# Two different definitions of ellipticity

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

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.
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