I'm currently trying to understand a step of a proof in a paper.
Suppose that we have two sequences $(n_k), (V_k)\subset H^1(\Omega)\cap L^\infty(\Omega)$ that converge weakly in $H^1(\Omega)$ to some limit elements $n,V\in H^1(\Omega)$, where $\Omega\subset\mathbb{R}^d$ is a bounded Lipschitz domain $(d\in\{1,2,3\})$. Further, assume that the sequences are uniformly bounded in $H^1(\Omega)$ and $L^\infty(\Omega)$ as well as weak-* convergent in $L^\infty(\Omega)$ to $n,V$. The authors now argue that this is more than sufficient to show
$$
\lim_{k\to\infty}\int_\Omega n_k\nabla V_k\cdot\nabla \varphi\;\text{d}x=\int_\Omega n\nabla V\cdot\nabla\varphi\;\text{d}x\quad\forall\varphi\in H^1(\Omega)
$$
without giving more details. Here, $\nabla u$ denotes the weak derivative of $u$. As I'm quite new to the topic, I tried to understand why this holds true.
My approach was to consider the following:
$$
\int_\Omega (n_k\nabla V_k-n\nabla V)\cdot\nabla\varphi\;\text{d}x=\int_\Omega n_k(\nabla V_k-\nabla V)\cdot\nabla\varphi\;\text{d}x+\int_\Omega (n_k-n)\nabla V\cdot\nabla\varphi\;\text{d}x.
$$
We have that $\nabla V\cdot\nabla\varphi\in L^1(\Omega)$ and therefore the second integral tends to $0$ by the weak-* convergence of $(n_k)$. However, the first integral is tricky for me.
If there wasn't the term $n_k$, one could see that the integral tends to $0$ because of the weak convergence of $V_k$ in $H^1(\Omega)$. I could apply the Hoelder inequality to estimate
$$
\left\vert\int_\Omega n_k(\nabla V_k-\nabla V)\cdot\nabla\varphi\;\text{d}x\right\vert\leq\Vert n_k\Vert_{L^\infty(\Omega)}\Vert \nabla V_k-\nabla V\Vert_{L^2(\Omega)}\Vert\nabla\varphi\Vert_{L^2(\Omega)}
$$
but I don't know if $(\nabla V_k)$ strongly converges in $L^2(\Omega)$.
Could someone provide me with a hint on how to tackle the problem? Thanks.
To give a little more context: I'm given a weakly convergent sequence $(n_k,V_k)$ that is a weak solution of a semilinear elliptic PDE system. I want to show that I can pass to the limit in the weak formulation where the term above appears. For the other terms in the weak formulation it was easy to verify that this is possible as they were linear.
Best Answer
The thesis is: $$\int_\Omega (n_k\nabla V_k-n\nabla V)\cdot\nabla\varphi\;\text{d}x \to 0$$ Observe first: $$\int_\Omega (n_k\nabla V_k-n\nabla V)\cdot\nabla\varphi\;\text{d}x = \int_\Omega (n_k-n)\nabla V_k\cdot\nabla\varphi\;\text{d}x + \int_\Omega (\nabla V_k-\nabla V)\cdot n \nabla\varphi\;\text{d}x $$ The second term goes to zero thanks to $V_k\to V$ weakly in $H^1$. This is because the map: $$\psi \mapsto \int_\Omega \nabla \psi\cdot n \nabla\varphi\;\text{d}x$$ is linear and continuous on $H^1$.
For the first term, the argument is a bit longer and uses two key ideas: the compact embedding of $H^1$ in $L^2$ and the density of $C^{\infty}(\mathbb{R}^d)$ in $H^1$.
$H^1$ embeds compactly in $L^2$. For this reason $n_k\to n$ weakly in $H^1$ entails $n_k\to n$ strongly in $L^2$. This can be used to show:
$$\int_\Omega (n_k-n)\nabla V_k\cdot\nabla\phi\;\text{d}x \to 0$$
for every $\phi \in C^{\infty}(\mathbb{R}^d)$.
In what follows it is explained in details how to use the density argument. Observe that:
$$\bigg| \int_\Omega (n_k-n)\nabla V_k\cdot\nabla\psi\;\text{d}x \bigg| \leq 2 A B \|\psi\|_{H^1} \qquad \forall \psi \in H^1$$ where $A,B$ are positive constants satisfying $\|n_k\|_\infty , \|n\|_\infty \leq A$ and $\|V_k\|_{H^1} \leq B$ for every $k$. Such constants existence is granted by boundedness of $n_k$ and $V_k$ in $L^{\infty}$ and $H^1$ respectively. Therefore, for every $\phi \in C^{\infty}(\mathbb{R}^d)$ we have:
$$\bigg| \int_\Omega (n_k-n)\nabla V_k\cdot\nabla\varphi\;\text{d}x \bigg| \leq \bigg| \int_\Omega (n_k-n)\nabla V_k\cdot\nabla\phi\;\text{d}x \bigg| + 2 A B \|\varphi - \phi\|_{H^1}$$ which yields: $$ \limsup_{k\to \infty}\bigg| \int_\Omega (n_k-n)\nabla V_k\cdot\nabla\varphi\;\text{d}x \bigg| \leq 2 A B \|\varphi - \phi\|_{H^1}$$ Since $\|\varphi - \phi\|_{H^1}$ can be chosen arbitrarily small thanks to $C^{\infty}(\mathbb{R}^d)$ density in $H^1$, this concludes the proof.