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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?

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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.

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  • $\begingroup$ Thanks. I forgot to clarify that $Y$ is finite-dimensional (have now fixed this) which I believe means that it is Chebyshev. $\endgroup$ – SecretlyAnEconomist Feb 2 at 21:22
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    $\begingroup$ Well, $Y$ is definitely proximinal., since it is boundedly compact. However, since $\mathcal{B}(X)$ is not strictly convex, it's very possible that $Y$ is not Chebyshev. For example, extend a line segment on the sphere of $\mathcal{B}(X)$ to a $1$-dimensional subspace. Then, the metric projection will not be single-valued. $\endgroup$ – Theo Bendit Feb 3 at 5:37
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    $\begingroup$ If $Y$ is Chebyshev and finite-dimensional, then yes, the metric projection will be continuous. This follows from the metric being locally bounded, $Y$ being boundedly compact, and the closed graph theorem. However, it won't be linear in general. $\endgroup$ – Theo Bendit Feb 3 at 5:41

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