# A note on Hölder regularity of weak solutions to linear elliptic equations

Karthik Adimurthi111Supported by the Department of Atomic Energy, Government
of India, under project no. 12-R&D-TFR-5.01-0520<EMAIL_ADDRESS>and
<EMAIL_ADDRESS>Tata Institute of Fundamental Research, Centre for
Applicable Mathematics,Bangalore, Karnataka, 560065, India

###### Abstract

In this paper, we show that weak solutions of

$\displaystyle-\operatorname{div}\mathbb{A}(x)\nabla
u=0\qquad\text{where}\qquad
A(x)=A(x)^{T}\,\,\text{and}\,\,\lambda|\zeta|^{2}\leq\left\langle
A(x)\zeta,\zeta\right\rangle\leq\Lambda|\zeta|^{2},$

and $\displaystyle\mathbb{A}(x)\equiv\mathbb{A}$ is a constant matrix are
Hölder continuous $\displaystyle u\in C^{\alpha}_{\operatorname{loc}}$ with
$\displaystyle\alpha\geq\tfrac{1}{2}\left(-(n-2)+\sqrt{(n-2)^{2}+\tfrac{4(n-1)\lambda}{\Lambda}}\right)$.
This implies that the example constructed by Piccinini - Spagnolo is sharp in
the class of constant matrices $\displaystyle\mathbb{A}(x)\equiv\mathbb{A}$.
The proof of Hölder regularity does not go through a reduction of oscillation
type argument and instead is achieved through a monotonicity formula.

In the case of general matrices $\displaystyle\mathbb{A}(x)$, we obtain the
same regularity under some additional hypothesis.

###### keywords:

Hölder regularity, Pohožaev identity, De Giorgi - Nash - Moser theory

###### MSC:

[2020]: 35B45 , 35B65 , 35J15

###### Contents

  1. 1 Introduction
  2. 2 Ideas of Piccinini-Spagnolo revisited
  3. 3 Proof of Theorem 1.3 when $\displaystyle\mathbb{A}(x)=\mathbb{I}$ is identity matrix - case of harmonic functions
    1. 3.1 Some discussion
    2. 3.2 Discussion of the two proofs
    3. 3.3 First proof
    4. 3.4 Second proof
  4. 4 Proof of Theorem 1.3 when $\displaystyle\mathbb{A}(x)\equiv\mathbb{A}$ has constant entries

##  1 Introduction

In this paper, we shall consider weak solutions $\displaystyle u\in
W^{1,2}_{\operatorname{loc}}(\mathbb{R}^{n})$ solving

$-\operatorname{div}\mathbb{A}(x)\nabla
u=0\qquad\text{with}\qquad\lambda|\xi|^{2}\leq\left\langle\mathbb{A}(x)\xi,\xi\right\rangle\leq\Lambda|\xi|^{2},$
(1.1)

where $\displaystyle\mathbb{A}(x)$ is assumed to be symmetric and
$\displaystyle\lambda,\Lambda\in(0,\infty)$ are any two fixed positive
constants.

The Hölder regularity results were first proved in [3, 8] and subsequently, a
different proof was given in [5, 6, 7]. All three approaches provided the
tools to prove a much stronger Harnack inequality and Hölder regularity was
deduced from this. In [1], a sharp form of Harnack inequality was proved which
implied that the Hölder exponent had exponential dependence on
$\displaystyle\tfrac{\lambda}{\Lambda}$. Since Harnack inequality obtained in
[1] was sharp, this implied that the approaches developed in [3, 8, 5] could
at best give the Hölder exponent that had exponential dependence on
$\displaystyle\tfrac{\lambda}{\Lambda}$.

In $\displaystyle\mathbb{R}^{2}$, a new proof using a monotonicity formula
which is an optimized version of the ‘hole filling’ technique was given in [9]
and they obtained the sharp Hölder exponent to be
$\displaystyle\sqrt{\tfrac{\lambda}{\Lambda}}$ and conjectured that in higher
dimensions, the exponent should have the form

$\displaystyle\alpha\geq\frac{1}{2}\left(-(n-2)+\sqrt{(n-2)^{2}+\frac{4(n-1)\lambda}{\Lambda}}\right).$

In this regard, we also mention [2, 12] and references therein where the ‘hole
filling’ technique was extended to higher dimensional linear elliptic and
parabolic equations using Green’s function estimates.

In this paper, we prove the following theorem:

######  Theorem 1.1.

Any local weak solutions of Eq. 1.1 with
$\displaystyle\mathbb{A}(x)\equiv\mathbb{A}$ is Hölder continuous
$\displaystyle u\in C_{\operatorname{loc}}^{\alpha}(\mathbb{R}^{n})$ with

$\displaystyle\alpha\geq\frac{1}{2}\left(-(n-2)+\sqrt{(n-2)^{2}+\frac{4(n-1)\lambda}{\Lambda}}\right).$

######  Corollary 1.2.

Any local weak solutions of Eq. 1.1 with a general
$\displaystyle\mathbb{A}(x)$ and
$\displaystyle\operatorname*{\mathbb{E}_{r}}\geq 0$ is Hölder continuous
$\displaystyle u\in C_{\operatorname{loc}}^{\alpha}(\mathbb{R}^{n})$ with

$\displaystyle\alpha\geq\frac{1}{2}\left(-(n-2)+\sqrt{(n-2)^{2}+\frac{4(n-1)\lambda}{\Lambda}}\right).$

The quantity $\displaystyle\operatorname*{\mathbb{E}_{r}}$ is as obtained in
Lemma 4.1 and the regularity is obtained in Remark 4.5.

It suffices to prove the following theorem instead:

######  Theorem 1.3.

Let $\displaystyle S_{1}$ be the sphere of unit radius in
$\displaystyle\mathbb{R}^{n}$ and $\displaystyle B_{1}$ be the unit ball in
$\displaystyle\mathbb{R}^{n}$. Then, any weak solution of Eq. 1.1 satisfies

$\displaystyle\int_{S_{1}}\left\langle\mathbb{A}\nabla u,\nabla
u\right\rangle\,d\sigma\geq\sqrt{(n-2)^{2}+\frac{4(n-1)\lambda}{\Lambda}}\int_{B_{1}}\left\langle\mathbb{A}\nabla
u,\nabla u\right\rangle\,dx.$

######  Remark 1.4.

In the rest of the paper, we will assume $\displaystyle\mathbb{A}(x)$ and
$\displaystyle u$ are smooth and obtain a priori estimates. Such an assumption
can be made to hold by a standard approximation procedure, see [4, Section 4]
for the details. The proof also shows that for general matrices
$\displaystyle\mathbb{A}(x)$, we can obtain the same regularity as Theorem 1.1
provided $\displaystyle\operatorname*{\mathbb{E}_{r}}\geq 0$, where
$\displaystyle\operatorname*{\mathbb{E}_{r}}$ is as obtained in Lemma 4.1.

