Let $X$ be a normed linear space over $\mathbb C$ such that $||x-y|| \ge \dfrac 12 (||x||+||y||)\bigg|\bigg| \dfrac x{||x||}- \dfrac y {||y||} \bigg|\bigg| , \forall 0\ne x, y \in X$ , then is it true that the norm on $X$ comes from an inner-product ? ( I can show that for a complex inner-product space , the inequality is true ) If not true in general , what if we moreover assume $X$ is Banach or finite dimensional ?

  • $\begingroup$ IIRC, a norm comes from an inner product if and only if it satisfies the parallelogram law $\lVert x + y \rVert^2 + \lVert x + iy \rVert^2 + \lVert x - y \rVert^2 + \lVert x - iy \rVert^2 = 4 (\lVert x \rVert^2 + \lVert y \rVert^2)$, and in that case the inner product formula is found from the polarization identity $\langle x, y \rangle = \frac{1}{4} (\lVert x + y \rVert^2 - i \lVert x+iy \rVert^2 - \lVert x-y \rVert^2 + i \lVert x-iy \rVert^2)$. $\endgroup$ Oct 11 '17 at 16:59
  • $\begingroup$ @DanielSchepler : I know that ... but how does that help here ? $\endgroup$
    – user
    Oct 11 '17 at 17:07
  • 2
    $\begingroup$ The two-dimensional case is the general case, because having an inner product is determined by 2D subspaces (via the parallelogram law). My impression is that the statement is true. $\endgroup$
    – user357151
    Oct 12 '17 at 2:44
  • $\begingroup$ wouldn't the space $\mathbb{C}^2$ with the norm $||(x, y)|| = |x| + |y|$ satisfy that? $\endgroup$ Oct 13 '17 at 20:40
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    $\begingroup$ @ПетяНарышкин No. The vectors $a=(0,1)$ and $b=(1,1)$ have $\|a-b\|=1$, compared to $\frac12(\|a\|+\||b\|)\|(0,1)-(1/2,1/2)\| = 3/2$. $\endgroup$
    – user357151
    Oct 14 '17 at 2:36

Initially, this was proved in

W.A. Kirk and M.F. Smiley, Mathematical Notes: Another characterization of inner product spaces, Amer. Math. Monthly 71 (1964), no. 8, 890–891.

but I don't have access to this paper. If you are interested in the complete history on this question see

F. Dadipour, A. Maric, M. S. Moslehian, and R. Rajic, A glimpse at the Dunkl-Williams inequality Banach J. Math. Anal. Volume 5, Number 2 (2011), 138-151.

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    $\begingroup$ To be precise, the one-page note of Kirk and Smiley doesn't prove much; they observe that the condition (trivially) implies $\|tx-t^{-1}y\|\ge \|x-y\|$ for all unit vectors $x,y$ and all $t>0$. And the latter was shown to characterize inner product spaces by E. R. Lorch: "Certain Implications Which Characterize Hilbert Space", Annals of Mathematics, Vol. 49, No. 3 (Jul., 1948), pp. 523-532. If I manage to extract a not-too-boring proof of the latter, I'll post it as an answer. $\endgroup$
    – user357151
    Oct 16 '17 at 22:31

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