I am reading a version of the Blackwell's Theorem for an operator to be a contraction mapping when given a metric space.
The Blackwell's Theorem:
Consider $X\subset\mathbb{R}^k$ and let $C(X)$ be a space of bounded functions, $f:X\rightarrow\mathbb{R}$, with the $sup$-metric. Let $B:C(X)\rightarrow C(X)$ be an operator satisfying two condition:
(1) Monotonicity: if $f,g\in C(X)$ and $f(x)\leq g(x)$ $\forall x\in X$, then $(Bf)(x)\leq(Bg)(x)$ for all $x$.
(2) Discounting: $\exists\delta\in(0,1)$ such that $[B(f+a)](x)\leq(Bf)(x)+\delta a$ for all $f\in C(X), a\geq0,x\in X$.
This is the part of the proof I don't understand:
For any $f,g\in C(X)$, $f(x)\leq g(x)+d(f,g)$ for all $x$ where $d$ is a metric.
By (1) and (2),
$Bg\leq B(f+d(f,g))\leq Bf+\delta d(f,g)$
$Bf\leq B(g+d(f,g))\leq Bg+\delta d(f,g)$
How do we obtain the first line with $Bg$?
Reference: https://projects.iq.harvard.edu/files/econ2010c/files/lecture_02_2010c_2014.pdf
Slide 20
If for any $f,g\in C(X)$ you have that $f(x)\leq g(x)+d(f,g)$, then you also have $g(x)\leq f(x)+d(f,g)$.
Just swap the symbols $f$ and $g$ and use the fact that $d(f,g) =d(g,f)$.