##  2 Ideas of Piccinini-Spagnolo revisited

In [9], the authors proved the following result:

######  Theorem 2.1.

When $\displaystyle n=2$, weak solutions of Eq. 1.1 are $\displaystyle
C^{\alpha}_{\operatorname{loc}}$ regular with
$\displaystyle\alpha=\sqrt{\tfrac{{\lambda}}{{\Lambda}}}$. In particular, the
following estimate holds for any $\displaystyle r>0$:

$\int_{B_{r}}\left\langle\mathbb{A}(x)\nabla u,\nabla
u\right\rangle\,dx\leq\frac{r{\sqrt{\frac{\Lambda}{\lambda}}}}{2}\int_{S_{r}}\left\langle\mathbb{A}(x)\nabla
u,\nabla u\right\rangle\,d\sigma,$ (2.1)

where $\displaystyle B_{r}$ is the ball of raidus $\displaystyle r$ and
$\displaystyle S_{r}=\partial B_{r}$ is the boundary of $\displaystyle B_{r}$.

###### Proof.

We shall change to polar co-ordinates and use the following notation:

$\displaystyle\begin{bmatrix}x\\\
y\end{bmatrix}=\begin{bmatrix}r\cos(\theta)\\\
r\sin(\theta)\end{bmatrix},\quad\mathbb{J}(\theta)=\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\
\sin(\theta)&\cos(\theta)\end{bmatrix},\quad\mathbb{P}(x)=\mathbb{J}(\theta)^{\top}\mathbb{A}(x)\mathbb{J}(\theta),\quad\underbrace{\begin{bmatrix}u_{x}\\\
u_{y}\end{bmatrix}}_{:=\nabla
u}=\mathbb{J}(\theta)\underbrace{\begin{bmatrix}u_{r}\\\
\tfrac{1}{r}u_{\theta}\end{bmatrix}}_{:=\bm{\bar{}}{\nabla}u}:=\begin{bmatrix}u_{N}\\\
u_{T}\end{bmatrix}.$

Following [9], we have

$\begin{array}[]{rcl}g(r)&:=&\int_{B_{r}}\left\langle\mathbb{A}(x)\nabla
u,\nabla
u\right\rangle\,dx=\int_{S_{r}}(u-k)\left\langle\mathbb{P}(x)\bm{\bar{}}{\nabla}u,\vec{e_{1}}\right\rangle\,d\sigma=\int_{S_{r}}(u-k)\left(p_{11}u_{\rho}+p_{12}\tfrac{1}{r}u_{\theta}\right)d\sigma,\\\
g^{\prime}(r)&:=&\int_{S_{r}}\left\langle\mathbb{P}(x)\bm{\bar{}}{\nabla}u,e_{1}\right\rangle\,d\sigma,\end{array}$
(2.2)

where to obtain the first equality in the definition of $\displaystyle g(r)$,
we made use of the fact that $\displaystyle u$ solves Eq. 1.1.

With the choice $\displaystyle k=\fint_{S_{r}}u\,d\sigma$ and $\displaystyle
L:=\tfrac{\Lambda}{\lambda}$, we shall estimate $\displaystyle g(r)$ as
follows:

$\begin{array}[]{rcl}g(r)&\overset{\hypertarget{a2.3a}{{\color[rgb]{0,0,0}(}{\color[rgb]{1,0,0}a}{\color[rgb]{0,0,0})}}}{\leq}&\left(\int_{S_{r}}p_{11}(u-k)^{2}\,d\sigma\right)^{\frac{1}{2}}\left(\int_{S_{r}}\left(\sqrt{p_{11}}u_{r}+\frac{p_{12}}{\sqrt{p_{11}}}\frac{1}{r}u_{\theta}\right)^{2}\,d\sigma\right)^{\frac{1}{2}}\\\
&\overset{\hypertarget{a2.3b}{{\color[rgb]{0,0,0}(}{\color[rgb]{1,0,0}b}{\color[rgb]{0,0,0})}}}{\leq}&r\sqrt{L}\left(\lambda\int_{S_{r}}u_{T}^{2}\,d\sigma\right)^{\frac{1}{2}}\left(\int_{S_{r}}\left(\sqrt{p_{11}}u_{N}+\frac{p_{12}}{\sqrt{p_{11}}}u_{T}\right)^{2}\,d\sigma\right)^{\frac{1}{2}}\\\
&\overset{\hypertarget{a2.3c}{{\color[rgb]{0,0,0}(}{\color[rgb]{1,0,0}c}{\color[rgb]{0,0,0})}}}{\leq}&\frac{r\sqrt{L}}{2}\left[\int_{S_{r}}\left(p_{22}-\frac{p_{12}^{2}}{p_{11}}\right)u_{T}^{2}\,d\sigma+\int_{S_{r}}\left(p_{11}u_{N}^{2}+\frac{p_{12}^{2}}{p_{11}}u_{T}^{2}+2p_{12}u_{N}u_{T}\right)\,d\sigma\right]\\\
&=&\frac{r\sqrt{L}}{2}\int_{S_{r}}\left\langle
P\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,d\sigma=\frac{r\sqrt{L}}{2}g^{\prime}(r),\end{array}$
(2.3)

where to obtain (a), we applied Hölder’s inequality, to obtain (b), we applied
Proposition 3.1 along with the bound $\displaystyle\lambda\leq
p_{11}(x)\leq\Lambda$ and finally to obtain (c), we made use of the matrix
inequality $\displaystyle\lambda\leq
p_{22}(x)-\frac{p_{12}^{2}(x)}{p_{11}(x)}\leq\Lambda$ along with Young’s
inequality. Recalling Eq. 2.2, the estimate in Eq. 2.3 is exactly the claim in
Eq. 2.1 and this completes the proof of the theorem.

