Elliptic estimates for self-adjoint operators

Question

Let $A$ be a symmetric matrix in $\mathbb R^n$ such that $A$ is positive definite and hence satisfies $0< \lambda \le A \le \Lambda < \infty.$

Let $T$ be a densely defined and closed operator from some Hilbert space $H$ into $H^n$. It is a classical theorem by John von Neumann that $T^*T$ is self-adjoint with domain $D(T^*T).$

I wonder whether it is true that for some $C>0$

$$\Vert \langle AT,T \rangle x \Vert \le C (\Vert T^*T x \Vert + \Vert x \Vert) \text{ and all } x \in D(T^*T).$$

Similarly, it seems natural to ask whether we also have that

$$\Vert T^*T x \Vert \le C (\Vert \langle AT,T \rangle x \Vert + \Vert x \Vert) \text{ and all } x \in D(\langle AT,T \rangle)?$$

Answer 1

As I understand your question now, after edit, in dimension two you ask whether $$ \|T_2^* T_1 x\| \leqslant C(\|T_1^* T_1 x + T_2^* T_2 x\| + \|x\|) $$ whenever $T_1$, $T_2$ are densely defined closed operators; $C$ can depend on $T_1$ and $T_2$.

This need not be the case. Let $T_1$ be the identity operator acting on $\ell^2$, and let $T_2$ be given by the matrix $$ T_2 = \pmatrix{0&1&0&0&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&0&2&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&0&0&0&3&\cdots\\0&0&0&0&0&0&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots} . $$ Then $$ T_1^* T_1 + T_2^* T_2= \pmatrix{1&0&0&0&0&0&\cdots\\0&1 + 1&0&0&0&0&\cdots\\0&0&1&0&0&0&\cdots\\0&0&0&1 + 4&0&0&\cdots\\0&0&0&0&1&0&\cdots\\0&0&0&0&0&1 + 9&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots} , $$ and $$ T_2^* T_1 = T_2^* = \pmatrix{0&0&0&0&0&0&\cdots\\1&0&0&0&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&2&0&0&0&\cdots\\0&0&0&0&0&0&\cdots\\0&0&0&0&3&0&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots} . $$ In particular, if $e_k$ is the $k$-th vector of the canonical basis of $\ell^2$, then $$\|T_2^* T_1 e_{2n-1}\| = \|n e_{2n}\| = n,$$ but $$\|T_1^* T_1 e_{2n-1} + T_2^* T_2 e_{2n-1}\| = \|e_{2n-1} + 0\| = 1 .$$ Therefore a constant $C$ with the desired property does not exist.