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

How do IIs 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.

Let $$u_m \rightharpoonup u$$ (weak convergence) in $X$, a Hilbert space. We are given $f:X \to X$, a continuous invertible map which is monotone $$(f(x)-f(y),x-y) \geq 0\quad\text{for all $x, y$}$$ and we have $$f(u_m) \rightharpoonup f(v)$$ in $X$ for some $v \in X$.

How do I 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.

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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Let $$u_m \rightharpoonup u$$ in $X$ (weak convergence) in $X$, a Hilbert space. We are given $f:X \to X$, a continuous invertible map which is monotone $$(f(x)-f(y),x-y) \geq 0\quad\text{for all $x, y$}$$ and we have $$f(u_m) \rightharpoonup f(v)$$ in $X$ for some $v \in X$.

How do I 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.

Let $$u_m \rightharpoonup u$$ in $X$ (weak convergence). We are given $f:X \to X$, a continuous invertible map which is monotone $$(f(x)-f(y),x-y) \geq 0\quad\text{for all $x, y$}$$ and we have $$f(u_m) \rightharpoonup f(v)$$ in $X$ for some $v \in X$.

How do I 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.

Let $$u_m \rightharpoonup u$$ (weak convergence) in $X$, a Hilbert space. We are given $f:X \to X$, a continuous invertible map which is monotone $$(f(x)-f(y),x-y) \geq 0\quad\text{for all $x, y$}$$ and we have $$f(u_m) \rightharpoonup f(v)$$ in $X$ for some $v \in X$.

How do I 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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If $u_m \rightharpoonup u$, how to show using monotonicity that $f(u_m) \rightharpoonup f(u)$?

Let $$u_m \rightharpoonup u$$ in $X$ (weak convergence). We are given $f:X \to X$, a continuous invertible map which is monotone $$(f(x)-f(y),x-y) \geq 0\quad\text{for all $x, y$}$$ and we have $$f(u_m) \rightharpoonup f(v)$$ in $X$ for some $v \in X$.

How do I 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.