∎

##  3 Proof of Theorem 1.3 when $\displaystyle\mathbb{A}(x)=\mathbb{I}$ is
identity matrix - case of harmonic functions

Let us first recall the Poincaré inequality on spheres (see [11, Chapter IV,
Section 2] for the details):

######  Proposition 3.1.

Let $\displaystyle u\in W^{1,2}_{\operatorname{loc}}(S_{r})$ where
$\displaystyle S_{r}$ is the unit sphere in $\displaystyle\mathbb{R}^{n}$ with
$\displaystyle n\geq 2$ and $\displaystyle\bm{\bar{}}{\nabla}u=(u_{N},u_{T})$
be the tangential and normal derivative of $\displaystyle u$. Then the
following sharp form of Poincaré’s inequality holds:

$\displaystyle\int_{S_{r}}(u-k)^{2}\,d\sigma\leq\left(\frac{r^{2}}{n-1}\right)\int_{S_{r}}|u_{T}|^{2}\,d\sigma,$

where $\displaystyle k=\fint_{S_{r}}u\,d\sigma$.

We shall also recall the well known Pohožaev identity for harmonic functions
obtained in [10].

######  Proposition 3.2.

Let $\displaystyle u$ to be a smooth, weak solution of $\displaystyle-\Delta
u=0$ (i.e., $\displaystyle u$ is locally harmonic), then the following
identity holds:

$\int_{S_{1}}|u_{T}|^{2}\,d\sigma=\int_{S_{1}}|u_{N}|^{2}\,d\sigma+(n-2)\int_{B_{1}}|\nabla
u|^{2}\ dx.$

###### Proof.

Simple calculations implies

$\Delta u\left\langle\nabla u,x\right\rangle=\operatorname{div}(\nabla
u\left\langle\nabla u,x\right\rangle)-\frac{1}{2}\operatorname{div}(|\nabla
u|^{2}x)+\frac{n-2}{2}|\nabla u|^{2}.$ (3.1)

Using $\displaystyle-\Delta u=0$ along with integrating Eq. 3.1 over
$\displaystyle B_{1}$, we have the following identity

$\begin{array}[]{rcl}0=\int_{B_{1}}\Delta u\left\langle\nabla
u,x\right\rangle\,dx&=&\int_{S_{1}}\left\langle\nabla
u,x\right\rangle^{2}\,d\sigma-\frac{1}{2}\int_{S_{1}}|\nabla
u|^{2}\left\langle x,x\right\rangle\,d\sigma+\frac{n-2}{2}\int_{B_{1}}|\nabla
u|^{2}\,dx\\\
&=&\int_{S_{1}}|u_{N}|^{2}\,d\sigma-\frac{1}{2}\int_{S_{1}}(|u_{T}|^{2}+|u_{N}|^{2})\,d\sigma+\frac{n-2}{2}\int_{B_{1}}|\nabla
u|^{2}\,dx,\end{array}$

which completes the proof. ∎

###  3.1 Some discussion

If we consider solutions of $\displaystyle-\Delta u=0$ in
$\displaystyle\mathbb{R}^{n}$ (i.e., $\displaystyle\mathbb{A}(x)=I$) for
$\displaystyle n\geq 3$ and try to follow the calculations of Theorem 2.1,
then we get

$\int_{B_{r}}\left\langle\nabla u,\nabla
u\right\rangle\,dx\leq\frac{r}{2\sqrt{n-1}}\int_{S_{r}}\left\langle\nabla
u,\nabla u\right\rangle\,d\sigma,$ (3.2)

which fails to prove Lipschitz regularity (i.e., the constant in Eq. 3.2
should be $\displaystyle\tfrac{r}{n}$) when $\displaystyle n=3$ and for
$\displaystyle n\geq 7$, we have $\displaystyle 2\sqrt{n-1}<n-2$ and thus no
regularity follows.

On the other hand, if we consider the simple example of harmonic functions
which satisfy $\displaystyle|\nabla u|=C$ for some constant $\displaystyle C$,
then we have

$\displaystyle\frac{\int_{B_{r}}\left\langle\nabla u,\nabla
u\right\rangle\,dx}{\int_{S_{r}}\left\langle\nabla u,\nabla
u\right\rangle\,d\sigma}=\frac{\omega_{n}r^{n}}{n\omega_{n}r^{n-1}}=\frac{r}{n},$

which implies Lipschitz regularity. But the above calculations shows that the
factor $\displaystyle\tfrac{1}{n}$ is coming from the ratio of
$\displaystyle|B_{r}|$ and $\displaystyle|S_{r}|$ which is a property of the
fact that we are in $\displaystyle\mathbb{R}^{n}$ and not necessarily from the
property of the equation and solution.

This suggest that some additional cancellations should hold that can further
improve Eq. 3.2 and these cancellations should be obtained from the fact that
we are in $\displaystyle\mathbb{R}^{n}$. We shall formalize this observation
for harmonic functions in the following theorem:

######  Theorem 3.3.

Let $\displaystyle u$ be a local weak solution of $\displaystyle-\Delta u=0$,
then we have

$\displaystyle\int_{B_{r}}\left\langle\nabla u,\nabla
u\right\rangle\,dx\leq\frac{r}{n}\int_{S_{r}}\left\langle\nabla u,\nabla
u\right\rangle\,d\sigma,$

holds for all $\displaystyle r>0$ and in particular $\displaystyle u\in
C^{0,1}_{\operatorname{loc}}$.

###  3.2 Discussion of the two proofs

We shall give two proofs of Theorem 3.3, both very similar, but have some
crucial differences which are detailed in the following observations:

  1. (i)

In the first proof, we need to make use of Eq. 3.6 to obtain Eq. 3.7. This
does not require equality to hold in Eq. 3.6, but it suffices if the following
inequality holds:

$\int_{S_{1}}|u_{T}|^{2}\,d\sigma\leq\int_{S_{1}}|u_{N}|^{2}\,d\sigma+(n-2)\int_{B_{1}}|\nabla
u|^{2}\ dx.$ (3.3)

  2. (ii)

In the first proof, we need to make use of Eq. 3.6 a second time in Eq. 3.9.
This again does not require equality to hold in Eq. 3.6, but it suffices if
the following inequality holds:

$\int_{S_{1}}|u_{T}|^{2}\,d\sigma\geq\int_{S_{1}}|u_{N}|^{2}\,d\sigma+(n-2)\int_{B_{1}}|\nabla
u|^{2}\ dx.$ (3.4)

  3. (iii)

In particular, the first proof requires Eq. 3.4 and Eq. 3.3 to both hold,
which is the equality form of the identity obtained in Proposition 3.2.

