A compactness result: if $f_n(u_n) \rightharpoonup w$ in $L^2(0,T;L^2)$, then $f_n(u_n) \to w$ in $L^2(s,T;H^{-1})$ for all $s > 0$. Let $f_n \to f$ on compact subsets of the real line.

If $u_m \rightharpoonup u$ in $L^2(0,T;H^1) \cap L^p(0,T;L^p)$ and $f_n(u_n) \rightharpoonup w$ in $L^2(0,T;L^2)$, then $f_n(u_n) \to w$ in $L^2(s,T;H^{-1})$ for all $s > 0$.

What result is this? Please can you refer me?
 A: There can be no such result, because the statement is fundamentally false. It will stay false in case  $f_n=f\;\,\forall\,n\,$, and it will stay false even if $f_n=f=I\;\,\forall\,n$ with the identity mapping $I\colon\,\mathbb{R}\to\mathbb{R}$. To see this, consider the following counterexample. Take a sequence 
$$
u_m(t,x)=\varphi(x)\sin{\frac{m\pi t}{T}}\,,\quad m\geqslant 1,
$$ 
with some fixed nonzero $\varphi\in H^1\cap L^p$. It is clear that $\,u_m\rightharpoonup\,0=u=w\,$ in $\,L^2(0,T;L^2)$, as well as in $\,L^2(0,T;H^1) \cap L^p(0,T;L^p)$,  while strong convergence $\,u_m\to\,0\,$ in $\,L^2(s,T;H^{-1})\,$ for any $\,s\in (0,T)\,$ is excluded by the fact that
$$
\begin{align}
\|u_m\|^2_{L^2(s,T;H^{-1})}=\frac{1}{2}\Bigl[T-s+\frac{T}{2\pi m}
\sin{\Bigl(\frac{2\pi ms}{T}\Bigr)}\Bigr]\|\varphi\|^2_{H^{-1}}\\
\geqslant \frac{1}{2}\Bigl[T-s-\frac{T}{2\pi m}\Bigr]\|\varphi\|^2_{H^{-1}}
\geqslant \frac{1}{2}\Bigl[T-s-\frac{T}{2\pi N_s}\Bigr]
\|\varphi\|^2_{H^{-1}}\\
\geqslant \frac{1}{2}\Bigl[T-s-\frac{T-s}{2}\Bigr]
\|\varphi\|^2_{H^{-1}}=\frac{T-s}{4}\|\varphi\|^2_{H^{-1}}>0
\quad\forall\,s\in (0,T),\;\forall m\geqslant N_s\,, 
\end{align}
$$
with $N_s$ defined as the least natural number such that
$$
N_s\geqslant\frac{T}{\pi(T-s)}\,. 
$$ 
To achieve what you want, you need some positive fractional order smoothness, no mattter how small, w.r.t. variable $t\in (0,T)$, or  its generic substitute of the form $\frac{\partial\,}{\partial t}u_m\rightharpoonup
\frac{\partial}{\partial t}u\,$ in $\,L^2(0,T;H^{-1})$, to say nothing of the more significant assumptions on $f_n$ and $f$.
