Directional derivative in a Sobolev-like inequality

Question

I am trying to do the following problem:

Let $\Omega \subset \subset \overline{\mathbb{R}_{+}^{n+1}} = \{ (x, x_{n+1}); \, x_{n+1}\geq 0\}$ (i.e., $\Omega$ is bounded inside the closed upper half-space). (By definition, smooth functions on $\overline{\mathbb{R}_{+}^{n+1}}$ are functions which are restrictions of smooth functions on an open neighborhood of $\overline{\mathbb{R}_{+}^{n+1}}$ in $\mathbb{R}^{n+1}$.) Let $u \in C_{0}^{\infty}(\Omega)$. Show that for any $p$ such that $1\leq p \leq \infty$, there is a constant $C_{\Omega}$ independent of $u$ so that, $$\|u(\cdot,0)\|_{L^{p}(\mathbb{R}^{n})}\leq C_{\Omega}\|\partial_{n+1}u\|_{L^{p}(\Omega)}.$$

Now, I've been told that the notation $\partial_{n+1}u$ means "the directional derivative of $u$ in the direction of $n+1$". What does that even mean/look like? I feel like I won't be able to do this problem until I find out what this actually means, so if anyone could shed some light on the subject, it would be much appreciated!

Answer 1

Assume that $\overline{\Omega}\subset K\times [a,b]$, where $0<a<b$ and $K\subset \mathbb R^n$, bounded. Then, clearly, for every $\boldsymbol{x}=(x_1,\ldots,x_{n+1})\in \Omega$, $$ u(\boldsymbol{x})=u(x_1,\ldots,x_{n+1})=\int_a^{x_{n+1}} \partial_{n+1} u(x_1,\ldots,x_n,t)\,dt $$ and hence, for $q=p/(p-1)$, using Holder inequality we obtain \begin{align} \lvert u(x_1,\ldots,x_n,x_{n+1})\rvert &\le \int_a^{x_{n+1}} \lvert \partial_{n+1} u(x_1,\ldots,x_n,t) \rvert \,dt \\ &\le \Big(\int_a^{x_{n+1}} \lvert \partial_{n+1} u(x_1,\ldots,x_n,t)\rvert^p\,dt\Big)^{1/p} \Big(\int_a^{x_n+1}1\,dt\Big)^{1/q} \\ &\le (b-a)^{1/q}\Big(\int_a^b \lvert\partial_{n+1} u(x_1,\ldots,x_n,t)\rvert^p\,dt\Big)^{1/p}. \end{align} Hence, using Fubini's Theorem \begin{align} \int_{\Omega}\lvert u(\boldsymbol{x})\rvert^p \,d\boldsymbol{x}&\le (b-a)^{p/q} \int_{K\times [a,b]}\left(\int_a^b \lvert\partial_{n+1} u(x_1,\ldots,x_n,t)\rvert^p\,dt\right)\,dx_1\cdots dx_n\,dx_{n+1} \\ &= (b-a)^{1+p/q} \int_{K}\left(\int_a^b \lvert\partial_{n+1} u(x_1,\ldots,x_n,t)\rvert^p\,dt\right)\,dx_1\cdots dx_n \\ &=(b-a)^{1+p/q} \int_{\Omega} \lvert\partial_{n+1} u(\boldsymbol{x})\rvert^p\,d\boldsymbol{x}. \end{align} Therefore $$ \|u\|_{L^p(\Omega)}\le (b-a)\, \|\partial_{n+1}u\|_{L^p(\Omega)}. $$

Answer 2

Hint:

The problem is actually independent of the first $n$-variables. You might want to deal with this baby version first: Let $u\in C^\infty_0([0,d))$. Then show that for $1\leq p< \infty$,

$$|u(0)|^p \leq C(d, p) \int_0^d |u'|^p dx . $$