# Positive Monotonic Transformations and Nested Functions

Suppose there is an economic agent with the utility function $u(x,y)$. A second agent has the utility function $h(g(f(u(x,y))))$.

Am I correct in thinking that if $f'(x)>0$, $g'(x)>0$, and $h'(x)>0$, then $h(g(f(u(x,y))))$ must also be positive and therefore $h(g(f(u(x,y))))$ is a positive monotonic transformation of $u(x,y)$?

If not, how come?

The value of $h(g(f(u(x,y))$ is not necessarily positive but the transformation $u \rightarrow h \circ g \circ f \circ u$ is indeed a positive monotone transformation.

To see this, take any $(x,y,z,w)$. We have the following chain of equivalences: \begin{align} u(x,y) \geq u(z,w) & \Leftrightarrow f(u(x,y)) \geq f(u(z,w)) \text{ since } f \text{ is increasing} \\ & \Leftrightarrow g(f(u(x,y)) \geq g(f(u(z,w)) \text{ since } g \text{ is increasing} \\ & \Leftrightarrow h(g(f(u(x,y))) \geq h(g(f(u(z,w))) \text{ since } h \text{ is increasing} \end{align} Thus, the transformation $u \rightarrow h \circ g \circ f \circ u$ preserves the ordering, i.e. it is a positive monotone transformation.

Another way to see it is to assume that all functions are differentiable and notice that \begin{equation*} \frac{d (h \circ g \circ f )(x)}{dx}= f'(x) g'(f(x)) h'(g(f(x)) \end{equation*} which is positive since $f'>0,g'>0,h'>0$.