Poincare type inequality along the boundary Let the $C^1$ domain $\Omega\subset \mathbb{R}^n$ have connected boundary. Assume $\vec{F}:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is a sufficiently smooth vector field and $\int_{\partial \Omega} \vec{F}=0$, show the inequality
$$\int_{\partial \Omega} |\vec{F}|^2\leq C\int_{\partial \Omega} |\nabla_T\vec{F}|^2$$
where $$\nabla_T \vec{F}=\nabla \vec{F}-(\nabla \vec{F}\cdot N)N=\left( \partial_j F^k-\frac{\partial F^k}{\partial N}N^j \right)_{1\leq j,k\leq n}$$
$N$ is the outer normal vector.
I have two questions:


*

*How to intuitively understand $\nabla _T F$ is the 'matrix of tangential derivatives'.

*How to prove the inequality using classical poincare inequality.


Any help will be appreciated.
 A: To have a Poincare inequality the domain (or manifold) has to be necessarily bounded. Let us assume $\varphi$ be a $C^1$ defining function for $\Gamma \, (= \partial \Omega)$ i.e., $\Gamma = \{ \varphi = 0 \}$ and furthermore we can normalize $\varphi$ to have $\left.|\nabla \varphi|\right|_\Gamma = 1$.
It suffices to see the inequality for scalar functions $f \in C^{1}(\Gamma)$. Let us consider a $C^1$ extension of $f$ to a $\epsilon$-nbd of $\Gamma$ say $\Omega_\epsilon = \{x: \operatorname{dist}(x,\Gamma) < \epsilon\}$ (which we continue to denote by $f$) which satisfies the inequality $$|\nabla f| \le A|\nabla_\tau f| \text{ on } \Gamma \tag{0}$$ i.e., the tangential gradient on $\Gamma$ controls the total gradient. 
N.B.: If the hypersurface $\Gamma$ has slightly higher regularity $C^{1,1}$ (instead of just $C^1$) then signed distance to $\Gamma$ is a good candidate for $\varphi$ and we can also take advantage of a uniform Fermi (normal) coordinate. In this case we actually have $\nabla f = \nabla_{\tau} f$ (in normal coordinates) by extending $f$ by constant along normal directions to $\Gamma$.
Now, by coarea formula we have $$\int_{V_r} f \,dx = \int_{-r}^{r} \int_{\Gamma_t} \frac{f}{|\nabla \varphi|} \,d\mathcal{H}^{n-1}\llcorner \Gamma_t\,dt \tag{1}$$ where, $\Gamma_t := \{\varphi = t\}$ denotes the level sets of $\varphi$ and $r > 0$ is small enough s.t. $V_r := \{x: |\phi(x)| < r\} \subset \Omega_\epsilon$. Note that $\Gamma_0 = \Gamma$.
Since, $\varphi \in C^1$ and since $\left.|\nabla \varphi|\right|_\Gamma = 1$ then from $(1)$ we have $$\left|\frac{1}{2r}\int_{V_r} f\,dx - \int_{\Gamma} f \,d\mathcal{H}^{n-1}\llcorner \Gamma\right| \le o(1) \tag{2}$$ with $o(1) \to 0$ as $r \to 0^+$.
Now, let us assume $f$ satisfies $\displaystyle \int_\Gamma f \,d\mathcal{H}^{n-1}\llcorner \Gamma = 0$ and prove the Poincare Inequality. Let us denote the average integral $\displaystyle \overline{f}_{V_r} = \frac{1}{|V_r|}\int_{V_r} f\,dx$. Then from $(2)$ we have $$\overline{f}_{V_r}^2 \le \frac{o(r^2)}{|V_r|^2}. \tag{3}$$
Using $f^2$ (instead of $f$) in the inequality $(2)$ and rearranging the triangle inequality we have 
\begin{align*} \int_\Gamma |f|^2\,d\mathcal{H}^{n-1}\llcorner \Gamma - o(1) &\le \frac{1}{2r}\int_{V_r} f^2 \,dx \\ &\le \frac{1}{r} \left(\int_{V_r} \left(f - \overline{f}_{V_r}\right)^2\,dx + |V_r|\overline{f}_{V_r}^2\right) \tag{4}\\ &\le \frac{C_1}{r}\int_{V_r} |\nabla f|^2\,dx + \frac{o(r)}{|V_r|} \tag{5}\end{align*} where, we used the Poincare inequality for $\mathbb{R}^n$ to the first term in line $(4)$ to the function $(f - \overline{f}_{V_r})$ and inequality $(3)$ to the second term. I.e., we have the inequality
\begin{align*} \int_\Gamma |f|^2\,d\mathcal{H}^{n-1}\llcorner \Gamma \le \frac{C_1}{r}\int_{V_r} |\nabla f|^2\,dx + o(1) + \frac{o(r)}{|V_r|}. \tag{6}\end{align*}
Now, $|V_r| \sim \mathcal{H}^{n-1}(\Gamma) \times r$ (since, $|\nabla \varphi||_\Gamma = 1$) and from inequality $(2)$ with $|\nabla f|^2$ instead of $f$ we see that $$\frac{1}{2r}\int_{V_r} |\nabla f|^2 \,dx \to \int_{\Gamma} |\nabla f|^2\,d\mathcal{H}^{n-1}\llcorner \Gamma, \, \text{ as } r \to 0^+. \tag{7}$$ 
Therefore, letting $r\to 0^+$ in $(6)$ combined with $(0)$ we see that $$\int_\Gamma |f|^2\,d\mathcal{H}^{n-1}\llcorner \Gamma \le 2AC_1 \int_\Gamma |\nabla_\tau f|^2\,d\mathcal{H}^{n-1}\llcorner \Gamma.$$
