# Unique continuation for a gradient inequality with $L^{n}$ potential

Adam Coffman, Yifei Pan, and Yuan Zhang  Department of Mathematical Sciences,
Purdue University Fort Wayne, 2101 East Coliseum Boulevard, Fort Wayne, IN
46805, USA<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We establish a unique continuation property for solutions of the differential
inequality $|\nabla u|\leq V|u|$, where $V$ is locally $L^{n}$ integrable on a
domain in $\mathbb{R}^{n}$. A stronger uniqueness result is obtained if in
addition the solutions are locally Lipschitz. One application is a finite
order vanishing property in the $L^{2}$ sense for the exponential of $W^{1,n}$
functions. We further discuss related results for the Cauchy-Riemann operator
$\bar{\partial}$ and characterize the vanishing order for smooth extension of
holomorphic functions across the boundary.

###### Key words and phrases:

unique continuation, divergent integrals.

###### 2020 Mathematics Subject Classification:

Primary 35R45. Secondary 35A02, 26B35, 32A40.

##  1\. Introduction and Results

Let $\Omega$ be a connected open subset of $\mathbb{R}^{n}$. We investigate
solutions to the following differential inequality concerning the gradient
operator $\nabla$:

$|\nabla u|\leq V|u|\ \ \text{on}\ \ \Omega,$ (1.1)

with the potential $V\in L^{n}_{loc}(\Omega)$.

A function $u\in L^{2}_{loc}(\Omega)$ is said to vanish to infinite order (or
to be flat) at a point $x_{0}\in\Omega$ (in the $L^{2}$ sense) means that for
all $m\geq 0$,

$\lim_{r\rightarrow 0}r^{-m}\int_{|x-x_{0}|<r}|u(x)|^{2}dv=0,$

where $dv$ is the Lebesgue measure element in $\mathbb{R}^{n}$. Otherwise, $u$
vanishes to finite order at $x_{0}$ in the $L^{2}$ sense. We say a
differential (in)equality satisfies the (strong) unique continuation property
to mean that every $H^{1}_{loc}(\Omega)\left(=W^{1,2}_{loc}(\Omega)\right)$
solution that vanishes to infinite order at a point in the $L^{2}$ sense must
vanish identically. Here for $p\geq 1$, $W^{1,p}_{loc}(\Omega)$ is the
standard Sobolev space of $L^{p}_{loc}(\Omega)$ functions whose first order
weak derivatives exist and belong to $L_{loc}^{p}(\Omega)$. While studying the
unique continuation property of the Cauchy-Riemann operator $\bar{\partial}$
in several complex variables:

$|\bar{\partial}u|\leq V|u|\ \ \ \text{on}\ \ \Omega\subset\mathbb{C}^{n},$
(1.2)

we observe that (1.2) is reduced to (1.1) when the solutions are real-valued.
This motivates us to study the following unique continuation property of
$H^{1}_{loc}(\Omega)$ solutions to (1.1).

######  Theorem 1.1.

Let $\Omega$ be a domain in $\mathbb{R}^{n},n\geq 2$ and $V\in
L_{loc}^{n}(\Omega)$. Suppose
$u=(u_{1},\ldots,u_{M}):\Omega\rightarrow\mathbb{R}^{M}$ with $u\in
H^{1}_{loc}(\Omega)$ and satisfies $|\nabla u|\leq V|u|$ a.e. on $\Omega$. If
$u$ vanishes to infinite order at some $x_{0}\in\Omega$, then $u\equiv 0$.

The $n=2$ case in Theorem 1.1 is due to a unique continuation property result
in [PZ] concerning the $\bar{\partial}$ operator. For higher dimensions, the
proof makes use of a Hardy-type inequality, along with the Gagliardo-
Nirenberg-Sobolev inequality. When the potential is no longer in $L^{n}$, one
can still get the unique continuation property for some special types of
potentials, see Theorem 2.6. However, as shown in Example 2.5, the property
fails in general for $V\notin L^{n}$. On the other hand, Theorem 2.7 states
that the weak unique continuation property always holds for (1.1) as long as
$V\in L^{2}$.

As a consequence of Theorem 1.1, we obtain the following property of vanishing
to finite order for the exponential of $W^{1,n}$ functions. Note that the
$W^{1,n}$ space is the critical Sobolev space where the Sobolev embedding
theorem fails, and instead is substituted by the Moser-Trudinger inequality.

######  Theorem 1.2.

Let $\Omega$ be an open set in $\mathbb{R}^{n},n\geq 2$. Suppose
$\phi:\Omega\to\mathbb{R}$ with $\phi\in W_{loc}^{1,n}(\Omega)$. Then the
exponential $e^{\phi}$ of $\phi$ vanishes to finite order in the $L^{2}$ sense
at each point in $\Omega$.

In the second part of the paper, we focus on locally Lipschitz solutions to
(1.1). Under the context of this more restricted function space, we are able
to prove a uniqueness result below by just assuming the vanishing of the first
jets. A similar uniqueness result was discussed in [PW] for higher order
differential operators on smooth functions of one variable ($n=1$). It is
worth pointing out that the Lipschitz assumption on the solutions cannot be
dropped here when $n\geq 2$, see Remark 3.9.

######  Theorem 1.3.

Let $\Omega$ be a domain in $\mathbb{R}^{n},n\geq 1$ and $V\in
L^{n}_{loc}(\Omega)$. Suppose $u:\Omega\to\mathbb{R}$ is a locally Lipschitz
function on $\Omega$ satisfying $|\nabla u|\leq V|u|$ a.e. on $\Omega$. If
$u(x_{0})=0$ at some $x_{0}\in\Omega$, then $u\equiv 0$.

Theorem 1.3 can be readily applied to study the uniqueness of some types of
nonlinear differential systems, as indicated in Corollary 3.11. In Section 4,
we discuss further applications under the Lipschitz setting. In particular,
Theorem 4.2 shows that the logarithm of a positive Lipschitz function cannot
fall in $W^{1,n}$ near every zero point of the function. On the other hand, we
prove that if in addition $e^{\phi}$ in Theorem 1.2 is Lipschitz, then
$e^{\phi}$ must be nowhere zero, see Corollary 4.5.

In the last section, we discuss related results for the $\bar{\partial}$
operator on domains in $\mathbb{C}^{n}$. To start with, we construct Example
5.1 to show that the gradient operator $\nabla$ in Theorem 1.3 cannot be
replaced by the $\bar{\partial}$ operator even for real analytic functions. On
the other hand, we give finer characterizations in terms of an $L^{2}$
divergence for holomorphic functions that are extended smoothly across the
boundary.

##  2\. Unique continuation for $H^{1}$ solutions

Let $\Omega$ be a domain (by which we mean a connected open set) in
$\mathbb{R}^{n}=\\{(x_{1},\ldots,x_{n})\\}$. For scalar valued
$u:\Omega\to\mathbb{R}$, $u\in W^{1,p}_{loc}(\Omega)$, the gradient of $u$ is
the vector of first order weak partial derivatives:

$\nabla u=(\partial_{x_{1}}u,\ldots,\partial_{x_{n}}u),$

defined on $\Omega$. The norm of a vector $x\in\mathbb{R}^{n}$ is
$|x|=\sqrt{x_{1}^{2}+\cdots+x_{n}^{2}}$, and in particular the norm of the
gradient is defined on the domain of the gradient by

$|\nabla u|^{2}=(\partial_{x_{1}}u)^{2}+\cdots+(\partial_{x_{n}}u)^{2}.$

In this section, we prove Theorem 1.1, the unique continuation property for
vector valued $H^{1}$ solutions $u:\Omega\to\mathbb{R}^{M}$, where the
inequality (1.1) reads as

$|\nabla
u|=\left(\sum_{j=1}^{n}\sum_{k=1}^{M}|\partial_{x_{j}}u_{k}|^{2}\right)^{\frac{1}{2}}\leq
V\left(\sum_{k=1}^{M}|u_{k}|^{2}\right)^{\frac{1}{2}}=V|u|.$ (2.1)

We first prove the following Hardy-type inequality for $\nabla$. Denote by
$B_{r}$ the ball in $\mathbb{R}^{n}$ of radius $r$ with center at the origin.

######  Lemma 2.1.

Let $u\in H^{1}(\mathbb{R}^{n})$ with support outside a neighborhood of $0$.
Then for any $\lambda>\frac{n}{2}$,

$\int_{\mathbb{R}^{n}}\frac{|u(x)|^{2}}{|x|^{2\lambda}}dv\leq\frac{4}{(2\lambda-n)^{2}}\int_{\mathbb{R}^{n}}\frac{|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv.$ (2.2)

###### Proof.

We first show the inequality when $u\in
C_{c}^{\infty}(\mathbb{R}^{n}\setminus\\{0\\})$. Let

$F(x):=\sum_{j=1}^{n}\frac{|u(x)|^{2}x_{j}}{|x|^{2\lambda}}dx_{1}\wedge\cdots\wedge
dx_{j-1}\wedge dx_{j+1}\wedge\cdots\wedge dx_{n}.$

Then $F$ is a smooth $(n-1)$ form with compact support. Note for each
$j=1,\ldots,n$,

$\sum_{j=1}^{n}\partial_{x_{j}}\left(\frac{x_{j}}{|x|^{2\lambda}}\right)=\sum_{j=1}^{n}\left(\frac{1}{|x|^{2\lambda}}-\frac{2\lambda
x_{j}^{2}}{|x|^{2\lambda+2}}\right)=\frac{n-2\lambda}{|x|^{2\lambda}}.$

Applying Stokes’ theorem on $F$, we have

$0=\int_{\mathbb{R}^{n}}dF=\int_{\mathbb{R}^{n}}\frac{(n-2\lambda)|u(x)|^{2}}{|x|^{2\lambda}}dv+\int_{\mathbb{R}^{n}}\frac{2u(x)\langle\nabla
u(x),x\rangle}{|x|^{2\lambda}}dv.$

Thus

$\int_{\mathbb{R}^{n}}\frac{|u(x)|^{2}}{|x|^{2\lambda}}dv=\frac{1}{(2\lambda-n)}\int_{\mathbb{R}^{n}}\frac{u(x)\langle\nabla
u(x),x\rangle}{|x|^{2\lambda}}dv.$

By the Cauchy-Schwarz inequality, one further gets

$\begin{split}\int_{\mathbb{R}^{n}}\frac{|u(x)|^{2}}{|x|^{2\lambda}}dv\leq&\frac{2}{(2\lambda-n)}\int_{\mathbb{R}^{n}}\frac{|u(x)||\nabla
u(x)|}{|x|^{2\lambda-1}}dv\\\
\leq&\frac{2}{(2\lambda-n)}\left(\int_{\mathbb{R}^{n}}\frac{|u(x)|^{2}}{|x|^{2\lambda}}dv\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^{n}}\frac{|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv\right)^{\frac{1}{2}}.\end{split}$

Dividing both sides by
$\left(\int_{\mathbb{R}^{n}}\frac{|u(x)|^{2}}{|x|^{2\lambda}}dv\right)^{\frac{1}{2}}$
and then squaring both sides, we obtain (2.2) for $u\in
C_{c}^{\infty}(\mathbb{R}^{n}\setminus\\{0\\})$.

