Why isn't the theorem of approximation applicable in Banach spaces? Let X be a Hilbert space, A a convex, closed subset of X. Then there exists for every x in X exactly one best approximation in A, that is there exists a y in A such that || x-y || = d(x,A) = inf { || x-z || : z in A}.
Why isn't that theorem true for Banach spaces?
 A: EDIT: since I'm not an expert on Banach spaces, I feel I shouldn't say anything more, but anyway; an essential ingredient is an exact formula in Hilbert spaces for $\|x + y\|^2$. Just an idea (perhaps I am being stupid): maybe if you have a Banach space where $\| x + y \| = F(\|x\|, \|y \|, g(x,y))$ for some reasonably simple functions $F$, $g$ then something can be said.
If you examine the proof for Hilbert spaces, it makes essential use of the scalar product; so it's not really surprising that it doesn't work for general Banach spaces. The norm is a nice enough structure to do a lot of things, but not that nice.
It also demonstrates that Banach spaces have far more detailed structure than just ordinary vector spaces, but that Hilbert spaces have even more structure again. In fact, there are many properties Hilbert spaces have which general Banach spaces don't (and many which even characterise Hilbert space uniquely).
A: Happened to have this slide in one of my lectures.  The green circle and green dot correspond to the Euclidean norm, the purple square and purple line to the sup norm.  With the Euclidean norm, the green dot is the closest point in the square to the blue dot; with the sup norm, the purple line is the set of closest points in the square to the blue dot.

A: A subset in a Banach space for which every point in the space has a unique element of best approximation is called a Chebyshev set. In a Banach space every closed convex set is a Chebyshev set if and only if your space is strictly convex and reflexive (so any space that fails to have these properties is a counterexample). See here for some references.
