Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this theorem true? Relationship between linear functionals and inner products

  1. Aug 11, 2012 #1
    Is this "theorem" true? Relationship between linear functionals and inner products

    Suppose we have a finite dimensional inner product space V over the field F. We can define a map from V to F associated with every vector v as follows:
    [tex]\underline{v}:V\rightarrow \mathbb{F}, \ w \mapsto \langle w,v\rangle[/tex]
    Clearly this is a linear functional.

    My question is whether all linear functionals from V to F are of this form. That is, is it true that for every f in V*, there exists a unique v such that f = v?

    I have a felling that it is, but I can't prove it.
     
  2. jcsd
  3. Aug 11, 2012 #2

    pwsnafu

    User Avatar
    Science Advisor

    Re: Is this "theorem" true? Relationship between linear functionals and inner product

    It is true for any Hilbert space (including infinite dimensional). The important thing is completeness. This is called the Riesz representation theorem.
     
  4. Aug 11, 2012 #3

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Re: Is this "theorem" true? Relationship between linear functionals and inner product

    It is quite easy to see for finite dimensional spaces. If [itex]\{e_1,...,e_n\}[/itex] are a basis for V, then we can define

    [tex]\varepsilon_i:V\rightarrow \mathbb{F}:v\rightarrow <e_i,v>[/tex]

    The [itex]\varepsilon_i[/itex] are easily seen to be linearly independent. Indeed, if [itex]\alpha_i[/itex] are such that

    [tex]\sum_{i=1}^n \alpha_i\varepsilon_i=0[/tex]

    then for all v in V holds that

    [tex]0=\sum_{i=1}^n \alpha_i<e_i,v>=<\sum_i \alpha_ie_i,v>.[/tex]

    Since this is true for all v, it is in particular true for [itex]\sum_i\alpha_ie_i[/itex]. And thus
    [itex]\sum_i \alpha_ie_i=0[/itex]. Since [itex]\{e_1,...,e_n\}[/itex] is a basis, it follows that [itex]\alpha_1=...=\alpha_n=0[/itex]. Thus linear independence holds.

    The [itex]\{\varepsilon_1,...,\varepsilon_n\}[/itex] also span [itex]V^*[/itex]. Indeed, if [itex]\varphi:V\rightarrow \mathbb{F}[/itex] is an arbitrary functional, then we define

    [tex]\alpha_i=\varphi(e_i)[/tex]

    For an arbitrary v holds that we can write [itex]v=\sum_i <e_i,v>e_i[/itex]. Thus

    [tex]\varphi(v)=\varphi(\sum_i <e_i,v> e_i)=\sum_i\varphi(e_i) <e_i,v>[/tex]

    Since this holds for all v, we have

    [tex]\varphi=\sum_i \alpha_i \varepsilon_i[/tex]

    So this proves the result for finite dimensional spaces. The result in infinite dimensions is false, since [itex]V^*[/itex] can really be huge.

    However, if we restrict our attention to complete inner-product spaces and to continuous functionals, then the result is true. The proof is not as easy as the one I just gave though.

    This Riesz representation theorem forms the justification for bra-ket notation (if you're familiar with that).
     
  5. Aug 11, 2012 #4
    Re: Is this "theorem" true? Relationship between linear functionals and inner product

    Thank you pwsnafu and micro mass.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Is this theorem true? Relationship between linear functionals and inner products
Loading...