# A characterization of inner product spaces ?

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 ?

• 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 $\langle x, y \rangle = \frac{1}{4} (\lVert x + y \rVert^2 - i \lVert x+iy \rVert^2 - \lVert x-y \rVert^2 + i \lVert x-iy \rVert^2)$. Oct 11 '17 at 16:59
• @DanielSchepler : I know that ... but how does that help here ?
– user
Oct 11 '17 at 17:07
• 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.
– user357151
Oct 12 '17 at 2:44
• wouldn't the space $\mathbb{C}^2$ with the norm $||(x, y)|| = |x| + |y|$ satisfy that? Oct 13 '17 at 20:40
• @ПетяНарышкин No. The vectors $a=(0,1)$ and $b=(1,1)$ have $\|a-b\|=1$, compared to $\frac12(\|a\|+\||b\|)\|(0,1)-(1/2,1/2)\| = 3/2$.
– user357151
Oct 14 '17 at 2:36

• 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.