Blackwell's condition for a contraction: Why is boundedness neccessary?
(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.
In this post, I am more concerned about the definition of $T: B(X)\to B(X)$.
$B(X)$ is the space of bounded functions, so the elements in $B(X)$ should be functions, each element\function is uniquely determined by how ALL $x\in X$ is associated to $\mathbb{R}$, meaning it should in space $\mathbb{R}^X$.
However, $f(x),g(x)\in\mathbb{R}$ but not in $\mathbb{R}^X$, why $T$ define like this ?
why isn't $T$ defined as from $\mathbb{R}$ to $\mathbb{R}$.
[see warning of composition function][2]
[2]: https://proofwiki.org/wiki/Definition:Composition_of_Mappings requires codomain equals to the domain of anthother.
You're not composing $f$ by a function $T : \mathbb{R} \to \mathbb{R}$ here, you're applying an operator $T : B(X) \to B(X)$ to a function $f \in B(X)$, and the image is a function $T(f) \in B(X)$.
You can check that it's coherent with the conditions of the theorem as you're looking at $(Tf)(x) = (T(f))(x)$ and not at $T(f(x))$.