# $u\in W^{1,p}(B)$ implies $u\in W^{1,p}(\partial B_t)$ for almost $t\in (0,1]$?

Suppose that $u\in W^{1,p}(B)$ where $B=B(0,1)\subset\mathbb{R}^N$ and $p\geq 1$. It was showed on this post (as an application of Fubini's theorem) that for almost $t\in (0,1]$ $$u_{|\partial B_t},(\nabla u)_{|\partial B_t}\in L^p(\partial B_t)$$

where $B_t=B(0,t)$. My question is: Is there any simple way to see that $u\in W^{1,p}(\partial B_t)$ for those balls where $u_{|\partial B_t},(\nabla u)_{\partial B_t}$ are defined and if $v\in L^p(\partial B_t)^N$ is the gradient of $u$ in the distributional sense, then $v=(\nabla u)_{|\partial B_t}$?

Update: I have an idea. Let $u_n\in C^1(\overline{B})$ such that $u_n\to u$ in $W^{1,p}(B)$. Note that byt Fatou's Lemma: $$\int_0^1\liminf\int_{\partial B}(|u_n(r\omega)-u(r\omega)|^p+|\nabla u_n(r\omega)-\nabla u(r\omega)|^p)r^{N-1}drd\Gamma\leq \\ \lim\int_B(|u_n(x)-u(x)|^p+|\nabla u_n(x)-\nabla u(x)|^p)dx \to 0$$

Hence, for almost $t\in (0,1]$, there exist a subsequence of $u_n$ not relabeled such that $$\int_{\partial B_t}(|u_n(y)-u(y)|^p+|\nabla u_n(y)-\nabla u(y)|^p)d\Gamma_t\to 0$$

Can I conclude what I have estated from the last convergence?

Thank you

• So you wanna show something like $\nabla(u\big|_{\partial B_t}) = (\nabla u)\big|_{\partial B_t}$? If this is the case, then the gradient on a codimension 1 surface has to be delicately defined to match the restriction of gradient on this surface. – Shuhao Cao May 27 '13 at 20:37
• I think they will differ only by a set of null measure @ShuhaoCao . Also, I have updated the question, please take a look on it. – Tomás May 27 '13 at 20:46

To avoid unnecessary complication, let me straigthen up the spheres. Then the problem becomes the following: $\newcommand{\R}{\mathbb R}$
If $u(x,y) \in W^{1,p}(\R^n \times \R^k)$, then $v_x := u(x,\cdot) \in W^{1,p}(\R^k)$ for a.e. $x \in \R^n$, with weak gradient $\nabla v_x (y) = \nabla_y u(x,y)$.
To see this, we first test the definition of weak gradient with functions of the form $\varphi(x)\eta(y)$, where $\varphi \in C_c^\infty(\R^n)$, $\eta \in C_c^\infty(\R^k)$, and consider only derivatives in the last $k$ directions: $$\int_{\R^{n+k}} \nabla_y u(x,y) \varphi(x) \eta(y) \, dx dy = - \int_{\R^{n+k}} u(x,y) \varphi(x) \nabla \eta(y) \, dx dy$$ Let me rephrase it as follows: the function $$I_\eta(x) := \int_{\R^k} \left( \nabla_y u(x,y) \eta(y) + u(x,y) \nabla \eta(y) \right) \, dy$$ satisfies $$\int_{\R^n} I_\eta(x) \varphi(x) \, dx = 0 \quad \text{for every } \varphi \in C_c^\infty(\R^n).$$ By the fundamental lemma of Calc. Var., this means that for each $\eta \in C_c^\infty(\R^k)$ we have $I_\eta(x) = 0$ for a.e. $x \in \R^n$.
We would like to reverse the order of quantifiers. To this end we choose a countable subset $\eta_j \in C_c^\infty(\R^k)$ dense in the $C^1$ norm. Then for a.e. $x \in \R^n$ we have $$\int_{\R^k} \nabla_y u(x,y) \eta_j(y) \, dy = - \int_{\R^k} u(x,y) \nabla \eta_j(y) \, dy \quad \text{for } j=1,2,\ldots$$ By density, we obtain this identity for the same set of $x$'s and all possible $\eta$'s.