Let $f\_n \to f$ on compact subsets of the real line (these are functions defined on the real line) satisfying some conditions: $f$ has linear growth (but is nonlinear function) and is continuous and increasing and passes through the origin. The $f\_n$ are smooth, satisfy the same linear growth condition, and $f\_n'$ are bounded above and below positively.



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> If $u\_n \rightharpoonup u$ 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 $s > 0$.
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I saw this on page 103 on [this paper](http://link.springer.com/article/10.1007/BF01049490#page-1). The author obtains uniform (in $m$) bounds on 
$|u\_m|\_{L^2(0,T;H^1)}$ and $|u\_m|\_{L^\infty(s,T;L^\infty)}$ for every $s$. 


And then the claims the highlighted text above. He cites "a result of Aubin (see Temam, 1984)" but I was unable to find such a statement. I posted this on [MSE](https://math.stackexchange.com/questions/814837/a-compactness-result-if-f-nu-n-rightharpoonup-w-in-l20-tl2-then-f) and was told what I had thought: that we need a bound on time derivative. But the author makes no reference to this and obtains no such bounds AFAIK.


**EDIT**: I think what the author does is rearrange $\frac{d}{dt}f\_n(u\_n) - \Delta u\_n = g$ and uses the above bounds to obtain a bound on $\frac{d}{dt}f\_n$ in the space $L^2(0,T;H^{-1})$. Then he applies the Lions-Aubin lemma with $L^2 \subset H^{-1} \subset H^{-1}$ to extract strong convergence in $H^{-1}$. Somebody agrees with this?



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 $s > 0$
I agree with your edited question.


In fact, the information about $u\_n$ seems to be unnecessary. 


Consider $g\_n := f\_n(u\_n)$; then your assumptions give 



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> $g\_n\rightharpoonup w$ in $L^2(0,T;L^2)$ and $\partial\_t g\_n$ is bounded in $L^2(0,T;H^{-1})$.
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Indeed by the Lions-Aubin Lemma, for which I always use the more general reference of Simon (DOI:10.1007/BF01762360; see Theorem 5), $g\_n$ is strongly compact in $L^2(0,T;H^{-1})$. The weak convergence $g\_n\rightharpoonup w$ in $L^2(0,T;L^2)$ implies weak convergence in $L^2(0,T;H^{-1})$, and therefore uniquely characterizes the strong limit as $w$.


This argument does not give $w=f(u)$, however ...


