In my textbook, there is a problem in which I'm given $V$, a vector space of infinitely differentiable (smooth) functions vanishing outside the certain interval.
Formally, I'm aware that $\exists a, b \in \mathbb{R}, I=[a, b]$ (closed interval) such that $\forall f \in V, f(x)=0, \forall x \notin I$. But besides this straightforward definition, what are other general properties of such functions?
Examples
Are there any constraints for vanishing intervals?
If $I=[a, b]$ and $f \in V$ is a smooth function, are there any constraints for choice of $a$ and $b$ on which $f$ vanishes? Is there any way such bounds can be found? Shall such interval $I$ contain the origin (0)?
Are there any constraints for derivatives and integrals of such functions?
For example, if $f(x)=0, \forall x \notin I$, what does this imply about $D(f(x)), \forall x \notin I$ or $D(f(x)), \forall x \in I$? May it imply that all of its derivatives vanish at origin?
Furthermore, does this imply something about $\int_a^b f(t) \, dt, \forall t \in I$ besides the fact that it is not zero?
If smooth functions vanish outside the certain interval, are they necessarily non-analytic? Concordantly, are they considered smooth transition functions?
This very interesting Wikipedia article, shows examples of smooth functions that can not be approximated by convergent power series – but that example contains a function which has derivative that contains the origin its vanishing interval (as stated in my second example).
In this case, where the definition of $f$ is not explicit, does this say something about its analytic/non-analytic property? The reason for my interest in this property is because such functions seem very similar to smooth transition functions.
In Short
Is there any explicit name for smooth functions that vanish outside the interval? If not, what are the properties that make them "special"?
Note:
To be more explicit about the definition of "special", this is the problem from my textbook (Serge Lang, Linear Algebra):
Let $V$ be a finite dimensional space over $\mathbb{R}$ of infinitely
differentiable functions vanishing outside some interval. Let
the scalar product be defined as usual by:$$\langle f, g \rangle = \int_0^1 {f(t)g(t)} \, dt$$
Let $D$ be the derivative. Show that one can define $D^T$ as
before, and $D^T=-D$.
Best Answer
The functions you are talking about are called bump functions and are incredibly important in the theory of distributions.
Bump functions can have any close interval as their support; as an example, the function
$$f(x)=\begin{cases} e^{\frac{-1}{(x-a)^2(x-b)^2}}\ \ x\in [a,b]\\ 0\end{cases}$$
is a smooth function with compact support $[a,b]$.
Actually, more is true: given any compact set $K$ and an open set $U$ containing $K$ there is a bump function that has value $1$ inside $K$ and $0$ outside $U$ (for the construction see the linked Wikipedia page).
The derivative of a bump function is still a bump function vanishing outside the same set $K$
Proof: Let $A=\mathbb{R}-K$. Then $A$ is an open set on which $f=0$. For every point $x$ in $A$ we have, for $h$ sufficiently small such that $x+h$ is still in $A$ $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{0}{h}=0$$
Integration is possible, but since it depends on a constant the integral is not assured to be $0$ outside $K$.
Yes. Every non zero bump function is smooth but is not analytic: this easily follows from a theorem known as the identity theorem, which states that two analytic functions defined on an open and connected set that are equal on a set of points $S$ such that $S'≠0$ are equal on all the domain. Thus, if a non zero bump function was to be analytic, it would have to be zero everywhere, which is not the case.