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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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    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.2011-06-03

2 Answers 2

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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 $u \in C^\infty_c(0, 1)$.2011-06-04
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Generally speaking, the word "elliptic" is used in many different contexts to indicate various positivity or nonvanishing conditions.

This is the case with Definition 2, where "H-ellipticity" (I presume H stands for Hilbert) is defined as a positivity property of an operator in a Hilbert space.

On the other hand, without any predicate, the "ellipticity" of a differential (or pseudo-differential) operator is a more standard concept, of which Definition 1 in the question is a special case.

First of all, the ellipticity of operators has nothing to do with their divergence forms, it is merely the positivity property of the operator's principal symbol. In the context of Definition 1, $A(x)\xi\cdot\xi$ is the principal symbol.

Regardless of any divergence form, a linear scalar-valued differential operator of order $2$, $Lu = a^{ij}(x) u_{x_i x_j}+\ldots,$ has symbol $\sigma(\xi) = a^{ij}\xi_i \xi_j$, and is elliptic whenever $\sigma(\xi)>0$ for $\xi\ne0$.

More generally, ellipticity is defined for any differential operator between sections of vector bundles as the property of the symbol being injective for all nonzero cotangent vectors. See for instance, page 289 in the book of Demailly, which is freely available online.