This is a special case of "why do we need Sobolev spaces". There seem to be good general answers to the question here (I believe I read some), but too high-level for me (I didn't really get the point yet). So here's a more specific question.
In my understanding, we are using Sobolev spaces because we want to find a "less regular" solution to a PDE that has no classical solution (that is, one with sufficiently many well-behaved derivatives).
However, if we have a problem in weak formulation, represented as B[u,v] = (f,v) $\forall v \in H$ where H is a Hilbert space, and the requirements of the Lax-Milgram theorem fulfilled , can we not choose both a fitting Sobolev Space (e.g. $H = H^1((0,1))$) OR also a fitting "truly" differentiable space, e.g.:
- $H = F$ := {$f \in C^0((0,1))$ | $f' \in L^2((0,1))$}?
Doesn't Lax-Milgram also guarantee a solution lying in $F$ ? So why make use of Sobolev spaces here? (We cannot use $C^\infty$ since it is not complete, and Lax-Milgram won't apply.)