# Exercise 4.7 in SLP (dynamic programming)

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: Let $$\lambda y \in \Gamma(\lambda x)$$ be the solution.
Let $$\delta = \frac{1}{\lambda}$$.
$$\lambda y \in\Gamma(\lambda x)$$ implies $$\delta \lambda y \in \Gamma(\delta\lambda x)$$ in other words $$y \in \Gamma(x)$$