# Upper bound for integral over boundary in terms of integral over interior

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?

## 1 Answer

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!

• 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? – Adriaan Dec 3 '14 at 16:46
• So $n$ is a fixed vector? $D_r(p)=\{ x \in \mathbb{R}^2: |x-p| \leq r\}$ – GGG Dec 3 '14 at 17:00
• No not fixed, it is the unit vector normal to the boundary $\Gamma$ and thus a function of $\bf{x}\in$$\Gamma$. – Adriaan Dec 3 '14 at 19:12
• The function $v$ has gradient parallel to $n$. So is there any problem with my counterexample? – GGG Dec 3 '14 at 21:11