Global Analysis — Global elliptic regularity

3.2 Global elliptic regularity

Let MM be a compact manifold, E0,E1E_0, E_1 complex vector bundles, and LL an elliptic differential operator from E0E_0 to E1E_1 of order kk. By patching local results together, we conclude that

Theorem (Global elliptic regularity)
  1. There is a constant AA such that for all s0s \geq 0 and uHs+k(M;E0)u \in H^{s + k}(M; E_0), we have

    us+kA(us+Lus). \| u\| _{s + k} \leq A(\| u\| _ s + \| Lu\| _ s).
  2. If LuHsLu \in H^ s, then uHs+ku \in H^{s + k}. In particular, if Lu=0Lu = 0, then uC(M)u \in C^\infty (M).

This implies a lot of very nice properties about elliptic operators. First observe the following result:


Let U,V,WU, V, W be Hilbert spaces and L:UVL: U \to V bounded, K:UWK: U \to W compact. If there is an AA such that

uUC(LuV+KuW), \| u\| _ U \leq C (\| L u\| _ V + \| K u\| _ W),

then kerL\ker L is finite-dimensional and imL\operatorname{im}L is closed.


We show that the unit ball of kerL\ker L is compact. If (un)(u_ n) is a sequence in the unit ball of kerL\ker L, then

unumUAKunKum. \| u_ n - u_ m\| _ U \leq A \| Ku_ n - Ku_ m\| .

Since KK is compact, there is a subsequence uniu_{n_ i} such that KuniK u_{n_ i} is Cauchy. So uniu_{n_ i} is Cauchy. So we are done.

To show imL\operatorname{im}L is closed, by restricting to the complement of the kernel, we may assume LL is injective. We will show that there is a cc such that

uUcLuV. \| u\| _ U \leq c \| L u\| _ V.

If not, pick a sequence unu_ n with unU=1\| u_ n\| _ U = 1 but LuV0\| L u\| _ V \to 0. By compactness, we may assume that KunKu_ n is Cauchy. Then we see that unu_ n must also be Cauchy, and the limit uu must satisfy uU=1\| u\| _ U = 1 and Lu=0Lu = 0, a contradiction.


L:Hs+kHsL: H^{s + k} \to H^ s has finite-dimensional kernel and closed image.

The compactness of the manifold MM is crucial for the inclusion Hs+kHsH^{s + k} \to H^ s to be compact.

We want to show that in fact LL is Fredholm, i.e. both the kernel and the cokernel are compact. Here we simply use duality. We need to show that

K={vHs:Lu,v=0 for all uHs+k} K = \{ v \in H^{-s} : \langle Lu, v\rangle = 0\text{ for all }u \in H^{s + k}\}

is finite-dimensional. But this is exactly the kernel of LL^*. So we deduce


LL is Fredholm, and in fact

Hs=kerL+im(L:Hs+kHs). H^ s = \ker L^* + \operatorname{im}(L: H^{s + k} \to H^ s).

The first factor is independent of ss (since all elements are CC^\infty ), and taking the limit ss \to \infty , we get

Γ(M,E1)=kerLimL. \Gamma (M, E_1) = \ker L^* \oplus \operatorname{im}L.\fakeqed

A bit of care is actually needed to take the limit ss \to \infty , but we leave that for the reader.