<p>Let <span class="math-container">$X$</span> be a nonempty set, <span class="math-container">$B(X)$</span> be the space of bounded functions from <span class="math-container">$X$</span> to <span class="math-container">$\mathbb{R}$</span>, and <span class="math-container">$\|\cdot\|_\infty$</span> the sup-norm on <span class="math-container">$B(X)$</span>. Write <span class="math-container">$f\leq g$</span> for <span class="math-container">$f,g\in B(X)$</span> if <span class="math-container">$f(x)\leq g(x)$</span> for all <span class="math-container">$x\in X$</span>. Let <span class="math-container">$\mathbf{1}\in B(x)$</span> be the function with constant value <span class="math-container">$1$</span>.</p>
<p>Here is the result of Blackwell:</p>
<blockquote>
<p><strong>Theorem:</strong> Let <span class="math-container">$T:B(X)\to B(X)$</span> be a function such that for some <span class="math-container">$\beta\in(0,1)$</span></p>
<ol>
<li>Whenever <span class="math-container">$f,g\in B(X)$</span> satisfy <span class="math-container">$f\leq g$</span>, then <span class="math-container">$T(f)\leq T(g)$</span> (monotonicity).</li>
<li>For all <span class="math-container">$f\in B(X)$</span> and <span class="math-container">$\alpha\geq 0$</span> one has <span class="math-container">$T(f+\alpha\mathbf{1})\leq T(f)+\alpha\beta\mathbf{1}$</span> (discounting).</li>
</ol>
<p>Then, <span class="math-container">$\|T(f)-T(g)\|_\infty\leq \beta\|f-g\|_\infty$</span> for all <span class="math-container">$f,g\in B(X)$</span>.</p>
</blockquote>
<p>The following argument should make things a bit more transparent. Take <span class="math-container">$\alpha\geq 0$</span> large enough that both <span class="math-container">$f\leq g+\alpha\mathbf{1}$</span> and <span class="math-container">$g\leq f+\alpha\mathbf{1}$</span> holds. Since both <span class="math-container">$f$</span> and <span class="math-container">$g$</span> are bounded, this is certainly possible.</p>
<p>By monotonicity,
<span class="math-container">$$T(f)\leq T\big(g+\alpha\mathbf{1}\big).$$</span>
By discounting,
<span class="math-container">$$T(f)\leq T(g)+\beta\alpha\mathbf{1}.$$</span>
Therefore, <span class="math-container">$$T(f)-T(g)\leq \beta\alpha\mathbf{1}.$$</span>
By switching <span class="math-container">$f$</span> and <span class="math-container">$g$</span>, we get
<span class="math-container">$$T(g)-T(f)\leq \beta\alpha\mathbf{1}.$$</span>
It follows that
<span class="math-container">$$|T(g)(x)-T(f)(x)|\leq \beta\alpha$$</span>
for all <span class="math-container">$x\in X$</span>,
which implies
<span class="math-container">$$\|T(f)-T(g)\|_\infty\leq \beta\alpha.$$</span>
To finish the proof, simply note that we can choose <span class="math-container">$\alpha=\|f-g\|_\infty$</span>.</p>
<p>Here, monotonicity guarantees that <span class="math-container">$T$</span> preserves the pointwise ordering of functions, and discounting guarantees that if we increase every coordinate of a function by the same amount, the resulting value of <span class="math-container">$T$</span> increases at most <span class="math-container">$\beta$</span> times this amount.</p>
<p>We want to use discounting even for two functions in which neither equals the other plus a constant amount. Monotonicity guarantees us that if one function is bounded above by another function plus a constant value, a similar bound can be used for the transformed functions under <span class="math-container">$T$</span>. This makes discounting much more powerful.</p>

CC BY-SA 4.0
2021-02-22T16:50:49.650Z
Thank you Michael, much appreciated.
EB3112
CC BY-SA 4.0
2021-02-22T15:58:58.047Z
2021-02-22T15:58:58.047Z
Michael Greinecker
<p>I wondered if anyone could throw some intuition on Blackwell's sufficiency conditions:</p>
<p>(1) Monotononicty;
(2) Discounting;</p>
<p>Would be genuinely appreciated to hear both terms expressed in plain English.</p>

<mathematical-economics>
Blackwell's Sufficiency Conditions for a Contraction:
EB3112
# Blackwell's Sufficiency Conditions for a Contraction:

(Asked by: username_0 on 2021-02-21)

I wondered if anyone could throw some intuition on Blackwell's sufficiency conditions:

(1) Monotononicty;
(2) Discounting;

Would be genuinely appreciated to hear both terms expressed in plain English.

**Tags:** mathematical-economics


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## 1 answers

### Answer by username_1 on 2021-02-22. Score: 5 | Accepted answer

Let $X$ be a nonempty set, $B(X)$ be the space of bounded functions from $X$ to $\mathbb{R}$, and $\|\cdot\|\_\infty$ the sup-norm on $B(X)$. Write $f\leq g$ for $f,g\in B(X)$ if $f(x)\leq g(x)$ for all $x\in X$. Let $\mathbf{1}\in B(x)$ be the function with constant value $1$.

Here is the result of Blackwell:

> **Theorem:** Let $T:B(X)\to B(X)$ be a function such that for some $\beta\in(0,1)$
>
> 1. Whenever $f,g\in B(X)$ satisfy $f\leq g$, then $T(f)\leq T(g)$ (monotonicity).
> 2. For all $f\in B(X)$ and $\alpha\geq 0$ one has $T(f+\alpha\mathbf{1})\leq T(f)+\alpha\beta\mathbf{1}$ (discounting).
>
> Then, $\|T(f)-T(g)\|\_\infty\leq \beta\|f-g\|\_\infty$ for all $f,g\in B(X)$.

The following argument should make things a bit more transparent. Take $\alpha\geq 0$ large enough that both $f\leq g+\alpha\mathbf{1}$ and $g\leq f+\alpha\mathbf{1}$ holds. Since both $f$ and $g$ are bounded, this is certainly possible.

By monotonicity,
$$T(f)\leq T\big(g+\alpha\mathbf{1}\big).$$
By discounting,
$$T(f)\leq T(g)+\beta\alpha\mathbf{1}.$$
Therefore, $$T(f)-T(g)\leq \beta\alpha\mathbf{1}.$$
By switching $f$ and $g$, we get
$$T(g)-T(f)\leq \beta\alpha\mathbf{1}.$$
It follows that
$$|T(g)(x)-T(f)(x)|\leq \beta\alpha$$
for all $x\in X$,
which implies
$$\|T(f)-T(g)\|\_\infty\leq \beta\alpha.$$
To finish the proof, simply note that we can choose $\alpha=\|f-g\|\_\infty$.

Here, monotonicity guarantees that $T$ preserves the pointwise ordering of functions, and discounting guarantees that if we increase every coordinate of a function by the same amount, the resulting value of $T$ increases at most $\beta$ times this amount.

We want to use discounting even for two functions in which neither equals the other plus a constant amount. Monotonicity guarantees us that if one function is bounded above by another function plus a constant value, a similar bound can be used for the transformed functions under $T$. This makes discounting much more powerful.

> **Comments:**
> **username_0**: Thank you Michael, much appreciated.  

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