Differences between $C_c^\infty[0,T]$ and $C_c^\infty(0,T)$ I believe it is true that:


*

*If $f \in C_c^\infty(0,T)$, then $f(T)=f(0)=0$.

*$C_c^\infty(0,T) \subset C_c^\infty[0,T]$ 

*$C^\infty(0,T) \subset C_c^\infty[0,T]$

*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.

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?).
