What can I assume in this proof?

Question

I know that in this question I'm supposed to let f, g, a, and b all be arbitrary. Can I assume that $f''$ and $g''$ exist?

Answer 1

$f,g\in A \implies f'',g''$ exist, and $(f'' + f) = (g''+g) = 0$

$q = af+bg$

Differentiation is a "linear operator"

$q'' = af'' + bg''\\ q + q'' = a(f + f'') + b(g+g'')\\ q+ q'' = 0\\ q+ q'' \in A$