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


1 Answer 1


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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