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Let $V$ be a vector space over $\mathbb{C}$. If $V$ is an inner product space, then $V$ is normed (where the norm is defined as $\|x\|=\sqrt{(x,x)}\,\,$). Now if $V$ is normed, does it follow that $V$ is an inner product space ? I suspect no. I would like to see an example.

Thank you.


After reading my question again, I think it needs some clarification:

Suppose that $V$ is normed with norm $||\,||$. Can $V$ be given an inner product space structure such that $(x,x)=||x||^2$ ?

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No. See this. –  David Mitra Oct 16 '13 at 18:58
    
@DavidMitra Thanks. –  Amr Oct 16 '13 at 19:01

3 Answers 3

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For an example of a norm that is not induced by an inner product, consider Euclidean space $\Bbb R^n$ (where $n\ge 2$) with the norm $$\lVert \vec x\rVert_1:=\sum_{k=1}^n |x_k|.$$

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Is it clear that no inner product induces this norm ? I don't see it immediately. –  Amr Oct 16 '13 at 19:47
    
Recall that in David's link, it was mentioned that a norm is induced by an inner product if and only if the parallelogram law holds. It does not hold for $\lVert\cdot\rVert_1.$ See if you can prove that in the two-dimensional case. –  Cameron Buie Oct 16 '13 at 19:51

Inner product spaces satisfy the parallelogram law.

If you can find a counterexample to the parallegram law in a space with the sup norm, then you've found a normed linear space that is not an inner product space.

Even more interestingly, inner product spaces are the ONLY normed vector spaces that satisfy the parallelogram law.

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Normed vector spaces with an inner product are, in fact, quite rare. For example $L^p(X)$ is normed by $$\Vert f\Vert_p := \left( \int_X |f|^p d\mu \right)^{\frac 1 p}$$ for all $1\leq p < \infty$ (and $\mathop {{\rm ess}\, \sup}_{x\in X} |f(x)|$ for $p=\infty$) but only has an inner product for $p=2$ $$\langle f,g \rangle_2 = \int_X f \bar{g} d\mu$$

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