# A Pythagorian inequality characterization of inner-product spaces

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.)

• 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"? Dec 19 '19 at 21:44
• (My point being that closest points will exist by compactness but I don't see why they should be unique) Dec 19 '19 at 21:45
• 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 Dec 19 '19 at 21:47
• @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. Dec 19 '19 at 22:29
• 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) Dec 19 '19 at 23:24