Linear Functionals & Inner Products: Is This Theorem True?

[dEdt]

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:
\underline{v}:V\rightarrow \mathbb{F}, \ w \mapsto \langle w,v\rangle
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.

 

[pwsnafu]

It is true for any Hilbert space (including infinite dimensional). The important thing is completeness. This is called the Riesz representation theorem.

 

[micromass]

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

\varepsilon_i:V\rightarrow \mathbb{F}:v\rightarrow <e_i,v>

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

\sum_{i=1}^n \alpha_i\varepsilon_i=0

then for all v in V holds that

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

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

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

\alpha_i=\varphi(e_i)

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

\varphi(v)=\varphi(\sum_i <e_i,v> e_i)=\sum_i\varphi(e_i) <e_i,v>

Since this holds for all v, we have

\varphi=\sum_i \alpha_i \varepsilon_i

So this proves the result for finite dimensional spaces. The result in infinite dimensions is false, since V^* 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).

 

[dEdt]

Thank you pwsnafu and micro mass.