The space $\mathcal{D}((0,T);V)$ and its norm/embeddings? 
Let $V$ be a Hilbert space. Define $\mathcal{D}((0,T);V)$ to be the set of functions $u:(0,T) \to V$ such $u$ is compactly supported on $(0,T)$ and is a $C^\infty$ test function.

What is the norm on this space? I want to know if $V \subset H$ is continuous embedded, and if $u \in \mathcal{D}((0,T);V)$ then $u \in \mathcal{D}((0,T);H)$ and other properties like that.
I guess it is a subspace of $C^\infty((0,T);V)$. But this does not possess a norm?
 A: As Jochen said, the topology of $\mathscr{D}$ is not induced by a norm. Consider test functions supported on a fixed compact interval $[a,b]$. On this subspace, the topology is defined by the countable collection of norms, namely 
$$\|f\|_{C^k } = \max_{0\le j\le k}\sup_{x\in (a,b)} |f^{(j)}(x)|$$
Suppose this topology can be given by a norm. Let $B$ be the open unit ball for this norm. Then for every open set $U$ containing $0$ there is $r>0$ such that $rB\subset U$. (This is what happens in normed spaces). On the other hand, since $B$ is open in our topology, there is an integer $k$ and  $\rho>0$ such that $\{f: \|f\|_{C^k}<\rho\}\subset B$. Let $U=\{f:\|f\|_{C^{k+1}} <1\}$. By the above, there is a constant $c>0$ such that
$$ \|f\|_{C^k} < c \implies \|f\|_{C^{k+1}} <1 \tag{1}$$
But (1) is false. Take any $f$ with $\sup  |f^{(k+1)}|>1   $ and scale it as $n^{-k-1} f(nx)$, where $n$ is large; this scaling preserves the supremum of $(k+1)$th derivative but can make the $C^k$ norm arbitrarily small. 
