# Proving that function F is a Lipschitz function

So i have $(X,d)$, a metric space. And function $F:X \times X \to \mathbb{R}$ given as $F(x,y)=d(x,y)$. So i want to prove that $F$ is Lipschitz on $X \times X$ if that space gets a metric $D_{1}$ defined as: $D_{1}=d(x_1,x_2)+d(y_1,y_2)$ So i know that i have to find $c>0$ such that it will work like this: $$\lvert F(x_1,x_2)-F(y_1,y_2)\rvert\le c \cdot D_1(x,y)$$ for any $(x,y)\in X\times X$

Any help would be appreciated.

• The condition should have been $\lvert F(x_1,x_2)-F(y_1,y_2)\rvert\le c \cdot D_1(x,y)$, not what you wrote, because: 1) Lipschitz condition is about inequalities. 2) $F(x,y)$ is in $\Bbb R$, not in $X$. – user228113 Sep 3 '16 at 10:46
• @G.Sassatelli i fixed it, was a little mistaken when writing it here... – MathIsTheWayOfLife Sep 3 '16 at 11:19
• Use $\lvert a-b\rvert\le \lvert a\rvert +\lvert -b\rvert$ to see that $c=1$ works. – user228113 Sep 3 '16 at 11:25