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What is the relation (if any) between dual spaces and inner product? As far as I understand the dual space of a vector space is the set of all linear mappings from the vector set to the field over which the space is defined. But the definition of the inner product is a bilinear mapping of two vectors to a scalar. It sounds to me like if we had defined the same thing twice, in two different ways, is that so?

If the answer is yes, and given that every space has a dual space, does that mean that every vector space is automatically an inner product space? Moreover, if the polarization identity can be used to define a norm from an inner product, are all vector spaces inner normed spaces?

I am sure I'm misunderstanding some definition, but I'm totally lost here. Any help?

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2 Answers 2

Not every vector space is an inner product space because not every norm satisfies the polarization identity. As classic counterexamples, consider the the spaces $\mathbb{R}^n$ under the $1$-norm (aka taxicab norm) and the $\infty$-norm (aka maximum/supremum norm).

If you are given an inner-product space (aka Hilbert space), then there is indeed a strong connection between the dual space and the inner product. This result is known as the Riesz representation theorem. Note, however, that his does not mean that we've "defined the same thing twice". The dual space is the set of all linear mappings to the scalar field, whereas the inner product space is a particular (bilinear) map on two vectors to the scalar field.

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So there is not necessarily a connection between the two of them? I mean, could the linear mappings in the dual be defined as something different from L(U) = <L|U>? (where L is a linear mapping in V*, U is a vector in V and <·|·> is the inner product) –  carllacan Dec 8 '13 at 17:43
    
Presumably, you mean that for an inner product space $(V,\langle \cdot,\cdot \rangle)$ and any $\phi \in V^*$, we can find some vector $x \in V$ for which $\phi(y) = \langle y, x \rangle$ (note that the inner product isn't explicitly defined on functionals). This is precisely what is stated by the Riesz representation theorem. While it is intuitive that this should hold in the finite-dimensional case, extending this notion further takes a little bit more subtlety. –  Omnomnomnom Dec 8 '13 at 17:49
    
Ok, it makes sense now. However: "(note that the inner product isn't explicitly defined on functionals)" you lost me there. I was talking about abstract vector spaces, is there anything special with function spaces? –  carllacan Dec 8 '13 at 17:58
    
The word "functional" is another term for a linear mapping. Remember that an inner product is something that takes in two vectors and gives you a scalar. Trying to apply the inner product to a linear mapping (a functional) is like trying to fit a square peg into a round hole. –  Omnomnomnom Dec 8 '13 at 18:02
    
Aaah, right, now has everything clicked. The inner product is a bi-linear mapping from the vector space, not from the dual space, that is where I was confused, I think. I guess it is because I'm encountered these concepts while studying quantum mechanics, and the dual space if only introduced in order to define the Dirac notation, in a manner in which elements of the dual vector space and vectors are pretty much "the same", only transposed so that we can calculate their inner product. I realise now that this is not generally the case, i.e dual vectors can be radically different from vectors. –  carllacan Dec 8 '13 at 18:08

If you have an inner product, then you have an isomorphism from $V$ to its dual $V^*$ given by $v\mapsto \langle v,-\rangle$. But, there are vector spaces that are not inner product spaces. For that see here.

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Always false if $V$ is of infinite dimension. –  Alexander Grothendieck Dec 8 '13 at 17:49
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@000 If $V$ is a Hilbert space, then it's not false. –  Joe Johnson 126 Dec 8 '13 at 17:54
    
It's still false because in this case you have to say topological dual space. –  Alexander Grothendieck Dec 8 '13 at 19:00

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