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For a normed linear space $(X, \|\cdot\|)$, the Jordan-von Neumann theorem specifies when exactly the norm is induced via an inner product, namely when the parallelogram law is satisfied.

I would like to know if there is a reference for a normed linear space that violates the parallelogram law, but in only one direction. That is, for any non-zero $x \in X$ and $y \in X$, such that $x \neq \pm y$, the following holds: $$ \frac{1}{2}\frac{\|x+y\|^2 + \|x-y\|^2}{\|x\|^2 + \|y\|^2} < 1.$$

In other words, the parallelogram law holds only when the vectors are equal or opposite. Specifically, I'd like to know if such spaces, although they aren't Hilbert spaces, have nice properties that might be useful in their study.

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    $\begingroup$ This is equivalent to the full law: en.wikipedia.org/wiki/… $\endgroup$ Commented Aug 14 at 9:21
  • $\begingroup$ @EmilJeřábek Yes $x+y=a, x-y=b$ mplies equality $\endgroup$ Commented Aug 14 at 9:33
  • $\begingroup$ @EmilJeřábek I added a clarification here. Sorry about the previous draft. $\endgroup$
    – Hikaru
    Commented Aug 14 at 9:36
  • $\begingroup$ How is this not even stronger (and therefore just inconsistent for any nontrivial space) than your previous assumption? $\endgroup$
    – David Gao
    Commented Aug 14 at 9:45

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If the above inequality holds for all nonzero $x,y$, then if all of $x,y,x+y,x-y$ are nonzero, we also have (applying your inequality to $x+y$ and $x-y$): $$ \frac{1}{2} \frac{\|2x\|^2 + \|2y\|^2}{\|x+y\|^2+\|x-y\|^2}\leq 1, $$ so $\|x+y\|^2+\|x-y\|^2 \leq 2(\|x\|^2+\|y\|^2) \leq \|x+y\|^2+\|x-y\|^2$, and we have the parallelogram identity. Of course, if $x-y=0$ or $x+y=0$, the parallelogram identity is trivial.

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  • $\begingroup$ Ah, I overlooked this. I will edit the question, because I have something else in mind. $\endgroup$
    – Hikaru
    Commented Aug 14 at 9:25
  • $\begingroup$ Now the same argument goes through to simply show that no such normed spaces exist. $\endgroup$ Commented Aug 14 at 9:46
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    $\begingroup$ @AchimKrause Technically, if we’re being extremely pedantic, and if we count the zero space as a normed space, it does satisfy the requirement. $\endgroup$
    – David Gao
    Commented Aug 14 at 9:47
  • $\begingroup$ @DavidGao I’d call that slightly pedantic, if at all. Or does it mean that I am an extreme pedant? $\endgroup$ Commented Aug 15 at 19:12
  • $\begingroup$ @EmilJeřábek I mean, I felt like I was being overly pedantic when I wrote down my comment, given that I routinely ignored “trivial” objects when writing all sorts of arguments and it has worked well for me. So I felt I was being a bit hypocritical, which was why I added that phrase. It does seem too much an exaggeration now though. Depending on the field, ignoring trivial objects could be a quite bad idea, so I suppose it wouldn’t be pedantic at all if you’re in such fields. $\endgroup$
    – David Gao
    Commented Aug 15 at 20:03

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