For general $u\in H^{1}(\mathbb{R}^{n})$ with support, say, away from
$B_{r},r>0$, we use the standard density argument. In detail, let $u^{(j)}\in
C_{c}^{\infty}(\mathbb{R}^{n}\setminus B_{r})\rightarrow u$ in $H^{1}$ norm.
Then

$\begin{split}\left(\int_{\mathbb{R}^{n}}\frac{|u(x)|^{2}}{|x|^{2\lambda}}dv\right)^{\frac{1}{2}}\leq&\left(\int_{\mathbb{R}^{n}\setminus
B_{r}}\frac{|u(x)-u^{(j)}(x)|^{2}}{|x|^{2\lambda}}dv\right)^{\frac{1}{2}}+\left(\int_{\mathbb{R}^{n}}\frac{|u^{(j)}(x)|^{2}}{|x|^{2\lambda}}dv\right)^{\frac{1}{2}}\\\
\leq&\frac{1}{r^{\lambda}}\left(\int_{\mathbb{R}^{n}}|u(x)-u^{(j)}(x)|^{2}dv\right)^{\frac{1}{2}}+\frac{2}{(2\lambda-n)}\left(\int_{\mathbb{R}^{n}}\frac{|\nabla
u^{(j)}(x)|^{2}}{|x|^{2\lambda-2}}dv\right)^{\frac{1}{2}}.\end{split}$

Here we used (2.2) for $u^{(j)}\in
C_{c}^{\infty}(\mathbb{R}^{n}\setminus\\{0\\})$. Thus

$\displaystyle\left(\int_{\mathbb{R}^{n}}\frac{|u(x)|^{2}}{|x|^{2\lambda}}dv\right)^{\frac{1}{2}}$
$\displaystyle\leq$
$\displaystyle\frac{1}{r^{\lambda}}\left(\int_{\mathbb{R}^{n}}|u(x)-u^{(j)}(x)|^{2}dv\right)^{\frac{1}{2}}+\frac{2}{(2\lambda-n)}\left(\int_{\mathbb{R}^{n}\setminus
B_{r}}\frac{|\nabla u^{(j)}(x)-\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv\right)^{\frac{1}{2}}$ $\displaystyle\
+\frac{2}{(2\lambda-n)}\left(\int_{\mathbb{R}^{n}}\frac{|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv\right)^{\frac{1}{2}}$ $\displaystyle\leq$
$\displaystyle\frac{1}{r^{\lambda}}\left(\int_{\mathbb{R}^{n}}|u(x)-u^{(j)}(x)|^{2}dv\right)^{\frac{1}{2}}+\frac{2}{(2\lambda-n)r^{\lambda-1}}\left(\int_{\mathbb{R}^{n}}|\nabla
u^{(j)}(x)-\nabla u(x)|^{2}dv\right)^{\frac{1}{2}}$ $\displaystyle\
+\frac{2}{(2\lambda-n)}\left(\int_{\mathbb{R}^{n}}\frac{|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv\right)^{\frac{1}{2}}$ $\displaystyle\leq$
$\displaystyle\left(\frac{1}{r^{\lambda}}+\frac{2}{(2\lambda-n)r^{\lambda-1}}\right)\|u-u^{(j)}\|_{H^{1}(\mathbb{R}^{n})}+\frac{2}{(2\lambda-n)}\left(\int_{\mathbb{R}^{n}}\frac{|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv\right)^{\frac{1}{2}}.$

Letting $j\rightarrow\infty$, we have the desired inequality (2.2). ∎

######  Lemma 2.2.

Let $u\in H^{1}(\mathbb{R}^{n})$ with support outside a neighborhood of 0.
Then there exists a constant $C_{0}>0$ such that for any
$\lambda>>\frac{n}{2}$,

$\int_{\mathbb{R}^{n}}\left|\nabla\left(\frac{u(x)}{|x|^{\lambda-1}}\right)\right|^{2}dv\leq
C_{0}\int_{\mathbb{R}^{n}}\frac{|\nabla u(x)|^{2}}{|x|^{2\lambda-2}}dv.$

###### Proof.

Since $|\nabla|x||=1$, we have

$\begin{split}\int_{\mathbb{R}^{n}}\left|\nabla\left(\frac{u(x)}{|x|^{\lambda-1}}\right)\right|^{2}dv\leq&2\int_{\mathbb{R}^{n}}\frac{|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv+2(\lambda-1)^{2}\int_{\mathbb{R}^{n}}\frac{|u(x)|^{2}}{|x|^{2\lambda}}dv\\\
\leq&2\int_{\mathbb{R}^{n}}\frac{|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv+\frac{2(\lambda-1)^{2}}{(2\lambda-n)^{2}}\int_{\mathbb{R}^{n}}\frac{|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv\\\
=&\left(2+\frac{2(\lambda-1)^{2}}{(2\lambda-n)^{2}}\right)\int_{\mathbb{R}^{n}}\frac{|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv.\end{split}$

Here in the second inequality we used Lemma 2.1. The lemma thus follows from
the fact that
$\displaystyle{\lim_{\lambda\rightarrow\infty}\frac{2(\lambda-1)^{2}}{(2\lambda-n)^{2}}=\frac{1}{2}}$.
∎

Throughout the rest of the paper, we occasionally use the notation $a\lesssim
b$ for two quantities $a$ and $b$, to mean that there exists a universal
constant $C$ (dependent only possibly on $n$) such that $a\leq Cb$. To prove
Theorem 1.1 in the case when $n=2$, we will use the following unique
continuation property established in [PZ] for $\bar{\partial}$. Note that
identifying $z\in\mathbb{C}$ with $(x_{1},x_{2})\in\mathbb{R}^{2}$, then for a
function $u$ on $\Omega$,
$\bar{\partial}_{z}u=\frac{1}{2}\left(\partial_{x_{1}}u+i\partial_{x_{2}}u\right)$.
It would be interesting to have a real-variable approach for this case, but we
currently do not.

######  Proposition 2.3.

[PZ] Let $\Omega$ be a domain in $\mathbb{C}$. Suppose
$u=(u_{1},\ldots,u_{N}):\Omega\rightarrow\mathbb{C}^{N}$ with $u\in
H^{1}_{loc}(\Omega)$ and satisfies $|\bar{\partial}u|\leq V|u|$ a.e. on
$\Omega$ for some $V\in L_{loc}^{2}(\Omega)$. If $u$ vanishes to infinite
order at $z_{0}\in\Omega$, then $u$ vanishes identically.

###### Proof of Theorem 1.1:.

The $n=2$ case follows from Proposition 2.3 and the trivial fact that
$|\bar{\partial}u|\lesssim|\nabla u|$. When $n\geq 3$, without loss of
generality assume $x_{0}=0$. Fix $r\in(0,1)$ so that

$\left(\int_{B_{r}}|V(x)|^{n}dv\right)^{\frac{2}{n}}<\frac{1}{2C_{0}C_{1}^{2}},$
(2.3)

where $C_{0}$ is the constant in Lemma 2.2, and $C_{1}$ is the constant in the
Gagliardo-Nirenberg-Sobolev inequality:

$\|f\|_{L^{\frac{2n}{n-2}}(\mathbb{R}^{n})}\leq C_{1}\|\nabla
f\|_{L^{2}(\mathbb{R}^{n})},\ \ \text{for all}\ \ f\in H^{1}(\mathbb{R}^{n}).$

We shall show that $u=0$ in $B_{\frac{r}{2}}$. Thus, applying a standard
propagation argument we obtain $u\equiv 0$ on $\Omega$.

Choose $\eta\in C_{c}^{\infty}(\mathbb{R}^{n})$ such that $0\leq\eta\leq 1$,
$\eta=1$ on $B_{r}$, $\eta=0$ outside $B_{2r}$, and
$|\nabla\eta|\leq\frac{2}{r}$ on $B_{2r}\setminus B_{r}$. Let $\psi\in
C^{\infty}(\mathbb{R}^{n})$ be such that $0\leq\psi\leq 1$, $\psi=0$ in
$B_{1}$, $\psi=1$ outside $B_{2}$, and $|\nabla\psi|\leq 2$ on
$B_{2}\setminus{B_{1}}$. For each $k\geq\frac{4}{r}$ (then
$\frac{2}{k}\leq\frac{r}{2}$), let $\psi_{k}(x)=\psi(kx),x\in\mathbb{R}^{n}$.
Defining $u^{(k)}=\psi_{k}\eta u$, note that $u^{(k)}\in
H^{1}(\mathbb{R}^{n})$ and is supported inside $B_{r}\setminus
B_{\frac{1}{k}}$. Then for each $k\geq\frac{4}{r}$ and $\lambda>\frac{n}{2}$,

$\displaystyle\int_{B_{2r}}\frac{|\nabla u^{(k)}(x)|^{2}}{|x|^{2\lambda-2}}dv$
(2.4) $\displaystyle\lesssim$
$\displaystyle\int_{B_{2r}}\frac{|\psi_{k}(x)\eta(x)|^{2}|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv+\int_{B_{r}}\frac{|\nabla\psi_{k}(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv+\int_{B_{2r}\setminus
B_{r}}\frac{|\nabla\eta(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv$
$\displaystyle\leq$
$\displaystyle\int_{B_{2r}}\frac{|V(x)|^{2}|\psi_{k}(x)\eta(x)u(x)|^{2}}{|x|^{2\lambda-2}}dv+\int_{B_{r}}\frac{|\nabla\psi_{k}(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv+\int_{B_{2r}\setminus
B_{r}}\frac{|\nabla\eta(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv$
$\displaystyle\leq$
$\displaystyle\left(\int_{B_{2r}}|V(x)|^{n}dv\right)^{\frac{2}{n}}\left(\int_{\mathbb{R}^{n}}\left(\frac{|u^{(k)}(x)|}{|x|^{\lambda-1}}\right)^{\frac{2n}{n-2}}dv\right)^{\frac{n-2}{n}}+\int_{B_{r}}\frac{|\nabla\psi_{k}(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv$
$\displaystyle\ +\int_{B_{2r}\setminus
B_{r}}\frac{|\nabla\eta(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv.$

