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Normally, we can see lots of example of Hilbert space, I am wondering if there is a space we can't define an inner product on it?

89085731
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    Using the axiom of choice, you can always select a basis $(x_i)i$ of a vector space and define $\langle \sum \alpha_i x_i, \sum \beta_i x_i \rangle := \sum \alpha_i \overline{\beta_i}$. But in the infinite-dimensional case, your space will not be complete w.r.t. this inner product. Take e.g. an infinite sequence $(i_n)_n$ of pairwise different indices. Then in the completion, we would have an element $\sum_n 1/n \cdot x{i_n}$, but as $(x_i)_i$ is a basis, every element of $V$ has only finitely many nonvanishing coefficients. – PhoemueX Jul 22 '14 at 09:09
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    @PhoemueX I suspect that the question is a bit more difficult. The OP is probably asking for a vector space on which no inner product can be defined. – Siminore Jul 22 '14 at 09:11

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Let me make my comment into a full-fledged answer:

Using the axiom of choice, you can always select a basis $(x_i)_i$ of a vector space and define

$$\langle \sum \alpha_i x_i, \sum \beta_i x_i \rangle := \sum \alpha_i \overline{\beta_i}.$$

But in the infinite-dimensional case, your space will not be complete w.r.t. this inner product. Take e.g. an infinite sequence $(i_n)_n$ of pairwise different indices. Then in the completion, we would have an element

$$\sum_n 1/n \cdot x_{i_n},$$

but as $(x_i)_i$ is a basis, every element of $V$ has only finitely many nonvanishing coefficients.

There are also spaces on which no inner product can be defined that makes the space complete. Take e.g. $\Bbb{R}[X]$. Then there is no norm on this space that makes it complete, because we have

$$ \Bbb{R}[X] = \bigcup_n \langle 1, x, \dots, x^n\rangle, $$

a union of strict, finite-dimensional subspaces. Every finite-dimensional subspace is closed and every strict subspace has nonempty interior (both w.r.t. any norm).

This shows that Baire's category theorem does not hold for $\Bbb{R}[X]$, so that it does not admit a complete metric, hence no complete inner product.

PhoemueX
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  • I can't see how u use axiom of choice to choose a basis. From the axiom of choice,we can form a set of elements which we choose one from each subset, then how can we choose the basis? – 89085731 Jul 22 '14 at 09:33
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    See here http://en.wikipedia.org/wiki/Zorn%27s_lemma#Equivalent_forms_of_Zorn.27s_lemma . What I said was somewhat imprecise. The axiom of choice implies Zorn's Lemma. Using Zorn's Lemma, choose a maximal linearly independent set. It is then not too hard to see that this is a basis. – PhoemueX Jul 22 '14 at 09:38
  • If u arrange the set well-ordered, how can u get an upper bound for each subset? – 89085731 Jul 22 '14 at 10:30
  • See here: http://math.stackexchange.com/questions/189693/every-vector-space-has-a-basis-proof-using-zorns-lemma-in-linear-algebra . You don't arrange the set well-ordered. For Zorn's Lemma, you have to show that every chain has an upper bound. For this you have to show that if $(M_i)_i$ is a family of linearly independent sets with $M_i \subset M_j$ or $M_j \subset M_i$ for all $i,j$, then $\bigcup_i M_i$ is also linearly independent. – PhoemueX Jul 22 '14 at 10:51