Weak convergence in Bochner space What does $u_n$ converges weakly to $u$ in $H = L^2(0,T,H^1_0)$ mean explicitly? My thinking is; the dual space of $H$ is $H^* = L^2(0,T,H^{-1})$. So this means for every $f \in H^*$, $f(u_n) \to f(u)$. How to write this out in inner products? Apparently it should be
$$\int_0^T \langle v, u_n \rangle \to \int_0^T \langle v, u \rangle$$ but what are the inner products over and how to get there?
I was working along the lines: since $f$ is in a dual space of a Hilbert space, Riesz rep. theorem gives $f$ is associated to a function $v \in H$. So $$\langle v, u_n \rangle_H \to \langle v, u \rangle_H$$ is the statement. But I'm not sure how to proceed. thanks
 A: What you have said so far is correct. For a measure space $X$ and Banach space $V$, so long as $V$ is "nice" (precisely, if its dual $V'$ has the Radon-Nikodym property), then the dual of the Bochner space $L^p(X,V)$ is $L^q(X,V')$, where $1/p + 1/q = 1$. This dual is realized by
$$\langle \varphi, f \rangle_{L^p(X,V)} = \int_X \langle \varphi(x), f(x) \rangle \, d\mu(x) $$
where $f(x) \in V$ and $\varphi(x) \in V'$, so the pairing $\langle \cdot, \cdot \rangle$ inside the integral is simply the pairing of $V'$ with $V$. 
In your case, $X = [0,T]$, $p = q = 2$, and $V$ is the Sobolev space $H_0^1$. Therefore the dual of $L^2([0,T],H_0^1)$ should be $L^2([0,T],H^{-1})$. Therefore, for any $u \in L^2([0,1],H_0^1)$ and $v \in L^2([0,1],H^{-1})$, the appropriate pairing of $v$ and $u$ is
$$\langle v, u \rangle = \int_0^T \langle v(t), u(t) \rangle \, dt$$
Here, $u(t)$ and $v(t)$ are themselves functions, in the spaces $H_0^1$ and $H^{-1}$ respectively; $H^{-1}$ is the dual of $H_0^1$. Now, your question remains, how do we characterize the pairing $\langle v(t), u(t) \rangle$? If you look in a PDE book, such as Partial Differential Equations by Evans, you'll find that functions $f \in H^{-1}(U)$ where $U \subset \mathbb{R}^n$ can be characterized by being a tuple $f_0, \ldots, f_n \in L^2(U)$, such that for all $g \in H_0^1$,
$$ \langle f, g \rangle = \int_U \left[f_0 g + \sum_{i=1}^n f_i D_{x_i}g\right] \, dx$$
Therefore this is the appropriate pairing of $H_0^1$ and $H^{-1}$ to use in the corresponding pairing of $L^2([0,1], H_0^1)$ and $L^2([0,1], H^{-1})$, so then weak convergence is precisely $\langle v, u_n \rangle \rightarrow \langle v, u \rangle$.