Here we have used Hölder’s inequality in (2.4). Since
$\frac{|u^{(k)}(x)|}{|x|^{\lambda-1}}\in H^{1}(\mathbb{R}^{n}),n\geq 3$,
making use of the Gagliardo-Nirenberg-Sobolev inequality and Lemma 2.2, we get

$\begin{split}\left(\int_{\mathbb{R}^{n}}\left(\frac{|u^{(k)}(x)|}{|x|^{\lambda-1}}\right)^{\frac{2n}{n-2}}dv\right)^{\frac{n-2}{n}}\leq&C_{1}^{2}\int_{\mathbb{R}^{n}}\left|\nabla\left(\frac{|u^{(k)}(x)|}{|x|^{\lambda-1}}\right)\right|^{2}dv\leq
C_{0}C_{1}^{2}\int_{B_{2r}}\frac{|\nabla
u^{(k)}(x)|^{2}}{|x|^{2\lambda-2}}dv.\end{split}$

This combined with (2.4) and (2.3) for each $k\geq\frac{4}{r}$ and
$\lambda>\frac{n}{2}$ leads to

$\int_{B_{2r}}\frac{|\nabla u^{(k)}(x)|^{2}}{|x|^{2\lambda-2}}dv\leq
2\int_{B_{r}}\frac{|\nabla\psi_{k}(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv+2\int_{B_{2r}\setminus
B_{r}}\frac{|\nabla\eta(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv.$ (2.5)

Now suppose toward a contradiction that $\nabla u\not\equiv 0$ on
$B_{\frac{r}{2}}$. Then there exists $k_{1}>0$ such that

$M_{1}=\int_{B_{\frac{r}{2}}\setminus B_{\frac{2}{k_{1}}}}|\nabla
u(x)|^{2}dv>0.$ (2.6)

Consequently for each fixed $\lambda>\frac{n}{2}$,

$M_{\lambda}=\int_{B_{\frac{r}{2}}\setminus B_{\frac{2}{k_{1}}}}\frac{|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv>0.$

Noting that $\nabla u^{(k)}=\nabla u$ on $B_{\frac{r}{2}}\setminus
B_{\frac{2}{k_{1}}}$ for any $k\geq k_{1}$ by construction of $u^{(k)}$, we
further have for any $k\geq k_{1}$,

$\int_{B_{\frac{r}{2}}}|\nabla
u^{(k)}(x)|^{2}\geq\int_{B_{\frac{r}{2}}\setminus B_{\frac{2}{k_{1}}}}|\nabla
u(x)|^{2}dv=M_{1}>0,$ (2.7)

and

$\int_{B_{\frac{r}{2}}}\frac{|\nabla
u^{(k)}(x)|^{2}}{|x|^{2\lambda-2}}dv\geq\int_{B_{\frac{r}{2}}\setminus
B_{\frac{2}{k_{1}}}}\frac{|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv=M_{\lambda}>0.$ (2.8)

On the other hand, by flatness of $u$ at $0$,

$\int_{B_{r}}\frac{|\nabla\psi_{k}(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv=\int_{B_{\frac{2}{k}}\setminus
B_{\frac{1}{k}}}\frac{|\nabla\psi_{k}(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv\leq
4k^{2\lambda}\int_{B_{\frac{2}{k}}}|u(x)|^{2}dv\rightarrow 0$ (2.9)

as $k\rightarrow\infty$. In particular, by (2.8) one can get some
$k_{\lambda}>k_{1}$ such that

$\int_{B_{r}}\frac{|\nabla\psi_{k_{\lambda}}(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv\leq\frac{M_{\lambda}}{4}\leq\frac{1}{4}\int_{B_{r}}\frac{|\nabla
u^{(k_{\lambda})}(x)|^{2}}{|x|^{2\lambda-2}}dv.$

Thus (2.5) with $k=k_{\lambda}$ becomes

$\begin{split}&\int_{B_{2r}}\frac{|\nabla
u^{(k_{\lambda})}(x)|^{2}}{|x|^{2\lambda-2}}dv\leq 4\int_{B_{2r}\setminus
B_{r}}\frac{|\nabla\eta(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv.\end{split}$
(2.10)

Since

$\int_{B_{2r}}\frac{|\nabla
u^{(k_{\lambda})}(x)|^{2}}{|x|^{2\lambda-2}}dv\geq\int_{B_{\frac{r}{2}}}\frac{|\nabla
u^{(k_{\lambda})}(x)|^{2}}{|x|^{2\lambda-2}}dv\geq\left(\frac{2}{r}\right)^{2\lambda-2}\int_{B_{\frac{r}{2}}}|\nabla
u^{(k_{\lambda})}(x)|^{2}dv$

and

$\int_{B_{2r}\setminus
B_{r}}\frac{|\nabla\eta(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv\leq\frac{1}{r^{2\lambda-2}}\int_{B_{2r}\setminus
B_{r}}|\nabla\eta(x)|^{2}|u(x)|^{2}dv,$

we obtain from (2.10) that

$2^{2\lambda-4}\int_{B_{\frac{r}{2}}}|\nabla
u^{(k_{\lambda})}(x)|^{2}dv\leq\int_{B_{2r}\setminus
B_{r}}|\nabla\eta(x)|^{2}|u(x)|^{2}dv.$

Letting $\lambda\rightarrow\infty$ and making use of the fact that $u\in
H^{1}_{loc}(\Omega)$, we see that

$\int_{B_{\frac{r}{2}}}|\nabla u(x)|^{2}dv=0.$

But this contradicts (2.6)! We thus have $\nabla u\equiv 0$ on
$B_{\frac{r}{2}}$. By flatness of $u$ at $0$, $u$ must be zero on
$B_{\frac{r}{2}}$. ∎

######  Example 2.4.

Given $0<\varepsilon<\frac{n-1}{n},n\geq 2$, consider the differential
equation

$|\nabla u|=V|u|\ \ \text{on}\ \ B_{\frac{1}{2}},$

where

$V=\frac{\varepsilon(-\log|x|)^{\varepsilon-1}}{|x|}\ \ \text{on}\ \
B_{\frac{1}{2}}.$

It is straightforward to verify that $V\in L^{n}(B_{\frac{1}{2}})$. As a
consequence of Theorem 1.1, every nonconstant $H^{1}$ solution must vanish to
finite order in the $L^{2}$ sense at each point in $B_{\frac{1}{2}}$.

On the other hand, the function

$u_{0}(x)=\exp\left({-{(-\log|x|)^{\varepsilon}}}\right)$

(extended to the origin by $u_{0}(0)=0$) is continuous on $B_{\frac{1}{2}}$,
and smooth on $B_{\frac{1}{2}}\setminus\\{0\\}$. Moreover, $u_{0}\in
H^{1}(B_{\frac{1}{2}})$ and is a solution of $|\nabla u|=Vu$ a.e. on
$B_{\frac{1}{2}}$. Note that there is no contradiction with Theorem 1.1 since
$u_{0}$ vanishes to finite order in the $L^{2}$ sense everywhere in
$B_{\frac{1}{2}}$.

When $V\in L^{p},p<n$, the unique continuation property fails in general as
seen below.

######  Example 2.5.

For each $1\leq p<n$, and $0<\epsilon<\frac{n-p}{p}$ (so that
$(\epsilon+1)p<n$),

$u(x)=\exp\left({-\frac{1}{|x|^{\epsilon}}}\right)$

(extended to the origin by $u(0)=0$) is a smooth function on $B_{1}$ and
vanishes to infinite order at $0$. Moreover, the function $u$ satisfies
$|\nabla u|\leq V|u|$ on $B_{1}$ with

$V=\frac{\epsilon}{|x|^{\epsilon+1}}\in L^{p}(B_{1}).$

On the other hand, the following theorem states that for some special
potentials in the form of multiples of $\frac{1}{|x|}$, the unique
continuation property can still hold. Note that $\frac{1}{|x|}\notin
L^{n}_{loc}$.

######  Theorem 2.6.

Let $\Omega$ be a domain in $\mathbb{R}^{n},n\geq 1$. Suppose
$u=(u_{1},\ldots,u_{M}):\Omega\rightarrow\mathbb{R}^{M}$ with $u\in
H^{1}_{loc}(\Omega)$ and satisfies $|\nabla u|\leq\frac{C}{|x|}|u|$ a.e. for
some constant $C>0$. If $u$ vanishes to infinite order at some
$x_{0}\in\Omega$ in the $L^{2}$ sense, then $u$ vanishes identically.

###### Proof.

Assume $x_{0}=0$ and consider $u^{(k)}=\psi_{k}\eta u$, where $\psi_{k}$ and
$\eta$ are defined as in the proof of Theorem 1.1. Then by Lemma 2.1,

$\displaystyle\int_{B_{2r}}\frac{|u^{(k)}(x)|^{2}}{|x|^{2\lambda}}dv$
$\displaystyle\leq$
$\displaystyle\frac{4}{(2\lambda-n)^{2}}\int_{B_{2r}}\frac{|\nabla
u^{(k)}(x)|^{2}}{|x|^{2\lambda-2}}dv$ $\displaystyle\lesssim$
$\displaystyle\frac{4}{(2\lambda-n)^{2}}\int_{B_{2r}}\frac{|\psi_{k}(x)\eta(x)|^{2}|\nabla
u(x)|^{2}}{|x|^{2\lambda-2}}dv+\frac{4}{(2\lambda-n)^{2}}\int_{B_{r}}\frac{|\nabla\psi_{k}(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv$
$\displaystyle\ +\frac{4}{(2\lambda-n)^{2}}\int_{B_{2r}\setminus
B_{r}}\frac{|\nabla\eta(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv$
$\displaystyle\leq$
$\displaystyle\frac{4C^{2}}{(2\lambda-n)^{2}}\int_{B_{2r}}\frac{|u^{(k)}(x)|^{2}}{|x|^{2\lambda}}dv+\frac{4}{(2\lambda-n)^{2}}\int_{B_{r}}\frac{|\nabla\psi_{k}(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv$
$\displaystyle\ +\frac{4}{(2\lambda-n)^{2}}\int_{B_{2r}\setminus
B_{r}}\frac{|\nabla\eta(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv.$

