## First Dirichlet eigenvalue is simple for connected domains

Suppose is a connected open set and consider the first two eigenvalues of the Laplace operator with Dirichlet boundary conditions . Then .

**Proof:** First step is to prove that we can choose , the first eigenfunction, to be positive. Using the definition and the weak formulation we get

On the other hand if we denote and the positive and negative parts of we have

This means that we can choose the positive or negative part of and still get an eigenfunction (because at least one of them is non-zero). Because the first eigenfunction is analytic and non identically zero, it must be positive in the interior of (here we have used the fact that is connected).

Suppose now, that , and choose an eigenfunction for which is orthogonal to . Using the same trick as above, and because is characterized here by the same minimization problem as , we can apply the same reasoning as above, and see that we can choose non-negative.

Therefore we have found two orthogonal functions in one which is strictly positive and the other one non-negative. This cannot happen unless one of them is zero, and this contradicts the fact that are eigenfunctions.

As a consequence we deduce that if is connected.

I’m still a little confused. The positive/negative part of might not be in ?

Yes, the positive and negative parts of a function which is are again . In fact, a more general theorem holds: If is a Lipschitz function with then is if is . For a reference and proofs, look in A. Henrot, M. Pierre, Variation et Optimisation de Formes, Proposition 3.1.11.