Sobolev space-exercice

Question

Let $\Omega = \mathbb{R}^2_+$. My question is: how we prove that if $v \in H^2(\Omega)$ such as $v(x,0)=0$, then $\dfrac{\partial v}{\partial x} \in H^1_0(\Omega)$ ?

Answer 1

Define $T:C_0^2(\overline{\mathbb{R}^2_+})\to C_0^1(\mathbb{R})$ by $$(Tu)(x)=u(x,0),\ \forall\ x\in \mathbb{R}$$

By using the calculations in Theorem 4.1. here, we have that $$\|Tu\|_{H^1}\le C\|u\|_{H_2},\ \forall\ u\in C_0^2(\overline{\mathbb{R}^2_+}) \tag{1}$$

Moreover, note that $$\frac{d}{dx}(Tu)(x)=T\left(\frac{\partial }{\partial x}u\right)(x)\ \forall u\in C_0^2(\overline{\mathbb{R}^2_+})\tag{2}$$

Now we use the denseness of $C_0^2(\overline{\mathbb{R}^2_+})$ in $H^2(\mathbb{R}^2_+)$, $(1)$, $(2)$ and the continuity of the differential operator $\frac{d}{dx}$ to conclude that

$$\frac{d}{dx}(Tu)(x)=T\left(\frac{\partial }{\partial x}u\right)(x)\ \forall u\in H^2(\mathbb{R}^2_+)\tag{3}$$

To conclude, if $v$ satisfies $v(x,0)=0$ then $Tv=0$, hence, by using $(3)$ we conclude that $T\left(\frac{\partial }{\partial x}v\right)=0$ which is the same to say that $$\frac{\partial v}{\partial x}\in H_0^1(\mathbb{R}^2_+)$$