I came across the following statement: If $V=\{v \in C^1[0,1])~ |~ v(0)=0\}, ~~f \in C([0,1])$ and $u \in C^2([0,1])$ then any solution to
\begin{equation} u(0)=0 ~~\text{and}~~ \text{for all}~~ v \in V \int_0^1u'(t)v'(t)dt=\int_0^1 f(t)v(t)dt \end{equation} is a solution to \begin{equation} -u''(t)=f(t)~~ \text{for all}~ t \in [0,1] ~~~~~~\text{with}~~~~~~~~ u(0)=0 ~~~~~~~~~~\text{and} ~~~~~~~~ u'(1)=0 \end{equation} and vice versa.
I can see why any strong solution is a weak solution since we can multiply the equation \begin{equation} -u''(t)=f(t) \end{equation} with a test function $v \in V$ and integrate over the intervall $[0,1]$. Using the boundary conditions we obtain \begin{equation} -\int_0^1u''(t)v(t)dt=-u'(t)v(t)|_{[0,1]}+\int_0^1u'(t)v'(t)=\int_0^1u'(t)v'(t)dt. \end{equation} Hence, any strong solution is a weak solution. I'm aware that the condition $u\in C^2[0,1]$ is crucial for the other direction but I still fail to understand how to prove it properly. Could anyone show me why our weak solution is a strong solution (in this particular case)?
Let $ u$ be a weak solution such that $ {u} \in {\mathcal{C}}^{1} \left(\left[0 , 1\right]\right)$. Let
\begin{equation} F \left(t\right) = \int_{t}^{1}f \left(s\right) d s\end{equation}
Let $ {\phi} \in {\mathcal{C}}_{0}^{0}$ be an arbitrary continuous function such that $ {\phi} \left(0\right) = {\phi} \left(1\right) = 0$ and let
\begin{equation}v \left(t\right) = \int_{0}^{t}{\phi} \left(s\right) d s \in V\end{equation}
We have $ {v'} = {\phi}$. By the weak solution property we have
\begin{equation}\int_{0}^{1}\left({u'} \left(t\right)-F \left(t\right)\right) {\phi} \left(t\right) d t = \int_{0}^{1}{u'} \left(t\right) {v'} \left(t\right) d t-{\left[F \left(t\right) v \left(t\right)\right]}_{0}^{1}-\int_{0}^{1}f \left(t\right) v \left(t\right) = 0\end{equation}
because $ F \left(1\right) = 0$ and $v \left(0\right) = 0$. The value of this integral is zero for every function $ {\phi} \in {\mathcal{C}}_{0}^{0}$ and this proves that
\begin{equation}{u'} \left(t\right) = F \left(t\right)\end{equation}
Note in particular that $ {\lim }_{t \rightarrow 1} {u'} \left(t\right) = 0$.
It follows that $ {u'} \in {\mathcal{C}}^{1}$ hence $ u \in {\mathcal{C}}^{2}$ and $ u$ is a strong solution.