<p>It follows from Assumption 4.10.</p>
<p>Let <span class="math-container">$\lambda y \in \Gamma(\lambda x)$</span> be the solution.</p>
<p>Let <span class="math-container">$\delta = \frac{1}{\lambda}$</span>.</p>
<p>By Assumption 4.10,</p>
<p><span class="math-container">$$
\lambda y \in\Gamma(\lambda x) 
$$</span>
implies
<span class="math-container">$$
\delta \lambda y \in \Gamma(\delta\lambda x)
$$</span>
in other words
<span class="math-container">$$
 y \in \Gamma(x)
$$</span></p>

CC BY-SA 4.0
2021-01-13T16:15:26.780Z
2021-01-13T16:15:26.780Z
Walrasian Auctioneer
<blockquote>
<p>Exercise 4.7 (b) : Show that under Assumptions 4.10 and 4.11, <span class="math-container">$T:H(X) \to H(X)$</span>.</p>
</blockquote>
<p><span class="math-container">$H(X)$</span> is the set of continuous and homogeneous of degree one functions and <span class="math-container">$Tf(x) = \sup_{y \in \Gamma(x)} \{F(x,y) + \beta f(y)\}$</span>. The continuity follows from the theorem of maximum. I am a little confused with showing <span class="math-container">$Tf$</span> is homogeneous of degree one. According to the solution,
<span class="math-container">\begin{equation}
\begin{split}
Tf(\lambda x)&amp; = \sup_{\lambda y \in \Gamma(\lambda x)} \{F(\lambda x,\lambda y) + \beta f(\lambda y)\}\\
&amp; = \lambda \sup_{y \in \Gamma(x)} \{F(x,y) + \beta f(y)\} \\
&amp; = \lambda Tf(x).
\end{split}
\end{equation}</span>
<span class="math-container">$\lambda$</span> can be taken out of functions because they are homogeneous of degree one. But, how can we justify the change from <span class="math-container">$\Gamma(\lambda x)$</span> to <span class="math-container">$\Gamma(x)$</span>? More precisely,
<span class="math-container">\begin{equation}
\begin{split}
Tf(\lambda x)&amp; = \sup_{\lambda y \in \Gamma(\lambda x)} \{F(\lambda x,\lambda y) + \beta f(\lambda y)\}\\
&amp; = \lambda \sup_{\lambda y \in \Gamma(\lambda x)} \{F(x,y) + \beta f(y)\} \\
&amp; = \lambda \sup_{y \in \Gamma( x)} \{F(x,y) + \beta f(y)\} .
\end{split}
\end{equation}</span>
I do understand the second equality, but I do not understand the third equality.</p>
<p>Assumption 4.10 and 4.11 are as follows:
<a href="https://i.sstatic.net/mEgvz.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/mEgvz.png" alt="enter image description here" /></a></p>

<macroeconomics><dynamic-programming>
Exercise 4.7 in SLP (dynamic programming)
shk910
# Exercise 4.7 in SLP (dynamic programming)

(Asked by: username_0 on 2021-01-13)

> Exercise 4.7 (b) : Show that under Assumptions 4.10 and 4.11, $T:H(X) \to H(X)$.

$H(X)$ is the set of continuous and homogeneous of degree one functions and $Tf(x) = \sup\_{y \in \Gamma(x)} \{F(x,y) + \beta f(y)\}$. The continuity follows from the theorem of maximum. I am a little confused with showing $Tf$ is homogeneous of degree one. According to the solution,
\begin{equation}
\begin{split}
Tf(\lambda x)& = \sup\_{\lambda y \in \Gamma(\lambda x)} \{F(\lambda x,\lambda y) + \beta f(\lambda y)\}\\
& = \lambda \sup\_{y \in \Gamma(x)} \{F(x,y) + \beta f(y)\} \\
& = \lambda Tf(x).
\end{split}
\end{equation}
$\lambda$ can be taken out of functions because they are homogeneous of degree one. But, how can we justify the change from $\Gamma(\lambda x)$ to $\Gamma(x)$? More precisely,
\begin{equation}
\begin{split}
Tf(\lambda x)& = \sup\_{\lambda y \in \Gamma(\lambda x)} \{F(\lambda x,\lambda y) + \beta f(\lambda y)\}\\
& = \lambda \sup\_{\lambda y \in \Gamma(\lambda x)} \{F(x,y) + \beta f(y)\} \\
& = \lambda \sup\_{y \in \Gamma( x)} \{F(x,y) + \beta f(y)\} .
\end{split}
\end{equation}
I do understand the second equality, but I do not understand the third equality.

Assumption 4.10 and 4.11 are as follows:
[![enter image description here](https://i.sstatic.net/mEgvz.png)](https://i.sstatic.net/mEgvz.png)

**Tags:** macroeconomics, dynamic-programming

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## 1 answers

### Answer by username_1 on 2021-01-13. Score: 3

It follows from Assumption 4.10.

Let $\lambda y \in \Gamma(\lambda x)$ be the solution.

Let $\delta = \frac{1}{\lambda}$.

By Assumption 4.10,

$$
\lambda y \in\Gamma(\lambda x)
$$
implies
$$
\delta \lambda y \in \Gamma(\delta\lambda x)
$$
in other words
$$
y \in \Gamma(x)
$$


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