# Reverse fixed point conclusion

If I have a function such as:

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

where $$M$$ is any metric space denoted by :

$$(M,d)$$

$$f(x) =d(x,y)$$ 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 $$K=1$$.