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I know that generally, an inner product defines a norm on an inner product space, But, generally speaking, If I have a normed space (on purpose I do not say which) with the norm 2 does it mean that I have an inner product for sure? Thanks

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No. Not every normed space has an inner product which gives rise to the given norm.

A normed space $(V, \|\cdot\|)$ is an inner product space if and only if it satisfies the parallelogram law: $\|x+y\|^2 + \|x - y\|^2 = 2(\|x\|^2 + \|y\|^2)$ for all $x, y \in V$.

An example of a normed space which does not satisfy the parallelogram law, and is therefore not an inner product space, is $(\mathbb{R}^2, \|\cdot\|_1)$ where $\|(x, y)\|_1 = |x| + |y|$. Note that

\begin{align*} \|(1, 0) + (0, 1)\|^2_1 + \|(1, 0) - (0, 1)\|^2_1 &= \|(1, 1)\|^2_1 + \|(1, -1)\|^2_1\\ &= (|1| + |-1|)^2 + (|1| + |-1|)^2\\ &= (1 + 1)^2 + (1 + 1)^2\\ &= 8, \end{align*}

but

\begin{align*} 2(\|(1, 0)\|^2_1 + \|(0, 1)|_1^2) &= 2((|1| + |0|)^2 + (|0| + |1|)^2)\\ &= 2((1 + 0)^2 + (0 + 1)^2)\\ &= 2(1 + 1)\\ &= 4 \end{align*}

so $\|(1, 0) + (0, 1)\|^2_1 + \|(1, 0) - (0, 1)\|^2_1 \neq 2(\|(1, 0)\|^2_1 + \|(0, 1)|_1^2)$.

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  • $\begingroup$ Awsome, thanks! assuming I have an abstract normed space with a norm $\|\|_2$ does it still mean that I would not know whether I have an inner product? (It does not specify the vectors so I do now know what Im dealing with, but I do know that the norm is of the type $\|\|_2$) $\endgroup$ – user3921 Jan 21 '15 at 22:49
  • $\begingroup$ What do you mean by a norm "of the type $\| \|_2$"? $\endgroup$ – Michael Albanese Jan 22 '15 at 0:55

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