Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Assume that $P \in \Psi^{m}(X)$ (X is a $C^{\infty}$ manifold)is properly supported and has a real principal part p which is homogeneous of degree m.I'm interested in the existence theorem(at least locally) for the equation $Pu=f$,according to the abstract functional analysis,we should first learn the kernel of the adjoint $P^{\ast}$,that is $$N(K)=\lbrace v \in \epsilon'(K),P^{\ast}v=0 \rbrace$$

Here K is a compact subset of X(since only locally existence is concerned)such that no complete bicharacteristic curve is contained in K.Under this condition,is N(K) a finite dimensional subspace of $C_{0}^{\infty}(K)$ ? If it is, then how to prove this ? Or are there other conditions (about the domain or the operator) to make sure the finiteness of the kernel ?

*EDIT*According to the condition on the domain,it can be shown that actually $N(K)\subset C^{\infty}(K)$,but then I don't know how to prove that the dimension is finite(i.e. the unit ball is precompact).

share|improve this question
add comment

1 Answer 1

If $\epsilon'(K)$ is the space $\mathscr E'(K)$ of distributions with support in $K$ and you know that $N(K)$ is contained in $C^\infty(X)$ then it is finite dimensional: Since $N(K)$ is closed in the bigger space $\mathscr E'(X)$ (the dual of the Frechet-Schwartz space $C^\infty(X)$) it is also closed in the Frechet space $C^\infty(X)$. On the other hand, closed subspaces of DFS-spaces (= strong duals of Frechet-Schwartz spaces) are again DFS and we have an open mapping theorem which yields that $N(K)$ is a Frechet space as well as a DFS-space. But such spaces are finite-dimensional (because they have a relatively compact $0$-neighbourhood).

The question when $N(K)$ is actually contained in $C^\infty(X)$ is treated in Hörmander's Propagation of Singularities and Semi-Global Existence Theorems for (Pseudo)-Differential Operators of Principal Type, Annals of Mathematics, 108 (1978),569-609.

EDIT: That Frechet-Schwartz spaces $X$ which are closed subspaces of the (strong) dual $Y'$ of a Frechet space are finite dimensional is a combination of Baire's theorem, the open mapping theorem, and a compactness argument: If $p_n$ is a sequence of semi-norms describing the topology of $Y$ one has $Y'= \bigcup_n (Y,p_n)'$ (just because continuity of a linear functional on $Y$ is continuity with respect to some $p_n$), and then Baire's theorem implies that some of the Banach spaces $Z_n=(Y,p_n)'$ (endowed with the dual norm $p_n'(\phi)=\sup \lbrace |\phi(x)|: p_n(x)\le 1\rbrace$) contains a set of second category of $X$. The open mapping theorem (as in Rudin's Functional Analysis) for $\lbrace (x,\phi)\in X\times Z_n:x=\phi \rbrace \to X$, $(x,\phi)\mapsto x$ then implies $X\subseteq Z_n$. (This is essentially the proof of Grothendieck's factrization theorem.) Since $X$ is closed even in $Y'$ (endowed with the topology of uniform convergence on bounded subsets of $Y$) it is closed in $Z_n$ and therefore, $X$ is a Banach space (the Frechet space topology coincides with the Banach topology by the open mapping theorem). But in a Frechet-Schwartz space all bounded sets (in our case, in particular the unit ball of the norm) are relatively compact, and a Hausdorff topological vector space having a relatively compact neighbourhood of zero is finite-dimensional (Rudin's book, theorem 1.22).

I wrote up this proof because (despite their usefulness and simplicity) such things are not covered by most text books on functional analysis. An exception is the one of Meise and Vogt (Oxford University Press, 1997).

share|improve this answer
    
@Jochen Wengenroth.Thanks very much.Did you mean that a frechet and DFS-subspace of a frechet space is finite-dimensional ?Since i don't know much about the topology of Frechet spaces,I'm also very appreciated if you can show me some references where I can find results on such topics. –  user23078 Aug 29 '12 at 12:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.