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Let $$u_m \rightharpoonup u \quad \text{(weakly) in $L^\infty(0,T;L^2(\Omega)) \cap L^2(0,T;H^1(\Omega))$}.$$ We are given $f:\mathbb R \to \mathbb R$, a Lipschitz continuous invertible map which is monotone $$(f(x)-f(y))(x-y) \geq 0\quad\text{for all $x, y$}$$ and we have $$f(u_m) \rightharpoonup f(v) \quad \text{in $L^2(0,T;H^1(\Omega))$}$$ for some $v \in L^2(0,T;H^1(\Omega))$ (assume that for $u \in L^2(0,T;H^1)$, $f(u)$ and $f^{-1}(u)$ are in $L^2(0,T;H^{1}(\Omega))$.)

Is it possible to show that indeed $v=u$, i.e. $f(u_m) \rightharpoonup f(u)$? I can't seem to do it by using the monotonicity method.

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    $\begingroup$ What is $X$? What is $(,)$? If $f$ it continuous, $f(u_m)\to u$ by definition. Thus, $f(u)=f(v)$. If $f$ is also invertible, you've got to have $u=v$. $\endgroup$ Commented Feb 22, 2014 at 16:48
  • $\begingroup$ @AlexDegtyarev Apologies, $X$ is a Hilbert space. All convergences are weak convergences, not strong. $\endgroup$
    – maximumtag
    Commented Feb 22, 2014 at 17:12
  • $\begingroup$ Not sure why this got downvoted. If the question is trivial please post the answer! $\endgroup$
    – maximumtag
    Commented Feb 22, 2014 at 17:17
  • $\begingroup$ Dear @maximumtag, please understand that if the question is trivial, it is likely to be off-topic at MathOverflow. $\endgroup$ Commented Feb 22, 2014 at 19:53
  • $\begingroup$ @RicardoAndrade I agree, but at least the downvoter should say that it it is trivial. $\endgroup$
    – maximumtag
    Commented Feb 22, 2014 at 20:49

2 Answers 2

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Do you really mean weak convergence in $L^\infty$ or weak-$*$ convergence? If it is the latter, this does not seem correct. Forget about the spatial dependence and consider functions depending only on t. Now let $u_m(t)=1+\sin(mt)$, and let $f$ be some monotone function which agrees with $f(x)=x+x^3$ on the interval $[0,2]$. Then $u_m$ converges to 1, but $f(u_m)$ converges to 7/2.

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Check Proposition 2.1(i) in the book Nonlinear Differential Equations of Monotone Types in Banach Spaces, Springer Verlag, 2010, by Viorel Barbu.

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  • $\begingroup$ But we don't have $y_n \to y$, only $y_n \rightharpoonup y$ here. Unless I miss something about my case. $\endgroup$
    – maximumtag
    Commented Mar 6, 2014 at 17:55

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