This question is simply applying a theorem. But I do not understand how . One can treat most of the content as black box. I will provide the definitions.
The context:
I want to show
If $M$ is a compact $n$-dimensional manifold, $P$ an ellipitic differential operator of order $k$, then $P:W^{k+l} \rightarrow W^l$ has kernel whose dimension is independent of $l$.
Black box terminology explanation:
Definition: Let $E_i \rightarrow M$ be two vector bundles, $P:\Gamma(M,E_0) \rightarrow \Gamma(M,E_1)$ is an elliptic differential operator of order $k$, if locally $P$ can be written as
$$ Pf = \sum_{|\alpha| \le k } A^\alpha(y) \frac{\partial^\alpha}{\partial x_\alpha} f(y).$$
Definition 2: Let $E \rightarrow M$ be a complex vector bundle over a compact $n$-dimensional manifold. Then we can give the space of sections $\Gamma(M,E)$ a Sobolev norm. We denote $W^k$ be the completion of $\Gamma(M,E)$ with respect to this norm.
Let us suppose $P:W^{k+l} \rightarrow W^k$ is well defined and:
Theorem: Let $Pu=f$, $f \in W^l$, $u \in W^r$ for some integer $r$, then $u \in W^{l+k}$.
It is claimed that then we have the kernel is independent of $l$.
How does this follow?
Reference: Pg 48-49.
Best Answer
Suppose $u \in W^{k+r}$ lies in the kernel of $P,$ so $Pu = 0.$ Then as $0 \in W^{\ell}$ for each $\ell,$ so the elliptic regularity theorem gives $u \in W^{k+\ell}$ for all $\ell.$ So if $N_r \subset W^{k+r}$ is the null space of $P$ viewed as an operator $W^{k+r} \rightarrow W^r,$ we get $N_r \subset N_{\ell}$ for each $\ell.$ As $r$ was also arbitrary, we see that $N_r$ is independent of $r$ and hence its dimension also is.