Let $O\subset\mathbb{R}^d$ be a bounded domain of the class $C^{1,1}$ (or $C^2$ for simplicity). Let the operator $A_D$ be formally given by the differential expression $A=-\mathrm{div}g(x)\nabla$ with the Dirichlet boundary condition. ($A$ means the expression, and $A_D$ is the corresponding operator.) Here $g(x)>0$, $g,g^{−1}\in L_\infty$. The precise definition is given via the quadratic form on $H^1_0(O)$.

If the matrix of coefficients $g$ and the boundary are smooth enough, we can define $A_D$ by the differential expression $-\mathrm{div}g(x)\nabla$ on $H^2(O)\cap H^1_0(O)=\mathrm{Ran}A_D^{-1}$, so for the expression $A$ acting on the resolvent $A^{−1}_D$ we have $$\Vert A A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}=\Vert A_D A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}=1.$$

Question:

Is there are any hope to prove the estimate $$\Vert A A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}\leqslant \mathrm{const}$$ for the operator with non-smooth coefficients in the domain of class $C^{1,1}$ (or $C^2$ )? (Here the expression $A$ is initially considered as operator from $H^1$ to $H^{−1}$ .) E. g., by using of mollification of coefficients.

(I am also interested in this question for strongly elliptic systems in a divergent form: $A=b(D)^∗g(x)b(D)$ , $b=\sum _{l=1}^d b_lD_l$ , $b_l$ are constant matrices, the symbol of $b(D)$ has maximal rank.)

I am trying to do the following. Write $A=D^*gD$, $D=-i\nabla$. Let $f\in H^1_0(O)$, $\phi\in L_2(O)$. We have $(AA_D^{-1}\phi ,f)_{L_2}=(g^{1/2}D A_D^{-1}\phi,g^{1/2}Df)_{L_2}$. Functions $A_D^{-1}\phi$ and $f$ are in $H^1_0(O)$ that is in the domain of the quadratic form of $A_D$. So, we can continue our equality as $(AA_D^{-1}\phi ,f)_{L_2}=(A_D^{1/2}A_D^{-1}\phi,A_D^{1/2}f)_{L_2}=(A_D^{-1/2}\phi,A_D^{1/2}f)_{L_2}=(\phi,f)_{L_2}$. So, $(AA_D^{-1})^*=A_D^{-1}A$ coincides with the identity operator $I$ on $H^1_0(O)\subset L_2(O)$. And I can extend it by continuity to the whole space $L_2(O)$, right?

Let $O\subset\mathbb{R}^d$ be a bounded domain of the class $C^{1,1}$ (or $C^2$ for simplicity). Let the operator $A_D$ be formally given by the differential expression $A=-\operatorname{div}g(x)\nabla$ with the Dirichlet boundary condition. ($A$ means the expression, and $A_D$ is the corresponding operator.) Here $g(x)>0$, $g,g^{−1}\in L_\infty$. The precise definition is given via the quadratic form on $H^1_0(O)$.

If the matrix of coefficients $g$ and the boundary are smooth enough, we can define $A_D$ by the differential expression $-\operatorname{div}g(x)\nabla$ on $H^2(O)\cap H^1_0(O)=\operatorname{Ran}A_D^{-1}$, so for the expression $A$ acting on the resolvent $A^{−1}_D$ we have $$\Vert A A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}=\Vert A_D A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}=1.$$

Question:

Is there are any hope to prove the estimate $$\Vert A A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}\leqslant \mathrm{const}$$ for the operator with non-smooth coefficients in the domain of class $C^{1,1}$ (or $C^2$ )? (Here the expression $A$ is initially considered as operator from $H^1$ to $H^{−1}$ .) E. g., by using of mollification of coefficients.

(I am also interested in this question for strongly elliptic systems in a divergent form: $A=b(D)^∗g(x)b(D)$ , $b=\sum _{l=1}^d b_lD_l$ , $b_l$ are constant matrices, the symbol of $b(D)$ has maximal rank.)

