Proof monotonicity on Blackwell sufficient conditions

Question

I need to prove monotonicity assumption on Blackwell’s sufficient conditions for a contraction, that is:

Given the operator T defined as

$(Tf)(x) = sup [F(x,y)+bf(y)]$

I need to show that $f\leq g\Rightarrow Tf\leq Tg$

My first approach is:

$F(x,y) + bf(y)\leq F(x,y) + bg(y)$

$sup F(x,y) + bf(y)\leq sup F(x,y) + bg(y)$

$T(f)(y)\leq T(g)(y)$

Am I missing something? Is this proof too simple to fully demonstrate the statement above?

Appreciate your help.

Is your $sup$ over $x$ or $y$? Notations in your first eq. are different from the last eq. — Bertrand, Mar 18 '21 at 09:11

score 1 · Answer 1 · answered May 11 '22 at 17:48

Blackwell's sufficiency theorem requires (1) Monotonicity and (2) Discounting.

Checked my notes from the first year macro, and this is what I have:

Theorem Blackwell's Sufficient Conditions for a contraction: Let $X\subseteq R^l$ and $B(X)$ be the space of bounded functions $f:X \rightarrow R$ with metric $d(f,g) = \sup_{x\in X} |f(x)-g(x)|$.

Let $T: B(X)\rightarrow B(X)$ be an operator satisfying:

Monotonicity: If $f,g\in B(X)$ and $f(x) \leq g(x)\ \forall x\in X$, then $(Tf) (x) \leq (\geq) (Tg)(x) \ \forall x \in X$.
Discounting: Let the function $f+a$, for $f\in B(X)$ and $a\in R_{+}$ be defined by $(f+a)(x)=f(x)+a$ (i.e., for all $x$ the number $a$ is added to $f(x)$). There exists $\beta\in(0,1)$ such that for all $f\in B(X)$, $a \geq 0$ and $x\in X$, $$[T(f+a)](x)\leq [Tf](x)+\beta a$$

If 1 and 2 hold, then the operator $T$ is a contraction with modulus $\beta$.

Proof If $f(x)\leq g(x)$ for all $x\in X$, we write $f\leq g$. For any $f\in B(X)$ and any $x\in X$

$$f(x)-g(x) \leq \sup_{x\in X}|f(x)-g(x)|=d(f,g)$$

since it holds for all $x$ (which also holds by boundness of $f$)

$$f\leq g +d(f,g)$$

By monotonicity

$$Tf \leq T(g+d(f,g))$$

and notice that $d(f,g)=a\geq 0$ is a constant.

By discounting

$$T(g+d(f,g)) \leq Tg+\beta d(f,g).$$

So,

$$Tf \leq Tg +\beta d)f,g) \implies Tf-Tg \leq \beta d(f,g).$$

Reversing the roles of $f$ and $g$, we have

$$Tg \leq Tf + \beta d(f,g) \implies Tg - Tf \leq \beta d(f,g).$$

Combining gives

$$(Tf)(x) - (Tg)(x) \leq \beta d(f,g)$$ $$(Tg)(x) - (Tf) (x) \leq beta d(f,g)$$

for all $x\in X$. Thus,

$$\sup_{x\in X} | (Tf)(x) - (Tg)(x)| = d(Tf, Tg) \leq \beta d(f,g).$$

QED

Proof monotonicity on Blackwell sufficient conditions

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