Let $(E, | \cdot|)$ be a Banach space and $F$ a closed subset of $E$. Fix $a \in E$. In general, there does not necessarily exist $b \in F$ such that $$ |b-a| = \inf_{x\in F}|x-a|. $$
Are there some conditions on the set $F$ to guarantee that such projection exists for every $a\in E$?
Update: I have recently solved this question. It suggests that in case of the dual space $E'$, if $M \subseteq E'$ is closed in the weak$^\star$ topology $\sigma(E', E)$, then the projection into $M$ exists for any $f\in E'$.
Such sets that admit best approximations are often called proximinal. It can be shown that if $E$ is reflexive, and $F$ is closed and convex then $F$ is proximinal. (The converse is also true; namely, if every closed and convex subset of $E$ is proximinal then $E$ is reflexive). Furthermore, if $E$ is reflexive and strictly convex, then there exists exactly one best approximation. In that case, $F$ is said to be Chebyshev. (In fact, $E$ is strictly convex and reflexive iff each of its closed and convex subsets is Chebyshev). The (nonlinear) map that sends each $x \in E$ to its best approximation in $F$ is called the metric projection of $F$.