Differences between $C_c^\infty[0,T]$ and $C_c^\infty(0,T)$

Question

I believe it is true that:

1. If $f \in C_c^\infty(0,T)$, then $f(T)=f(0)=0$.
2. $C_c^\infty(0,T) \subset C_c^\infty[0,T]$
3. $C^\infty(0,T) \subset C_c^\infty[0,T]$
4. If $f \in C_c^\infty[0,T]$, it doesn't necessarily mean that $f(T)=f(0)=0$.

Finally, what happen if I define weak derivative as $$\int_0^T u\phi' = -\int_0^T u'\phi$$ for all $\phi \in C_c^\infty[0,T]$, instead of $C_c^\infty(0,T)$.

I will think more on this but maybe somebody has already studied it. Anyway, more information about these differences would be appreciated.

Answer 1

First let's assume some definitions (see for example Gilbard-Trudinger page 10): $C^\infty([0,T])$ is the space of all functions $f\in C^\infty((0,T))$ such that $f$ and all it's derivatives, can be continuously extended up to the boundary of $(0,1)$.

With this definition we have that all $f\in C^\infty([0,T])$ does have compact support, hence $$C^\infty([0,T])=C_c^\infty([0,T])$$

The last equality shows that $3$ is not true and $4$ is true.

On the other hand if $f\in C_c^\infty ((0,T))$, then you can find a $\delta >0$ (try to prove it), such that $f(x)=0$ for $x\in (0,\delta)\cup (T-\delta,T)$, hence $1$ and $2$ is true.

To conclude, note that if you define weak derivative (as you have formulated) by using the space $C^\infty([0,T])$ as a test space, then you have some problems with integration by parts (can you see this?).

Answer 2

I think the correct definition for $C_c^\infty[0,T]$ should be

$$ C_c^\infty[0,T]=\{ f\in C_c^\infty(\mathbb{R}): \text{supp}(f) \subset [0,T]\}, $$

since this is the natural analog of the definition for $C_c^\infty(0,T)$. This gives raise to an equivalent definition

$$ C_c^\infty[0,T]=\{ f\in C^\infty[0,T] : f^{(n)}(0)=f^{(n)}(T)=0, \text{ for all } n \geq0\}. $$

In this case 4 is clearly false and the definition of weak derivatives remains unchanged.

Edit: The first definition can be found, for example, in Trèves' "Topological vector spaces, distributions and kernels" on page 130 (where he gives the example of test functions as an LF space).

As I see you are interested in Lions-Magenes' definition of $\mathcal{D}[a,b]$, let me note that their definition is equivalent to $C^\infty[a,b]$ by Borel's lemma (I consider $V=\mathbb{R}$): They define

$$ \mathcal{D}[a,b]:= \{ f|_{[a,b]} : f\in C_c^\infty(\mathbb{R}) \}. $$ From this obviously $\mathcal{D}[a,b] \subset C^\infty[a,b]$ and for the other direction take a function $f\in C^\infty[a,b]$ and take $g\in C^\infty[b,\infty)$ so that $g^{(n)}(b)=f^{(n)}(b)$ for all $n$. This gives a function $C^\infty[a,\infty)$ which you can then "cut-off" to have compact support. Doing the same for $a$ we get that $f\in \mathcal{D}[a,b]$ so they are indeed the same.

Notice that they $\textbf{don't}$ define weak derivatives with respect to this space, but with respect to $\mathcal{D}]a,b[$ which is the usual space of test functions.

On the other hand this answers another question of yours on why is $\mathcal{D}[a,b]$ dense in $W(a,b)$: This assertion is the analog of the theorem that says $C^\infty(\bar{\Omega})$ is dense in $W^{1,2}(\Omega)$ when $\Omega$ has nice boundary.