I am reading a paper that deals with the solution for the Sturm-Liouville problem's uniqueness of solutions by defining an operator $K$:
$$K:L^2([0,1]) \to L^2([0,1])$$ $$f \hspace{0.5cm} \longrightarrow u$$
Where u is the solution to the ODE $u'' + \rho u=f$ (With the boundary conditions $u(0)=u(1)=0$). It is then shown that the operator $K$ is injective, so that for a given $f \in L^2([0,1])$ there is a unique solution $u \in L^2([0,1])$.
Also, if $f \in C([0,1])$, as u verifies $u'' + \rho u=f$ then $u''$ is also continuous so that $u \in C^2([0,1])$.
So my question is:
We are dealing with equivalence classes in $L^2([0,1])$, so the uniqueness that comes from the fact that $K$ is injective is for the equivalence class $[u]$. So precisely what I need to know is whether in the set $[u]$ the representative in $C^2([0,1])$ is also unique.
Continuous representatives are unique whenever they exist. That is just because if you modify a continuous function at one point, to have the result also be a continuous function, you need to make a modification on an interval, which will leave the equivalence class you started in (since intervals have positive measure).
Another way of putting it is that the complement of a measure zero set is always dense, and two continuous functions that agree on a dense set agree everywhere.