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EDIT: I need to think more about the question I want to ask given comments in the answer below. Please close the thread if required. I leave it undeleted because answer is useful.

Let $V \subset H \subset V^*$ be separable Hilbert spaces with continuous and dense embeddings. Define the Hilbert space $$W(0,T) = \{u \in L^2(0,T;V) : u' \in L^2(0,T;V^*)\}$$ with inner product $$(u,v)_W = \int_0^T (u(t),v(t))_{L^2(0,T;V)} + \int_0^T (u'(t), v'(t))_{L^2(0,T;V^*)}.$$

I want to know whether the set of functions of the form $$w(t) = \sum_j \phi_j w_j, \qquad\text{where $\phi_j \in C_c^\infty(0,T)$ and $w_j \in V$}$$ are dense in $W(0,T).$

We know from Lions and Magenes that $\mathcal{D}([0,T];V) \subset W(0,T)$ is dense, so the above should hopefully be true. According to a book, the set of functions $$f(t) = \sum_j t^j w_j \quad \text{where $w_j \in V$}$$ are indeed dense in $W(0,T)$.

Does this imply the result I want? Can I approximate the $t^j$ by $C_c^\infty(0,T)$ functions or something like that? (I don't think so). Or is there another way to do this? I guess I may need to replace $C_c^\infty(0,T)$ by $C_c^\infty[0,T]$..

I posted this at math.stackexchange.com but got no answers.

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We do not know that $C^\infty_c(0,T;V)$ is dense in $W(0,T)$. Note that $W(0,T)$ embeds into $C([0,T],V^*)$ (actually even into $C([0,T],H)$). Since $W(0,T)$ clearly contains functions which do not vanish at the endpoints, no set of functions which are required to vanish at the endpoints can be dense.

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  • $\begingroup$ Thanks for replying. But is not $C_c^\infty(0,T;V) = \mathcal{D}(0,T;V)$? The latter (defined as infinitely differentiable compactly-supported $V-$valued functions) is dense in $W(0,T)$ by a theorem of Lions and Magenes. $\endgroup$
    – aere
    Commented Jul 18, 2013 at 18:16
  • $\begingroup$ Lions and Magenes use the notation ${\cal D}([a,b])$ to denote $C^\infty$ functions with compact support in the closed interval $[a,b]$. The condition of compact support is of course redundant in this case unless the interval is infinite. This should not be confused with ${\cal D}(a,b)$, which is a set of functions which have support which is compact in the open interval $(a,b)$. $\endgroup$
    – Michael Renardy
    Commented Jul 18, 2013 at 19:11
  • $\begingroup$ Ah, I see. I will edit my post then. Thanks for pointing it out. $\endgroup$
    – aere
    Commented Jul 18, 2013 at 21:07

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