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I've encountered quite some papers in which it is simply assumed that

$\exists C>0 : \left(\displaystyle{\int\limits_{\Gamma}}((\nabla v)\cdot \hat{\bf{n}})^2d\Gamma\right)^{\dfrac{1}{2}}\leq C\left(\displaystyle{\int\limits_{\Omega}}(\nabla v)\cdot(\nabla v)d\Omega\right)^{\dfrac{1}{2}}\quad\forall v:\bar{\Omega}\rightarrow\mathbb{R},\quad\Gamma:=\partial\Omega,\bar{\Omega}:=\Omega\cup\Gamma$

instead of proven. The $\hat{\bf{n}}$ is unit vector normal to the boundary $\Gamma$. How do you prove this or where do you find a proof of this? Is there a name for such a type of inequality?

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I think that the inequality is false. For $\Omega=D_1(0) \subset \mathbb{R}^2$ and $v(x)=|x|^n$, then $\nabla v(x)= n|x|^{n-2}x$, so the left term of the inequality gives $\sqrt{2 n^2 \pi} $, but the right one gives $C \sqrt{n \pi}$. I hope not to be wrong with integrals!

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  • $\begingroup$ I did not mention that the $\hat{\bf{n}}$ is the unit vector normal to the boundary $\Gamma$. So I think you misunderstood the meaning of the integrals. What do you mean with $D_1(0)$ btw? $\endgroup$ – Adriaan Dec 3 '14 at 16:46
  • $\begingroup$ So $n$ is a fixed vector? $ D_r(p)=\{ x \in \mathbb{R}^2: |x-p| \leq r\} $ $\endgroup$ – GGG Dec 3 '14 at 17:00
  • $\begingroup$ No not fixed, it is the unit vector normal to the boundary $\Gamma$ and thus a function of $\bf{x}\in$$\Gamma$. $\endgroup$ – Adriaan Dec 3 '14 at 19:12
  • $\begingroup$ The function $ v$ has gradient parallel to $n$. So is there any problem with my counterexample? $\endgroup$ – GGG Dec 3 '14 at 21:11

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