Density in $H^1_0$

Question

Let $\Omega = \mathbb{R}^2_+=\{(x,y)\in \mathbb{R}^2; y>0\}$ and let $v \in H^1_0(\Omega)$. For $h \neq 0$ we define $D_h v = \dfrac{v(x+h,y)-v(x,y)}{h}$ such as $\forall \varphi \in \mathcal{D}(\Omega)$, we have

1. $\displaystyle\iint D_h v \varphi = - \iint v D_{-h} \varphi$

2. $\iint |D_h v|^2 dx dy \leq \iint \Big|\dfrac{\partial v}{\partial x} (x,y)\Big|^2 dx dy$

Prove that for all $v \in H^1_0(\Omega)$, we have $D_h \Big(\dfrac{\partial v}{\partial x}\Big) = \dfrac{\partial}{\partial x} (D_h v) \quad\text{et}\quad D_h \Big(\dfrac{\partial v}{\partial y}\Big) = \dfrac{\partial}{\partial y} (D_h v)$

So, I write a proof for the functions $C^{\infty}(\Omega)$:$D_h(\dfrac{\partial v}{\partial x}) = \dfrac{\partial}{\partial x} (D_h v)$ But how we use the density to obtain the result for the functions $v \in H^1_0(\Omega)$?

Thank's for help.

Answer 1

You've shown that $D_h$ commutes with $\dfrac{\partial}{\partial x}$ when applied to smooth functions. Now use the definition of weak derivative. If $v \in H^1(\Omega)$ then \begin{align*} \int_\Omega D_h v \frac{\partial \psi}{\partial x} \, dx dy &= - \int_\Omega v D_{-h} \frac{\partial \psi}{\partial x} \, dxdy \\ &= - \int v \frac{\partial (D_{-h} \psi)}{\partial x} v\, dxdy \\ &= \int_\Omega \frac{\partial v}{\partial x} D_{-h} \psi \, dxdy \\ &= - \int_\Omega D_h \left( \frac{\partial v}{\partial x} \right) \psi \, dxdy \end{align*} for all $\psi \in C_0^\infty(\Omega)$. Thus the weak derivative $\dfrac{\partial D_h v}{\partial x}$ equals $D_h \left( \dfrac{\partial v}{\partial x} \right)$.

Answer 2

(1) You should prove the identity for smooth functions.

(2) Prove that $\frac{\partial}{\partial x}:H^1_0(\Omega)\to L^2(\Omega)$ and $D_h:L^2(\Omega)\to L^2(\Omega)$ are continuous.

(3) Approximate $v\in H^1_0(\Omega)$ by $v_\epsilon\in C_0^\infty(\Omega)$ with $\|v-v_\epsilon\|_{H^1(\Omega)}\le \epsilon$.

(4) Conclusion: Use (1)-(3) to prove the claim with the following inequality $$ \left\|\left(D_h\frac{\partial}{\partial x} - \frac{\partial}{\partial x}D_h\right)v\right\|_{L^2} \le \left\|\left(D_h\frac{\partial}{\partial x} - \frac{\partial}{\partial x}D_h\right)v_\epsilon\right\|_{L^2} + \left\|\left(D_h\frac{\partial}{\partial x} - \frac{\partial}{\partial x}D_h\right)(v-v_\epsilon)\right\|_{L^2} $$