MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

3 added 393 characters in body

Clearly, what you call $\newcommand{\bn}{\boldsymbol{n}}$ $\nabla u\cdot \bn$ is the normal derivative $\frac{\partial u}{\partial \bn}$. The trace theorem (see e.g. Lions and Magenes, Non-Homogeneous Boundary Value Problems and Applications. I, Thm. 9.4, Chap 1) shows that for $s> \frac{3}{2}$ the restriction map $\newcommand{\pa}{\partial}$

$$R: C^\infty(B_1)\ni u\mapsto (u|_{\pa B_1}, \frac{\pa u}{\pa \bn})\in C^\infty(\pa B_1)\times C^\infty(\pa B_1)$$

extends to a continuous split surjective map

$$H^s(B_1)\to H^{s-\frac{1}{2}}(\pa B_1) \times H^{s-\frac{3}{2}}(\pa B_1).$$

The result is optimal because Theorem 9.5, Chap. 1 op. cit. shows that if $s\leq \frac{3}{2}$, then for any $\phi\in C^\infty(\pa B_1)$ the linear functional

$$C^\infty( B_1)\ni u\mapsto \int_{\pa B_1} \frac{\pa u (x)}{\pa \bn} \phi (x) dA(x)\in\mathbb{R}$$

is not continuous in the topology induced by $H^s(B_1)$. In particular, if $s\leq \frac{3}{2}$ there cannot exist a constant $C>0$ such that

$$\left|\int_{\pa B_1} \frac{\pa u}{\pa \bn} dA\right| \leq C\Vert u\Vert_{H^s(B_1)},\;\;\forall u\in C^\infty(B_1),$$

so that there exists a sequence $u_k\in C^\infty(B_1)$ such that $\Vert u_k\Vert_{H^1(B_1)}\leq 1$ and

$$\left|\int_{\pa B_1} \frac{\pa u_k}{\pa \bn} dA\right| \to \infty.$$

2 added 43 characters in body

Clearly, what you call $\newcommand{\bn}{\boldsymbol{n}}$ $\nabla u\cdot \bn$ is the normal derivative $\frac{\partial u}{\partial \bn}$. The trace theorem (see e.g. Lions and Magenes, Non-Homogeneous Boundary Value Problems and Applications. I, Thm. 9.4, Chap 1) shows that for $s> \frac{3}{2}$ the restriction map $\newcommand{\pa}{\partial}$

$$R: C^\infty(B_1)\ni u\mapsto (u|_{\pa B_1}, \frac{\pa u}{\pa \bn})\in C^\infty(\pa B_1)\times C^\infty(\pa B_1)$$

extends to a continuous split surjective map

$$H^s(B_1)\to H^{s-\frac{1}{2}}(\pa B_1) \times H^{s-\frac{3}{2}}(\pa B_1).$$

The result is optimal because Theorem 9.5, Chap. 1 op. cit. shows that if $s\leq \frac{3}{2}$, then for any $\phi\in C^\infty(\pa B_1)$ the linear functional

$$H^sC^\infty( B_1)\ni u\mapsto \int_{\pa B_1} \frac{\pa u (x)}{\pa \bn} \phi (x) dA(x)\in\mathbb{R}$$

is not continuous .in the topology induced by $H^s(B_1)$.

1

Clearly, what you call $\newcommand{\bn}{\boldsymbol{n}}$ $\nabla u\cdot \bn$ is the normal derivative $\frac{\partial u}{\partial \bn}$. The trace theorem (see e.g. Lions and Magenes, Non-Homogeneous Boundary Value Problems and Applications. I, Thm. 9.4, Chap 1) shows that for $s> \frac{3}{2}$ the restriction map $\newcommand{\pa}{\partial}$

$$R: C^\infty(B_1)\ni u\mapsto (u|_{\pa B_1}, \frac{\pa u}{\pa \bn})\in C^\infty(\pa B_1)\times C^\infty(\pa B_1)$$

extends to a continuous split surjective map

$$H^s(B_1)\to H^{s-\frac{1}{2}}(\pa B_1) \times H^{s-\frac{3}{2}}(\pa B_1).$$

The result is optimal because Theorem 9.5, Chap. 1 op. cit. shows that if $s\leq \frac{3}{2}$, then for any $\phi\in C^\infty(\pa B_1)$ the linear functional

$$H^s( B_1)\ni u\mapsto \int_{\pa B_1} \frac{\pa u (x)}{\pa \bn} \phi (x) dA(x)\in\mathbb{R}$$

is not continuous.