Global Analysis — Global elliptic regularity

## 3.2 Global elliptic regularity

Let $M$ be a compact manifold, $E_0, E_1$ complex vector bundles, and $L$ an elliptic differential operator from $E_0$ to $E_1$ of order $k$. By patching local results together, we conclude that

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

$\| u\| _{s + k} \leq A(\| u\| _ s + \| Lu\| _ s).$
2. If $Lu \in H^ s$, then $u \in H^{s + k}$. In particular, if $Lu = 0$, then $u \in C^\infty (M)$.

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

Proposition

Let $U, V, W$ be Hilbert spaces and $L: U \to V$ bounded, $K: U \to W$ compact. If there is an $A$ such that

$\| u\| _ U \leq C (\| L u\| _ V + \| K u\| _ W),$

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

Proof

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

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

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

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

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

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

Corollary

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

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

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

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

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

Corollary

$L$ is Fredholm, and in fact

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

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

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

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