How do fixed point arguments for PDE work? My understanding (which I'm not altogether sure of) is that a a "fixed point" argument often used in PDE goes something like this:
If we have some PDE like
$$u=F(u),$$
we consider a sequence of functions $u^{(n)}$ recursively defined so
$$u^{(n+1)}=F(u^{(n)}).$$
Then you compute an estimate like
$$\|u^{(n+1)}\|_X\leq g(\|u^{(n)}\|_X)$$
(where the mathematician knows what function space $X$ should be) and then you solve the recurrence equation to find say
$$\|u^{(n)}\|_X\leq c$$
and then use some kind of compactness argument to say the $u^{(n)}$ have a particular kind of limit in $X$ which by construction solves the desired equation.
Is my understanding of this strategy broadly correct? There are some details I don't understand. For instance, what is the particular kind of compactness argument we need? Say $X=H^s$. A ball in $H^s$ is not compact in the norm topology, right? So how does one make this work?
I ask this question because it seems that people who do PDE use things like this all the time so they skim over the technical details.
 A: Your understanding is correct. The compactness part comes from the nice and workhorse theorem in these areas: Rellic Compactness theorem:
If you have a bounded sequence $\{f_n\}$ in $H^s$ it sure is not guaranteed to have a convergent subsequence (in $H^s$). But notice that for $t<s$, $H^s \subset H^t$, ie. Sobolev spaces are nested, and $L^2=H^0$ is the biggest, say.
Now this beautiful theorem says that the inclusion
$$ H^s \longrightarrow H^t$$ is a compact operator-- which exactly means that the sequence above does have a convergent subsequence in $H^t$, in particular in $L^2$.
It is therefore necessary and easier to work with mixed norms. Look at this example from differential geometry. You define an operator $D$ on the smooth functions on a manifold $M$ (compact and oriented). $D^2=\Delta$ is the Laplacian on $M$. Now as operator on Sobolev spaces, we have this $$\| u\|_s \leq C(\| u \| _{s-1} + \|Du \| _{s-1}).$$
Now assume $u \in $L^2  is harmonic, i.e. $\Delta u=0$, which in fact implies that $Du=0$, then, the norm inequality above with $s=1$, shows that it is in $H^1$, since its $H^1$ norm is bounded then. Then use the same inequality with $s=2$ to see that $u$ is in $H^2$, and inductively prove that it is in all Sobolev spaces. But this means that $u$ is in fact a smooth function, and a classical harmonic function, not just an $L^2$ one.
