# why isn't “functional operator” a contradiction?

Consider the term "functional operator". My understanding was that:

(a) An operator in this context refers to a mapping from one vector space to another vector space.

(b) A functional is a mapping from a vector space to its underlying scalar field.

(c) Scalars are fundamentally different objects than vectors, and cannot, for instance, be thought of as little 1x1 vectors.

So if a functional is a mapping from a vector space into a scalar field, how can it be an operator? And I know that "operator" can sometimes have a much more general meaning ("something that does something") but I'm pretty sure in the context of a phrase like "functional operator" we're clearly in the more specific conversation about vector spaces.

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... the scalars form a vector space ... –  Olivier Bégassat Jul 20 '12 at 5:05
I think point (c) is off the mark. Any field satisfies the axioms of a vector space over that field. There are some instances, like Calc III where we make a big deal of distinguishing vectors from scalars, but what we mean in that context is that $1$-vectors should not be confused with $2$-vectors or $3$-vectors. –  Grumpy Parsnip Jul 20 '12 at 5:14
@Lucas The general notion of a vector is somewhat different. A vector is simply an element of a vector space, which can be very strange objects. For example, $\mathbb R$ is an infinite-dimensional vector space over $\mathbb Q$. –  Alex Becker Jul 20 '12 at 5:45
Do you know the definition of a vector space? –  Henning Makholm Jul 20 '12 at 12:25
(B) is the definition of "functional (noun)". The expression in the title involves "functional (adjective)". Dictionaries teach us that the meaning if two can be different. –  user31373 Jul 21 '12 at 12:12

## 1 Answer

Assumption (c) is wrong. The scalar field can and is usually seen as a one-dimensional vector space (or Banach space, or Hilbert space).

When one considers the continuous dual of a normed space over $\mathbb C$, say, functionals are seen as the bounded linear operators $X\to\mathbb C$.

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