Let $g(t)=t^{\beta-1}$, where $\beta\in (0,1)$. I am pondering, if there exists a weak derivative of $g$ in space $L^1(0,T)$, $T>0$.
Firstly, we see that $g\in L^1(0,T)$. Now, we are looking for function $v\in L^1(0,T)$, so that for every $\phi\in C_c^\infty[0,T]$ it holds $$\int_0^T g(t)\phi'(t)dt=-\int_0^T v(t)\phi(t)dt.$$
I thought, it is very easy, one needs just to count it. So, let's count, using per partes we are getting $$\int_0^T t^{\beta-1}\phi'(t)dt=[t^{\beta-1}\phi(t)]_0^T-\int_0^T(\beta-1)t^{\beta-2}\phi(t)dt=$$$$=-lim_{t\rightarrow 0+} t^{\beta-1}\phi(t)-\int_0^T(\beta-1)t^{\beta-2}\phi(t)dt=-\int_0^T(\beta-1)t^{\beta-2}\phi(t)dt$$
To count the limit we can use l'Hospital. So, we've got $v(t)=(\beta-1)t^{\beta-2}$, but it is not in $L^1(0,T)$. From this I think, we can say that $g\in L^1(0,T)$ has not a weak derivative at all. Have I done everything good? Have I missed something?
There have already been little bit similar questions, but I was not satisfied with the answers, therefore I am asking separately. It is my first question, I hope it is well written. If something, please correct me.
It seems to me that you proved that $g$ does not belong to the Sobolev space $W^{1,1}(0,T)$. Many books give a definition of weak derivative as follows.
Definition. A function $u \in L^1_{\mathrm{loc}}(a,b)$ has a weak derivative in $(a,b)$ if there exists a function $v \in L^1_{\mathrm{loc}}(a,b)$ such that $$\int_a^b u \varphi' = - \int_a^b v \varphi$$ for every $\varphi \in C_c^\infty(a,b)$. If such a $v$ exists, then we write $\partial u=v$.
Here you are forced to saty away from $a$ and $b$, since you are interested only on integrability in the interior.