Here we used the inequality $|\nabla u|\leq\frac{C}{|x|}|u|$ in the first term
of the last inequality. When $\frac{4C^{2}}{(2\lambda-n)^{2}}\leq\frac{1}{2}$
(equivalently, when $\lambda>\frac{n}{2}+\sqrt{2}C$), one can move this first
term to the left hand side and get

$\int_{B_{2r}}\frac{|u^{(k)}(x)|^{2}}{|x|^{2\lambda}}dv\leq\frac{8}{(2\lambda-n)^{2}}\int_{B_{r}}\frac{|\nabla\psi_{k}(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv+\frac{8}{(2\lambda-n)^{2}}\int_{B_{2r}\setminus
B_{r}}\frac{|\nabla\eta(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv.$

Letting $k\rightarrow\infty$ and making use of the flatness of $u$ with a
similar argument as in (2.9), we obtain

$\int_{B_{2r}}\frac{|u(x)|^{2}}{|x|^{2\lambda}}dv\leq\frac{16}{(2\lambda-n)^{2}}\int_{B_{2r}\setminus
B_{r}}\frac{|\nabla\eta(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv.$

Since

$\int_{B_{2r}}\frac{|u(x)|^{2}}{|x|^{2\lambda}}dv\geq\int_{B_{\frac{r}{2}}}\frac{|u(x)|^{2}}{|x|^{2\lambda}}dv\geq\left(\frac{2}{r}\right)^{2\lambda}\int_{B_{\frac{r}{2}}}|u(x)|^{2}dv$

and

$\int_{B_{2r}\setminus
B_{r}}\frac{|\nabla\eta(x)|^{2}|u(x)|^{2}}{|x|^{2\lambda-2}}dv\leq\frac{1}{r^{2\lambda-2}}\int_{B_{2r}\setminus
B_{r}}|\nabla\eta(x)|^{2}|u(x)|^{2}dv,$

we have

$\int_{B_{\frac{r}{2}}}|u(x)|^{2}dv\leq\frac{r^{2}}{(2\lambda-n)^{2}2^{2\lambda-4}}\int_{B_{2r}\setminus
B_{r}}|\nabla\eta(x)|^{2}|u(x)|^{2}dv.$

Letting $\lambda\rightarrow\infty$, we see $u\equiv 0$ on $B_{\frac{r}{2}}$. ∎

Although the unique continuation property for (1.1) fails for general $L^{n}$
potentials as demonstrated in Example 2.5, the following theorem shows that
the weak continuation property holds if the potential is in $L^{2}$. Recall
that weak unique continuation for a differential (in)equality is the property
that every solution that vanishes in an open subset vanishes identically.

######  Theorem 2.7.

Let $\Omega$ be a domain in $\mathbb{R}^{n},n\geq 2$, and let $V\in
L_{loc}^{2}(\Omega)$. Suppose
$u=(u_{1},\ldots,u_{M}):\Omega\rightarrow\mathbb{R}^{M}$ with $u\in
H^{1}_{loc}(\Omega)$ and satisfies $|\nabla u|\leq V|u|$ on $\Omega$. If $u$
vanishes in an open subset of $\Omega$, then $u$ vanishes identically.

###### Proof.

The $n=2$ case is a direct consequence of Theorem 1.1, since the (strong)
unique continuation implies the weak unique continuation property. We shall
show below when $n=3$, for any two domains $D_{1},D_{2}$ in $\mathbb{R}^{2}$
with $D_{1}\subset D_{2}$, and $s>0$, if $u$ satisfies (1.1) on the product
domain $D_{2}\times(-s,s)$ and $u=0$ on $D_{1}\times(-s,s)$, then $u=0$ on
$D_{2}\times(-s,s)$. If so, then $u\equiv 0$ with a standard propagation
argument. The proof for $n\geq 3$ cases follows from an induction.

Since $V\in L_{loc}^{2}(D_{2}\times(-s,s))$, by Fubini’s theorem, for almost
every $x_{3}\in(-s,s)$, $V(\cdot,x_{3})\in L^{2}_{loc}(D_{2})$, and similarly
$u(\cdot,x_{3})\in H^{1}_{loc}(D_{2})$. Restricting (2.1) at each such
$x_{3}=c_{3}\in(-s,s)$, we have $v=u(\cdot,c_{3})$ satisfies

$|\nabla
v|=\left(\sum_{k=1}^{M}|\partial_{x_{1}}u_{k}(\cdot,c_{3})|^{2}+|\partial_{x_{2}}u_{k}(\cdot,c_{3})|^{2}\right)^{\frac{1}{2}}\leq|\nabla
u(\cdot,c_{3})|\leq V(\cdot,c_{3})|u(\cdot,c_{3})|=V(\cdot,c_{3})|v|$

on $D_{2}$ and $v=0$ on $D_{1}$. Applying the $n=2$ case we have $v=0$ on
$D_{2}$. Thus $u=0$ on $D_{2}\times(-s,s)$. ∎

##  3\. Uniqueness for Lipschitz functions

In this section, we focus on locally Lipschitz functions whose definition is
given below.

######  Definition 3.1.

A function $u:\Omega\to\mathbb{R}$ is said to be locally Lipschitz on $\Omega$
means that for any point $p\in\Omega$, there is some neighborhood $p\in
U_{p}\subseteq\Omega$ and some constant $C_{p}$ so that for all $x,y\in
U_{p}$, $|u(y)-u(x)|<C_{p}|y-x|$. The function $u$ is Lipschitz on $\Omega$
means that there exists a constant $C$ such that for all $x,y\in\Omega$,
$|u(y)-u(x)|<C|y-x|$.

According to Rademacher’s Theorem, if $u$ is locally Lipschitz on $\Omega$,
then $\nabla u$ is defined a.e. on $\Omega$. See, for instance, [E, pp. 296].
Moreover,

######  Proposition 3.2.

[E, pp. 294] Let $\Omega$ be a domain in $\mathbb{R}^{n}.$ Then $u$ is locally
Lipschitz on $\Omega$ if and only if $u\in W^{1,\infty}_{loc}(\Omega)$.

Following the convention of [E], even for $u\in W^{1,\infty}_{loc}(\Omega)$
defined a.e. in $\Omega$ or with some measure zero set of discontinuities,
there is a unique continuous function agreeing with $u$ a.e., which we will
also denote $u$.

To prove Theorem 1.3, we begin with a uniqueness property of Lipschitz
functions in one real variable on an interval, making use of the following
fundamental theorem of calculus for Lipschitz functions.

######  Proposition 3.3.

[R, Theorem 7.20, Fundamental Theorem of Calculus] If $u:[0,1]\to\mathbb{R}$
is Lipschitz on $[0,1]$, then for any $0\leq a<b\leq 1$,

$u(b)-u(a)=\int_{a}^{b}u^{\prime}(t)dt.$

######  Lemma 3.4.

Let $\varphi:[0,1]\to\mathbb{R}$ be Lipschitz on $[0,1]$, with $\varphi(0)=0$.
If there exist $p\geq 1$ and a non-negative function
$\lambda:[0,1]\to\mathbb{R}$ with $\lambda\in L^{p}([0,1])$ such that for a.e.
$x\in(0,1)$,

$|\varphi^{\prime}(x)|\leq\lambda(x)\,|\varphi(x)|\,x^{\frac{1-p}{p}},$ (3.1)

then $\varphi\equiv 0$ in $[0,1]$.

###### Proof.

We note first that we can assume without loss of generality that $\lambda$ is
non-vanishing. Indeed, if that is not the case, then we can just replace
$\lambda$ with $1+\lambda\in L^{p}([0,1])$ and (3.1) still holds.

Let $\delta=\sup\\{d\in[0,1]\,|\,\varphi\equiv 0\text{ in }[0,d]\\}$. By
continuity, $\varphi(\delta)=0$, and by Proposition 3.3 (which uses the
Lipschitz hypothesis), for all $x\in(0,1]$,

$|\varphi(x)|=|\varphi(x)-\varphi(\delta)|=\left|\int_{\delta}^{x}\varphi^{\prime}(t)dt\right|\leq\int_{\delta}^{x}\left|\varphi^{\prime}(t)\right|dt.$
(3.2)

The existence of the RHS integral is from Proposition 3.2, with
$L^{\infty}([\delta,x])\subseteq L^{p}([\delta,x])\subseteq
L^{1}([\delta,x])$.

For $p>1$, let $q$ be the conjugate exponent so that
$\frac{1}{p}+\frac{1}{q}=1$. By Hölder’s inequality, we have

$\int_{\delta}^{x}|\varphi^{\prime}(t)|\,dt\leq\bigg{(}\int_{\delta}^{x}|\varphi^{\prime}(t)|^{p}\,dt\bigg{)}^{\frac{1}{p}}\bigg{(}\int_{\delta}^{x}1^{q}\,dt\bigg{)}^{\frac{1}{q}}\leq\bigg{(}\int_{\delta}^{x}|\varphi^{\prime}(t)|^{p}\,dt\bigg{)}^{\frac{1}{p}}x^{\frac{p-1}{p}}.$
(3.3)

It follows from (3.2) and (3.3) that for $p\geq 1$,

$|\varphi(x)|^{p}\leq x^{p-1}\int_{\delta}^{x}|\varphi^{\prime}(t)|^{p}\,dt.$
(3.4)

We multiply both sides of (3.4) by the function $x^{1-p}\lambda^{p}(x)$, to
obtain:

$\lambda^{p}(x)\,|\varphi(x)|^{p}x^{1-p}\leq\lambda^{p}(x)\int_{\delta}^{x}|\varphi^{\prime}(t)|^{p}\,dt.$
(3.5)

Suppose toward a contradiction that $\delta<1$, and let $s\in(\delta,1)$.
Integrating in the variable $x$ on both sides of (3.5) gives

$\int_{\delta}^{s}\lambda^{p}(x)\,|\varphi(x)|^{p}\,x^{1-p}\,dx\leq\int_{\delta}^{s}\bigg{(}\lambda^{p}(x)\int_{\delta}^{x}|\varphi^{\prime}(t)|^{p}\,dt\bigg{)}\,dx.$
(3.6)

Note that the Lipschitz property of $\varphi$ on $[0,1]$ and $\varphi(0)=0$
imply there is some constant $C$ so that $|\varphi(x)|\leq C|x|$, so
$|\varphi(x)|^{p}\,x^{1-p}$ is continuous and bounded as a function of $x$ on
$(0,1]$. Then the hypothesis $\lambda\in L^{p}([0,1])$ applies, so that both
the LHS and RHS integrals in (3.6) exist. The inequality (3.6) then implies,
first using $x\leq s$, and then the hypothesis (3.1):

