}\text{rot}\left(\tilde{\alpha}\right)q(\left|x\right|)^{2}dx.$

Returning to the original potential
$\alpha=\frac{B}{2}\left(-x_{2},x_{1}\right)$, the lemma follows from Theorem
A.1, the above calculation and that $\text{rot}\left(\alpha\right)=B$. ∎

Moreover, Erdős proved the following estimates. See Proposition A.1 in [10].

######  Lemma A.3.

There are universal constants $C_{1},C_{2}$ such that

$B+\frac{C_{1}}{R^{2}}e^{-\frac{3}{4}BR^{2}}\leq\lambda\left(B,D_{R}\right)\leq
B+C_{2}B\left(\frac{1}{BR^{2}}+BR^{2}\right)e^{-\frac{1}{8}BR^{2}}.$

Improving these estimates is an ongoing area of research [2],[9],[16] & ref.
therein. In the absence of a magnetic field,
$\lambda(0,D_{R})=j_{0,1}^{2}R^{-2}$ where $j_{0,1}\approx 2.4048$ is the
first zero of the Bessel function of order zero.

###  Appendix B Asymmetry of Large Subsets

If a subset is large enough, its asymmetry is comparable to the asymmetry of
the whole domain [7],[15].

######  Lemma B.1.

Let $U\subseteq\Omega$ with $\left|U\right|=\pi r^{2}$ and
$\left|\Omega\right|=\pi R^{2}$. If
$\left|U\right|\geq\left|\Omega\right|\left(1-\frac{1}{2}\mathcal{A}\left(\Omega\right)\right)$,
then
$r\mathcal{A}\left(U\right)\geq\frac{1}{2}R\mathcal{A}\left(\Omega\right)$.

###### Proof.

First we consider the interior asymmetry. From our assumption on the area of
$U$, we have
$\left|U\right|\geq\left|\Omega\right|\left(1-\frac{1}{2}\mathcal{A}_{I}\left(\Omega\right)\right)^{2}$
and thus $r\geq
R\left(1-\frac{1}{2}\mathcal{A}_{I}\left(\Omega\right)\right)$. We then deduce
that $r-\rho_{-}(U)\geq
r-\rho_{-}(\Omega)\geq\frac{1}{2}\left(R-\rho_{-}(\Omega)\right)$, which
yields the lemma.

Now we turn to the Fraenkel asymmetry. Let $D_{U}$ and $D_{\Omega}$ denote two
concentric balls such that $\left|D_{U}\right|=\left|U\right|$ and
$\left|D_{\Omega}\right|=\left|\Omega\right|$. Then,
$\left|D_{\Omega}\triangle\Omega\right|\leq\left|D_{U}\triangle
U\right|+2\left(\left|\Omega\right|-\left|U\right|\right).$ Using this
inequality and our assumption on the area of $U$, we deduce

$\frac{|D_{U}\Delta
U|}{2|U|}\geq\frac{|D_{\Omega}\Delta\Omega|}{2|U|}-\frac{|\Omega|-|U|}{|U|}\geq\frac{1}{2}\mathcal{A}_{F}(\Omega)\frac{|\Omega|}{|U|}\geq\frac{1}{2}\frac{R}{r}\mathcal{A}_{F}(\Omega).$

Taking the infimum over all translations of $D_{U}$ concludes the proof. ∎

### References

  * [1] Avron, J.E., I.W. Herbst and B. Simon. Schrödinger operators with magnetic fields. III. Atoms in homogeneous magnetic field. Comm. Math. Phys. 79(4), 529-572 (1981).
  * [2] Barbaroux, J.-M, L. Le Treust, N. Raymond and E. Stockmeyer. On the semiclassical spectrum of the Dirichlet-Pauli operator. J. Eur. Math. Soc. 23(10), 3279-3321 (2021).
  * [3] Bhattacharya, T. Some observations on the first eigenvalue of the $p-$Laplacian and its connections with asymmetry. Electron. J. Differ. Equat. 35, 1-15 (2001).
  * [4] Bhattacharya, T. and A. Weitsman. Estimates for Green’s function in terms of asymmetry. AMS Contemporary Math. Series 221, 31-58 (1999).
  * [5] Bonheure, D., J. Dolbeault, M.J. Esteban, A. Laptev and M. Loss. Symmetry Results in Two-Dimensional Inequalities for Aharonov-Bohm Magnetic Fields. Comm. Math. Phys. 375, 2071-2087 (2020).
  * [6] Bonnesen, T. Über die isoperimetrische Defizit ebener Figuren. Math. Ann. 91, 252-268 (1924).
  * [7] Brasco, L. and G. de Philipis. “Spectral inequalities in quantitative form.” Shape Optimization and Spectral Theory, edited by Antoine Henrot, De Gruyter Open, Warsaw, 2017, 201-281.
  * [8] Brasco, L., G. de Phillipis and B. Velichkov. Faber-Krahn inequalities in sharp quantitative form. Duke Math. J. 164(9), 1777-1831 (2015).
  * [9] Ekholm, T., H. Kovařík and F. Portmann. Estimates for the lowest eigenvalue of magnetic Laplacians. J. Math. Anal. Appl. 439(1), 330-446 (2016).
  * [10] Erdös, L. Rayleigh-type isoperimetric inequality with a homogeneous magnetic field. Calc. Var. Partial Differ. Equ. 4, 283-292 (1996).
  * [11] Erdős, L. Lifschitz tail in a magnetic field: the nonclassical regime. Probab. Theory Relat. Fields 112, 321-371 (1998).
  * [12] Erdős, L. Lifschitz tail in a magnetic field: coexistence of classical and quantum behavior in the borderline case. Probab. Theory Relat. Fields 121, 219-236 (2001).
  * [13] Fusco, N., F. Maggi and A. Pratelli. The sharp quantitative isoperimetric inequality. Ann. of Math. 168(3), 941-980 (2008).
  * [14] Fusco, N., F. Maggi and A. Pratelli. Stability estimates for certain Faber-Krahn, isocapacitory and Cheeger inequalities. Ann. Scuola Norm. Sup. Pisa Cl. Sci. VIII(5), 51-71 (2009).
  * [15] Hansen, W. and N. Nadirashvili. Isoperimetric inequalities in potential theory. Potential Analysis 3, 1-14 (1994).
  * [16] Helffer, B. and A. Morame. Magnetic Bottles in Connection with Superconductivity. J. Funct. Anal. 185(2), 604-680 (2001).
  * [17] Nadirashvili, N. Conformal maps and isoperimetric inequalities for eigenvalues of the Neumann problem. Proceedings of the Ashkelon Workshop of Complex Function Theory, 197-201, Israel Math. Conf. Proc. 11, Bar-Ilan Univ., Ramat Gan (1997).
  * [18] Osserman, R. Bonnesen-Style Isoperimetric Inequalities. Amer. Math. Monthly 86(1), 1-29 (1979).
  * [19] Pólya, G. and G. Szegő. Isoperimetric Inequalities in Mathematical Physics. Annals of Mathematics Studies 27, Princeton University Press, Princeton (1951).
  * [20] Talenti, G. Best constant in Sobolev inequality. Ann. Mat. Pura. Appl. 110, 353-372 (1976).