<p>I am by no means an expert on this, but maybe this helps. Here is a simple example for a bellman equation</p>

<p>$V(y) = \max_x u(x,y) + \beta V(y')$</p>

<p>$s.t. \, y' = f(x,y)$</p>

<p>This is a functional equation in an unknown function V. A solution to this problem is a function V that satisfies the equation above. If you look at the equation, it's pretty clear that the solution has to be a fixed point of the operator on the RHS of the bellman equation: if you take the correct V and an arbitrary y and calculate</p>

<p>$\max_x u(x,y) + \beta V(y')$</p>

<p>$s.t. \, y' = f(x,y)$</p>

<p>you will get $V(y)$. The operator that is the RHS of the Bellman equation operates on functions, and the solution is a fixed point in some space of functions.</p>

<p>It's a different question whether this fixed point exists and how to find it. Here, you appeal to the contraction mapping theorem: under typical assumptions on u and provided $\beta&lt;1$, the maximization step above is a contraction mapping for any guess of V. This means that there exists a unique fixed point V, and you can find it by successive iteration.</p>

CC BY-SA 3.0
2017-01-26T00:51:27.463Z
Thanks, after reading your post I see that it it indeed a fixed point by definition. Yes, I just assumed, that utility function is bounded, etc., by default. I have found an [article](https://editorialexpress.com/cgi-bin/conference/download.cgi?db_name=AMES2014&paper_id=390)  that explains the point in more detail
dpd
CC BY-SA 3.0
2017-01-28T01:33:57.403Z
@due_to_revision if you're happy with the answer, you should consider accepting it, so the site doesn't keep your question in the 'Unanswered' queue.
Theoretical Economist
CC BY-SA 3.0
2017-01-26T00:33:38.763Z
2017-01-28T02:30:44.107Z
2017-01-28T02:30:44.107Z
Tobias
<p>I have recently started studying dynamic optimization. I cannot quite wrap my head around the fact that the value function of the Bellman equation is a fixed point of a contraction mapping.
As far my understanding is rather naive: if the problem is finite, say:
$$\sum_{t=0}^T \beta^tu(c_t)$$
we construct the Bellman equation from the end, as if we knew the maximum possible value of the sequence in advance. Starting from the last period $T$, we just repeat the maximization by adding an optimal term reflecting current period utility $u(c_t)$, until we arrive to the period $0$. From here I can clearly see how contraction mapping works.
But the infinite case is not so easy for me to comprehend: I can only suppose, that, by iteration of the Bellman operator $(Bv)(x)$, we perform a "calibration" of the policy function until we find the value function (i.e. the maximum possible utility given our transversality conditions) $(Bv)(x)=v(x)$. Am I, at least, thinking in the right direction, or this idea should be understood in a different way?
Thank you in advance.
(Also, this is my first question on .stackexchange ever, and if there are any issues with presentation of my question, please, let me know)</p>

<dynamic-programming><bellman-equations>
Solution to the Bellman equation is a fixed point
dpd
# Solution to the Bellman equation is a fixed point

(Asked by: username_0 on 2017-01-25)

I have recently started studying dynamic optimization. I cannot quite wrap my head around the fact that the value function of the Bellman equation is a fixed point of a contraction mapping.
As far my understanding is rather naive: if the problem is finite, say:
$$\sum\_{t=0}^T \beta^tu(c\_t)$$
we construct the Bellman equation from the end, as if we knew the maximum possible value of the sequence in advance. Starting from the last period $T$, we just repeat the maximization by adding an optimal term reflecting current period utility $u(c\_t)$, until we arrive to the period $0$. From here I can clearly see how contraction mapping works.
But the infinite case is not so easy for me to comprehend: I can only suppose, that, by iteration of the Bellman operator $(Bv)(x)$, we perform a "calibration" of the policy function until we find the value function (i.e. the maximum possible utility given our transversality conditions) $(Bv)(x)=v(x)$. Am I, at least, thinking in the right direction, or this idea should be understood in a different way?
Thank you in advance.
(Also, this is my first question on .stackexchange ever, and if there are any issues with presentation of my question, please, let me know)

**Tags:** dynamic-programming, bellman-equations

---

## 1 answers

### Answer by username_1 on 2017-01-26. Score: 1 | Accepted answer

I am by no means an expert on this, but maybe this helps. Here is a simple example for a bellman equation

$V(y) = \max\_x u(x,y) + \beta V(y')$

$s.t. \, y' = f(x,y)$

This is a functional equation in an unknown function V. A solution to this problem is a function V that satisfies the equation above. If you look at the equation, it's pretty clear that the solution has to be a fixed point of the operator on the RHS of the bellman equation: if you take the correct V and an arbitrary y and calculate

$\max\_x u(x,y) + \beta V(y')$

$s.t. \, y' = f(x,y)$

you will get $V(y)$. The operator that is the RHS of the Bellman equation operates on functions, and the solution is a fixed point in some space of functions.

It's a different question whether this fixed point exists and how to find it. Here, you appeal to the contraction mapping theorem: under typical assumptions on u and provided $\beta<1$, the maximization step above is a contraction mapping for any guess of V. This means that there exists a unique fixed point V, and you can find it by successive iteration.

> **Comments:**
> **username_0**: Thanks, after reading your post I see that it it indeed a fixed point by definition. Yes, I just assumed, that utility function is bounded, etc., by default. I have found an [article](https://editorialexpress.com/cgi-bin/conference/download.cgi?db_name=AMES2014&paper_id=390)  that explains the point in more detail  
> **username_2**: @due_to_revision if you're happy with the answer, you should consider accepting it, so the site doesn't keep your question in the 'Unanswered' queue.  

---