$\displaystyle\int_{\delta}^{s}\lambda^{p}(x)\,|\varphi(x)|^{p}\,x^{1-p}\,dx$
$\displaystyle\leq$
$\displaystyle\bigg{(}\int_{\delta}^{s}\lambda^{p}(x)\,dx\bigg{)}\bigg{(}\int_{\delta}^{s}|\varphi^{\prime}(x)|^{p}\,dx\bigg{)}$
(3.7) $\displaystyle\leq$
$\displaystyle\bigg{(}\int_{\delta}^{s}\lambda^{p}(x)\,dx\bigg{)}\,\bigg{(}\int_{\delta}^{s}\lambda^{p}(x)\,|\varphi(x)|^{p}\,x^{1-p}\,dx\bigg{)}.$

By the construction of $\delta$ as the supremum of a set where
$\varphi(x)\equiv 0$, we can find a sequence of points $s_{j}\in(\delta,1)$ so
that $s_{j}$ is decreasing, $\displaystyle{\lim_{j\to\infty}s_{j}=\delta}$,
and $\varphi(s_{j})\neq 0$. By the continuity of $|\varphi(x)|^{p}\,x^{1-p}$
and the property that $\lambda^{p}\geq 1$, the integrand
$\lambda^{p}(x)\,|\varphi(x)|^{p}\,x^{1-p}$ is strictly positive in some
neighborhood of $s_{j}$. So, for all $j=1,2,3,\ldots$,

$\int_{\delta}^{s_{j}}\lambda^{p}(x)\,|\varphi(x)|^{p}\,x^{1-p}\,dx>0.$

The inequality (3.7) then yields, for all $j$,

$1\leq\int_{\delta}^{s_{j}}\lambda^{p}(x)\,dx.$ (3.8)

Since $\lambda\in L^{p}(0,1)$, letting $s_{j}\to\delta$ in (3.8) leads to a
contradiction. ∎

Recalling Rademacher’s theorem that a Lipschitz function is differentiable
almost everywhere, the following simple, but useful, Lemma gives a set of
points where the square of a Lipschitz function is known to be differentiable.

######  Lemma 3.5.

Let $u$ be a locally Lipschitz function on an open set
$\Omega\subseteq\mathbb{R}^{n}$. Then the square function $g=u^{2}$ is also
locally Lipschitz on $\Omega$. Moreover, $g$ is differentiable wherever $u$
vanishes and in fact $\nabla g(x)=0$ there.

###### Proof.

The locally Lipschitz property of $g$ follows from the well-known fact that
the product of locally Lipschitz functions is locally Lipschitz; this is
easily checked as an elementary consequence of Definition 3.1. The second
claim is also elementary; let $x_{0}\in\Omega$ be such that $u(x_{0})=0$. The
properties that $g=u^{2}$ is differentiable at $x_{0}$ and $\nabla g(x_{0})=0$
follow from the definition of differentiability,

$\lim_{x\to x_{0}}\frac{g(x)-g(x_{0})-0\cdot(x-x_{0})}{|x-x_{0}|}=\lim_{x\to
x_{0}}\frac{u^{2}(x)}{|x-x_{0}|}=0,$

where we have used the Lipschitz property of $u(x)=u(x)-u(x_{0})=O(|x-x_{0}|)$
in a neighborhood of $x_{0}$. ∎

######  Lemma 3.6.

For $n\geq 2$, let $A$ be a set of measure zero in the unit ball in
$\mathbb{R}^{n}$. Then for almost all points $\omega$ in the unit sphere
$S^{n-1}$, the set of intersection of $A$ with the radius segment
$\\{r\omega:0\leq r\leq 1\\}$ is of measure zero in the line measure.

###### Proof.

Let $|K|$ denote the $d$-dimensional measure of a measurable set
$K\subseteq\mathbb{R}^{d}$, and let $\chi_{A}:\mathbb{R}^{n}\to\mathbb{R}$ be
the characteristic function of the set $A$. We have

$0=|A|=\int_{|x|<1}\chi_{A}(x)dv=\int_{S^{n-1}}\int_{0}^{1}\chi_{A}(r\omega)r^{n-1}drd\omega.$

By Fubini’s theorem, we conclude that for a.e. $\omega\in S^{n-1}$,
$\int_{0}^{1}\chi_{A}(r\omega)r^{n-1}dr=0$, which is the desired result:
$|A\cap\\{r\omega\\}|=0$. ∎

Given a locally Lipschitz function $u$ on $\Omega$, denote by $Z_{u}$ be the
zero set of $u$ in $\Omega$, that is, $Z_{u}=\\{x\in\Omega\,|\,u(x)=0\\}$.
Theorem 1.3 will be a consequence of the following general result concerning
Lipschitz functions.

######  Theorem 3.7.

Let $\Omega$ be a domain in $\mathbb{R}^{n},n\geq 2$, and $u$ be a locally
Lipschitz function on $\Omega$. If the zero set ${Z_{u}}$ of $u$ is neither
$\emptyset$ nor $\Omega$, then

$\int_{\Omega\setminus{Z_{u}}}\bigg{|}\frac{\nabla
u(x)}{u(x)}\bigg{|}^{n}\,dv=\infty.$ (3.9)

###### Proof.

First we make the observation that in order to prove the theorem it suffices
to prove it for $g=u^{2}$. In fact, since

$\int_{\Omega\setminus{Z_{u}}}\bigg{|}\frac{\nabla
g(x)}{g(x)}\bigg{|}^{n}dv=2^{n}\int_{\Omega\setminus{Z_{u}}}\bigg{|}\frac{\nabla
u(x)}{u(x)}\bigg{|}^{n}\,dv,$

if the conclusion (3.9) is true for $g=u^{2}$, then it is also true for $u$.
Hence we only need to prove (3.9) for functions that are the square of a
locally Lipschitz function. By Lemma 3.5, the gradient of $g$ is $0$ at every
point where $g$ is $0$ (the same set ${Z_{u}}$ where $u$ is $0$), and $g$ also
satisfies the locally Lipschitz assumption. For the rest of the proof we
assume (by replacing $u$ with $g$) that $\nabla u(x)=0$ wherever $u(x)=0$. Let

$V(x)=\begin{cases}\Big{|}\frac{\nabla u(x)}{u(x)}\Big{|}\quad
x\in\Omega\setminus{Z_{u}}\\\ 0\quad\quad\quad\,\,\,x\in{Z_{u}}.\end{cases}$
(3.10)

Note that $V$ is a measurable function in $\Omega$. The zero set ${Z_{u}}$ is
closed in $\Omega$, and by the assumptions that ${Z_{u}}\neq\emptyset$,
${Z_{u}}\neq\Omega$, and $\Omega$ is connected, there is some boundary point
$x_{0}\in\partial{Z_{u}}\subseteq{Z_{u}}\subseteq\Omega$, and a ball
$B(x_{0},r_{0})$ of radius $r_{0}>0$ centered at $x_{0}$ such that the closure
$\overline{B(x_{0},r_{0})}\subset\Omega$, and $u$ is Lipschitz on
$\overline{B(x_{0},r_{0})}$. We can assume, after a translation and scaling,
that $x_{0}$ is the origin and the radius $r_{0}$ is equal to $1$. Because
$B(0,1)\setminus{Z_{u}}$ is open and non-empty, there is some
$B(x_{1},r_{1})\subseteq B(0,1)$ where $u$ is nonvanishing.

Now suppose toward a contradiction that (3.9) is false:

$\int_{\Omega}V^{n}(x)\,dv=\int_{\Omega\setminus{Z_{u}}}\bigg{|}\frac{\nabla
u(x)}{u(x)}\bigg{|}^{n}\,dv<\infty,$ (3.11)

and therefore $V\in L^{n}(\Omega)$. Hence in polar coordinates,

$\int_{B(0,1)}V^{n}(x)\,dv=\int_{S^{n-1}}\int_{0}^{1}V^{n}(r\omega)r^{n-1}\,drd\omega.$
(3.12)

Since the integral (3.12) is finite, Fubini’s theorem implies that for a.e.
$\omega\in S^{n-1}$ we have

$\int_{0}^{1}V^{n}(r\omega)r^{n-1}\,dr<\infty.$ (3.13)

From Rademacher’s theorem, let $A\subseteq\Omega$ be the set of measure zero
where $\nabla u(x)$ does not exist at $x$. Choose $\omega_{0}\in S^{n-1}$ such
that (3.13) holds, that is, $V(r\omega_{0})r^{\frac{n-1}{n}}\in L^{n}([0,1])$
and at the same time, by Lemma 3.6 the same $\omega_{0}\in S^{n-1}$ can be
chosen such that $\nabla u(x)$ exists a.e. on the radius segment
$\\{r\omega_{0}\\}$. Define $\varphi$, for $t\in[0,1]$, by

$\varphi(t)=u(t\omega_{0}).$

It is evident that $\varphi(t)$ is Lipschitz on $[0,1]$ from the Lipschitz
property of $u$. Then applying the chain rule at points $t$ such that $u$ is
differentiable at $t\omega_{0}$, we have

$\varphi^{\prime}(t)=\nabla u(t\omega_{0})\cdot\omega_{0},$

which implies, for a.e. $t\in[0,1]$,

$|\varphi^{\prime}(t)|\leq|\nabla u(t\omega_{0})|.$ (3.14)

By the definition of $V$, we have

$|\varphi^{\prime}(t)|\leq
V(t\omega_{0})|u(t\omega_{0})|=V(t\omega_{0})|\varphi(t)|\quad\text{for
}\,u(t\omega_{0})\neq 0.$ (3.15)

However, when $u(t\omega_{0})=0$, we have, by the observation at the beginning
of the proof, $\nabla u(r\omega_{0})=0$ and therefore $\varphi^{\prime}(t)=0$.
Hence we have shown that

$|\varphi^{\prime}(t)|\leq
V(t\omega_{0})|\varphi(t)|=V(t\omega_{0})t^{\frac{n-1}{n}}|\varphi(t)|t^{-\frac{n-1}{n}}$

holds for a.e. $t\in[0,1]$. By Lemma 3.4, with
$\lambda(t)=V(t\omega_{0})t^{\frac{n-1}{n}}$ and $p=n$, $\varphi(t)\equiv 0$.
So $u\equiv 0$ on all the radius segments $\\{t\omega_{0}\\}$ for a.e.
$\omega_{0}\in S^{n-1}$, but this contradicts the fact that $u$ has no zeros
in the ball $B(x_{1},r_{1})$. ∎

######  Remark 3.8.

The proof of Theorem 3.7 actually leads to the following stronger conclusion:
under the same assumption as in Theorem 3.7, on every neighborhood
$U\subset\Omega$ of a point $a\in\Omega\cap\partial{Z_{u}}$, one has

$\int_{U\setminus{Z_{u}}}\bigg{|}\frac{\nabla
u(x)}{u(x)}\bigg{|}^{n}\,dv=\infty.$

###### Proof of Theorem 1.3.

For the one-dimensional case $n=1$, $\Omega$ is an open interval $(a,b)$ and
Lemma 3.4 can be used directly. For any $s\in(x_{0},b)$, let
$\varphi(x)=u((s-x_{0})x+x_{0})$ so that $\varphi(0)=u(x_{0})=0$,
$\varphi(1)=u(s)$, and $\varphi$ is Lipschitz on $[0,1]$. Lemma 3.4 applies to
$\varphi$ with $p=1$ and $\lambda(x)=V((s-x_{0})x+x_{0})\cdot|s-x_{0}|\in
L^{1}([0,1])$, to show $\varphi(1)=0=u(s)$. Similarly, $u(t)=0$ for any
$a<t<x_{0}$.