  4. (iv)

In the second proof, we only need to use the Pohožaev identity Eq. 3.11 once
to obtain Eq. 3.12. In particular, the second proof works if we only had the
following one sided inequality:

$\int_{S_{1}}|\nabla u|^{2}\,d\sigma\geq
2\int_{S_{1}}|u_{N}|^{2}\,d\sigma+(n-2)\int_{B_{1}}|\nabla u|^{2}\ dx.$

###  3.3 First proof

We shall use the notation $\displaystyle\bm{\bar{}}{\nabla}u=(u_{N},u_{T})$
where $\displaystyle
u_{N}=\left\langle\bm{\bar{}}{\nabla}u,\vec{e_{1}}\right\rangle$ and
$\displaystyle
u_{T}=(\left\langle\bm{\bar{}}{\nabla}u,\vec{e_{2}}\right\rangle,\ldots,\left\langle\bm{\bar{}}{\nabla}u,\vec{e_{n}}\right\rangle)$
are the normal and tangential derivatives on the sphere respectively in polar
coordinates. We can also take $\displaystyle r=1$ since the required estimate
can be obtained by scaling.

######  Remark 3.4.

We will be deliberately careless in switching between $\displaystyle\nabla u$
and $\displaystyle\bm{\bar{}}{\nabla}u$, i.e., cartesian and polar coordinates
in the proof, so as to better present the ideas of the proof, noting that this
switch is just a transformation by orthogonal matrices and hence does not
affect any of the calculations.

The proof will now proceed in several steps:

Step 1:

Since $\displaystyle u$ is harmonic, we see that

$\begin{array}[]{rcl}\int_{B_{1}}\left\langle\nabla u,\nabla
u\right\rangle\,dx&=&\int_{S_{1}}(u-k)\left\langle\nabla
u,\vec{n}\right\rangle\,d\sigma=\int_{S_{1}}(u-k)\left\langle\bm{\bar{}}{\nabla}u,\vec{e_{1}}\right\rangle\,d\sigma\\\
&\overset{\hypertarget{a3.2a}{{\color[rgb]{0,0,0}(}{\color[rgb]{1,0,0}a}{\color[rgb]{0,0,0})}}}{\leq}&\left(\frac{1}{2(n-1)}\int_{S_{1}}|u_{T}|^{2}\,d\sigma+\frac{1}{2}\int_{S_{1}}|u_{N}|^{2}\,d\sigma\right),\\\
\end{array}$ (3.5)

where to obtain (a), we applied Young’s inequality followed by Proposition 3.1
with $\displaystyle k=\fint_{S_{1}}u\,d\sigma$.

Step 2:

We recall Pohožaev identity from Proposition 3.2:

$\int_{S_{1}}|u_{T}|^{2}\,d\sigma=\int_{S_{1}}|u_{N}|^{2}\,d\sigma+(n-2)\int_{B_{1}}|\nabla
u|^{2}\ dx.$ (3.6)

Step 3:

We shall substitute Eq. 3.6 into Eq. 3.5 and obtain

$\int_{B_{1}}|\nabla u|^{2}\
dx\leq\frac{1}{2(n-1)}\left(\int_{S_{1}}|u_{N}|^{2}\,d\sigma+(n-2)\int_{B_{1}}|\nabla
u|^{2}\,dx\right)+\frac{1}{2}\int_{S_{1}}|u_{N}|^{2}\,d\sigma,$ (3.7)

which after simplification, becomes

$\int_{B_{1}}|\nabla u|^{2}\ dx\leq\int_{S_{1}}|u_{N}|^{2}\,d\sigma.$ (3.8)

Step 4:

Let us add $\displaystyle\int_{S_{1}}|u_{N}|^{2}\,d\sigma$ to both sides of
Eq. 3.6 and then make use of Eq. 3.8 to get

$\begin{array}[]{rcl}\int_{S_{1}}|\nabla
u|^{2}\,d\sigma=\int_{S_{1}}(|u_{T}|^{2}+|u_{N}|^{2})\,d\sigma&\overset{\lx@cref{creftype~refnum}{a3.7}}{=}&2\int_{S_{1}}|u_{N}|^{2}\,d\sigma+(n-2)\int_{B_{1}}|\nabla
u|^{2}\ dx\\\
<EMAIL_ADDRESS>u|^{2}\
dx+(n-2)\int_{B_{1}}|\nabla u|^{2}\ dx\\\ &=&n\int_{B_{1}}|\nabla
u|^{2}\,dx.\end{array}$ (3.9)

This completes the proof of the theorem.

###  3.4 Second proof

Using the same notation as the first proof, we will give a slightly modified
proof of Theorem 3.3 which will proceed in several steps:

Step 1:

Since $\displaystyle u$ is harmonic, we see that

$\begin{array}[]{rcl}\int_{B_{1}}\left\langle\nabla u,\nabla
u\right\rangle\,dx&=&\int_{S_{1}}(u-k)\left\langle\nabla
u,\vec{n}\right\rangle\,d\sigma=\int_{S_{1}}(u-k)\left\langle\bm{\bar{}}{\nabla}u,\vec{e_{1}}\right\rangle\,d\sigma\\\
&\overset{\hypertarget{a3.8ba}{{\color[rgb]{0,0,0}(}{\color[rgb]{1,0,0}a}{\color[rgb]{0,0,0})}}}{\leq}&\left(\frac{1}{2(n-1)}\int_{S_{1}}|u_{T}|^{2}\,d\sigma+\frac{1}{2}\int_{S_{1}}|u_{N}|^{2}\,d\sigma\right)\\\
&\overset{\hypertarget{a3.8bb}{{\color[rgb]{0,0,0}(}{\color[rgb]{1,0,0}b}{\color[rgb]{0,0,0})}}}{=}&\left(\frac{1}{2(n-1)}\int_{S_{1}}|\nabla
u|^{2}\,d\sigma+\left(\frac{1}{2}-\frac{1}{2(n-1)}\right)\int_{S_{1}}|u_{N}|^{2}\,d\sigma\right),\end{array}$
(3.10)

where to obtain (a), we applied Young’s inequality followed by Proposition 3.1
with $\displaystyle k=\fint_{S_{1}}u\,d\sigma$ and to obtain (b), we add and
subtract $\displaystyle\frac{1}{2(n-1)}\int_{S_{1}}|u_{N}|^{2}\,d\sigma$.

