Metric projection from space of bounded functions to finite-dimensional linear space Apologies if the answer is obvious or should be easy to find, but so far I've had no luck.
Let $X$  be a subspace of $\mathbb{R^k}$ for a finite $k$ and let $\mathcal{B}(X)$ be the Banach space of bounded real-valued functions on $X$. Let $Y$ be a finite-dimensional linear subspace of $\mathcal{B}(X)$. I know that there must exist a bounded linear projection from $\mathcal{B}(X)$ to $Y$, and I know that the  metric projection (i.e., projection that achieves to minimum distance from $Y$) exists. What I want to know is a. whether the metric projection is continuous, and b. whether it is linear. And, if not, can one find a bounded linear projection that achieves a distance arbitrarily close to that of the metric projection?
 A: I just want to say first up, you'd need to know from the outset that $Y$ is a Chebyshev (meaning: admits unique metric projections) subspace, as in general, $Y$ may not be Chebyshev. If $X$ is an infinite set, then $\mathcal{B}(X)$ is not a reflexive space, which means (by James' theorem) that $\mathcal{B}(X)$ admits an antiproximinal (meaning no metric projections exist from points outside the set) hyperplane. You'd need some extra assumptions, such as $Y$ is finite-dimensional.
If you assume $Y$ is Chebyshev, unfortunately no, there is no guarantee that the metric projection is continuous or linear. If I remember correctly, metric projections onto a closed subspace always being a linear map is a property exclusively of Hilbert Spaces. As for continuity, Lindenstrauss published an example of a Chebyshev subspace of a Banach space that is discontinuous. If I remember correctly, it was a subspace of $C[0, 1]$, so not quite $\mathcal{B}(X)$, but I think it's highly dubious to expect the projection map to be continuous.