I am trying to do the following. Write $A=D^*gD$, $D=-i\nabla$. Let $f\in H^1_0(O)$, $\phi\in L_2(O)$. We have $(AA_D^{-1}\phi ,f)_{L_2}=(g^{1/2}D A_D^{-1}\phi,g^{1/2}Df)_{L_2}$. Functions $A_D^{-1}\phi$ and $f$ are in $H^1_0(O)$ that is in the domain of the quadratic form of $A_D$. So, we can continue our equality as $(AA_D^{-1}\phi ,f)_{L_2}=(A_D^{1/2}A_D^{-1}\phi,A_D^{1/2}f)_{L_2}=(A_D^{-1/2}\phi,A_D^{1/2}f)_{L_2}=(\phi,f)_{L_2}$. So, $(AA_D^{-1})^*=A_D^{-1}A$ coincides with the identity operator $I$ on $H^1_0(O)\subset L_2(O)$. And I can extend it by continuity to the whole space $L_2(O)$, right?

Let $O\subset\mathbb{R}^d$ be a bounded domain of the class $C^{1,1}$ (or $C^2$ for simplicity). Let the operator $A_D$ be formally given by the differential expression $A=-\mathrm{div}g(x)\nabla$ with the Dirichlet boundary condition. ($A$ means the expression, and $A_D$ is the corresponding operator.) Here $g(x)>0$, $g,g^{−1}\in L_\infty$. The precise definition is given via the quadratic form on $H^1_0(O)$.

If the matrix of coefficients $g$ and the boundary are smooth enough, we can define $A_D$ by the differential expression $-\mathrm{div}g(x)\nabla$ on $H^2(O)\cap H^1_0(O)=\mathrm{Ran}A_D^{-1}$, so for the expression $A$ acting on the resolvent $A^{−1}_D$ we have $$\Vert A A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}=\Vert A_D A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}=1.$$

Question:

Is there are any hope to prove the estimate $$\Vert A A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}\leqslant \mathrm{const}$$ for the operator with non-smooth coefficients in the domain of class $C^{1,1}$ (or $C^2$ )? (Here the expression $A$ is initially considered as operator from $H^1$ to $H^{−1}$ .) E. g., by using of mollification of coefficients.

(I am also interested in this question for strongly elliptic systems in a divergent form: $A=b(D)^∗g(x)b(D)$ , $b=\sum _{l=1}^d b_lD_l$ , $b_l$ are constant matrices, the symbol of $b(D)$ has maximal rank.)

Let $O\subset\mathbb{R}^d$ be a bounded domain of the class $C^{1,1}$ (or $C^2$ for simplicity). Let the operator $A_D$ be formally given by the differential expression $A=-\mathrm{div}g(x)\nabla$ with the Dirichlet boundary condition. ($A$ means the expression, and $A_D$ is the corresponding operator.) Here $g(x)>0$, $g,g^{−1}\in L_\infty$. The precise definition is given via the quadratic form on $H^1_0(O)$.

If the matrix of coefficients $g$ and the boundary are smooth enough, we can define $A_D$ by the differential expression $-\mathrm{div}g(x)\nabla$ on $H^2(O)\cap H^1_0(O)=\mathrm{Ran}A_D^{-1}$, so for the expression $A$ acting on the resolvent $A^{−1}_D$ we have $$\Vert A A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}=\Vert A_D A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}=1.$$

Question:

Is there are any hope to prove the estimate $$\Vert A A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}\leqslant \mathrm{const}$$ for the operator with non-smooth coefficients in the domain of class $C^{1,1}$ (or $C^2$ )? (Here the expression $A$ is initially considered as operator from $H^1$ to $H^{−1}$ .) E. g., by using of mollification of coefficients.

(I am also interested in this question for strongly elliptic systems in a divergent form: $A=b(D)^∗g(x)b(D)$ , $b=\sum _{l=1}^d b_lD_l$ , $b_l$ are constant matrices, the symbol of $b(D)$ has maximal rank.)

I am trying to do the following. Write $A=D^*gD$, $D=-i\nabla$. Let $f\in H^1_0(O)$, $\phi\in L_2(O)$. We have $(AA_D^{-1}\phi ,f)_{L_2}=(g^{1/2}D A_D^{-1}\phi,g^{1/2}Df)_{L_2}$. Functions $A_D^{-1}\phi$ and $f$ are in $H^1_0(O)$ that is in the domain of the quadratic form of $A_D$. So, we can continue our equality as $(AA_D^{-1}\phi ,f)_{L_2}=(A_D^{1/2}A_D^{-1}\phi,A_D^{1/2}f)_{L_2}=(A_D^{-1/2}\phi,A_D^{1/2}f)_{L_2}=(\phi,f)_{L_2}$. So, $(AA_D^{-1})^*=A_D^{-1}A$ coincides with the identity operator $I$ on $H^1_0(O)\subset L_2(O)$. And I can extend it by continuity to the whole space $L_2(O)$, right?