TITLE: On the linearity of metric projections
QUESTION [7 upvotes]: Let $X$ be a reflexive and strictly convex Banach space. If $V$ is a closed subspace of $X$, then for each $x \in X$ there exists a unique vector $P_V(x) \in V$ that solves the feasibility problem $ \inf_{v \in V} ||x-v|| $.  The map $P_V \colon X \to X ,  ~ x \mapsto P_V(x) $ is called the   metric projection onto $V$. In general, such a map need not be linear. However, in the case of Hilbert spaces, the metric projections coincides with the orthogonal projection onto $V$, which is linear.
I was wondering if Hilbert spaces are the only ones with the property that each of its metric projection onto closed subspaces is linear. To be more exact, is the following statement true?

Let $X$ be a reflexive and strictly convex Banach space. Suppose  that
each of its
metric  projections onto closed subspaces is linear. Then $X$ is a Hilbert  space.

Any help is appreciated!

REPLY [4 votes]: The answer is $\textbf{yes}$.
Notice that if $P$ is a metric projection onto a closed subspace $V$, then
$$\|Px\|\:\leq\: \|Px -x\| +\|x\| \:=\: d(x,V) +\|x\|\:\leq\: 2\|x\|\,,$$
i.e., it is bounded.  If every metric projection is linear, then every closed subspace is complemented (as there exists a linear bounded projection onto it). Hence, it is isomorphic to a Hilbert space by the
Lindenstrauss-Tzafriri theorem.