Let us begin with a discussion of the geometric question. If is an open set in (bounded or unbounded), let denote the lowest eigenvalue of in with Dirichlet boundary conditions. if is empty. Intuitively, if is small then must be large in some sense. One well known result in this direction is the Faber-Krahn inequality which states that among all domains with a given volume , the ball has the smallest . Thus,
where is the lowest eigenvalue of a ball of unit volume. This inequality clearly does not tell the whole story. If is small then must not only have a large volume, it must also be “fat” in some sense.
Let us place here a very beautiful result due to Lieb among other big contributions. This result was published in Invent. Math. during 1983. The proof relies upon the Rayleigh quotient and a very clever choice of a trial function for the variational characterization of . For the whole paper, we refer the reader to here.
Theorem. Let and be non-empty open sets in (), and and be the lowest eigenvalue of with Dirichlet boundary conditions. Let denote translated by . Let . Then there exists an such that
If and are both bounded then there is an such that
Before proving the theorem, let us recall the so-called Rayleigh quotient. Precisely
We turn now to the proof of the theorem.
Part 1. There exist , such that
Here, we think of as defined on all of with for and similarly for . Without loss and can be assumed to be real valued. The normalization
is imposed. The functions given by
satisfy . Let
We now compute
The last term can be written as
Thus the integral (over ) of this term vanishes and
on a set of positive measure. The proof then follows from the Rayleigh quotient.
Part 2. Since and are bounded there exist and such that
This is a simple consequence of the Rellich-Kondrachov compactness theorem. (Again, we extend to all of with , , and similarly for ; it is easy to see that .) Define
Since , . Likewise, since and are functions,
It is easy to see (by approximating by functions) that in the sense of distributions.
It is not a-priori clear that or . (They are, in fact, in because and . However, this latter fact is not elementary and we prefer to avoid using it. For our purpose it suffices to show that and for almost all , and this can be done by the following elementary argument.) We note that (by Fubini) satisfies . Thus, a.e. and a.e. Likewise, repeating the argument for we get
(Here, one has to note as above that a.e. and a.e. By the Schwartz inequality,
Finally, in the distributional sense, whence .). Thus, for almost all , and . We have that . The remainder of the proof is as before, except that in order to prove the strict inequality, i.e. we must show that
cannot hold a.e. To see this let (resp. ) be the characteristic function of (resp. ). Then
is a continuous function of compact support. For there is an open set such that for . By the strong maximum principle, both and are positive in the open set , . (Here we assume that and are connected; if not, it suffices to replace and by connected components on which .) Thus, , . If , , then , but this is impossible for sufficiently small by the Faber-Krahn inequality.