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  1. I was wondering what differences and relations are between a mapping and an operator generally?

    For topological vector spaces or functional analysis, it seems like an operator and a mapping are the same concept, doesn't it?

  2. What differences are between an operator and an operation? Is an operation a mapping from $X^n$ to $X$ for some set $X$ and some $n \in \mathbb{N}$?

Thanks and regards!

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Mapping is generally another name for function, and an operator is often used to mean a binary function (one with two arguments) but occasionally (especially in functional analysis) used in place of the word function. –  Alex Becker Feb 23 '12 at 23:58
    
@AlexBecker: Thanks! (1) what about an operation? (2) Do you mean that an operator and a binary operation are the same concept, generally except some special cases such as in functional analysis? –  Tim Feb 24 '12 at 0:00
    
I've seen operator/operation used interchangeably, but I don't know if its proper. –  Alex Becker Feb 24 '12 at 1:41

1 Answer 1

up vote 3 down vote accepted

Historically, "function" meant something like an element of the vector space $C(\mathbb{R})$ of continuous functions $\mathbb{R} \to \mathbb{R}$, "functional" meant something like a linear functional $C(\mathbb{R}) \to \mathbb{R}$ (that is, a thing which takes functions as input and returns numbers), and "operator" meant something like a linear transformation $C(\mathbb{R}) \to C(\mathbb{R})$ (that is, a thing which takes functions as input and returns functions).

Of course, from the modern point of view there's no real reason not to call operators functions, since we recognize sets of functions as after all just sets so we can talk about functions in and out of them. It's just convenient in functional analysis to invoke a certain context by using the word "operator."

You're more or less correct about what an operation is.

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+1 Thanks! In functional analysis or topological vector spaces, does "operator" imply its domain and codomain being same? –  Tim Feb 24 '12 at 0:23
    
@Tim: not necessarily. –  Qiaochu Yuan Feb 24 '12 at 0:43

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