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 $P_V \colon X \to V , ~ x \mapsto P_V(x) ,$ is called the metric projection onto $V$. In general, such a map need not be linear. It is linear if $X$ is a Hilbert space, since it coincides with the orthogonal projection on $V$ (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!
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.