Function which doesn't increase distance Let $(X, d)$ be a metric space, $A \subseteq X$ with $A \neq \varnothing$, and $$f: A \rightarrow \mathbb{R}\quad\text{such that}\quad \left|f(x)-f(y)\right|\leq d(x,y),\ x,y\in A.\tag{$\ast$}$$
Let $$f_1, f_2 : X \rightarrow \mathbb{R}$$be functions such that $$f_1 (x) = \sup_{a \in A}(f(a)-d(x, a))\quad \text{and}\quad f_2 (x) = \inf_{a \in A}(f(a)+d(x, a)).$$
Prove that $f_1$ and $f_2$ have the same property as $f$; I marked it with $(*)$.
I've tried proving it: $$|f_1(x)-f_2(y)|= \left|\sup_{a \in A}(f_1(a)-d(x, a)) - \sup_{a \in A}(f(a)-d(y, a))\right|.$$ I don't know how to show that it's less than or equal to $d(x,y)$. I suppose I should use the triangle inequality and some properties of $\sup / \inf$, but I don't know how.
Could you help me?
 A: 1- First, we should check that $f_1(x)$ is a finite number. Since $f$ is $1$-Lipschitz, we find 
$$
f(a)-d(x,a)\leq f(x)\quad\Rightarrow\quad f_1(x)\leq f(x)<\infty.
$$
By  triangular inequality of the distance $d$, we have $-d(x,a)\leq d(x,y)-d(y,a)$ so
$$
f(a)-d(x,a)\leq f(a)-d(y,a)+d(x,y)\leq f_1(y)+d(x,y)
$$
for every $x,y\in X$.
Now take the sup on the lhs to get
$$
f_1(x)\leq f_1(y)+d(x,y)\qquad\Rightarrow\qquad f_1(x)-f_1(y)\leq d(x,y)
$$
for every $x,y\in X$. For symmetry reasons, the same holds when we exchange $x$ and $y$ hence
$$
|f_1(x)-f_1(y)|\leq d(x,y)\quad\forall x,y\in X.
$$
2- You can handle $f_2$ in a similar fashion. First, since $f$ is $1$-Lipschitz,
$$
f(a)+d(x,a)\geq f(x)\quad \Rightarrow \quad f_2(x)\geq f(x)>-\infty
$$
so $f_2(x)$ is a finite number. 
Now by triangular inequality of $d$, we have $d(x,a)\geq d(y,a)-d(x,y)$ whence
$$
f(a)+d(x,a)\geq f(a)+d(y,a)-d(x,y)\geq f_2(y)-d(x,y).
$$
Taking the inf of the lhs, this yields
$$
f_2(x)\geq f_2(y)-d(x,y)\quad\Rightarrow\quad f_2(y)-f_2(x)\leq d(x,y).
$$
By symmetry, the same holds when we exchange $x$ and $y$, so 
$$
|f_2(x)-f_2(y)|\leq d(x,y)\quad\forall x,y\in X.
$$
