So, on the linear algebra class it is mentioned that function space is an example of the vector space.
By the definition of the vector space, any vector space $(V,+,\cdot)$ both an abelian group and a module. That means the function space has two operators working on a set. So far is correct, right?
One is the scalar multiplication $c\cdot f$ and one is the addition of functions $f+g$.
But in calculus we have seen that there are also many other operators that can work on function, for instance, the multiplication of functions $f\times g$ and the function composition $f\circ g$, also there are unary operators such as differential operator $\nabla$ and integral operator $\int$ and integral transformations $T f$, those are all linear, right?
[Main question] Does this means there are some more complex algebra structure for the function space? If not, how does the function composition enrolled into the discussion.
Decomposing the main question:
Part 1: If I obtain a function space with certain basis. It is closed under scalar multiplication and function addition because it is a vector space. Now, if I attach function composition to this structure, make it $$ (F, \cdot, +, \circ)=(\text{set of function, scalar product, addition, function composition}), $$ is the set still closed? Is there a proof or counter example for this.
[The question is explained by @RyanHowe in comment] Also, in wiki it is mentioned that function space is also a topological space idea. What's the difference when we treat the function space differently? Is the definition changed?
Does that means we only care scalar multiplication $c\cdot f$ and addition of functions $f+g$ in linear algebra. And, differential operator $\nabla$ and integral operator $\int$ are considered in Calculus?
Thank you guys very much.
In general, if we have some algebraic structure $R$, the set of functions $f : A \to R$ inherit all the algebraic structure that $R$ has. For example, if $R$ is an Abelian group, the function space $\{f : A \to R\}$ inherits the group operations from $R$.
So $\{f : A \to \mathbb{R}\}$ in particular will inherit the structure of $\mathbb{R}$ as an $\mathbb{R}$-algebra. This means that not only can we multiply a function by a scalar, but we can also multiply a function by another function.
As for function composition, this is only a useful operation when $A = \mathbb{R}$. And it doesn't play very nice with the other operations for the most part. So you could come up with an "algebraic structure" by saying things like $(f \cdot g) \circ h = (f \circ h) \cdot (g \circ h)$, for instance. But there isn't a nice name for it.
Finally, consider that differentiation (and even integration) aren't total operators. They rely on the function being continuous. But if you're interested in looking into the algebraic structure of differentiation, you should look into differential algebra.