Let $a, b \in \mathbb R^n$ and $f, g \in L^1 [0,1]$. Assume for all $h \in AC[0,1 ]$ (space of absolutely continuous functions) following integral equality holds 


$$ \int\_{0}^{1} \langle f(t) , h(t) \rangle \; dt + \int\_{0}^{1} \langle g(t) , h' (t) \rangle \; dt + \langle h(0) , a \rangle + \langle h(1) , b \rangle = 0 $$


My question : Can we simplify the above expression more? In the sense above can be equivalently written in the form of an ODE in terms of $f,g,a,b$? 


Note that all functions are from $[0,1]$ to $\mathbb R^n$. However we can assume they are just real valued function for simplicity. I was thinking about trying $h = \text{constant or exponential functions.}$



Converting an integral equation into a differential equation
This is elementary but let's state it. First one only considers smooth test functions $h$ with compact support (so that the last two terms disappear). If $F(x):=\int\_0^xf(t)dt$, integrating by parts the first term gives $\int\_0^1\langle (F-g), h'\rangle dt=0$ for all these $h$, whence $F-g$ is a constant, that is, $g$ is (has a representant which is) $AC$ and $f$ is its weak derivative. The initial relation then writes $\langle h(1),(b+g(1))\rangle +\langle h(0),( a-g(0))\rangle=0$ for all $h\in AC$, that is, $g(0)=a$, $g(1)=-b$. In conclusion, the initial weak relation (for a continuous representant of $g$) is equivalent to: $g$ is $AC$, $f=g'$, $g(0)=a$, $g(1)=-b$.


