This is a question originating in another mathematics forum, matematicamente.it (in Italian).
In literature one encounters the word elliptic in (at least) two different definitions. In what follows $\Omega$ is an open bounded subset of $\mathbb{R}^n$ and $\mathcal{D}(\Omega)$ denotes the space of smooth functions with compact support.
Definition 1 A differential operator (in divergence form)
$$L(u)(x)=-\mathrm{div} \big( A(x)Du(x) \big) u(x), \qquad x \in \Omega$$
is said to be (uniformly) elliptic (1) if there exists a $\theta >0 $ s.t. the matrix-valued function $A$ verifies
$$A(x)\xi \cdot \xi \ge \theta \lvert \xi \rvert^2, \qquad x, \xi \in \mathbb{R}^n.$$
Definition 2 A (densely defined) linear operator $(L, D(L))$ on a Hilbert space $H$ is said to be $H$-elliptic (2) if there exists a $c >0$ s.t.
$$(Lu, u) \ge c \lVert u \rVert^2, \qquad u \in D(L).$$
Question Let
$$L(u)(x)=-\mathrm{div} \big( A(x)Du(x) \big) u(x), \quad
D(L)=\mathcal{D}(\Omega),\quad
H=L^2(\Omega).$$Is it true that $L$ is elliptic as in
definition 1 if and only if it is
$H$-elliptic as in definition 2? Assume that $A$ depends continuously on $x$ and is symmetric everywhere.
It is straightforward to prove that definition 1 implies definition 2; I find it nontrivial to prove the converse (if true).
What do you think?
1) cfr. Evans, Partial differential equations, §6.1.1.
2) cfr. Kesavan, Topics in functional analysis, §3.1.1.
Best Answer
The answer is no. For instance, definition 2 can be satisfied by subelliptic operators which are not elliptic in the sense of definition 1. The sublaplacian on the Heisenberg group is an example.
I'll add some more details later when I have more time.
Edit: Actually, unless I am mistaken, even a fully degenerate operator can satisfy definition 2. Let $\Omega = (0,1)^2$ be the open unit square in $\mathbb{R}^2$, and let $L = - \frac{\partial^2} {\partial x^2}$. We know that $-\frac{d^2}{dx^2}$ is elliptic in all senses on $(0,1)$, so if $u \in C^\infty_c(\Omega)$, then for each $y \in (0,1)$ we have $$-\int_0^1 u_{xx}(x,y) u(x,y) dx \ge c \int_0^1 |u(x,y)|^2 dx$$ for a constant $c$ independent of $y$. Integrating with respect to $y$ now gives the result.