<p>Consider a contraction mapping on the space of (say) all bounded continuous functions. Let <span class="math-container">$A$</span> be the set bounded continuous and increasing functions: <span class="math-container">$V \in A$</span> if for all <span class="math-container">$w \le w', V(w) \le V(w')$</span> and let <span class="math-container">$B$</span> be the set of all bounded continuous and strictly increasing functions: <span class="math-container">$V \in B$</span> if for all <span class="math-container">$w &lt; w', V(w) &lt; V(w')$</span>. Note that the set <span class="math-container">$A$</span> is closed: a convergent sequence of increasing bounded and continuous functions converges to an increasing bounded and continuous function (here convergence is in the sup metric). Also note that <span class="math-container">$B \subseteq A$</span>. Note that <span class="math-container">$B$</span> is not closed.</p>
<p>The following holds:</p>
<p><strong>Lemma</strong> If for all <span class="math-container">$V \in A$</span>, <span class="math-container">$TV \in B$</span> then the fixed point of <span class="math-container">$T$</span> will be in <span class="math-container">$B$</span>.</p>
<blockquote>
<p><em>Proof:</em> As <span class="math-container">$T$</span> is a contraction mapping, the fixed point can be found |by taking any bounded and continuous function <span class="math-container">$V$</span> and generating the sequence <span class="math-container">$T V, T^2 V, T^3 V, \ldots$</span> By the contraction mapping theorem, this sequence converges (in the sup metric) to the unique fixed point of <span class="math-container">$T$</span>, i.e. the function <span class="math-container">$V^\ast$</span> whith <span class="math-container">$TV^\ast = V^\ast$</span>.</p>
<p>Now, start at any <span class="math-container">$V \in A$</span>, then <span class="math-container">$T V \in B \subseteq A$</span>, <span class="math-container">$T^2 V \in B \subseteq A$</span> and so on. So every element in the sequence <span class="math-container">$V, TV, T^2 V, \ldots$</span> belongs to <span class="math-container">$A$</span> and as <span class="math-container">$A$</span> is closed the limit, say <span class="math-container">$V^\ast$</span> (which is the fixed point) will also be in <span class="math-container">$A$</span>. But then <span class="math-container">$V^\ast = TV^\ast \in B$</span>.</p>
</blockquote>
<p>The Bellman operator is given by:
<span class="math-container">$$
T V(w) = w + \beta\left(\lambda_1(\int \max[V(w), V(\tilde x)] d F(\tilde x)) + \delta[V_0]\right).
$$</span>
Assume you already have shown that <span class="math-container">$T$</span> is a contraction mapping on (for example) the set of continuous and bounded functions from <span class="math-container">$\mathbb{R}$</span> to <span class="math-container">$\mathbb{R}$</span>).</p>
<p>Let <span class="math-container">$V$</span> be an increasing function <span class="math-container">$V$</span>, i.e. <span class="math-container">$w \le w'$</span> implies <span class="math-container">$V(w) \le V(w')$</span>. We need to show that <span class="math-container">$TV$</span> is strictly increasing: if <span class="math-container">$w &lt; w'$</span>, then <span class="math-container">$TV(w) &lt; TV(w')$</span>.</p>
<p>We have,
<span class="math-container">$$
\begin{align*}
TV(w) &amp;= \underbrace{w}_{&lt; w'} + \beta\left(\lambda_1 (\int\max[\underbrace{V(w)}_{\le V(w')}, V(\tilde x)] dF(\tilde x)) + \delta [V_0]\right),\\
&amp;&lt; w' + \beta\left(\lambda_1 (\int\max[V(w'), V(\tilde x)] dF(\tilde x)) + \delta [V_0]\right),\\
&amp;= TV(w')
\end{align*}
$$</span></p>

