<p>Blackwell's sufficiency theorem requires (1) Monotonicity and (2) Discounting.</p>
<p>Checked my notes from the first year macro, and this is what I have:</p>
<p><strong>Theorem Blackwell's Sufficient Conditions for a contraction</strong>:
Let <span class="math-container">$X\subseteq R^l$</span> and <span class="math-container">$B(X)$</span> be the space of bounded functions <span class="math-container">$f:X \rightarrow R$</span> with metric <span class="math-container">$d(f,g) = \sup_{x\in X} |f(x)-g(x)|$</span>.</p>
<p>Let <span class="math-container">$T: B(X)\rightarrow B(X)$</span> be an operator satisfying:</p>
<ol>
<li><em>Monotonicity</em>: If <span class="math-container">$f,g\in B(X)$</span> and <span class="math-container">$f(x) \leq g(x)\  \forall x\in X$</span>, then <span class="math-container">$(Tf) (x) \leq (\geq) (Tg)(x) \  \forall x \in X$</span>.</li>
<li><em>Discounting</em>: Let the function <span class="math-container">$f+a$</span>, for <span class="math-container">$f\in B(X)$</span> and <span class="math-container">$a\in R_{+}$</span> be defined by <span class="math-container">$(f+a)(x)=f(x)+a$</span> (i.e., for all <span class="math-container">$x$</span> the number <span class="math-container">$a$</span> is added to <span class="math-container">$f(x)$</span>). There exists <span class="math-container">$\beta\in(0,1)$</span> such that for all <span class="math-container">$f\in B(X)$</span>, <span class="math-container">$a \geq 0$</span> and <span class="math-container">$x\in X$</span>,
<span class="math-container">$$[T(f+a)](x)\leq [Tf](x)+\beta a$$</span></li>
</ol>
<p>If 1 and 2 hold, then the operator <span class="math-container">$T$</span> is a contraction with modulus <span class="math-container">$\beta$</span>.</p>
<p><em><strong>Proof</strong></em>
If <span class="math-container">$f(x)\leq g(x)$</span> for all <span class="math-container">$x\in X$</span>, we write <span class="math-container">$f\leq g$</span>. For any <span class="math-container">$f\in B(X)$</span> and any <span class="math-container">$x\in X$</span></p>
<p><span class="math-container">$$f(x)-g(x) \leq \sup_{x\in X}|f(x)-g(x)|=d(f,g)$$</span></p>
<p>since it holds for all <span class="math-container">$x$</span> (which also holds by boundness of <span class="math-container">$f$</span>)</p>
<p><span class="math-container">$$f\leq g +d(f,g)$$</span></p>
<p>By monotonicity</p>
<p><span class="math-container">$$Tf \leq T(g+d(f,g))$$</span></p>
<p>and notice that <span class="math-container">$d(f,g)=a\geq 0$</span> is a constant.</p>
<p>By discounting</p>
<p><span class="math-container">$$T(g+d(f,g)) \leq Tg+\beta d(f,g).$$</span></p>
<p>So,</p>
<p><span class="math-container">$$Tf \leq Tg +\beta d)f,g) \implies Tf-Tg \leq \beta d(f,g).$$</span></p>
<p>Reversing the roles of <span class="math-container">$f$</span> and <span class="math-container">$g$</span>, we have</p>
<p><span class="math-container">$$Tg \leq Tf + \beta d(f,g) \implies Tg - Tf \leq \beta d(f,g).$$</span></p>
<p>Combining gives</p>
<p><span class="math-container">$$(Tf)(x) - (Tg)(x) \leq \beta d(f,g)$$</span>
<span class="math-container">$$(Tg)(x) - (Tf) (x) \leq beta d(f,g)$$</span></p>
<p>for all <span class="math-container">$x\in X$</span>.
Thus,</p>
<p><span class="math-container">$$\sup_{x\in X} | (Tf)(x) - (Tg)(x)| = d(Tf, Tg) \leq \beta d(f,g).$$</span></p>
<p><em><strong>QED</strong></em></p>

CC BY-SA 4.0
2022-05-11T17:48:26.890Z
2022-05-11T17:48:26.890Z
Evangelos Con
<p>I need to prove monotonicity assumption on Blackwell’s sufficient conditions for a contraction, that is:</p>
<p>Given the operator T defined as</p>
<p><span class="math-container">$(Tf)(x) = sup [F(x,y)+bf(y)]$</span></p>
<p>I need to show that <span class="math-container">$f\leq g\Rightarrow Tf\leq Tg$</span></p>
<p>My first approach is:</p>
<p><span class="math-container">$F(x,y) + bf(y)\leq F(x,y) + bg(y)$</span></p>
<p><span class="math-container">$sup F(x,y) + bf(y)\leq sup F(x,y) + bg(y)$</span></p>
<p><span class="math-container">$T(f)(y)\leq T(g)(y)$</span></p>
<p>Am I missing something? Is this proof too simple to fully demonstrate the statement above?</p>
<p>Appreciate your help.</p>

CC BY-SA 4.0
2021-03-16T18:12:33.440Z
It seems fine assuming $b\geq 0$
Jesper Hybel
CC BY-SA 4.0
2021-03-18T09:11:36.627Z
Is your $sup$ over $x$ or $y$? Notations in your first eq. are different from the last eq.
Bertrand
<macroeconomics><mathematical-economics><recursive-macroeconomics>
Proof monotonicity on Blackwell sufficient conditions
vmp
# Proof monotonicity on Blackwell sufficient conditions

(Asked by: username_0 on 2021-03-16)

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.

**Tags:** macroeconomics, mathematical-economics, recursive-macroeconomics
> **Comments:**
> **username_1**: It seems fine assuming $b\geq 0$  
> **username_2**: Is your $sup$ over $x$ or $y$? Notations in your first eq. are different from the last eq.  

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## 1 answers

### Answer by username_3 on 2022-05-11. Score: 1

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:

1. *Monotonicity*: If $f,g\in B(X)$ and $f(x) \leq g(x)\ \forall x\in X$, then $(Tf) (x) \leq (\geq) (Tg)(x) \ \forall x \in X$.
2. *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***


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