Step 2:

We rewrite the Pohožaev identity from Proposition 3.2 as:

$\int_{S_{1}}|\nabla
u|^{2}\,d\sigma=2\int_{S_{1}}|u_{N}|^{2}\,d\sigma+(n-2)\int_{B_{1}}|\nabla
u|^{2}\ dx.$ (3.11)

Step 3:

We shall substitute Eq. 3.11 into Eq. 3.10 to get

$\int_{B_{1}}|\nabla u|^{2}\ dx\leq\frac{1}{2(n-1)}\int_{S_{1}}|\nabla
u|^{2}\,d\sigma+\left(\frac{1}{2}-\frac{1}{2(n-1)}\right)\frac{1}{2}\left(\int_{S_{1}}|\nabla
u|^{2}\,d\sigma-(n-2)\int_{B_{1}}|\nabla u|^{2}\,dx\right),$ (3.12)

which after simplification, becomes

$\frac{n^{2}}{4(n-1)}\int_{B_{1}}|\nabla u|^{2}\
dx\leq\frac{n}{4(n-1)}\int_{S_{1}}|\nabla u|^{2}\,d\sigma.$

This completes the proof of the lemma.

##  4 Proof of Theorem 1.3 when $\displaystyle\mathbb{A}(x)\equiv\mathbb{A}$
has constant entries

The first lemma we generalize is the Pohožaev identity from Proposition 3.2 to
variable coefficient operators of the form considered in Eq. 1.1. This is well
known in literature and we present all the details for sake of completeness.

######  Lemma 4.1.

Let $\displaystyle u$ be a weak solution to Eq. 1.1 and assume both
$\displaystyle u$ and $\displaystyle\mathbb{A}(x)$ are sufficiently smooth.
Then the following Pohožaev type identity holds:

$\begin{array}[]{rcl}\int_{S_{1}}\left\langle\mathbb{A}(x)\nabla u,\nabla
u\right\rangle\left\langle\mathbb{A}(x)x,x\right\rangle\,d\sigma&=&2\int_{S_{1}}\left\langle\mathbb{A}(x)\nabla
u,x\right\rangle^{2}\,d\sigma+\int_{B_{1}}\left(\sum_{i}^{n}a_{ii}(x)\right)\left\langle\mathbb{A}(x)\nabla
u,\nabla u\right\rangle\,dx\\\
&&+\sum_{i,j,k,l}\int_{B_{1}}x_{j}(\partial_{x_{i}}a_{ij})a_{kl}u_{x_{k}}u_{x_{l}}\,dx-2\int_{B_{1}}\left\langle\mathbb{A}(x)\nabla
u,\mathbb{A}(x)\nabla u\right\rangle\,dx\\\
&&-2\sum_{i,j,k,l}\int_{B_{1}}u_{x_{i}}a_{ij}u_{x_{k}}x_{l}(\partial_{x_{j}}a_{kl})\,dx\\\
&&+\sum_{i,j,m,n}\int_{B_{1}}a_{ij}x_{j}u_{x_{m}}u_{x_{n}}(\partial_{x_{i}}a_{mn})\,dx\\\
&=&2\int_{S_{1}}\left\langle\mathbb{A}(x)\nabla
u,x\right\rangle^{2}\,d\sigma+\int_{B_{1}}\left(\sum_{i}^{n}a_{ii}(x)\right)\left\langle\mathbb{A}(x)\nabla
u,\nabla u\right\rangle\,dx\\\ &&-2\int_{B_{1}}\left\langle\mathbb{A}(x)\nabla
u,\mathbb{A}(x)\nabla
u\right\rangle\,dx+\operatorname*{\mathbb{E}_{r}},\end{array}$ (4.1)

where we have denoted

$\displaystyle\operatorname*{\mathbb{E}_{r}}:=\sum_{i,j,k,l}\int_{B_{1}}x_{j}(\partial_{x_{i}}a_{ij})a_{kl}u_{x_{k}}u_{x_{l}}\,dx-2\sum_{i,j,k,l}\int_{B_{1}}u_{x_{i}}a_{ij}u_{x_{k}}x_{l}(\partial_{x_{j}}a_{kl})\,dx+\sum_{i,j,m,n}\int_{B_{1}}a_{ij}x_{j}u_{x_{m}}u_{x_{n}}(\partial_{x_{i}}a_{mn})\,dx.$

######  Remark 4.2.

Note that all the terms of $\displaystyle\operatorname*{\mathbb{E}_{r}}$
contain derivates of the entries of $\displaystyle\mathbb{A}$ and does not
contain any terms from the Hessian matrix $\displaystyle[D^{2}u]$. In
particular, Eq. 4.1 already captures the cancellations of the form
$\displaystyle\left\langle x,[D^{2}u]\nabla u\right\rangle=\left\langle\nabla
u,[D^{2}u]x\right\rangle$ which is crucially used in the case of harmonic
functions to prove Proposition 3.2.

###### Proof of Lemma 4.1.

Simple calculations give

$\begin{array}[]{l}\operatorname{div}(\mathbb{A}\nabla u\left\langle
x,\mathbb{A}\nabla u\right\rangle)=\operatorname{div}(\mathbb{A}\nabla
u)\left\langle x,\mathbb{A}\nabla u\right\rangle+\left\langle\mathbb{A}\nabla
u,\mathbb{A}\nabla u\right\rangle+\left\langle\mathbb{A}\nabla
u,[D(\mathbb{A}\nabla u)]x\right\rangle,\\\
\operatorname{div}(\mathbb{A}x\left\langle\mathbb{A}\nabla u,\nabla
u\right\rangle)=\operatorname{div}(\mathbb{A}x)\left\langle\mathbb{A}\nabla
u,\nabla u\right\rangle+\left\langle\mathbb{A}x,[D(\mathbb{A}\nabla u)]\nabla
u\right\rangle+\left\langle\mathbb{A}x,[D^{2}u]\mathbb{A}\nabla
u\right\rangle,\end{array}$ (4.2)

where $\displaystyle D(\mathbb{A}\nabla u)$ is a matrix of the form

$(D(\mathbb{A}\nabla
u))_{ij}=\sum_{k=1}^{n}(\partial_{x_{i}}a_{jk})u_{x_{k}}+(D^{2}u\mathbb{A})_{ij}=:(D(A)(\nabla
u))_{ij}+([D^{2}u]\mathbb{A})_{ij},$ (4.3)

and $\displaystyle([D^{2}u]\mathbb{A})$ is the matrix product of
$\displaystyle[D^{2}u]\cdot\mathbb{A}$. Using Eq. 4.3 in Eq. 4.2 and noting
that $\displaystyle\operatorname{div}(\mathbb{A}\nabla u)=0$, we get

