If I have a function such as:

$$f:M \rightarrow \mathbb{R} $$

where $M$ is any metric space denoted by :


$$f(x) =d(x,y) $$ where $y \in M$ is a fixed point.

I am trying to show that this function satisfies the Lipschitz condition. https://en.wikipedia.org/wiki/Lipschitz_continuity $$\frac{d(f(x),f(z))}{d(x,z)}\leq K$$ for $K \geq 0$

Currently I am stuck at two things.

First, the distance representation is not quite clear for me , I am not sure if I can use here $d(x,y) = |x-y|$?

I only learnt two fixed points theorems, the Banach fixed point and the Brouwer fixed point.

Secondly, If I have a fixed point $y$ : it could be a Banach case(unique fixed point), if $M$ is a complete metric space and If $f$ is a contraction i.e. $0 \leq K <1$

Otherwise, it is not unique fixed point.

I am not sure what approach should I use here? contradiction or attempt some sort of reverse fixed point iteration direct proof?


The triangle inequality implies that, for all $x_1,x_2\in M$, $$ d(x_1,x_2)+d(x_2,y)\ge d(x_1,y) \quad\text{and}\quad d(x_2,x_1)+d(x_1,y)\ge d(x_2,y) $$ hence $$ d(x_1,x_2)\ge d(x_1,y)-d(x_2,y) \quad\text{and}\quad d(x_1,x_2)\ge d(x_2,y)-d(x_1,y) $$ and therefore $$ |\,f(x_1)-f(x_2)|=\big|\,d(x_1,y)-d(x_2,y)\big|\le d(x_1,x_2). $$ So, $f$ is Lipschitz, with $K=1$.


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