# Density of a subspace in $\mathcal{D}(0,T;V)$ under $H^1$ norm

Let $V$ be Hilbert. Let $\mathcal{D}((0,T);V)$ be space of infinite differentiable functions with values in $V$ with compact support. Are functions of the form $$\sum_j \psi_n(t)w_n$$ where $\psi_n \in \mathcal{D}(0,T)$ and $w_n \in V$, dense under the norm $$\lVert {u}\rVert_{W} = \sqrt{\int_0^T \lVert {u}\rVert_{V}^2 + \lVert {u'}\rVert_{V}^2}$$ in $\mathcal{D}((0,T);V)$?

If $V=\mathbb{R}$, the question seems stupid.. Otherwise I don't know...

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If $(e_n,n\geqslant 1)$ is a Hilbert basis, why wouldn't you take $\psi_n(t):=\langle \psi(t),e_n\rangle_V$ and $w_n=e_n$? (In particular, finite combinations of $\psi_ne_n$ are dense in $\mathcal D((0,T);V)$). – Davide Giraudo Jul 19 '13 at 12:27
@DavideGiraudo I think you're right. I was just thinking of that after seeing this question math.stackexchange.com/questions/416543/… You can put your comment as answer if you like. – aere Jul 19 '13 at 12:31
But is it obvious that $\frac{d}{dt}\psi_n(t) = \langle \psi'(t), en \rangle$? – aere Jul 19 '13 at 12:32
I wouldn't say "it's obvious", but it follows from a not too hard computation. – Davide Giraudo Jul 19 '13 at 12:35

Let $(e_n,n\geqslant 1)$ be a Hilbert basis of $V$. On can show that the set of functions of the form $$\sum_{j=1}^N\psi_j(t)e_j,N\in\mathbb N, \psi_j\in \mathcal D(0,T),1\leqslant j\leqslant N$$ is dense in $W:=\mathcal D((0,T);V)$. Indeed, let $\psi\in W$. Let us define $\psi_n(t):=\langle \psi(t),e_n\rangle_V$. Then $\psi_n$ has compact support in $(0,T)$ and is smooth. We have $\psi'_n(t)=\langle \psi'(t),e_n\rangle_V$, and $$\int_0^T\left\lVert \psi(t)-\sum_{j=1}^N\psi_j(t)e_j\right\rVert_V^2\mathrm dt=\int_0^T \sum_{j\geqslant N+1}|\psi_j(t)|^2\mathrm dt.$$ Then we conclude by monotone convergence. We can do the same for the derivative.