Let $(\mathcal{X}, d)$ be a complete metric space.
For any subset set $C$ of $\mathcal{X}$, we say that $C$ is convex if for any two points $x, y, \in C$, there is a geodesic $\gamma (t)$ from $x$ to $y$, such that $\gamma (0) = x$, $\gamma (1) = y$ and $\gamma ([0, 1]) \subseteq C$.
We further define the projection operator in $\mathcal{X}$ as $P(x) := \arg\min_{y \in C} d(x, y)$.
The definition of the geodesic is provided as follows (c.f. a course in metric geometry).
Let $X$ be a length space. A curve $\gamma \colon I \to X$ is called a geodesic if for every $t \in I$ there exists an interval $J$ containing a neighborhood of $t$ in I such that $\gamma \mid J$ is a shortest path. In other words, a geodesic is a curve which is locally a distance minimizer (i.e., a shortest path).
We also assume that the minimum in the projection operator can be achieved and the minimizer is unique. For example, if $\mathcal{X}$ is a uniformly convex space, then the minimum exists.
My question is that,
For all $z \in C$, does the inequality $d(z, P(x)) \leq d(z, x)$ hold or not?
Best Answer
Let $\mathcal{X}$ be the disjoint union of the unit interval $[0,1]$ (with usual metric), and a point $x$, where $d(x,t)=\frac {4+t}6$ for $t\in [0,1]$. As $|\frac{d}{dt}\frac {4+t}6|\leq 1$ and $\frac {4+t}6>\frac12$, we know $\mathcal{X}$ is a metric space. It is clearly complete, as $[0,1]$ is.
We have $[0,1]$ a convex subset, and $P(x)=0$. However $$d(0,1)=1>\frac56=d(x,1).$$