The usual definition for a sequence $u_k $ converge weakly to $u$ in $W^{1,p} (U)$ is that $ u_k \rightharpoonup u$ in $L^p (U)$ and ${u_k}_ {x_i } \rightharpoonup u_{x_i }$ in $L^p (U)$ for all $i$.
But if I think $W^{1,p } (U)$ itself as a Banach space, weak convergence means for any bounded linear operator $f$ on it, $f(u_k ) \to f(u)$.
Are the two definitions equivalent? It's easy to get the first one from the second, but I have trouble seeing how the inverse hold.
Any help is appreciated. Many thanks in advance.
Best Answer
Any linear functional $f$ on $W^{1,p}(U)$, $1 \le p < \infty$ can be identified with $(v_0,\ldots,v_n) \in L^{q}(U)^{n+1}$, $1/q + 1/p = 1$, via $$f(u) = \int_U u \, v_0 + \sum_{i=1}^n u_{x_i} \, v_i \, \mathrm{d}x.$$ This shows the equivalency of your conditions.
And this identification can be shown by using $T : W^{1,p}(U) \to L^p(U)^{n+1}$, $$u \mapsto (u, u_{x_1}, \ldots, u_{x_n}).$$ This $T$ is isometric (if the norms are defined accordingly). Hence, a bounded, linear functional $f$ on $W^{1,p}(U)$ can be identified with bounded functional on a closed subspace of $L^p(U)^{n+1}$. Now, extend it by Hahn-Banach and identify with an element from $L^q(U)^{n+1}$.