Let $-\infty < a < b < \infty$ and set $U = (a,b)$. A weak solution of the Poisson's equation $\Delta u = f$ subject to $u = 0$ on the boundary with $f \in L^2(U)$, is a function $u \in H_0^1(U)$ that satisfies $$\int_a^b u' v' = - \int_a^b f v \, \quad \forall v \in H_0^1(U)\,.$$
Existence can be obtained via the Riesz representation theorem.
Since we know that $u \in L^2(U)$, $u'$ acts on tests functions $\varphi \in C_c^\infty(U)$, as $u'[\varphi] = \int_a^b u' \varphi$. The distributional derivative of $u'$, acts on test functions as $Du'[\varphi] = - u'[\varphi'] = - \int_a^b u' \varphi'$. From the existence argument we know that $Du'[\varphi] = f[\varphi]$ for all test functions $\varphi \in C_c^\infty$. Hence, the distributional derivative of $u'$, which is the second weak derivative of $u$, is a function in $L^2$, hence $u \in H^2$.
These are my questions:
Is this correct for the one-dimensional case?
Once we try the same argument for more dimensions the result is not that the second derivates are in $L^2$ but just that the Laplacian is in $L^2$, and the argument to the whole second derivatives requires some integration by parts not possible for functions in $H_0^1$, and that's why we need the interior and boundary regularity theorems, right?
Best.