Blackwell's condition for a contraction: Why is boundedness neccessary?

I'm trying to understand the proof that certain operators $T$ are a contraction if they fulfill Blackwell's sufficient conditions. In particular, I try to understand why the operator $T$ has to map the set of bounded functions into itself because the definition of a contraction does not presume the operator to map bounded functions into itself. According to Stokey, Lucas and Prescott (1989) a contraction is defined as:

Definition. Let $(S,\rho)$ be a metric space and $T: S \rightarrow S $ be a function mapping S into itself. $T$ is a contraction mapping (with modulus $\beta)$ if for some $ \beta \in (0,1), \rho(Tx,Ty) \leq \beta \rho(x,y)$, for all $ x,y \in S$

Following the same authors, the theorem says:

(Theorem: Blackwell's sufficient condition for a contraction.) Let $X \subset \mathbf{R}^l $ and let $B(X)$ be a space of bounded functions $f: X \rightarrow \mathbf{R}$ with the sup norm. Let $T: B(X) \rightarrow B(X) $ be an operator satisfying:

• (monotonicity) $f,g \in B(X) $ and $f(x) \leq g(x) $, for all $ x \in X$, implies $(Tf)(x) \leq (Tg)(x) $ for all $x \in X$
• (discounting) there exists some $\beta \in (0,1) $ such that: $[T(f+a)](x) \leq (Tf)(x) + \beta a$, all $f \in B(X)$, $ a \geq 0, x \in X$

[Here $(f+a)(x)$ is the function defined by $(f+a)(x) = f(x) + a$]. Then $T$ is a contraction with modulus $\beta$.

The requirement of bounded functions should by utilized in the proof. The proof is:

Proof: If $f(x) \leq g(x)$ for all $ x \in X$, we write $f \leq g$. For any $f,g \in B(X) $ $f \leq g + || f - g||$. Then properties (a) and (b) imply that:

$Tf \leq T(g + || f - g||) \leq Tg + \beta || f - g||$.

Reversing the roles of $f$ and $g$ gives by the same logic

$T g \leq Tf + \beta || f - g ||$.

Combing these two inequalities, we find that $||Tf - Tg|| \leq \beta || f -g ||$, as was to be shown.

Can somebody point out which statement in the proof requires the functions $f,g$ to be bounded?

If $f$ or $g$ is unbounded the sup norm $\|f-g\|$ could be infinite.