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.