Does this property implies Lipschitz continuity?

Question

Let $f:\mathbb{R}^{n} \rightarrow \mathbb{R}$ be such that, for $x,y,z \in \mathbb{R}^{n}$, we have that

$$|f(z) - f(x)| \leq |f(z) - f(y)| \Rightarrow \|z-x\| \leq \|z-y\|$$

Can I say that this function is Lipschitz continuous?

Answer 1

Claim) There is no such function when $n>1$, and for $n=1$ only non-constant affine functions satisfy this property.

Let us begin with a general observation. The contrapositive says: $$\parallel z-y\parallel< \parallel z-x\parallel\Rightarrow |f(z)-f(y)|<|f(z)-f(x)|.$$ Writing this for all sides of a triangle in $\Bbb{R}^n$ implies:

• Given two distinct points $a,b\in\Bbb{R}^n$ and denoting $\parallel a-b\parallel$ by $r$, for any $c\in B_r(a)\cap B_r(b)$, $f(c)$ lies strictly between $f(a)$ and $f(b)$ on the real axis. ($B$ denotes the open ball).

But when $n>1$, there exist points $c_1,c_2\in B_r(a)\cap B_r(b)$ with $\parallel c_1-c_2\parallel>r$. Invoking the fact above, this time for $f(c_1)$, $f(c_2)$ and $f(a)$, the latter should be between $f(c_1)$, $f(c_2)$. But these two numbers belong to the interval $\left(\min{(f(a),f(b))},\max{(f(a),f(b))}\right)$, a contradiction. So such functions do not exist when $n>1$.

Now let us consider the case of $n=1$. Due to the general fact mentioned above, if $c\in (a,b)$, then $f(c)\in\left(\min{(f(a),f(b))},\max{(f(a),f(b))}\right)$. In particular, $f$ is strictly monotonic. WLOG, suppose it is strictly increasing. Indeed, $f$ cannot have any jumps and must be continuous: Otherwise, there exists a point $p$ and a number $\epsilon>0$ such that $f(y)-f(z)>\epsilon$ whenever $z<p<y$. But $y$ and $z$ can be arbitrarily close. So pick $x,z<p$ such that $|f(z)-f(x)|<\epsilon$ but $z$ is closer to $y$ than it is to $x$. This violates the inequality which $f$ is assumed to satisfy.

We now finish the proof. Let $a<b$ be arbitrary. If $c\in\left(a,\frac{a+b}{2}\right)$, then $|c-a|<|c-b|$, so $|f(c)-f(a)|=f(c)-f(a)$ must be smaller than $|f(c)-f(b)|=f(b)-f(c)$. In particular, $f(c)<\frac{f(a)+f(b)}{2}$. By the same logic, $c\in\left(\frac{a+b}{2},b\right)$ implies $f(c)>\frac{f(a)+f(b)}{2}$. Combining with continuity at $\frac{a+b}{2}$, we deduce $f\left(\frac{a+b}{2}\right)=\frac{f(a)+f(b)}{2}$. A continuous function $f:\Bbb{R}\rightarrow\Bbb{R}$ satisfies this functional equation iff it is of the form $f(x)=mx+h$ (here $m\neq 0$ due to monotonicity).