# Why is this function space dense in this Bochner-type space? (Rogers and Renardy)

Let $\phi(t) = \sum_{i=1}^N\beta_i(t)\phi_i(x)$, where $\phi_i$ is basis for space $V$, and $\beta_i \in C_c^\infty(0,T).$ Renardy and Rogers says that

1) Functions of the above form are dense in $H^1(0,T;V)$

2) Functions of the above form are dense in $C_c^\infty(0,T;V)$

Why is this true? How to prove it?

Edit: They didn't define $H^1(0,T;V).$ I guess it consists of functions $u(t) \in V$ for each $t$ with $u(t) \in L^2(0,T)$ such that the derivative $u'(t) \in V$ for each $t$ with $u'(t) \in L^2(0,T)$ too.

$C_c^\infty(0,T;V)$ consists of functions $f(\cdot):[0,T] \to V$, so $f(t) \in V$ for each $t$. Furthermore $f$ is $C_c^\infty$ around $[0,T]$ wrt. $t$.

-
Define the spaces? I think the second one is probably in error .... –  Ray Yang Feb 12 '13 at 13:29
@RayYang see my edit. –  george.s Feb 12 '13 at 22:07
What we need is a topology on $V$ to define the derivative, and also to define the notion of integral so as to make sense of a weak derivative for $V$-valued functions on $[0,T]$. –  A Blumenthal Feb 17 '13 at 20:07
@DavideGiraudo $V \subset H \subset V^*$ are Hilbert spaces (Hilbert triple) and all spaces separable. –  george.s Feb 18 '13 at 0:32