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Let $(X,\|\cdot\|)$ be a real normed space. For any points $A$ and $B$ in $X$, let $AB:=\|A-B\|$. Suppose that for any points $A$ and $B$ in $X$ and any straight line $\ell\subseteq X$ such that $B\in\ell$ we have $$AB^2\ge AA_\ell^2+BA_\ell^2 $$ for some point $A_\ell$ on $\ell$ that is at the smallest distance to $A$ among all points on $\ell$ (it is easy to see that such a point $A_\ell$ always exists). Does it then necessarily follow that the norm $\|\cdot\|$ is induced by an inner product?

Perhaps easier to answer is the version of this question with "for some point $A_\ell$ on $\ell$ that is at the smallest distance to $A$" replaced by "for all points $A_\ell$ on $\ell$ that are at the smallest distance to $A$".

(I was not sure if this question is appropriate for MO, but then saw this MO question, which seems to be of the same flavor.)

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  • $\begingroup$ When you say that $A_\ell$ is "the closest to $A$ among all points on $\ell$" do you just mean "a point on $\ell$ whose distance to $A$ is minimal" or "a unique point whose distance to $A$ is minimal"? $\endgroup$
    – Yemon Choi
    Dec 19, 2019 at 21:44
  • $\begingroup$ (My point being that closest points will exist by compactness but I don't see why they should be unique) $\endgroup$
    – Yemon Choi
    Dec 19, 2019 at 21:45
  • $\begingroup$ Final comment for now: it seems that it would suffice (for a positive or negative answer) to restrict to the case $\dim(X)=2$, unless I have made a silly error $\endgroup$
    – Yemon Choi
    Dec 19, 2019 at 21:47
  • $\begingroup$ @YemonChoi : Thank you for your comments. (i) I have replaced "the closest" by "at the smallest distance". (ii) I was cognizant of the nonuniqueness issue, which is why I offered two versions of the question. (iii) I too guess that the problem is probably mostly two dimensional, with perhaps some extra efforts needed to harmonize between the different two-dimensional subspaces. $\endgroup$ Dec 19, 2019 at 22:29
  • $\begingroup$ Thanks for the clarifications. Regarding (iii), here was my line of thought: if there is a 2-dimensional counterexample we are done. Conversely, suppose we know that in dimension 2 this property characterizes the Euclidean norm. Then for an arbitrary X satisfying your condition, I take two vectors x and y generating a 2-dim subspace V. Applying your condition to points and lines in V, I deduce that the subspace norm on V is Euclidean. But since this holds for all 2-dim subspaces of X, X is Euclidean/Hilbertian (parallelogram identity) $\endgroup$
    – Yemon Choi
    Dec 19, 2019 at 23:24

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