$\begin{array}[]{l}\operatorname{div}(\mathbb{A}\nabla u\left\langle
x,\mathbb{A}\nabla u\right\rangle)=\left\langle\mathbb{A}\nabla
u,\mathbb{A}\nabla u\right\rangle+\left\langle\mathbb{A}\nabla
u,[D(\mathbb{A})(\nabla u)]x\right\rangle+\left\langle\mathbb{A}\nabla
u,[D^{2}u]\mathbb{A}x\right\rangle,\\\
\operatorname{div}(\mathbb{A}x\left\langle\mathbb{A}\nabla u,\nabla
u\right\rangle)=\operatorname{div}(\mathbb{A}x)\left\langle\mathbb{A}\nabla
u,\nabla u\right\rangle+\left\langle\mathbb{A}x,[D(\mathbb{A})(\nabla
u)]\nabla u\right\rangle+2\left\langle\mathbb{A}x,[D^{2}u]\mathbb{A}\nabla
u\right\rangle,\end{array}$ (4.4)

Subtracting twice the first equation in Eq. 4.4 from the second equation in
Eq. 4.4 in order to cancel the terms containing $\displaystyle[D^{2}u]$ (since
$\displaystyle\left\langle\mathbb{A}x,[D^{2}u]\mathbb{A}\nabla
u\right\rangle=\left\langle\mathbb{A}\nabla
u,[D^{2}u]\mathbb{A}x\right\rangle$), we get

$2\operatorname{div}(\mathbb{A}\nabla u\left\langle x,\mathbb{A}\nabla
u\right\rangle)-\operatorname{div}(\mathbb{A}x\left\langle\mathbb{A}\nabla
u,\nabla u\right\rangle)\\\ =2\left\langle\mathbb{A}\nabla u,\mathbb{A}\nabla
u\right\rangle+2\left\langle\mathbb{A}\nabla u,[D(\mathbb{A})(\nabla
u)]x\right\rangle-\operatorname{div}(\mathbb{A}x)\left\langle\mathbb{A}\nabla
u,\nabla u\right\rangle-\left\langle\mathbb{A}x,[D(\mathbb{A})(\nabla
u)]\nabla u\right\rangle.$ (4.5)

We see that
$\displaystyle\operatorname{div}(\mathbb{A}x)=\sum_{i=1}^{n}a_{ii}+\sum_{i,j}x_{j}\partial_{x_{i}}a_{ij}$
which we substitute in Eq. 4.5 and integrate over $\displaystyle B_{1}$ to get

$\begin{array}[]{rcl}2\int_{B_{1}}\operatorname{div}(\mathbb{A}\nabla
u\left\langle x,\mathbb{A}\nabla
u\right\rangle)\,dx&=&\int_{B_{1}}\operatorname{div}(\mathbb{A}x\left\langle\mathbb{A}\nabla
u,\nabla u\right\rangle)\,dx+2\int_{B_{1}}\left\langle\mathbb{A}\nabla
u,\mathbb{A}\nabla u\right\rangle\,dx\\\
&&+2\int_{B_{1}}\left\langle\mathbb{A}\nabla u,[D(\mathbb{A})(\nabla
u)]x\right\rangle\,dx-\int_{B_{1}}\left\langle\mathbb{A}x,[D(\mathbb{A})(\nabla
u)]\nabla u\right\rangle\,dx\\\
&&-\int_{B_{1}}\left(\sum_{i}^{n}a_{ii}\right)\left\langle\mathbb{A}\nabla
u,\nabla
u\right\rangle\,dx-\int_{B_{1}}\left(\sum_{i,j}x_{j}\partial_{x_{i}}a_{ij}\right)\left\langle\mathbb{A}\nabla
u,\nabla u\right\rangle\,dx.\end{array}$

We can now apply the divergence theorem to complete the proof of the lemma. ∎

######  Corollary 4.3.

In the case $\displaystyle\mathbb{A}(x)\equiv\mathbb{A}$ is a constant matrix,
then $\displaystyle\operatorname*{\mathbb{E}_{r}}=0$ and Eq. 4.1 becomes

$\begin{array}[]{rcl}\int_{S_{1}}\left\langle\mathbb{A}\nabla u,\nabla
u\right\rangle\left\langle\mathbb{A}x,x\right\rangle\,d\sigma&=&2\int_{S_{1}}\left\langle\mathbb{A}\nabla
u,x\right\rangle^{2}\,d\sigma+\int_{B_{1}}\left(\sum_{i}^{n}a_{ii}\right)\left\langle\mathbb{A}\nabla
u,\nabla u\right\rangle\,dx\\\ &&-2\int_{B_{1}}\left\langle\mathbb{A}\nabla
u,\mathbb{A}\nabla u\right\rangle\,dx.\end{array}$ (4.6)

Let us now prove a matrix inequality using the Schur complement of a block
matrix that will be the higher dimensional replacement of the cancellation
used in (c) of Eq. 2.3.

######  Lemma 4.4.

Let $\displaystyle\mathbb{P}$ be the matrix similar to
$\displaystyle\mathbb{A}$ in polar coordinates which we write as

$\displaystyle\mathbb{P}=\left(\begin{array}[]{c|c}p_{11}&P_{12}\\\ \hline\cr
P_{21}&\mathbb{P}_{22}\end{array}\right),$

where $\displaystyle p_{11}$ is a scalar, $\displaystyle P_{12}=P_{21}^{\top}$
is $\displaystyle(n-1)$ vector and $\displaystyle\mathbb{P}_{22}$ is
$\displaystyle(n-1)\times(n-1)$ matrix. Then the following inequality holds:

$\lambda p_{11}|\xi|^{2}\leq
p_{11}\left\langle\mathbb{P}_{22}\xi,\xi\right\rangle-\left\langle
P_{12},\xi\right\rangle^{2}\qquad\text{for
every}\,\,\xi\in\mathbb{R}^{n-1}\setminus\\{0\\}.$ (4.7)

