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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.

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 $s > 0$.

I saw this on page 103 on this paper. 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 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?

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  • $\begingroup$ You're right, some control on time variation is necessary. But what is your question?} $\endgroup$ – Mark Peletier Jun 2 '14 at 17:00
  • $\begingroup$ @MarkPeletier My question is why is the stated strong convergence true? $\endgroup$ – LapLace Jun 2 '14 at 17:38
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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

$g_n\rightharpoonup w$ in $L^2(0,T;L^2)$ and $\partial_t g_n$ is bounded in $L^2(0,T;H^{-1})$.

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 ...

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  • $\begingroup$ Thanks. As for your last sentence, that is what is proved next in identification of the limit via a technique involving subdifferentials. $\endgroup$ – LapLace Jun 4 '14 at 21:01

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