I have trouble understanding the following proposition.
Proposition
$f,g\in L_{\text{loc}}^1(\Omega)$. Then $g=D^{\alpha}f$ iff. there exists $f_m\in C^{\infty}(\Omega)$ such that $f_m\to f$ in $L_{\text{loc}}^1(\Omega)$ and $D^{\alpha}f_m\to g$ in $L_{\text{loc}}^1(\Omega)$.
Here is the solution I received
Proof: ($\Longleftarrow$) $f_m\to f$ and $D^{\alpha}f_m\to g$ \begin{align*} \underbrace{\int_{\Omega}f_m\cdot D^{\alpha}\phi\,dx}_{\downarrow}= \underbrace{(-1)^{|\alpha|}\int_{\Omega}D^{\alpha}f_m\cdot\phi\,dx}_{\downarrow} \end{align*}
\begin{align*} \int_{\Omega}f\cdot D^{\alpha}\phi\,dx= (-1)^{|\alpha|}\int_{\Omega}D^{\alpha}g\phi\,dx \end{align*}
\begin{align*} \left|\int_{\Omega}f_m\cdot D^{\alpha}\phi\,dx-\int_{\Omega}f\cdot D^{\alpha}\phi\,dx\right|&\le\int_{\Omega}|f_m-f| \underbrace{|D^{\alpha}\phi|}_{C}\,dx\\ &\le C\int_{\text{sup}(\phi)}|f_m-f|dx\to 0 \end{align*}
Q1 I feel the solution is incomplete for the part $(\Longleftarrow)$. Also, base on the above equations, there is no explaination provided. I became quite lost on what is going on. Could any one help providing a complete solution for the part $(\Longleftarrow)$, or add some explaination on the above?
Q2 By the way, here is based don my understanding...but I am afraid I am wrong. Thanks for any corrections.
($\Longleftarrow$) Suppose there exists such a sequence $\{f_m\}$ then by assumptions, \begin{align*} f_m&\to u\quad\text{in}\quad\text{in}\quad L_{\text{loc}}^1(\Omega)\\ D^{\alpha}f_m&\to g\quad\text{in}\quad\text{in}\quad L_{\text{loc}}^1(\Omega) \end{align*} By the definition of weak derivatives, we have \begin{align*} \int_{\Omega}f D^{\alpha}\varphi\,dx&=(-1)^{|\alpha|}\int_{\Omega}D^{\alpha}f\cdot\varphi\,dx\\ \int_{\Omega}u D^{\alpha}\varphi\,dx&=(-1)^{|\alpha|}\int_{\Omega}g\varphi\,dx \end{align*} To match the rhs of above, we have For $p=1$, \begin{align*} &\ \left|\int_{\Omega}g(x)\varphi(x)dx-\int_{\Omega}D^{\alpha}f(x)\varphi(x)\right|\\ &\le \left|\int_{\Omega}(g(x)-D^{\alpha}f(x))\varphi(x)\right|\\ &\le \int_{\Omega}\left|g(x)-D^{\alpha}f(x)\right||\varphi(x)|dx\\ &\le C \int_{\Omega}\left|g(x)-D^{\alpha}f(x)\right|dx \end{align*} as $m\to\infty$, where $C=\sup_{\Omega}|\varphi(x)|<\infty.$
Now to match the lhs, we have For $p=1$, \begin{align*} &\ \left|\int_{\Omega}f(x)D^{\alpha}\varphi(x)dx-\int_{\Omega}u_m(x)D^{\alpha}\varphi(x)\right|\\ &\le \left|\int_{\Omega}(f(x)-u_m(x))D^{\alpha}\varphi(x)\right|\\ &\le \int_{\Omega}\left|f(x)-u_m(x)\right||D^{\alpha}\varphi(x)|dx\\ &\le C' \int_{\Omega}\left|f(x)-u_m(x)\right|dx \end{align*} as $m\to\infty$, where $C'=\sup_{\Omega}|D^{\alpha}\varphi(x)|<\infty.$
I understood that the downarrow mean D.C.T.. But I need more detail. Would be appreciate if someone could help filling up the gap. I added my own solution which I doubt there could contain lots of mistakes.