Density of $C\infty([0,T];V)$ in $W(0,T;V,V)$. Let $W=\{ u \in L^2(0,T;V) : u_t \in L^2(0,T;V)\}$ where $V$ is a Hilbert space in the Gelfand triple $V \subset H \subset V^*$ and $u_t$ is the weak time derivative.
Is it true that $C^\infty([0,T];V)$ is dense in $W$?
I would like a reference to this fact please??
 A: $\newcommand{\R}{\mathbb{R}}$Yes. The approximating argument that you use in the usual case of $C^\infty(0, T;\mathbb{R})\subset W^1(0, T;\mathbb{R})$ (that is, convolution against a mollifying family) works verbatim here. For the details you may look in Evans's book on PDEs (chapter on Sobolev spaces, look in the last paragraph "Spaces involving time"), or in Chapter 6 of Hunter's book. 
EDIT. Since you are asking, I'll write down some idea related to what I have in mind. Let us assume that $u\in L^2(\mathbb{R}; V)$ and that its weak derivative $u_t$ is in the same space. By this I mean that 
$$\int_{\R}u_t(x)\phi(x)\, dx= -\int_{\R}u(x)\phi_t(x)\, dx, \qquad \forall \phi\in C^{\infty}_c(\R;\R).$$
We take a mollifying family $\phi_\epsilon\in C^\infty_c(\R;\R)$. The convolution 
$$u_\epsilon=u\star \phi_\epsilon(x)=\int_\R u(y)\phi_\epsilon(x-y)\, dy$$
is smooth and it converges to $u$ in $L^2(\R;V)$ sense as $\epsilon\to 0$. Moreover, we have for its first derivative that
$$\frac{d}{dx}u_\epsilon= u_t\star \phi_\epsilon (x), $$
so that this derivative converges to $u_t$. All in all we have proved that $u_\epsilon \to u$ in $W^1(\R;V)$ sense. 
This is the core of the argument. The rest is fiddling with the support, so that the approximating family is compactly supported. Later you will have to do some other technicality to include the time boundary, that is, to deal with the $L^2(0, T; V)$ case. (Here we have dealt with $L^2(\R;V)$.) 
Anyway, as you can see, you do not really need any information on the Banach space $V$. This exactly same proof works in the case $V=\R$ as well.
