I'm working on a functional-analysis problem set, and the question is:
Let $g\in L^1(0,1)$, and define $f(x) = \int_0^x g(t)\,dt$. Show that $f\in W^{1,1}(0,1)$ and that the weak derivative of $f$ is $g$.
I (think I) solved the exercise correctly, although I don't understand why I can do some steps. Here is what I found out:
$f\in L^1(0,1)$ is obvious, as $g\in L^1(0,1)$. We can bound an estimate of $\int_0^1 f$ by $\|g\|_{L^1(0,1)} < \infty$ (by assumption).
For showing, that the weak derivative of $f$ is in $L^1(0,1)$, I wanted to show that $f'=g$, then it is clear that $f'\in L^1(0,1)$ (by assumption $g\in L^1(0,1)$).
We know, that each $L^p(U)$ function on an open subset $U\subset\mathbb{R}^N$ can be approximated by a sequence in $C^\infty_0(U)$. Especially we can apply this to our function $g$ which is in $L^1(0,1)$:
$\exists \{g_n\}_{n}\subset C^\infty_0(0,1),\,g_n\to g.$
Obviously all $g_n$ are integrable.
They are continuous, so they even have an antiderivative, right?
If I would find a majorant of the sequence, then I could apply dominated convergence to get
$\int_0^x g_n(t)\,dt = \int_0^x \lim_{n\to\infty}g_n(t)\,dt = \lim_{n\to\infty} \int_0^x g_n(t)\,dt$.
Is this the right way? Why can I do this?
Using this and the definition of the weak derivative I showed
$-\int_0^1 \varphi(x)f'(x)\,dx = \int_0^1\varphi'(x)\lim_{n\to\infty}\int_0^xg_n(t)\,dt\,dx,\quad\forall\varphi\in C^\infty_0(0,1)$
Now I would use that $g_n$ is continuous, therefore an antiderivative $G_n$ exists. This means $\int_0^x g_n(t)\,dt = G_n(x)$.
The above equation for the weak derivative gets
$ \begin{eqnarray*} -\int_0^1 \varphi(x)f'(x)\,dx &=& \int_0^1 \varphi'(x)\lim_{n\to\infty}\int_0^x g_n(t)\,dt\,dx\\ &=& \int_0^1\varphi'(x)\lim_{n\to\infty} G_n(x) \end{eqnarray*} $
Now I would like to again apply dominated convergence to get the limit out, i.e.
$\int_0^1\varphi'(x)\lim_{n\to\infty} G_n(x)\,dx = \lim_{n\to\infty}\int_0^1 \varphi'(x) G_n(x)\,dx$
Again, I have no clue why this is possible!
However, after partial integration I get the result I wanted, that means the weak derivative of $f$ is $g$. Therefore I think I can't be too wrong...
Some help/hints are appreciated!
Thank you :-)