CC BY-SA 4.0
2024-11-07T15:33:16.357Z
"To show that the fixed point of   is stictly increasing, it suffices to show that for all ∈ the function  is strictly increasing" How does this work? I want to show that V is strictly increasing. So how do we start off by saying take an increasing function V? Sorry but I am missing something here.
user584534
CC BY-SA 4.0
2024-11-07T16:24:28.500Z
I modified my response. Hope this helps.
tdm
CC BY-SA 4.0
2024-11-08T00:41:33.697Z
Thank you! Yes this is really helpful
user584534
CC BY-SA 4.0
2024-11-07T07:19:53.747Z
2024-11-07T16:24:09.930Z
2024-11-07T16:24:09.930Z
tdm
<p>I have a Bellman equation for employed worker in a setting where job to job transition is allowed. I want to show that V(w), the value function, is strictly increasing in w. In standard Bellman equation I know of a theorem that says if U and constraint correspondence G are monotone then V is also monotone. But what is the contraint correspondence here? Below is the BE. I can see that w is strictly increasing. But I do not know how to explain the second requirement.</p>
<p><span class="math-container">$V_1()=+\beta(_1(∫max[{_1(),_1(̃)}](̃))+[_0]]$</span></p>

<dynamic-programming><search-and-matching><bellman-equations>
Monotonicity of value function with on the job search
user584534
# Monotonicity of value function with on the job search

(Asked by: username_0 on 2024-11-07)

I have a Bellman equation for employed worker in a setting where job to job transition is allowed. I want to show that V(w), the value function, is strictly increasing in w. In standard Bellman equation I know of a theorem that says if U and constraint correspondence G are monotone then V is also monotone. But what is the contraint correspondence here? Below is the BE. I can see that w is strictly increasing. But I do not know how to explain the second requirement.

$V\_1()=+\beta(\_1(∫max[{\_1(),\_1(̃)}](̃))+[\_0]]$

**Tags:** dynamic-programming, search-and-matching, bellman-equations

---

## 1 answers

### Answer by username_1 on 2024-11-07. Score: 3 | Accepted answer

Consider a contraction mapping on the space of (say) all bounded continuous functions. Let $A$ be the set bounded continuous and increasing functions: $V \in A$ if for all $w \le w', V(w) \le V(w')$ and let $B$ be the set of all bounded continuous and strictly increasing functions: $V \in B$ if for all $w < w', V(w) < V(w')$. Note that the set $A$ is closed: a convergent sequence of increasing bounded and continuous functions converges to an increasing bounded and continuous function (here convergence is in the sup metric). Also note that $B \subseteq A$. Note that $B$ is not closed.

The following holds:

**Lemma** If for all $V \in A$, $TV \in B$ then the fixed point of $T$ will be in $B$.

> *Proof:* As $T$ is a contraction mapping, the fixed point can be found |by taking any bounded and continuous function $V$ and generating the sequence $T V, T^2 V, T^3 V, \ldots$ By the contraction mapping theorem, this sequence converges (in the sup metric) to the unique fixed point of $T$, i.e. the function $V^\ast$ whith $TV^\ast = V^\ast$.
>
> Now, start at any $V \in A$, then $T V \in B \subseteq A$, $T^2 V \in B \subseteq A$ and so on. So every element in the sequence $V, TV, T^2 V, \ldots$ belongs to $A$ and as $A$ is closed the limit, say $V^\ast$ (which is the fixed point) will also be in $A$. But then $V^\ast = TV^\ast \in B$.

The Bellman operator is given by:
$$
T V(w) = w + \beta\left(\lambda\_1(\int \max[V(w), V(\tilde x)] d F(\tilde x)) + \delta[V\_0]\right).
$$
Assume you already have shown that $T$ is a contraction mapping on (for example) the set of continuous and bounded functions from $\mathbb{R}$ to $\mathbb{R}$).

Let $V$ be an increasing function $V$, i.e. $w \le w'$ implies $V(w) \le V(w')$. We need to show that $TV$ is strictly increasing: if $w < w'$, then $TV(w) < TV(w')$.

We have,
$$
\begin{align\*}
TV(w) &= \underbrace{w}\_{< w'} + \beta\left(\lambda\_1 (\int\max[\underbrace{V(w)}\_{\le V(w')}, V(\tilde x)] dF(\tilde x)) + \delta [V\_0]\right),\\
&< w' + \beta\left(\lambda\_1 (\int\max[V(w'), V(\tilde x)] dF(\tilde x)) + \delta [V\_0]\right),\\
&= TV(w')
\end{align\*}
$$

> **Comments:**
> **username_0**: "To show that the fixed point of   is stictly increasing, it suffices to show that for all ∈ the function  is strictly increasing" How does this work? I want to show that V is strictly increasing. So how do we start off by saying take an increasing function V? Sorry but I am missing something here.  
> **username_1**: I modified my response. Hope this helps.  
> **username_0**: Thank you! Yes this is really helpful  

---

