Let $1<p<\infty$ and $f\_n$ be a sequence in $L^p(S^1)$ that converges weakly to some $f$. Here $S^1$ is the circle so we are dealing with periodic functions.


Let us see if $| f\_n |^p - | f |^p - | f\_n-f |^p$ converges to zero in distribution sense. This means that for every smooth function $\phi$ on $S^1$, $\int\_{S^1}(| f\_n |^p - | f |^p - | f\_n-f |^p)\phi$ goes to $0$.


Of course this is false because we can think of $f\_k(x)=1+sin(kx)$.


Now, what if we impose another condition that $f\_k$ converges almost everywhere to $f$?
I encountered this issue while studying calculus of variations and it seems much deeper than I thought. Is there any theorem or result for this problem? 



$| f_n |^p - | f |^p - | f_n-f |^p$ converges in distribution sense if $f_k$ converges almost everywhere and weakly to $f$?
Let $A\_{n,\epsilon}$ be the set where $|f|<\epsilon|f\_n|$. Now split the integral of $(|f\_n|^p-|f|^p-|f\_n-f|^p)\phi$ into an integral over $A\_{n,\epsilon}$ and an integral over its complement. Since the $L^p$ norm of $f\_n$ is bounded, the former part is bounded by a constant times $\epsilon$. The latter part converges to zero as $n\to\infty$ by the Lebesgue dominated convergence theorem.


