# Elliptic estimates for self-adjoint operators

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

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

• Is $A$ fixed, or is it a function of $x$? Do you really allow for $\Lambda = \infty$? Apr 7, 2020 at 7:02
• @MateuszKwaśnicki I don't want $\Lambda=\infty$, sorry. I am surprised it would make a difference whether $A$ depends on $x$. Does it hold only for non $x$-dependent $A$? Apr 7, 2020 at 7:35
• If $P(D)$ has constant coefficients, and all operators act on $L^2(\mathbb{R}^n)$, then the domain of $P(D)$ is indeed $W^{2,2}(\mathbb{R}^n)$: just have a look on the Fourier transform. However, if $A$ is a (non-smooth) function of $x \in \mathbb{R}^n$, then the domain of $P(D)$ can get quite unrelated to the domain of $\Delta$. Apr 7, 2020 at 7:39
• @MateuszKwaśnicki I see, is there also a proof that does not rely on the Fourier transform?-I am just wondering whether it is true in general for $T$ closed and densely-defined that $\langle A T,T\rangle$ is relatively $T^*T$ bounded under the above assumptions on $A$ (and assuming it is constant)-and such a proof could somehow give hints whether this is indeed true. Apr 7, 2020 at 8:08

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.
• thank you. What I don't quite understand at the moment is that your $A=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ no? This does not look positive-definite. Or is your answer assuming that without loss of generality it suffices to show you cannot control off-diagonal entries? Apr 7, 2020 at 23:53
• @KungYao: Yes this is what I meant; sort of "abstract second-order Riesz transforms need not be bounded". The same example shows that, say, $A = \pmatrix{1&1\\1&1}$ leads to an operator $\langle AT,T\rangle = 1 + T_2 + T_2^* + T_2^*T_2$ which fails to satisfy the desired inequality. Apr 8, 2020 at 0:08