For $n\geq 2$, suppose $u$ and $V$ satisfy $|\nabla u|\leq V|u|$ a.e. on
$\Omega$ with $V\in L^{n}_{loc}(\Omega)$. Let $B$ be a nonempty open ball
centered at $x_{0}$ such that $\overline{B}\subset\Omega$. Suppose toward a
contradiction that $B\not\subseteq{Z_{u}}$. Then $u$ is locally Lipschitz on
$B$, and the zero set $Z_{u}\cap B$ of $u$ in $B$ is neither $\emptyset$
(since $x_{0}\in Z_{u}\cap B$) nor $B$. However,

$\int_{B\setminus{Z_{u}}}\left|\frac{\nabla
u(x)}{u(x)}\right|^{n}dv\leq\int_{B\setminus{Z_{u}}}\left|V(x)\right|^{n}dv<\infty,$

contradicting Theorem 3.7. We can conclude $B\subseteq{Z_{u}}$. Thus ${Z_{u}}$
is both open and closed in the connected set $\Omega$ and $u\equiv 0$. ∎

######  Remark 3.9.

The Lipschitz condition cannot just be dropped in Theorem 1.3 when $n\geq 2$.
Indeed, Example 2.4 gives a nontrivial function $u_{0}$ that is locally
Lipschitz on $B_{\frac{1}{2}}\setminus\\{0\\}$, continuous on
$B_{\frac{1}{2}}$ with $u_{0}(0)=0$, and solves $|\nabla u|=V|u|$ on
$B_{\frac{1}{2}}$ for some $V\in L^{n}(B_{\frac{1}{2}})$. This indicates that
the zero set of solutions fail to propagate at a non-Lipschitz point in
general.

On the other hand, the hypothesis of Theorem 1.3 can be weakened as in the
following Corollary without contradicting Remark 3.9 — if $u$ is a continuous
$k^{th}$ root of a locally Lipschitz function then the uniqueness still holds.

######  Corollary 3.10.

Let $\Omega$ be a domain in $\mathbb{R}^{n}$ and $u:\Omega\to\mathbb{R}$ be
continuous on $\Omega$ with the zero set $Z_{u}\subseteq\Omega$. If there is
some integer $k\geq 1$ so that $v(x)=(u(x))^{k}$ is locally Lipschitz on
$\Omega$, then $\nabla u$ exists a.e. in $\Omega\setminus Z_{u}$. Further, if
there is some $V\in L^{n}_{loc}(\Omega)$, so that $\nabla u$ satisfies

$|\nabla u|\leq V|u|\mbox{ \ \ \ a.e.\ on $\Omega\setminus Z_{u}$,}$

and $Z_{u}\neq\emptyset$, then $u\equiv 0$.

###### Proof.

On the open set where $u(x)>0$, the partial derivatives of $v$ exist a.e. and
at each point where $\nabla v$ exists, by the chain rule, the partial
derivatives of $u(x)=(v(x))^{1/k}$ exist, with $\nabla
u(x)=\frac{1}{k}(v(x))^{\frac{1}{k}-1}\nabla v(x)$. Similarly, on the open set
where $u(x)<0$, the partial derivatives of $u=-((-1)^{k}v)^{1/k}$ exist a.e.,
establishing the first claim.

Consider $g(x)=(v(x))^{2}=(u(x))^{2k}$. By Lemma 3.5, $g$ is locally Lipschitz
on $\Omega$ and at every point where $v(x)=0$, which is the same set as
$Z_{u}$, the partial derivatives of $g$ exist with $\nabla g=0$. At every
point where $u(x)\neq 0$, if $\nabla u$ exists, then $\nabla g$ also exists
and is equal to $2ku^{2k-1}\nabla u$. So, at every point in $\Omega$ except
for a set of measure zero contained in $\Omega\setminus Z_{u}$, $\nabla g$
exists and satisfies:

$|\nabla g|=2k|u|^{2k-1}|\nabla u|\leq 2kV|u|^{2k}=2kV|g|.$

Applying Theorem 1.3 and the assumption $Z_{u}\neq\emptyset$, we have $g\equiv
0$, and thus $u\equiv 0$. ∎

We conclude the section with an application of Theorem 1.3 to a uniqueness
problem for a nonlinear system of differential equations.

######  Corollary 3.11.

For a domain $\Omega\subseteq\mathbb{R}^{n}$, $u:\Omega\to\mathbb{R}$,
$x_{0}\in\Omega$, $y_{0}\in\mathbb{R}$, let
$f:\mathbb{R}\rightarrow\mathbb{R}^{n}$ be Lipschitz on $\mathbb{R}$. Then
there exists at most one Lipschitz solution to $\nabla u=f(u)$ on $\Omega$
with $u(x_{0})=y_{0}$.

###### Proof.

Suppose there exists a pair of Lipschitz solutions $u_{1},u_{2}$ to $\nabla
u=f(u)$ on $\Omega$ with the same initial condition
$u_{1}(x_{0})=u_{2}(x_{0})=y_{0}$. Then $w=u_{1}-u_{2}$ is Lipschitz on
$\Omega$, $w(x_{0})=0$, and $w$ satisfies

$|\nabla w|=|\nabla u_{1}-\nabla u_{2}|=|f(u_{1})-f(u_{2})|\leq
C|u_{1}-u_{2}|=C|w|\ \ \text{on}\ \ \Omega,$

where $C>0$ is the Lipschitz constant for $f$. By Theorem 1.3, we have
$w\equiv 0$. ∎

##  4\. Further applications

######  Corollary 4.1.

Given any nonempty closed set $A\subsetneq\mathbb{R}^{n}$, there exists a
smooth function $F:\mathbb{R}^{n}\setminus A\to\mathbb{R}$ so that for any
point $a\in\partial A$, the boundary of $A$, $F$ cannot be extended to an
$L^{n}$ integrable function over any neighborhood of $a$.

###### Proof.

First, by a well-known theorem of Whitney ([W]), there exists a smooth
function $h:\mathbb{R}^{n}\to\mathbb{R}$ whose zero set is exactly $A$. Its
square, $u(x)=(h(x))^{2}$ is also smooth, has zero set exactly $A$, and
satisfies $\nabla u=0$ at every point of $A$ by Lemma 3.5. The quotient
$\frac{|\nabla u|}{|u|}$ is the claimed smooth function $F$ on the open set
$\mathbb{R}^{n}\setminus A$. If there were some ball $B$ centered at $a$ and a
function $V\in L^{n}(B)$ which agrees with $\frac{|\nabla u|}{|u|}$ on
$B\setminus A$, then $u$ and $V$ would satisfy (1.1) from Theorem 1.3 at every
point of $B\setminus A$ by construction of $V$, and at every point of $B\cap
A$, where $|\nabla u|=0$. By Theorem 1.3, $u\equiv 0$ on $B$, contradicting
the assumption that $a$ is a boundary point of the zero set. ∎

######  Theorem 4.2.

Let $\Omega$ be an open set in $\mathbb{R}^{n}$, and $u$ be a Lipschitz
function on $\Omega$. Then $\log|u(x)-u(a)|\notin
W_{loc}^{1,n}(\Omega\setminus Z_{u-u(a)})$ for every $a\in\Omega$. In
particular, if $\log|u(x)|\in W_{loc}^{1,n}(\Omega\setminus Z_{u})$, then $u$
is nowhere zero on $\Omega$. If in addition that $\Omega$ has Lipschitz
boundary, then the above results also hold true with $\Omega$ replaced by
$\overline{\Omega}$.

###### Proof.

Let $v=u-u(a)$ on $\Omega$. Then $v$ is Lipschitz on $\Omega$ with $v(a)=0$.
If $\log|v|\notin L_{loc}^{n}(\Omega)$, then we are done. If $\log|v|\in
L_{loc}^{n}(\Omega)$, then one further computes

$\left|\nabla\log|v|\right|=\frac{\left|\nabla v\right|}{|v|}$

wherever $v\neq 0$. If $a$ is an interior point of
$Z_{v}=\\{x\in\Omega\,|\,v(x)=0\\}$, the zero set of $v$, then the theorem is
trivially true. If $a\in\partial Z_{v}\cap\Omega$, then one can apply Remark
3.8 to conclude $\nabla\log|v|\notin L_{loc}^{n}(\Omega\setminus Z_{v})$.

Inr the case when $a\in\partial\Omega$ and $\Omega$ has Lipschitz boundary, if
$v(b)=0$ for some $b\in\Omega$, then it is reduced to $a\in\Omega$ case. Thus
we assume $a\in\partial\Omega$ and $v(x)\neq 0$ for all $x\in\Omega$. In
particular, this means there exists a cone $S_{a}\subset\Omega$ centered at
$a$ (which exists since $\Omega$ has Lipschitz boundary) such that $v\neq 0$
on $S_{a}$. Making use of a similar argument as in the proof of Theorem 3.7,
with $B(0,1)$ replaced by $S_{a}$, one can obtain $\nabla\log|v|\notin
L_{loc}^{n}(\Omega)$. ∎

A natural way to view Theorem 4.2 is as follows. Denote by $Lip(\Omega)$ the
set of all Lipschitz functions on $\Omega$. Theorem 4.2 implies that for each
$a\in{\Omega}$,

$T_{a}(Lip(\Omega))\cap W^{1,n}(\Omega)=\emptyset,$

where $T_{a}$ is a (non-linear) map on $Lip(\Omega)$ defined by
$T_{a}(u)=\log|u-u(a)|$, $u\in Lip(\Omega)$. The following Corollary is a
direct consequence of Theorem 4.2 as well.

