I'm trying to determine what weak derivatives the function $$ f(x)=\begin{cases} x&\mbox{if }0<x<1,\\ 1&\mbox{if }1\leq x<2, \end{cases} $$ has.
I already managed to prove that it has a first weak deriviative $$ g(x)=\begin{cases} 1&\mbox{if }0<x<1,\\ 0&\mbox{if }1\leq x<2. \end{cases} $$ I'm pretty confident that this deriviative does not have a weak deriviate of its own. How could I prove it?
So far I came to realise that if $h$ was a weak deriviative of $g$, it would need to fulfill the following equation:
$$\int h\cdot \psi = -\psi(1)$$ for every $\psi$ which is a function in $L^1_\mbox{loc}$. I got it by integration by parts, and using the fact that $\psi$ would need to be zero on $0$ and $2$.
Is it the right way to go? I could not find a combination of $\psi$'s that would lead me towards a contradiction.