Set-up:
Let $\Omega$ be an open subset of $\mathbb{R}^{n}$. We consider functions from $\Omega \times [0,T]$ into $\mathbb{R}^{n}$.
Let $u(x,t) \in L^{2}((0,T); H^{1}_{0, \sigma})$ such that there exists a sequence $\{u_{m}\}_{m=1}^{\infty}$ which converges to $u$ weakly in $L^{2}((0,T); H^{1}_{0, \sigma})$. Here, $H^{1}_{0, \sigma} = W^{1,2}_{0, \sigma} = \{ u \in W^{1,2}_{0} : \text{div}u = 0 \}$
Let $\Phi(x,t) := \sum^{N}_{i=1} \lambda_{i}(t) \phi_{i}(x)$, where $\lambda_{i}$ are arbitrary functions in $H^{1}((s,t) ; \mathbb{R}^{1})$ for some $0 \leq s \leq t \leq T$, and $\{\phi_{i}\}_{i=1}^{\infty} \in H^{1}_{0, \sigma} \cap L^{n}$ such that $\text{span}\{\phi_{i}\}_{i=1}^{\infty}$ is a complete orthonormal set dense in $L^{2}_{\sigma}$.
Now set $w_{i}(x,t) = u^{(i)}(x,t)\Phi(x,t)$, $(i=1,2,...,n)$ where $u^{(i)}$ is the $i^{\text{th}}$ component of $u$. Then $w_{i} \in L^{2}(\Omega \times(s,t))$, and hence there is a sequence $\{w_{i,k}\}_{k=1}^{\infty}$ in $C^{\infty}_{0}(\Omega \times(s,t))$ such that $w_{i,k} \rightarrow w_{i}$ in $L^{2}(\Omega \times(s,t))$.
Problem:
Our claim is that:
$|\int^{t}_{s} (u \cdot \nabla(u_{m} - u), \Phi) \text{d}t | \leq \sum^{n}_{i=1} \int^{t}_{s} |(u \cdot \nabla(u_{m} - u), \partial_{x_{i}}w_{i,k} )| \text{d}t \ + \ \sum^{n}_{i=1}(\int^{t}_{s} ||\nabla u_{m} - \nabla u||^{2} \text{d}t)^{1/2}(\int^{t}_{s} ||w_{i,k} - w_{i}||^{2} \text{d}t)^{1/2}$
Here $(\cdot,\cdot)$ is the $L^{2}(\Omega)$ inner-product, and $||\cdot ||$ is the $L^{2}(\Omega)$ norm.
My Attempt to Prove:
The paper I am reading claims that the above is a simple consequence of integration by parts...
$|\int^{t}_{s} (u \cdot \nabla(u_{m} - u), \Phi) \text{d}t |$
$\leq \int^{t}_{s} |(u \cdot \nabla(u_{m} - u), \Phi)| \text{d}t$ (Just moving the absolute value inside the integral...)
$\leq \sum^{n}_{i=1} \int^{t}_{s} |(u^{(i)}\partial_{x_{i}}(u_{m} - u), \Phi)| \text{d}t $ (Using Triangle Inequality...)
$= \sum^{n}_{i=1} \int^{t}_{s} | \int_{\Omega} u^{(i)}\partial_{x_{i}}(u_{m} - u) \cdot \Phi \ \text{d}x| \ \text{d}t $ (writing out the inner-product as an integral...)
$= \sum^{n}_{i=1} \int^{t}_{s} | \int_{\Omega} \partial_{x_{i}}(u_{m} - u) \cdot u^{(i)}\Phi \ \text{d}x| \ \text{d}t $ (Moving $u^{(i)}$ to other side of dot-product)
$= \sum^{n}_{i=1} \int^{t}_{s} | \int_{\Omega} \partial_{x_{i}}(u_{m} - u) \cdot w_{i} \ \text{d}x| \ \text{d}t $ (By definition of $w_{i}$)
This tells us nothing about the relation to $w_{i,k}$ however. So I tried to obtain the desired result by assuming the next two steps are possible:
$= \sum^{n}_{i=1} \int^{t}_{s} | \int_{\Omega} \partial_{x_{i}}(u_{m} - u) \cdot ^{\text{lim}}_{k \rightarrow \infty} w_{i,k} \ \text{d}x| \ \text{d}t $
$= \sum^{n}_{i=1} \ ^{\text{lim}}_{k \rightarrow \infty} \int^{t}_{s} | \int_{\Omega} \partial_{x_{i}}(u_{m} - u) \cdot w_{i,k} \ \text{d}x| \ \text{d}t $
However, using integration by parts here yields:
$\leq \sum^{n}_{i=1} \int^{t}_{s} |((u_{m} - u), \partial_{x_{i}}w_{i,k})| \text{d}t $
Which is clearly stronger than our actual claim. Thus I assume the abuse of limit I used above is incorrect. However, I cannot see how to arrive at the desired result. Can anyone help me? Thanks