###### Proof.

We have

$\displaystyle\underbrace{\begin{pmatrix}1&0\\\
-P_{21}p_{11}^{-1}&I\end{pmatrix}}_{:=\mathbb{S}}\underbrace{\begin{pmatrix}p_{11}&P_{12}\\\
P_{21}&P_{22}\end{pmatrix}}_{=\mathbb{P}}\underbrace{\begin{pmatrix}1&-p_{11}^{-1}P_{12}\\\
0&I\end{pmatrix}}_{=\mathbb{S}^{\top}}=\underbrace{\begin{pmatrix}p_{11}&0\\\
0&P_{22}-P_{21}p_{11}^{-1}P_{12}\end{pmatrix}}_{:=\mathbb{M}}.$

From this, we see that if we consider the $\displaystyle\mathbb{R}^{n}$ vector
$\displaystyle\begin{pmatrix}0\\\ \xi\end{pmatrix}$, then using Eq. 1.1, we
have

$\begin{array}[]{rcl}\lambda|\xi|^{2}\leq\lambda\frac{\left\langle
P_{12},\xi\right\rangle^{2}}{p_{11}^{2}}+\lambda|\xi|^{2}&\leq&\left\langle\mathbb{P}\begin{pmatrix}-p_{11}^{-1}\left\langle
P_{12},\xi\right\rangle\\\
\xi\end{pmatrix},\begin{pmatrix}-p_{11}^{-1}\left\langle
P_{12},\xi\right\rangle\\\ \xi\end{pmatrix}\right\rangle\\\
&=&\left\langle\mathbb{S}\mathbb{\mathbb{P}}\mathbb{S}^{\top}\begin{pmatrix}0\\\
\xi\end{pmatrix},\begin{pmatrix}0\\\
\xi\end{pmatrix}\right\rangle=\left\langle\mathbb{M}\begin{pmatrix}0\\\
\xi\end{pmatrix},\begin{pmatrix}0\\\ \xi\end{pmatrix}\right\rangle\\\
&=&\left\langle\begin{pmatrix}p_{11}&0\\\
0&P_{22}-P_{21}p_{11}^{-1}P_{12}\end{pmatrix}\begin{pmatrix}0\\\
\xi\end{pmatrix},\begin{pmatrix}0\\\ \xi\end{pmatrix}\right\rangle\\\
&=&\left\langle(\mathbb{P}_{22}-P_{21}p_{11}^{-1}P_{12})\xi,\xi\right\rangle\end{array}$

In particular, we have

$\lambda p_{11}|\xi|^{2}\leq
p_{11}\left\langle\mathbb{P}_{22}\xi,\xi\right\rangle-\left\langle
P_{12},\xi\right\rangle^{2}\qquad\text{for every}\,\,\xi\in\mathbb{R}^{n-1},$

which completes the proof of the lemma. ∎

We are now ready to prove Theorem 1.3 in the case
$\displaystyle\mathbb{A}(x)\equiv\mathbb{A}$ is a constant matrix:

###### Proof of Theorem 1.3.

We make a note about the notation used, we will use $\displaystyle\mathbb{A}$
and $\displaystyle\nabla u$ to denote quantities in Cartesian coordinates and
$\displaystyle\mathbb{P}$ and $\displaystyle\bm{\bar{}}{\nabla}u$ to denote
analogue quantities in Polar coordinates. It is easy to see that the matrices
$\displaystyle\mathbb{A}$ and $\displaystyle\mathbb{P}$ are similar to each
other via the spherical orthogonal change of variables, see the proof of
Theorem 2.1 for this in $\displaystyle\mathbb{R}^{2}$.

The proof follows in several steps:

Step 1:

Using integration by parts and noting that
$\displaystyle-\operatorname{div}\mathbb{A}(x)\nabla u=0$, we have

$\int_{B_{1}}\left\langle\mathbb{A}(x)\nabla u,\nabla
u\right\rangle\,dx=\int_{S_{1}}(u-k)\left\langle\mathbb{A}(x)\nabla
u,\vec{n}\right\rangle\,d\sigma,$

where we take $\displaystyle k=\fint_{S_{1}}u\,d\sigma$. This can be further
estimated after changing to polar coordinates to get

$\begin{array}[]{rcl}\int_{S_{1}}(u-k)\left\langle\mathbb{A}(x)\nabla
u,\vec{n}\right\rangle\,d\sigma&\leq&\frac{\varepsilon}{2}\int_{S_{1}}(u-k)^{2}\,d\sigma+\frac{1}{2\varepsilon}\int_{S_{1}}\left\langle\mathbb{P}(x)\bm{\bar{}}{\nabla}u,\vec{e_{1}}\right\rangle^{2}\,d\sigma\\\
&\overset{\hypertarget{a4.4a}{{\color[rgb]{0,0,0}(}{\color[rgb]{1,0,0}a}{\color[rgb]{0,0,0})}}}{\leq}&\frac{\varepsilon}{2(n-1)}\int_{S_{1}}|u_{T}|^{2}\,d\sigma+\frac{1}{2\varepsilon}\int_{S_{1}}\left\langle\mathbb{P}(x)\bm{\bar{}}{\nabla}u,\vec{e_{1}}\right\rangle^{2}\,d\sigma,\end{array}$
(4.8)

where to obtain (a), we made use of Proposition 3.1.

Step 2:

We will apply Eq. 4.7 with $\displaystyle\xi=u_{T}$ to get

$\begin{array}[]{rcl}p_{11}\lambda|u_{T}|^{2}&\leq&p_{11}\left\langle\mathbb{P}_{22}u_{T},u_{T}\right\rangle-\left\langle
P_{12},u_{T}\right\rangle^{2}\\\
&=&p_{11}\left\langle\mathbb{P}_{22}u_{T},u_{T}\right\rangle-\left\langle
P_{12},u_{T}\right\rangle^{2}+\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\vec{e_{1}}\right\rangle^{2}-\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\vec{e_{1}}\right\rangle^{2}\\\
&=&p_{11}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle-\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\vec{e_{1}}\right\rangle^{2}\\\
\end{array}$ (4.9)

Using Eq. 4.9 along with $\displaystyle\lambda\leq p_{11}\leq\Lambda$, we
estimate $\displaystyle\int_{S_{1}}|u_{T}|^{2}\,d\sigma$ in Eq. 4.8 to get

$\int_{B_{1}}\left\langle\mathbb{P}(x)\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,dx\leq\frac{\varepsilon}{2(n-1)\lambda}\int_{S_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,d\sigma+\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)\int_{S_{1}}\left\langle\mathbb{P}(x)\bm{\bar{}}{\nabla}u,\vec{e_{1}}\right\rangle^{2}\,d\sigma.$
(4.10)

