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Let $u \in H^{1}(\Omega)$ ($\Omega \subset R^n$ a bounded domain with smooth boundary). Suppose that there is a constant $C>0$ such that

$$ |u(x) - u(y)| \leq C |x-y|,$$

for every Lebesgue point $x,y$ of $u$. Can I conclude that $u $ is Lipschitz?

Thanks in advance!

math student
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1 Answers1

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Yes: Call $A$ the set of Lebesgue points of $u$ (which is dense in $\Omega$). Then $u$ restricted to $A$ is Lipschitz and therefore there is a unique Lipschitz extension $v$ to $\bar{A}=\Omega$. Since $|\Omega\setminus A|=0$ we get that $v$ is a representative of $u$, and therefore "$u$ is Lipschitz".

Jose27
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