Gårding's inequality for $\mathbb R^n$ implies that for bounded smooth domains? We are given (weak) Gårding's inequality for elliptic pseudodifferential operators:

Given $a\in S^m$ such that $\operatorname{Op}(a)$ is an elliptic operator, namely $\exists c,R>0$ such that for each $x,\xi\in\mathbb R^n$, we have $\operatorname{Re} a(x,\xi)\ge c(1+\lvert\xi\rvert)^m,\forall\lvert\xi\rvert\ge R$. Then for each $s\in\mathbb R$, there exists $A_s,B_s>0$ such that for each $u\in H^{m/2}$, we have $\operatorname{Re}\langle\operatorname{Op}(a)u,u\rangle\ge A_s\lVert u\rVert_{H^{m/2}}^2-B_s\lVert u\rVert_{H^s}^2$.

I wonder whether we can deduce the following version of Gårding-type inequality:

Suppose $\Omega\subseteq\mathbb R^n$ is a smooth bounded domain, and $L=\sum_{\lvert\alpha\rvert\le2k}a_\alpha\partial^\alpha$ is a real elliptic differential operator of order $2k$ on $\Omega$, namely the associated symbol $a(x,\xi)=\sum_\alpha a_\alpha(x)\xi^\alpha$ satisfies elliptic condition: there exists $c,R>0$ such that $a(x,\xi)\ge c(1+\lvert\xi\rvert)^{2k},\forall x\in\Omega,\lvert\xi\rvert\ge R$, and $a_\alpha\in C_b^\infty(\Omega,\mathbb R)$, the space of real smooth functions with all partial derivatives bounded. Then there exists $A_0,B_0>0$ such that for each $u\in H_0^k(\Omega)$, we have $\langle Lu,u\rangle\ge A_0\lVert u\rVert_{H^k}^2-B_0\lVert u\rVert_{L^2}^2$.

It seems to me that direct extension doesn't work, since we cannot extend $L$ to an elliptic operator on $\mathbb R^n$ (Whitney extension theorem needs compatible conditions near boundary). I wonder whether it's still a corollary of the inequality for that of $\mathbb R^n$ with a clever argument.
Any help is welcome.
 A: I think that can be solved in this way. Take $L= \sum a_{\alpha}(x)\partial^{\alpha}$, where $\alpha$ is the usual multi-index.
Extend each coefficient $a_{\alpha}(x)$ in an open neighborhood of $\Omega$, call it $\Omega_b$. By the fact that
$$
Re L(x,\xi) > c |\xi|^{2k} , for \;|\xi|>R, \; \forall x \in \Omega
$$
use the continuity to achieve that
$$
Re L(x,\xi) > \frac{c}{2} |\xi|^{2k} , for \;|\xi|>R, \; \forall x \in \Omega_b
$$
shrinking $\Omega_b$ if necessary. Now choose a cut-off, smooth function such that
$$
\varphi=1 \; on \;\;\Omega \;\; and \;\; \varphi=0 \;\; outside \;\;\Omega_b;
$$
the existence of such $\varphi$ is obvious. Now define
$$
a(x,\xi) = \varphi(x) L(x,\xi) + (1-\varphi(x))|\xi|^{2k}
$$
it is elliptic on $\mathbf{R}^n$. And we can apply the Garding inequality on $a(x,D)$, as you suggest. So we have
$$
Re (a(x,D)u,u) \geq C_1 || u||_{H^{k}}^2 - C_0|| u||_{L^2}^2
$$
for all $u \in H^{m/2}$. So notice that each element $v\in H^k_0(\Omega)$ can be approximated by $C^{\infty}_0(\Omega)$ element, so we have to deduce the Garding inequality for function in this space. But for definition
$$
a(x,D)v = \varphi(x)L(x,D)v + (1-\varphi(x))(-\Delta)^{k} v =  L(x,D)v 
$$ 
since the support of $v$ is conteined in $\Omega$. I wish I have included all details you need, but if you deserve I can make more clear the previous.
I want make you a question, that arise to me studying the inequality for pseudodifferential operator (maybe you can only suggest to me a book where the proof are made in this settings). The proof who I studied started with an elliptic pseudodifferential $Op(A)= A(x,D)$ operator of order $0$, so
$$ 
Re A(x,\xi) > c, for \;|\xi|\in \mathbf{R^n} \; \forall x \in \mathbf{R}^n
$$
the general statement can easily deduced by the same in this situation. And set 
$$
q(x,\xi) = (Re A(x,\xi)- \frac{c}{2})^{\frac{1}{2}}.
$$
Using the adjiont property he deduce
$$
q(x,D)q(x,D)^* = ReA(x,D) - \frac{c}{2} + r(x,D)
$$
where $r(x,D)$ is a remainder, but is not important for my question. Now from this, it seems to me that he compute in $u$ the previous operator equality, and take the duality pairing with $u$. But in this case he has a term like
$$(ReA(x,D)u,u) $$
rather than
$$Re(A(x,D)u,u) $$
Is not it? The proof is taken by "Partial Differential equation" author "Micheal E. Taylor" volume 2. Any suggestion?
P.S. I have I little different notation on pseudodifferntial operator, so you may read $A(x,D)$ for $Op(A)$ and similar.
