is it an equivalent definition of the Schwartz Space? all real smooth function that $ f^{(k)}(\infty)=0=f^{(k)}(-\infty)$

Question

Background:

1. I have only taken half semester Analysis course
2. It's in a german script from a linear algebra course on scalar products and self-adjoint endomorphisms, and so on. It mentions such a vector space $S(\mathbb{R})$ on $\mathbb{R}$:

Notation:

Let $C^\infty(\mathbb{R})$ be all infinitely differentiable real functions on $\mathbb{R}$.

Definition of $S(\mathbb{R})$ and contents in the script:

$ S(\mathbb{R}):= \{ f \in C^\infty(\mathbb{R}) ~|~ \exists \text{ a finite interval $ I_f \subsetneq \mathbb{R}$ such that $f$ and all derivatives of $f$ outside $I_f$ are $0$ } \} (\star)$

The script also provides such notation: $ f^{(k)}(\infty)=0=f^{(k)}(-\infty) $

The script then claims that

(1) We can define a scalar product as follows:

$\langle f,g\rangle := \int_{-\infty}^{\infty} f(t)g(t)dt~~ (f,g \in S(\mathbb{R})) $

(2)The second derivative (Laplace Operator)$\Delta:S(\mathbb{R}) \to S(\mathbb{R});~ f \mapsto f'' $ is self-adjoint

...

Question

1. I am wondering if this particular vector space $S(R)$ had a name and why letter $S$ is used here.

ChatGPT suggested the Schwartz Space.

But all the definition and alternation definitions for schwartz spaces' that I found look very different to definition $\star$, they usually involves usage of a Supermum (Supermum Norm?)and include notations that I am not familiar with.

for example

Equivalent definitions of Schwartz Space

https://www.math.ucdavis.edu/~hunter/m218a_09/ch5A.pdf

https://en.wikipedia.org/wiki/Schwartz_space

And sometimes $S(\mathbb{R}^{n}), C^\infty(\mathbb{R}^{n})$,not $S(\mathbb{R}), C^\infty(\mathbb{R})$, is used for schwartz space.

So is this $S(\mathbb{R})$ schwartz space or not?? If not, what is it? If it is, is there any material for schwartz space that contains this definition $\star$ ?

2. Is $S(\mathbb{R})$ an infinite-dimensional vector space on $\mathbb{R}$ ?

Answer 1

Shwartz is a space of rapidly decreasing functions.

Your space is a space of decreasing functions. So no it is not. It nowhere says how fast it decreases.

Long sentences: your space is a space of smooth functions whose derivatives (including the function itself) decay at infinity. Shwartz is a space of smooth functions whose derivatives (including the function itself) decay at infinity faster than any power.