I will for example that describe the problem: Let $f\in L^2(0,1)$ \begin{align}(1)\quad \begin{cases}-u''(x)=f(x),\forall x\in (0,1)\\ u(0)=u(1)=0\end{cases}\end{align}
Then $\forall \phi \in C^\infty _c(0,1)$, we get $$\int_0^1 u''(x)\phi(x)dx=\int^1_0f(x)\phi(x)dx\\ (2) \quad\Leftrightarrow \int^1_0u'(x)\phi'(x)dx=\int^1_0f(x)\phi(x)dx$$ Now, using Lax-Milgram theorem: $\exists ! u\in H^1_0(0,1)$ is solution (2), then use regularity theory, "we can prove $u\in H^2(0,1)$" (of course with (1) problem is easier because it is 1 dimesion ). The last, $u$ is (1) solution also (2) solution.
My quesion is "we can prove $u\in H^2(0,1)$" (i.e: $u''\in L^2(0,1)$ ), using regularity theory is fine, but before have $u''$ ( weak derivative ) or not ?. Clearly, in other problems, we need to have solution formula of $u''$ and then prove $u''\in L^2(0,1)$. I don't understand, if we don't sure $\exists u''$ is why we can continue to prove $u''\in L^2(0,1)$ (i.e: $||u''||_{L^2(0,1)}\leq C||f||_{L^2(0,1)}$ with $C>0$).