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Many simple vector spaces such as $\mathbb{R}^m$ have natural inner products, in this case the dot product. Even function spaces such as $C[0,1]$, continuous real-valued functions on the unit interval, lend themselves to nice inner products, e.g. $$ \langle f,g \rangle := \int^1_0 f(x)g(x) dx. $$

In trying to think of vector spaces $V$ without a sensible inner product, it seems easiest to put $V$ over a less "nice" field, say the finite field $\mathbb{F}_5$. If $V$ is the vector space of polynomials up to degree $n$ with coefficients in $\mathbb{F}_5$, what would a reasonable inner product on $V$ be? What's an example of a real vector space without an inner product?

More generally, given an $F$-vector space $V$, are there necessary and sufficient conditions to show $V$ has an inner product?

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    $\begingroup$ an inner product should be positive-definite, so in an ordered field $\endgroup$ Jun 7 '20 at 21:42
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Assuming the axiom of choice, every vector space has a basis. Using that, you can turn every real vector space into an inner product space by first defining the inner product on the basis vectors so as to make them orthonormal and then extending it.

If you don't assume choice, then this answer show that it is consistent with $\sf ZF+ DC$ that $\cal C(\Bbb R)$ has no norm. (Since every inner product gives a norm, it answers your question as well.)


Note that inner-product requires your base field to have an "order", that is, your field must be an ordered field. In particular, it must have characteristic $0$. Therefore, it doesn't make much sense to talk about finite fields.

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    $\begingroup$ Thanks! I suspected we needed $F$ to have characteristic $0$, since there is a natural connection between norms and inner products and so we need the triangle inequality to hold. $\endgroup$
    – ccroth
    Jun 8 '20 at 23:43

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