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Is it true that:

-an inner product satisfies the properties of a norm if and only if the norm satisfies the parallelogram equality

-a norm can be induced by a metric if and only if the metric satisfies $d(x+a,y+a)=d(x,y)$ and $d(ax,ay)=ad(x,y)$

or are the implications one way?

Dan Burrows
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    And what do you think? – TZakrevskiy Apr 20 '15 at 10:05
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    The wording seems to be a bit confused: The first statement should be something like, "a (vector space) norm is induced by an inner product iff it satisfies the parallelogram identity." In the second, a norm is a unary map $X \to \mathbb{R}$ but the notation here even indicates that a metric is a binary map $X \times X \to \mathbb{R}$, so a norm is never a metric---to ask a sensible question, we must first describe how to build a candidate metric in terms of a norm and/or vice versa. – Travis Willse Apr 20 '15 at 10:09
  • sorry yes my wording for the 2nd part was poor. What i meant was a norm can be induced by a metric by setting ||x-y||=d(x,y) iff ... – Dan Burrows Apr 20 '15 at 10:15
  • And I'm thinking that these are indeed iff statements, though I've never seen them described this way, hence the question. – Dan Burrows Apr 20 '15 at 10:18
  • Your inner product lines still have the problem that Travis pointed out. $\;$ –  Apr 20 '15 at 10:36
  • [Norms Induced by Inner Products and the Parallelogram Law](http://math.stackexchange.com/questions/21792/norms-induced-by-inner-products-and-the-parallelogram-law) – Martin Sleziak Apr 20 '15 at 11:12
  • [Not every metric is induced from a norm](http://math.stackexchange.com/questions/166380/not-every-metric-is-induced-from-a-norm), [When a metric space is a normed space?](http://math.stackexchange.com/questions/988770/when-a-metric-space-is-a-normed-space), [Was this metric induced by a norm?](http://math.stackexchange.com/questions/503542/was-this-metric-induced-by-a-norm) – Martin Sleziak Apr 20 '15 at 11:17

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The first statement is true both ways. Specifically, suppose $(X, ||\cdot||)$ is a normed linear space. Then the norm $||\cdot ||$ is induced by an inner product iff the parallelogram law holds in $(X,||\cdot||)$.

For the second statement, this is not true. Call the condition $d(x,y)=d(x+a,y+a)$ translation invariance, and the condition $d(x,y)=d(ax,ay)$ homogeneity. Consider the following variant of the characteristic function $$ \chi(x,y)=\left\{\begin{matrix} 1, & x\neq y, \\ 0, & \text{otherwise.} \end{matrix}\right. $$ Homogeneity fails for $\chi,~x\neq y$. Indeed, $$ \chi(x,y)=1\neq a=\chi(ax,ay). $$