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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?

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    $\begingroup$ I guess I don't quite get the notation/formalism: why isn't $A_D A^{-1}_D$ tautologically an identity? If the coefficients are not smooth, $A^{-1}_D$ will not have range $H^2 \cap H^1_0$, but that doesn't change what $A_D A^{-1}_D$ does. $\endgroup$
    – user378654
    Commented Dec 25, 2023 at 0:34
  • $\begingroup$ I have not the densely defined operator $A_D:L_2\rightarrow L_2$ acting on $A_D^{-1}$ but the differential expression (with not-smooth coefficients under differentiation): $A:H^1\rightarrow H^{-1}$. I want to say that $A$ coincides with $A_D$ on the range of $A_D^{-1}$, but I not see how to say it since I cannot define the operator by the differential expression with the Dirichlet boundary condition in the case of non-smooth coefficients. $\endgroup$
    – Yulia Meshkova
    Commented Dec 25, 2023 at 3:01

1 Answer 1

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From the identity $(AA_D^{-1}\phi,f)_{L_2}=(\phi,f)_{L_2}$ it follows that $$\vert (\phi, (AA_D^{-1})^* f)_{L_2}\vert \leqslant \Vert \phi\Vert _{L_2}\Vert f\Vert _{L_2}\quad \forall \phi\in L_2(O).$$ So, $$\Vert (AA_D^{-1})^* f\Vert _{L_2}\leqslant \Vert f\Vert _{L_2}.$$ Here $f$ is arbitrary function from $H^1_0(O)$. Since $H^1_0(O)$ is dense in $L_2(O)$, by continuity, this inequality is valid for all $f\in L_2(O)$. So, $$\Vert A_{D}^{-1}A\Vert _{L_2\rightarrow L_2}=\Vert (AA_D^{-1})^* \Vert _{L_2\rightarrow L_2}\leqslant 1.$$

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