I want to show the following statement:
Given $u \in H^1_0(Q_+)$ with $supp(u) \subset \{x \in \mathbb{R}^n\ | \ (\sum_{i=1}^{n-1} |x_i|^2)^{1/2}< 1 - \delta, \ 0 \leq x_n < 1-\delta \}$ y $h \in \mathbb{R}^{n-1} \times \{0\}$ such that $|h| \leq \delta$. Then: $$ \quad \quad \quad \quad ||D_hu||_{L^2(Q_+)} \leq ||\nabla u||_{L^2(Q_+)} $$ Where $D_hu= \frac{u(x+h)-u(x)}{|h|}$ and $Q_+=\{x=(x_1,...,x_n) \in \mathbb{R}^n \ | \ (\sum_{i=1}^{n-1} x_i^2)^{1/2} < 1 \ y \ 0 < |x_n|<1 \}$.
This is the proof:
For $u \in H_0^1(Q_+)$ there exists $(u_n) \subset C_c^{\infty}(Q_+)$ such that $u_n \to u $ in $H^1(Q_+)$. We define $v(t)=u_n(x+th)$ with $t \in [0,1]$, so $v'(t)=h \nabla \overline{u}(x+th)$ and we obtain: $$|u_n(x+h) - u_n(x)|^2 \leq \int_0^1 | h \nabla u_n(x+th)|^2 dt $$
(This is easily shown, it is not the part I want to check)
Integrating over $Q_+$: $$\int_{Q_+} |u_n(x+h) -u_n(x)|^2 dx \leq \int_{Q_+} \int_0^1 | h \nabla u_n(x+th)|^2 dt \ dx \leq \quad \quad \quad \quad \quad \quad$$ $$ \int_0^1 \int_{Q_+} | h \nabla u_n(x+th)|^2 dx \ dt = \int_0^1 \int_{Q_++th} | h \nabla u_n(y)|^2 dy \ dt = \int_{Q_+ +th} | h \nabla u_n(y)|^2 dy$$
(Using change of variables $y=x+th$). Then: $$||u_n(x+h) -u_n(x)||_{L^2(Q_+)} \leq ||h \nabla u_n||_{L^2{(Q_+ +th)}}$$
We take limits on both sides and due to the hypotesis over $h$ we have $sop(u(x+th)) \subset Q_+$, so we obtain:$$||u(x+h) -u(x)||_{L^2(Q_+)} \leq || h \nabla u||_{L^2{(Q_+ +th)}}= ||h \nabla u||_{L^2(Q_+)}$$