######  Corollary 4.3.

Let $\Omega$ be an open set in $\mathbb{R}^{n}$ and
$u:\Omega\rightarrow\mathbb{R}$ be locally Lipschitz on $\Omega$. If the zero
set $Z_{u}$ of $u$ is neither empty nor $\Omega$, and

$\int_{\Omega\setminus Z_{u}}|\nabla\log|u(x)||^{p}\,dv<\infty.$ (4.1)

then $p<n$.

######  Proposition 4.4.

Let $\Omega$ be an open set in $\mathbb{R}^{n},n\geq 2$, and
$\phi:\Omega\to\mathbb{R}$ with $\phi\in W_{loc}^{1,n}(\Omega)$. The following
statements hold for the exponential $e^{-|\phi|}$.

  1. (1)

$e^{-|\phi|}$ vanishes to finite order in the $L^{2}$ sense anywhere in
$\Omega$.

  2. (2)

If $e^{-|\phi|}$ is locally Lipschitz on $\Omega$, then $e^{-|\phi|}$ is
nowhere zero on $\Omega$. $e^{-|\phi|}$ is nowhere zero on $\overline{\Omega}$
if in addition $\Omega$ has Lipschitz boundary.

###### Proof.

Since $\phi\in W_{loc}^{1,n}(\Omega)$, we have $|\phi|\in
W_{loc}^{1,n}(\Omega)$ as well. The function $u=e^{-|\phi|}$ satisfies $|u|<1$
and

$|\nabla
u|=|\nabla|\phi||e^{-|\phi|}=\left|\nabla|\phi|\right||u|\leq\left|\nabla|\phi|\right|\in
L_{loc}^{n}(\Omega).$

See for instance [E, pp. 308]. Hence $u\in W_{loc}^{1,n}(\Omega)$ and
satisfies $|\nabla u|=V|u|$ with $V=|\nabla|\phi||\in L_{loc}^{n}(\Omega)$. By
Theorem 1.1, $u$ cannot vanish to infinite order in the $L^{2}$ sense anywhere
in $\Omega$.

If $e^{-|\phi|}$ is also Lipschitz on $\Omega$, and $e^{-|\phi|}$ is zero at
$x_{0}\in\Omega$, then $|\phi|=-\log|u-u(x_{0})|\notin W_{loc}^{1,n}(\Omega)$
by Theorem 4.2. Contradiction! ∎

Before proving Theorem 1.2, let us recall the Moser-Trudinger inequality: Let
$\Omega$ be a bounded domain in $\mathbb{R}^{n}$ with Lipschitz boundary, and
$\alpha_{n}=nw_{n-1}^{\frac{1}{n-1}}$ where $w_{n-1}$ is the surface area of
the unit sphere in $\mathbb{R}^{n}$. There exists a positive constant $C_{MT}$
depending only on $n$ such that

$\sup_{u\in W_{0}^{1,n}(\Omega),\ \ \|\nabla u\|_{L^{n}(\Omega)}\leq
1}\int_{\Omega}\exp\left({\alpha_{n}|u(x)|^{\frac{n}{n-1}}}\right)dv\leq
C_{MT}|\Omega|.$

Here $|\Omega|$ is the volume of $\Omega$. We shall use the Moser-Trudinger
inequality to prove that the exponential of $W^{1,n}$ functions is $L^{2}$
integrable.

###### Proof of Theorem 1.2: .

First we show that $e^{\phi}\in L^{2}_{loc}(\Omega)$. Given $x_{0}\in\Omega$,
let $B_{r}(x_{0})$ be the ball centered at $x_{0}$ with radius $r$. Pick $r$
small enough such that $B_{2r}(x_{0})\subset\Omega$. By Sobolev extension
theorem, there exists an extension $\tilde{\phi}\in
W_{0}^{1,n}(B_{2r}(x_{0}))$ of $\phi|_{B_{r}(x_{0})}$ such that

$a=\|\nabla\tilde{\phi}\|_{L^{n}(B_{2r}(x_{0})))}\leq
C\|\phi\|_{W^{1,n}(B_{r}(x_{0})))}$

for some constant $C$ dependent only on $r$ and $n$. In particular,
$\tilde{\phi}_{1}:=a^{-1}\tilde{\phi}\in W_{0}^{1,n}(B_{2r}(x_{0}))$ and
$\|\nabla\tilde{\phi}_{1}\|_{L^{n}(B_{2r}(x_{0}))}\leq 1.$ Thus one applies
the Moser-Trudinger inequality to obtain

$\int_{B_{2r}(x_{0})}\exp\left({\alpha_{n}|\tilde{\phi}_{1}(x)|^{\frac{n}{n-1}}}\right)dv\lesssim
1.$

Noting that $2\tilde{\phi}<\alpha_{n}|\tilde{\phi}_{1}|^{\frac{n}{n-1}}$ when
$|\tilde{\phi}|>2^{n-1}a^{n}\alpha_{n}^{1-n}$, we further have

$\int_{B_{2r}(x_{0})\cap\\{|\tilde{\phi}|>2^{n-1}a^{n}\alpha_{n}^{1-n}\\}}\exp\left({2\tilde{\phi}(x)}\right)\
dv\leq\int_{B_{2r}(x_{0})}\exp\left({\alpha_{n}|\tilde{\phi}_{1}(x)|^{\frac{n}{n-1}}}\right)\
dv\lesssim 1.$

The claim that $e^{\phi}\in L^{2}_{loc}(\Omega)$ is thus a consequence of the
following inequality

$\displaystyle\int_{B_{r}(x_{0})}e^{2\phi(x)}\ dv$ $\displaystyle\leq$
$\displaystyle\int_{B_{2r}(x_{0})\cap\\{|\tilde{\phi}|\leq
2^{n-1}a^{n}\alpha_{n}^{1-n}\\}}e^{2\tilde{\phi}(x)}\
dv+\int_{B_{2r}(x_{0})\cap\\{|\tilde{\phi}|>2^{n-1}a^{n}\alpha_{n}^{1-n}\\}}e^{2\tilde{\phi}(x)}\
dv$ $\displaystyle\lesssim$
$\displaystyle\exp\left({2^{n}a^{n}\alpha_{n}^{1-n}}\right)r^{n}+1.$

On the other hand, by Proposition 4.4 part (1), $e^{-|\phi|}$ vanishes to
finite order in the $L^{2}$ sense at $x_{0}$. Equivalently, there exists some
$m_{0}\geq 0$ such that

$\overline{\lim_{r\rightarrow 0}}\
r^{-m_{0}}\int_{|x-x_{0}|<r}|e^{-|\phi(x)|}|^{2}dv>0.$

Since $e^{\phi}\geq e^{-|\phi|}\geq 0$ and $e^{\phi}\in L^{2}_{loc}(\Omega)$,
we further have

$\overline{\lim_{r\rightarrow 0}}\
r^{-m_{0}}\int_{|x-x_{0}|<r}|e^{\phi(x)}|^{2}dv\geq\overline{\lim_{r\rightarrow
0}}\ r^{-m_{0}}\int_{|x-x_{0}|<r}|e^{-|\phi(x)|}|^{2}dv>0.$

Namely, $e^{\phi}$ vanishes to finite order in the $L^{2}$ sense at $x_{0}$. ∎

######  Corollary 4.5.

Let $\Omega$ be an open set in $\mathbb{R}^{n},n\geq 2$. Suppose
$\phi:\Omega\to\mathbb{R}$ with $\phi\in W_{loc}^{1,n}(\Omega)$. If $e^{\phi}$
is Lipschitz on $\Omega$, then $e^{\phi}$ is nowhere zero on $\Omega$.

###### Proof.

It is not hard to verify that for all $x_{1},x_{2}\in\Omega$,

$\left|e^{-|\phi(x_{2})|}-e^{-|\phi(x_{1})|}\right|\leq|e^{\phi(x_{1})}-e^{\phi(x_{2})}|.$

In particular, $e^{-|\phi|}$ is Lipchitz whenever $e^{\phi}$ is so. Applying
Proposition 4.4 part (2), we have $e^{-|\phi|}$, and thus $e^{\phi}$, is
nowhere zero on $\Omega$. ∎

##  5\. In relation to $\bar{\partial}$

On domains in $\mathbb{C}^{n}$, if the gradient operator $\nabla$ is replaced
by the $\bar{\partial}$ operator, then Theorem 1.3 fails, even for real
analytic functions. In fact, there are real analytic functions that vanish to
any given order at one point and satisfy $|\bar{\partial}u|\leq V|u|$ for some
$V\in L^{\infty}$.

######  Example 5.1.

Let $f$ be a holomorphic function on $B_{1}\subset\mathbb{C}^{n}$ that
vanishes to order $k$ at $0$, $k\geq 1$. Letting
$u(z)=\left(1+\frac{\bar{z}_{1}}{2}\right)f(z)$, then $u$ is real analytic on
$B_{1}$, vanishes to order $k$ at $0$ and satisfies $|\bar{\partial}u|\leq
4|u|$.

On the other hand, since $|\nabla u|^{2}=|\partial
u|^{2}+|\bar{\partial}u|^{2}$ for a Lipschitz $u$, by Theorem 1.3 we have near
any neighborhood $U$ of a zero point in $\partial Z_{u}$ of $u$,

$\int_{U}\frac{|\nabla u(z)|^{2}}{|u(z)|^{2}}\
dv=\int_{U}\frac{|\bar{\partial}u(z)|^{2}}{|u(z)|^{2}}+\frac{|\partial
u(z)|^{2}}{|u(z)|^{2}}\ dv=\infty.$

The following propositions discuss a finer property about the $L^{2}$
divergence of $\frac{\nabla u}{u}$ concerning the smooth extension of
holomorphic functions beyond the boundary. In particular, they exhibit an
intrinsic obstruction for holomorphic functions to be extended smoothly across
the boundary. We note that for smooth functions, the flatness in the $L^{2}$
sense at a point is equivalent to the vanishing of all jets at that point.

######  Proposition 5.2.

Let $\Omega$ be a domain in $\mathbb{C}$ and $z_{0}\in\partial\Omega$. Let $u$
be a nonconstant holomorphic function on $\Omega$. If $u$ can be extended
smoothly across $z_{0}$, still denoted by $u$, and $u(z_{0})=0$, then there
exists a neighborhood $U$ of $z_{0}$ such that one of the following holds.

