Let $f \in L^2$ be absolutely continuous on an open bounded set $(a,b)$. Now, take $\phi_n \rightarrow f$ in the $L^2$, sense.
I would like to understand if we have the following limits.
1.) Let $p \in C(a,b)$ so not necessarily bounded at the interval ends. and $pf'\in L^2$. Does this mean that we have $$p\phi_n' \rightarrow pf'$$ in the $L^2$-sense?
Now the other way around:
2.)Assume we know that the limit of $p \phi_n'$ exists in the $L^2$ sense, does this mean that $p \phi'_n \rightarrow pf$ in the $L^2$-sense(which of course implies that $pf'\in L^2$ exists)?
My attempts:
1.) Maybe we have from the $\phi_n \rightarrow f$ as $L^2$ convergence that a subsequence converges pointwise a.e. This could(I am not so sure) imply that $p \phi_n' \rightarrow pf'$ for a proper subsequence a.e. But I fail to see, how we can establish the $L^2$ convergence from this?
2.) By $\phi_n \rightarrow f$ we have that a subsequence $\phi_n \rightarrow f$ pointwise almost everywhere. Thus, we should also have $\phi_n' \rightarrow f'$ pointwise a.e for a subsequence. So we have pointwise a.e. convergence $p\phi_n' \rightarrow pf'$ and $p \phi_n'$ converges in the $L^2$ sense. So afais, both limits must be the same.
Could anyboy help me making this rigorous? One issue is that I am not sure, if we have $\phi_n \rightarrow f$ pointwise a.e. then the same holds true for the derivatives?
If anything is unclear, please let me know.