Convergence in dual of Sobolev space Hi please view the following question:
Consider Sobolev space $W^{1,p}(\Omega)$, where $\Omega \subset \mathbb{R}^{n}$ is bounded. We also have a mapping $a: \Omega \times \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$.  
Define the following operator $K(w,u) \in (W^{1,p}(\Omega))^{*}$ by $$\langle K(w,u), v \rangle := \int_{\Omega} a(x,w, \nabla u)\cdot \nabla vdx$$
I want to show that $K(u,u_{t}) \rightarrow K(u,u)$ in $W^{1,p}(\Omega)^{*}$ follows if it is given that $$u_{t} \rightarrow u~~\text{ in }~W^{1,p}(\Omega)~~~~~~\text{ and }~~~~~a(x,u, \nabla u_{t}) \rightarrow a(x,u, \nabla u)~~\text{ in}~~L^{p'}(\Omega; \mathbb{R}^{n}),$$ where $p' := \frac{p}{p-1}$.
Proposed Idea:
Use the fact that $(L^{p}(\Omega;\mathbb{R}^{n}))^{*} \simeq L^{p'}(\Omega; \mathbb{R}^{n})$ and $W^{1,p}(\Omega) \subset L^{p}(\Omega)$  [therefore $ (L^{p}(\Omega))^{*} \subset (W^{1,p}(\Omega))^{*}$]. Note also that $a(x,u, \nabla u_{t}) \rightarrow a(x,u, \nabla u)~~\text{ in}~~L^{p'}(\Omega; \mathbb{R}^{n})$.
But it doesn't seem that it can simply be stated that $K(u,u_{t}) \rightarrow K(u,u)$ in $(L^{p}(\Omega))^{*}$ follows due to the fact that the definition of the operator $K(w,u)$ contains $\nabla v$ in the integrand and not $v$. 
Can anyone see how the convergence $K(u,u_{t}) \rightarrow K(u,u)$ could be shown?
Any assistance would be appreciated. 
 A: You're definitely on the right track, but you at some point have to use the structure of $(W^{1,p}(Ω))^*$.
I would suggest a more elementary proof in the beginning: You want to prove convergence in $(W^{1,p}(Ω))^*$, i.e. $K(u, u_t) - K(u, u)$ must go to zero in that norm. Write down the definition of the dual norm, then apply the triangle inequality and Cauchy-Schwarz, then Hölder's inequality:
$$\begin{align}
\| K(u, u_t) - K(u, u) \|_{(W^{1,p}(Ω))^*} &= \sup_{\substack{v \in W^{1,p}(Ω)\\\| v \|_{W^{1,p}(Ω)} \leq 1}} \left| \int_Ω (a(x, u(x), \nabla u_t(x)) - a(x, u(x), \nabla u(x)) \cdot \nabla v(x) \right| dx \\
&\leq \sup_{\substack{v \in W^{1,p}(Ω)\\\| v \|_{W^{1,p}(Ω)} \leq 1}} \int_Ω \left| a(x, u(x), \nabla u_t(x)) - a(x, u(x), \nabla u(x)) \right| \left| \nabla v(x) \right| dx \\
&\leq \sup_{\substack{v \in W^{1,p}(Ω)\\\| v \|_{W^{1,p}(Ω)} \leq 1}} \| a(\cdot, u(\cdot), \nabla u_t(\cdot)) - a(\cdot, u(\cdot), \nabla u(\cdot)) \|_{L^{p'}(Ω; \mathbb R^n)} \| \nabla v \|_{L^{p}(Ω; \mathbb R^n)} \\
&\leq \| a(\cdot, u(\cdot), \nabla u_t(\cdot)) - a(\cdot, u(\cdot), \nabla u(\cdot)) \|_{L^{p'}(Ω; \mathbb R^n)} \to 0 \\
\end{align}$$
As for your more abstract argument: If you identify your $a(\ldots)$ functions by the natural isomorphism of $(L^p(Ω; \mathbb R^n))^* \cong L^{p′}(Ω; \mathbb R^n)$ with elements of $(L^p(Ω; \mathbb R^n))^*$ and call those $A$, $A_t$, i.e.
$$ \begin{align}
A\colon L^p(Ω; \mathbb R^n) &\to \mathbb R, & A(q) &= \int_Ω a(x, u(x), \nabla u(x)) \cdot q(x) dx,\\
A_t\colon L^p(Ω; \mathbb R^n) &\to \mathbb R, & A_t(q) &= \int_Ω a_t(x, u(x), \nabla u(x)) \cdot q(x) dx
\end{align}
$$
and fix the arguments of $K$, let's call those fixed versions $\tilde K$ and $\tilde K_t$, i.e.
$$ \begin{align}
\tilde K\colon W^{1,p}(Ω) &\to \mathbb R, & \tilde K(v) &= \int_Ω a(x, u(x), \nabla u(x)) \cdot \nabla v(x) dx,\\
\tilde K_t\colon W^{1,p}(Ω) &\to \mathbb R, & \tilde K(v) &= \int_Ω a(x, u(x), \nabla u_t(x)) \cdot \nabla v(x) dx
\end{align}
$$
there obviously holds $\tilde K(v) = A(\nabla v)$ and $\tilde K_t(v) = A_t(\nabla v)$ and by your requirements you know $A \to A_t$ in $(L^p(Ω; \mathbb R^n))^*$ (this follows with an argument as in the proof above). Now use the definition of the dual norm again to show $\tilde K_t \to \tilde K$.
