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.

 

[blue_raver22]

I cannot definitively say whether this "theorem" is true without further information or context. However, I can provide some insight into the relationship between linear functionals and inner products.

Firstly, it is important to note that linear functionals and inner products are two distinct mathematical concepts. A linear functional is a linear map from a vector space to its underlying field, while an inner product is a bilinear map from a vector space to its underlying field. They serve different purposes and have different properties.

However, there is a connection between the two. In a finite dimensional inner product space, we can define a map from the vector space to its underlying field by taking the inner product of a vector with a fixed vector. This map is a linear functional, as shown in the given example. This is known as the Riesz representation theorem.

So, in a sense, all linear functionals in a finite dimensional inner product space can be represented in this way. However, there may be other linear functionals that do not have a corresponding vector in the space. In fact, in infinite dimensional spaces, this is often the case.

In conclusion, the given "theorem" is not necessarily true in all cases and depends on the specific properties of the vector space and the linear functional in question. The relationship between linear functionals and inner products is complex and cannot be reduced to a simple yes or no statement.