Simple geometry question in a Banach space
Show that in a Banach space, three collinear points cannot all be the same distance from a fourth point. Or is this possible? I’m sure it isn’t in Hilbert spaces, but a proof eludes me in Banach spaces.
This came up as I’m trying to prove the convex projection/ best approximation theorem: given a closed convex set 
and a point 
, there exists a unique point in closest to .
As I was trying to recall anything about norms that might be useful in proving this, I realized that the law of cosines is a specific application of the polarization identity. In fact, if you write the law of cosines in the vector formulation (cf. the Wikipedia link) and replace the 
expression with 
, you get exactly the polarization identity. That was probably the motivation for the general Hilbert space result.
Another neat rephrasing of a more basic trig identity: 
.
Possibly relevant posts:
- Every finite spanning set is a frame (5/26/2006)
- Absorbing and balanced sets (1/28/2006)
- I’m stumped– you give it a try (4/1/2008)
It’s not true for a Banach space. The simplest examples I can think of are either the sup-norm or the L^1-norm in R^2. In each of these two cases, “circles” are really squares (although at different orientations in the two cases), so there can be (uncountably) infinitely many collinear points all the same distance from another point.
Comment by george — 8/2/2007 @ 9:40 am
oh, duh! thanks.
Comment by Alex — 8/2/2007 @ 1:04 pm