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When investigaing the regularity of certain functions, I encountered this problem:

if $u\in H^{3/2}([0,1]\times [0,1])$, what can we say about the continuity of $\nabla u\cdot\overrightarrow{n}$ across a line segment inside the domain, where $\overrightarrow{n}$ is the unit normal ?

Without loss of generality, we formulate the problem as below:

Let $\Omega$ be the unit square in two dimensions and let $\gamma$ denote a vertical line segment inside $\Omega$.

If $u\in H^{3/2}(\Omega)$, can we deduce that $$\frac{\partial u}{\partial x}|_{\gamma^+}=\frac{\partial u}{\partial x}|_{\gamma^-} \;\;\; ?$$

The subscripts $\gamma^+$ and $\gamma^-$ denote traces from left and right sides of $\gamma$, respectively.

Here we assume that $$ \frac{\partial u}{\partial x}|_{\gamma^+}\in L^2(\gamma),\quad \frac{\partial u}{\partial x}|_{\gamma^-}\in L^2(\gamma).$$ Note that this assumption is not redundant because trace theorem does not hold for $H^{1/2}(\Omega)$ with index $1/2$.

What if the assumption above is weakened into: $$ \frac{\partial u}{\partial x}|_{\gamma^+}\in H^{-1/2}(\gamma),\quad \frac{\partial u}{\partial x}|_{\gamma^-}\in H^{-1/2}(\gamma) \;\;\; ?$$

The question was asked in MSE last year but there is still no response, so I plan to ask here.

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  • $\begingroup$ If you can bound the trace, you can approximate H^3/2 by smooth functions and pass the identity to the limit. $\endgroup$
    – Fan Zheng
    Jan 10, 2017 at 3:29
  • $\begingroup$ Sounds true, so the property does not hold for $H^{3/2-\delta}$ in general for any $\delta>0$. $\endgroup$
    – booksee
    Jan 10, 2017 at 3:59
  • $\begingroup$ @FanZheng No, it is not true because trace theorem does not hold for $H^{1/2}(\Omega)$, so the question is still nontrivial. $\endgroup$
    – booksee
    Jan 10, 2017 at 4:13
  • $\begingroup$ Remark. For Lipschitz domain $\Omega$ and $H^s(\Omega)$, the index with trace theorem is $s\in (\frac{1}{2},\frac{3}{2})$. $\endgroup$
    – booksee
    Jan 10, 2017 at 4:17
  • $\begingroup$ Oh, I forgot about the corner case. But then how do you define the trace in the first place? $\endgroup$
    – Fan Zheng
    Jan 10, 2017 at 4:45

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