I'm sort of new to the notion of weak derivative and am learning more about it beyond the definition.
Consider the mapping f: $H^1(\mathbb{R})\to L^2(\mathbb{R})$ given by $f(u)=u'+u,$ with the derivative given in the weak sense. It seems to me like you could set up $u'+u=g\in L^2$ and solve for $u$ in the $H^1$ sense. In general, $u'+u=g$ a.e. implies $(u(z)e^z)'=g(z)e^z$ a.e. in the $L^2$ sense. We can throw these terms in integrals with test functions to be more precise. Something feels quite off about this analysis, but I can't see what.
Similarly, the mapping $u\mapsto u'$ seems like it may be invertible. Can anyone explain where I am incorrect?
The inverse of $u\to u+u'$ is $$v\to v \ast e^{-x}1_{x > 0}$$ which is bounded $L^2(\Bbb{R})\to H^1(\Bbb{R})$. Proof : $$u'\ast e^{-x}1_{x > 0}= u\ast (e^{-x}1_{x > 0})'=u\ast (\delta-e^{-x}1_{x > 0})=u-u\ast e^{-x}1_{x < 0}$$
The inverse of $u\to u',H^1(\Bbb{R})\to L^2(\Bbb{R})$ is unbounded densely defined on $\{ v\in L^1\cap L^2,\int_{-\infty}^\infty v(x)dx=0\}$ where it is $v\to v\ast 1_{x >0}$. Also the Fourier transform makes it correspond with $v\to \frac{v}{2i\pi \xi }$.