  1. (1)

If $u$ vanishes to finite order at $z_{0}$, then

$\int_{U}\frac{|\bar{\partial}u(z)|^{2}}{|u(z)|^{2}}\ dv<\infty\ \ \text{and}\
\ \int_{U}\frac{|\partial u(z)|^{2}}{|u(z)|^{2}}\ dv=\infty.$ (5.1)

  2. (2)

If $u$ vanishes to infinite order at $z_{0}$, then

$\int_{U}\frac{|\bar{\partial}u(z)|^{2}}{|u(z)|^{2}}\ dv=\infty.$ (5.2)

###### Proof.

Without loss of generality let $z_{0}=0$. In (1), since $u$ vanishes to finite
order at $0$ and is holomorphic on $\Omega$, $u=cz^{k}+O(|z|^{k+1})$ near $0$
for some constant $c\neq 0,k\in\mathbb{Z}^{+}$. With a direct computation we
have

$\frac{|\bar{\partial}u|}{|u|}=\frac{O(z^{k})}{|cz^{k}+O(|z|^{k+1})|}=O(1)\ \
\text{and}\ \ \frac{|\partial u|}{|u|}=\frac{k}{|z|}+O(1),$ (5.3)

from which (5.1) follows.

For (2), if not, then set $V=\frac{\bar{\partial}u}{u}$ where $u\neq 0$, and
$V=0$ otherwise on $U$, so that $V\in L^{2}(U)$ and $\bar{\partial}u=Vu$ on
$U$. According to Theorem 2.3, since $u$ is flat at $z_{0}$, we have $u\equiv
0$ on $U$. In particular, $u=0$ on the open set $U\cap\Omega$. By the
holomorphic property of $u$ on $\Omega$, we further have $u\equiv 0$ on
$\Omega$. This contradicts the assumption that $u$ is nonconstant on $\Omega$.
∎

The following two corollaries give alternative characterizations on the
vanishing order of smooth extension of holomorphic functions across the
boundary.

######  Corollary 5.3.

Let $\Omega$ be a domain in $\mathbb{C}$ and $z_{0}\in\partial\Omega$. Let $u$
be a nonconstant holomorphic function on $\Omega$, and extend smoothly across
$z_{0}$, still denoted by $u$, with $u(z_{0})=0$. Then the following
statements are equivalent to each other.

  1. (1)

$u$ vanishes to finite order at $z_{0}$.

  2. (2)

$\frac{|\bar{\partial}u|}{|u|}\in L^{\infty}$ near $z_{0}$.

  3. (3)

$\frac{|\bar{\partial}u|}{|u|}\in L^{2}$ near $z_{0}$.

######  Corollary 5.4.

Let $\Omega$ be a domain in $\mathbb{C}$ and $z_{0}\in\partial\Omega$. Let $u$
be a nonconstant holomorphic function on $\Omega$, and extend smoothly across
$z_{0}$, still denoted by $u$, with $u(z_{0})=0$. Then the following
statements are equivalent to each other.

  1. (1)

$u$ vanishes to infinite order at $z_{0}$.

  2. (2)

$\frac{|\bar{\partial}u|}{|u|}\notin L^{\infty}$ near $z_{0}$.

  3. (3)

$\frac{|\bar{\partial}u|}{|u|}\notin L^{2}$ near $z_{0}$.

###### Proof of Corollary 5.3 and 5.4:.

For Corollary 5.3, (2) $\Rightarrow$ (3) is trivial. (3) $\Leftrightarrow$ (1)
is a direct consequence of Proposition 5.2. (1) $\Rightarrow$ (2) follows from
(5.3) in the proof of Proposition 5.2. Corollary 5.4 can be proved similarly.
∎

######  Example 5.5.

Let $\mathbb{H}^{+}$ be the upper half plane in $\mathbb{C}$. The function

$u=\exp\left({\displaystyle{\frac{1}{i\sqrt{iz}}}}\right),\ \ \ \arg
iz\in(\frac{\pi}{2},\frac{3\pi}{2}),$

is holomorphic on $\mathbb{H}^{+}$ and vanishes to infinite order at
$z_{0}=0$. It allows for a smooth extension across $0$. By Proposition 5.2
(2), every smooth extension of $u$ on a neighborhood $U$ of $0$ should satisfy
(5.2). Note that $u$ cannot extend holomorphically across $0$.

For every $k\geq 1$, the function $u=z^{k}$ is holomorphic on $\mathbb{H}^{+}$
and vanishes to finite order $k$ at $0$. By Proposition 5.2 (1), every smooth
extension of $u$ on a neighborhood $U$ of $0$ should satisfy (5.1). For a less
trivial example towards Proposition 5.2 (1) without holomorphic extension
across $0$, one can consider $u=z^{k}+e^{\frac{1}{i\sqrt{iz}}}$ on
$\mathbb{H}^{+}$ instead, and obtain (5.1) for every smooth extension of $u$
across $0$.

######  Proposition 5.6.

Let $\Omega$ be a domain in $\mathbb{C}^{n}$ and $z_{0}\in\partial\Omega$. Let
$u$ be a nonconstant holomorphic function on $\Omega$. If $u$ can be extended
smoothly across $z_{0}$, still denoted by $u$, and $u(z_{0})=0$, then there
exists a neighborhood $U$ of $z_{0}$ such that one of the following holds.

  1. (1)

If $u$ vanishes to finite order at $z_{0}$, then there exists a complex line
$L$ passing through $z_{0}$ such that

$\int_{U\cap L}\frac{|\partial u(z)|^{2}}{|u(z)|^{2}}\ dv=\infty.$ (5.4)

  2. (2)

If $u$ vanishes to infinite order at $z_{0}$, then for all complex line $L$
passing through $z_{0}$,

$\int_{U\cap L}\frac{|\bar{\partial}u(z)|^{2}}{|u(z)|^{2}}\ dv=\infty.$ (5.5)

###### Proof.

For simplicity let $z_{0}=0$ and $n=2$. The higher dimensional cases can be
proved similarly. If $u$ vanishes to finite order at $0$, then after a
holomorphic change of coordinates, there exists some $k\in\mathbb{Z}^{+}$ such
that

$u=z_{1}^{k}+g_{k-1}(z_{2})z_{1}^{k-1}+\cdots+g_{0}(z_{2})+h(z)$

near $0$. Here for each $j=0,\ldots,k-1$, $g_{j}$ is smooth on $U$,
holomorphic on $\Omega\cap U$ and $g_{j}(0)=0$, and $h$ is a smooth function
on $U$ with $h=0$ on $\Omega\cap U$. In particular, $h$ is flat at $0$. Thus
on the complex line $L=\\{(z_{1},0)\in\mathbb{C}^{2}\\}$, we have $u|_{U\cap
L}=z_{1}^{k}+h(z_{1},0)$ and so

$\frac{|\partial_{z_{1}}u|}{|u|}=\frac{k}{|z_{1}|}+O(1).$

Hence (5.4) holds.

Assume $u$ vanishes to infinite order at $0$ and there exists a complex line
$L$ through $0$ such that $\frac{|\bar{\partial}u|}{|u|}\in L^{2}(U\cap L)$.
Applying a holomorphic change of coordinates if necessary, one can always
write $L=\\{(z_{1},0)\in\mathbb{C}^{2}\\}$. Then $v:=u|_{U\cap L}$ vanishes to
infinite order at $0$ and

$\frac{|\bar{\partial}_{z_{1}}u|}{|u|}\leq\frac{|\bar{\partial}u|}{|u|}\in
L^{2}(U\cap L).$

In particular, there exists some $W\in L^{2}(U\cap L)$ such that
$\bar{\partial}_{z_{1}}v=Wv$ on $U\cap L$. By Theorem 2.3, we have $v\equiv
0$. Thus (5.5) holds. ∎

######  Proposition 5.7.

Let $\Omega$ be a domain in $\mathbb{C}^{n}$ and $z_{0}\in\partial\Omega$. Let
$u$ be a function holomorphic on $\Omega$ and smooth on a neighborhood
$U\subset\mathbb{C}^{n}$ of $z_{0}$. Then either one of the following mutually
exclusive cases holds.

  1. (1)

$u$ is holomorphic on $\Omega\cup U$.

  2. (2)

$\int_{U\setminus
Z_{\bar{\partial}u}}\frac{\sum_{j,k=1}^{n}|\bar{\partial}^{2}_{z_{j}z_{k}}u(z)|^{2}}{\sum_{j=1}^{n}|\bar{\partial}_{z_{j}}u(z)|^{2}}\
dv=\infty,$ (5.6)

where $Z_{\bar{\partial}u}$ is the zero set of the vector function
$\bar{\partial}u$.

###### Proof.

Suppose $u$ is not holomorphic on $\Omega\cup U$ and (5.6) fails. Then
$Z_{\bar{\partial}u}\cap U\neq U$ and the function

$W=\begin{cases}\sqrt{\frac{\sum_{j,k=1}^{n}|\bar{\partial}^{2}_{z_{j}z_{k}}u|^{2}}{\sum_{j=1}^{n}|\bar{\partial}_{z_{j}}u|^{2}}},&\text{on}\
\ U\setminus Z_{\bar{\partial}u};\\\ 0,&\text{on}\ \
Z_{\bar{\partial}u}\end{cases}$

belongs to $L^{2}_{loc}(U)$. Let
$v=(\bar{\partial}_{z_{1}}u,\ldots,\bar{\partial}_{z_{n}}u)$. Then
$v:U\rightarrow\mathbb{C}^{n}$ satisfies $|\bar{\partial}v|=W|v|$ on $U$ and
vanishes on the nonempty open set $U\cap\Omega$. According to the weak unique
continuation property in [PZ, Theorem 1.2], we have $v\equiv 0$ on $U$,
contradicting the assumption that $Z_{\bar{\partial}u}\cap U\neq U$. ∎

We point out that in the case when $\Omega$ is pseudoconvex with smooth
boundary, there always exists a function which is holomorphic on $\Omega$ and
smooth on $\overline{\Omega}$, but does not extend holomorphically across a
boundary point $z_{0}$. Thus for every smooth extension of this function part
(2) of Proposition 5.7 always occurs.

######  Remark 5.8.

While all the propositions and corollaries in this section are formulated for
holomorphic functions with smooth extension across a boundary point, the same
reasoning and conclusions can be extended without effort to more general
settings, including formally holomorphic functions – smooth functions where
the Taylor expansion at that point does not contain $\bar{z}$ terms. See [FP]
for more discussion on formally holomorphic functions.

### Acknowledgments

Part of this work by the second named author was conducted while on sabbatical
leave visiting Huaqiao university in China in Spring 2024. He thanks Jianfei
Wang for his invitation, and the host institution for its hospitality and
excellent research environment.

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