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 $(x,x)=||x||^2$ ?

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

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 $$\lVert \vec x\rVert_1:=\sum_{k=1}^n |x_k|.$$

  • 2
    $\begingroup$ Is it clear that no inner product induces this norm ? I don't see it immediately. $\endgroup$ – Amr Oct 16 '13 at 19:47
  • $\begingroup$ 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. $\endgroup$ – 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.


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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