0
$\begingroup$

Please I dont understand this.

I have:

$ \parallel \nabla m_n \parallel_{L^{\infty}(\mathbb{R}^+, L^2(\Omega))}\leq C$

$ \parallel \frac{\partial m_n}{\partial z} \parallel_{L^{\infty}(\mathbb{R}^+, L^2(\Omega))}\leq C n $

$ \parallel \frac{\partial m_n}{\partial t} \parallel_{L^2(\mathbb{R}^+\times\Omega)}\leq C$

$ \parallel m_{3,n} \parallel_{L^{\infty}(\mathbb{R}^+, L^2(\Omega))}\leq C \sqrt{n}$

So we have for a subsequence,

$m_n \to m $ weakly * in $L^{\infty}(\mathbb{R}^+, H^1(\Omega))$

$\partial_z m_n \to 0 $ in $L^{\infty}(\mathbb{R}^+, L^2(\Omega))$

$m_{3,n} \to 0$ in $L^{\infty}(\mathbb{R}^+, L^2(\Omega))$

$\partial_t m_n \to \partial_t m$ in $ L^2([0,T],\times\Omega)$ for any $T>0$

my question why it follows that:

$m_n \to m $ in $L^{2}([0,T]\times \Omega)$ for any $T>0$ and in $L^{p}([0,T]\times \Omega)$ for any $p \geq 1$. and $m_n(0) \to m(0)$ weakly in $(L^2(\Omega))^3$ and $ \partial_t m_n,3 \to $ weakly in $L^2([0,T]\times \Omega)$ for any $T>0$

$\nabla m_{n,3} \to 0$ weakly in $L^2([0,T] \times \Omega) $ for any $T>0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.