# Parallelogram law for vectors of equal length

Does the parallelogram law for vectors of equal length imply the full parallelogram law? That is, if for all norm one vectors $$x$$ and $$y$$ in a Banach space $$X$$ it holds that $$\lVert x-y\rVert^2+\lVert x+y\rVert^2=4$$, does it follow that $$X$$ is isometric to a Hilbert space?

I suspect the answer is "no", but I cannot come up with an example. Of course, it is enough to consider the question in two dimensions.

I will give it a try, based on Day's idea. Let $$X$$ be a two-dimensional Banach space with the given property and denote by $$B_X$$ its unit ball. Consider the ellipsoid of maximal volume (denoted by $$B_2$$) contained in $$B_X$$ (the John's ellipsoid) and denote by $$\|\cdot\|_2$$ the induced Euclidean norm. The goal is to show that $$B_X=B_2$$.
From John's theorem concerning the ellipsoid of maximal volume (or Loewner's Lemma for two dimensions in Day's paper), it follows that $$B_X$$ and $$B_2$$ have at least four contact points. Unless $$B_X=B_2$$, the contact points cannot form a dense subset of $$B_X$$. Assuming $$B_X\neq B_2$$, we can find contact points $$x$$ and $$y$$ such that $$\displaystyle\frac{x+y}{\|x+y\|}\in B_X$$ is not a contact point. Hence $$\|x+y\|<\|x+y\|_2$$. Therefore:
$$4=\|x+y\|^2+\|x-y\|^2<\|x+y\|_2^2+\|x-y\|_2^2=4$$
This is a contradiction, hence $$B_X=B_2$$.