Distance from a weak* closed subspace is achieved by an element? Let $X$ denotes a separable Banach space and $X^{*}$ its dual space. Let $M\subseteq X^{*}$ be a weak* closed subspace of $X^{*}.$ Let $\varphi\in X^{*}\setminus M$ be given. I have the following question:
is it true that there exists a $\psi\in M$ such that $$\inf\{||\varphi-m||:m\in M\}=||\varphi-\psi||,$$
where $||.||$ denotes the operator norm of linear functionals on $X^{*}?$
 A: The separability of $X$ is irrelevant here. From the Banach-Alaoglu theorem, we know that for all $\varphi \in X^{\ast}$ and $r \geqslant 0$ the norm-closed ball $K_r(\varphi) := \{ \psi \in X^{\ast} : \lVert \varphi - \psi\rVert \leqslant r\}$ is weak$^{\ast}$-compact.
Then if $S$ is any weak$^{\ast}$-closed nonempty subset of $X^{\ast}$, the family $\{S\cap K_r(\varphi) : r > \operatorname{dist}(\varphi,S)\}$ is a family of nonempty weak$^{\ast}$-compact sets. Since the family is nested, it has the finite intersection property, and therefore
$$\bigcap_{r > \operatorname{dist}(\varphi,S)} S\cap K_r(\varphi) = S \cap K_{\operatorname{dist}(\varphi,S)}(\varphi) \neq \varnothing.$$
But $S \cap K_{\operatorname{dist}(\varphi,S)}(\varphi)$ is just the set of elements of $S$ that achieve the distance,
$$S \cap K_{\operatorname{dist}(\varphi,S)}(\varphi) = \{\psi \in S : \lVert \varphi - \psi\rVert = \operatorname{dist}(\varphi,S)\}.$$
So whatever the nonempty weak$^{\ast}$-closed subset, the distance is always achieved.
The set $S \cap K_{\operatorname{dist}(\varphi,S)}(\varphi)$ may or may not be a singleton. It is always a singleton if $\varphi\in S$ (trivially). It is also a singleton if the set $S$ is convex [e.g. a linear subspace] and the operator norm on $X^{\ast}$ is strictly convex (that is, the unit sphere contains no line segment), for if we have two distinct points $\psi_1,\psi_2\in S$ with $\lVert \varphi - \psi_1\rVert = \lVert \varphi - \psi_2\rVert$, and $S$ is convex, then $\psi_3 = \frac{1}{2}(\psi_1 + \psi_2) \in S$, and if the norm is strictly convex, then $\lVert \varphi - \psi_3\rVert < \lVert \varphi - \psi_1\rVert$, so $\lVert \varphi - \psi_1\rVert > \operatorname{dist}(\varphi,S)$ then.
If the operator norm is not strictly convex, the distance can be achieved in multiple points. E.g. if we take $X = \mathbb{R}^2$ endowed with the $\ell^1$-norm, then $X^{\ast}$ is (isometrically isomorphic to) $\mathbb{R}^2$ endowed with the maximum-norm. Then we can take $M = \{(x,0) : x \in \mathbb{R}\}$ and $\varphi = (0,1)$, and we see that the distance is achieved for all $(x,0)$ with $\lvert x\rvert \leqslant 1$. If on the other hand we take a subspace $M = \mathbb{R}\cdot (a,b)$ with $ab\neq 0$, then the point achieving the distance is unique.
A: Yes this is true and it can be proven along the following lines:


*

*find a sequence $\{\psi_n\} \subset M$, such that $\|\psi_n - \varphi\| \to \inf\{\|\varphi - m\| : m \in M\}$

*extract a weak-$*$ convergent subsequence with limit $\psi$

*conclude that $\psi$ has the desired properties

