If $(X,d)$ is a metric space and $x_0 \in X $ ,show that the function $f :X \to \mathbb R ; f(x)=d(x_0,x)$ is Lipschitz continuous

Question

I was writting $d(x_0,x_1) \le d(x_0,x_2)+d(x_2,x_1)$ Since $\displaystyle d(x_2,x_1) > 0;d(x_0,x_1) - d(x_2,x_1) \le d(x_0,x_2); |d(x_0,x_1) - d(x_2,x_1)| \le |d(x_0,x_2)|$

But I don't know how to keep going.

Answer 1

From $d(x_0,x_1) \leq d(x_0,x_2)+d(x_1,x_2)$ we get $d(x_0,x_1)-d(x_0,x_2) \leq d(x_1,x_2)$, and from $d(x_0,x_2) \leq d(x_0,x_1)+d(x_1,x_2)$ we get $-d(x_1,x_2) \leq d(x_0,x_1)-d(x_0,x_2)$. Thus $$-d(x_1,x_2) \leq d(x_0,x_1)-d(x_0,x_2) \leq d(x_1,x_2)$$ and this means $|d(x_0,x_1)-d(x_0,x_2)| \leq d(x_1,x_2)$, that is, $|f(x_1)-f(x_2)| \leq d(x_1,x_2)$.