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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.

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What are you assuming about $\Omega$? And what is $\mathcal{D}(\Omega)$? –  Nate Eldredge Jun 3 '11 at 12:21
    
@Nate: Sorry I forgot. $\Omega$ is open and bounded and $u \in \mathcal{D}(\Omega)$ means that $u$ is smooth and compactly supported. Also, there was an error in the definition of the differential operator, which is now fixed (thanks to the person who pointed this out - you know who you are! :-) ). –  Giuseppe Negro Jun 3 '11 at 12:40
    
Actually, is it obvious that $L=\frac{\partial^2}{\partial x^2}$ on, say, the unit disk or unit square in $\mathbb{R}^2$, doesn't satisfy definition 2? Somehow I'm not seeing it. –  Nate Eldredge Jun 3 '11 at 18:05

1 Answer 1

up vote 2 down vote accepted

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.

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Your example is really a relief: I had read your comment and was struggling with cutoff functions to prove that $-\frac{\partial^2}{\partial x^2}$ was not $L^2$-elliptic. Now I see why I couldn't do it: it was false! Thank you very much. To make things even simpler I would like to add that we can take $c=1$, because $$\int_0^1 \left( \frac{du}{dx} \right)^2 \, dx \ge \int_0^1 u(x)^2\, dx$$ for all $u \in C^\infty_c(0, 1)$. –  Giuseppe Negro Jun 4 '11 at 17:31

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