Step 3:

We recall Pohožaev identity from Eq. 4.6 noting that
$\displaystyle\sum_{i}^{n}a_{ii}=\sum_{i}^{n}p_{ii}$ to get

$\begin{array}[]{rcl}\int_{S_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\left\langle\mathbb{P}\vec{e_{1}},\vec{e_{1}}\right\rangle\,d\sigma&=&2\int_{S_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\vec{e_{1}}\right\rangle^{2}\,d\sigma+\int_{B_{1}}\left(\sum_{i}^{n}p_{ii}\right)\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,dx\\\
&&-2\int_{B_{1}}\left\langle\mathbb{P}\nabla u,\mathbb{P}\nabla
u\right\rangle\,dx+\operatorname*{\mathbb{E}_{r}}.\end{array}$ (4.11)

Step 4:

Let us restrict $\displaystyle\varepsilon\leq\sqrt{(n-1)\lambda\Lambda}$ which
is equivalent to
$\displaystyle\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)\geq
0$. Combining Eq. 4.11 and Eq. 4.10 gives

$\int_{B_{1}}\left\langle\mathbb{P}(x)\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,dx\leq\frac{\varepsilon}{2(n-1)\lambda}\int_{S_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,d\sigma+\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)\times\\\
\times\frac{1}{2}\left(\int_{S_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\left\langle\mathbb{P}\vec{e_{1}},\vec{e_{1}}\right\rangle\,d\sigma-\int_{B_{1}}\left(\sum_{i}^{n}p_{ii}\right)\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,dx+2\int_{B_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\mathbb{P}\bm{\bar{}}{\nabla}u\right\rangle-\operatorname*{\mathbb{E}_{r}}\right).$

In particular, after simplification, we get Simplifying, we get

$\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)\frac{1}{2}\operatorname*{\mathbb{E}_{r}}+\int_{B_{1}}\left(1-\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)\frac{(2\Lambda-\left(\sum_{i}^{n}p_{ii}\right))}{2}\right)\left\langle\mathbb{P}(x)\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,dx\leq\\\
\left(\frac{\varepsilon}{2(n-1)\lambda}+\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)\frac{\Lambda}{2}\right)\int_{S_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,d\sigma.$
(4.12)

Step 5:

Ignoring the contribution from $\displaystyle\operatorname*{\mathbb{E}_{r}}$
momentarily (see Remark 4.5), we proceed as follows: Denoting $\displaystyle
T:=\sum_{i}^{n}p_{ii}$, we see that $\displaystyle n\lambda\leq T\leq
n\Lambda$. So, let us maximize Eq. 4.12 over
$\displaystyle\varepsilon\leq\sqrt{(n-1)\lambda\Lambda}$ and $\displaystyle
n\lambda\leq T\leq n\Lambda$ to get

$\int_{B_{1}}\max_{T\in[n\lambda,n\Lambda]}\max_{\varepsilon\in[0,\sqrt{(n-1)\lambda\Lambda}]}\frac{\left(1-\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)\frac{(2\Lambda-T)}{2}\right)}{\left(\frac{\varepsilon}{2(n-1)\lambda}+\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)\frac{\Lambda}{2}\right)}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,dx\leq\int_{S_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,d\sigma.$

This maximum is achieved when $\displaystyle T=n\Lambda$ and

$\displaystyle\varepsilon=\frac{(2-n)\Lambda+\sqrt{(n-2)^{2}\Lambda^{2}+4(n-1)\lambda\Lambda}}{2}=\frac{\Lambda}{2}\left(-(n-2)+\sqrt{(n-2)^{2}+\frac{4(n-1)\lambda}{\Lambda}}\right).$

Furthermore, it is easy to see that
$\displaystyle\varepsilon\in[0,\sqrt{(n-1)\lambda\Lambda}]$.

Substituting the values of $\displaystyle\varepsilon$ and $\displaystyle T$
into Eq. 4.12 gives

$\left(\sqrt{(n-2)^{2}+\tfrac{4(n-1)\lambda}{\Lambda}}\right)\int_{B_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,dx\leq\int_{S_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,d\sigma,$

which completes the proof of the theorem. ∎

######  Remark 4.5.

If we were to keep track of $\displaystyle\operatorname*{\mathbb{E}_{r}}$,
then we would get

$\operatorname*{\mathbb{E}_{r}}\frac{\frac{1}{2}\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)}{\left(\frac{\varepsilon}{2(n-1)\lambda}+\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)\frac{\Lambda}{2}\right)}+\int_{B_{1}}\frac{\left(1-\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)\frac{(2\Lambda-T)}{2}\right)}{\left(\frac{\varepsilon}{2(n-1)\lambda}+\left(\frac{1}{2\varepsilon}-\frac{\varepsilon}{2(n-1)\lambda\Lambda}\right)\frac{\Lambda}{2}\right)}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,dx\leq\int_{S_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,d\sigma.$

which after substituting the values of $\displaystyle\varepsilon$ and
$\displaystyle T$, becomes

$\frac{(n-2)}{\Lambda\left(\sqrt{(n-2)^{2}+\tfrac{4(n-1)\lambda}{\Lambda}}\right)}\operatorname*{\mathbb{E}_{r}}+\left(\sqrt{(n-2)^{2}+\tfrac{4(n-1)\lambda}{\Lambda}}\right)\int_{B_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,dx\leq\int_{S_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,d\sigma.$

In particular, for general $\displaystyle\mathbb{A}(x)$ matrices, Theorem 1.3
holds provided $\displaystyle\operatorname*{\mathbb{E}_{r}}\geq 0$.

######  Question 4.6.

Does the following inequality hold for some $\displaystyle\mathbf{C}_{o}>0$?

$\displaystyle\mathbf{C}_{o}\operatorname*{\mathbb{E}_{r}}\geq\left(\sqrt{(n-2)^{2}+\tfrac{4(n-1)\lambda}{\Lambda}}\right)\int_{B_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,dx-\int_{S_{1}}\left\langle\mathbb{P}\bm{\bar{}}{\nabla}u,\bm{\bar{}}{\nabla}u\right\rangle\,d\sigma.$

If the above inequality holds, then in the case
$\displaystyle\operatorname*{\mathbb{E}_{r}}\leq 0$, we automatically get the
required estimate, thus completing the proof.

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