A unique continuation property for $\lvert\overline \partial u\rvert \leq V \lvert u\rvert$

Let $u\colon \Omega \subset \mathbb {C}^n \to \mathbb {C}^m$, for $n \geq 2$ and $m \geq 1$. Let $1 \leq p \leq 2$, and $2(2n)^2 -1 \leq q < \infty $ such that $\frac {1}{p} + \frac {1}{p'} = 1$ and $\frac {1}{p} - \frac {1}{p'} = \frac {1}{q}$. Suppose $\lvert \overline \partial u\rvert \leq V \lvert u\rvert $, where $V \in L^q_{\textrm {loc}}(\Omega )$. Then $u$ has a unique continuation property in the following sense: if $u \in W^{1,p}_{\textrm {loc}}(\Omega )$ and for some $z_0 \in \Omega $, $\lVert u \rVert _{L^{p'}(B(z_0,r))} $ decays faster than any powers of $r$ as $r \to 0$, then $u \equiv 0$. The same result holds for $q=\infty $ if $u$ is scalar-valued ($m=1$)

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Publication Details

Journal
MATHEMATICA SCANDINAVICA
Published
2025-03-25
DOI
https://doi.org/10.7146/math.scand.a-151062
Primary Topic
Advanced Mathematical Physics Problems
Type
article
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article

A unique continuation property for $\lvert\overline \partial u\rvert \leq V \lvert u\rvert$

Ziming Shi
MATHEMATICA SCANDINAVICA
Advanced Mathematical Physics Problems
article

A unique continuation property for $\lvert\overline \partial u\rvert \leq V \lvert u\rvert$

Ziming Shi
article en

Abstract

Let $u\colon \Omega \subset \mathbb {C}^n \to \mathbb {C}^m$, for $n \geq 2$ and $m \geq 1$. Let $1 \leq p \leq 2$, and $2(2n)^2 -1 \leq q < \infty $ such that $\frac {1}{p} + \frac {1}{p'} = 1$ and $\frac {1}{p} - \frac {1}{p'} = \frac {1}{q}$. Suppose $\lvert \overline \partial u\rvert \leq V \lvert u\rvert $, where $V \in L^q_{\textrm {loc}}(\Omega )$. Then $u$ has a unique continuation property in the following sense: if $u \in W^{1,p}_{\textrm {loc}}(\Omega )$ and for some $z_0 \in \Omega $, $\lVert u \rVert _{L^{p'}(B(z_0,r))} $ decays faster than any powers of $r$ as $r \to 0$, then $u \equiv 0$. The same result holds for $q=\infty $ if $u$ is scalar-valued ($m=1$)

MATHEMATICA SCANDINAVICAVol. 131(1)
Openalex Percentile: Top 90%
Advanced Mathematical Physics Problems
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