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Functionals / operators

  1. Nov 25, 2013 #1
    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. jcsd
  3. Nov 25, 2013 #2
    [Whenever we talk about vector spaces, there's some underlying field [itex]\mathbb F[/itex] in the background, and all the vector spaces under consideration are over the same field [itex]\mathbb F[/itex].]

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

    Depending who you ask, a linear operator could either mean an arbitrary linear transformation [itex]V\to W[/itex] (in which case a linear functional is indeed a special case with [itex]W=\mathbb F[/itex]), or it's the special case of a linear transformation [itex]V\to V[/itex] (in which case it's distinct from a linear functional).
     
  4. Nov 25, 2013 #3
    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.
     
  5. Nov 25, 2013 #4
    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)?
     
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