# Functionals / operators

1. Nov 25, 2013

### gentsagree

Are functionals a special case of operators (as written on Wiki)?

Operators are mappings between two vector spaces, whilst a functional is a map from a vector space (the space of functions, say) to a field [or from a module to a ring, I guess]. Now, the field is NOT NECESSARILY a vector space. It could be the field over which the vector space is defined.

Can somebody clear this up for me?

arigato.

2. Nov 25, 2013

### economicsnerd

[Whenever we talk about vector spaces, there's some underlying field $\mathbb F$ in the background, and all the vector spaces under consideration are over the same field $\mathbb F$.]

There's a natural way of viewing $\mathbb F$ itself as a vector space, with addition being field addition and scalar multiplication being field multiplication. Viewing $\mathbb F$ as a vector space in this way, a linear functional on the vector space $V$ is just a linear transformation $V\to\mathbb F$.

Depending who you ask, a linear operator could either mean an arbitrary linear transformation $V\to W$ (in which case a linear functional is indeed a special case with $W=\mathbb F$), or it's the special case of a linear transformation $V\to V$ (in which case it's distinct from a linear functional).

3. Nov 25, 2013

### R136a1

A field is always a vector space over itself (or can be seen as such). So $\mathbb{R}$ is canonically an $\mathbb{R}$-vector space. In that sense, a functional is always a special operator.

Note however that some authors tend to use operator and functional in a completely different way.

4. Nov 25, 2013

### gentsagree

Thank you for the good replies. One more question pops to mind:

What do we mean exactly when we say that we declass an operator to a function (say, in path integrals)?