Proof that composition of two unbounded functions is not a bounded function.

Question

I am supposed to answer the question if a composition of unbounded functions can be a bounded function. I have failed to find any counter example, so I should probably do a proof by contradiction but I don't know how. Could anyone help? Thanks!

Answer 1

The idea is to find functions $f:I\rightarrow \mathbf{R}$ and $g:J\rightarrow \mathbf{R}$ such that both are unbounded over their domain of definition but so that $g$ is bounded on $f(I)$. That is $g\vert_{f(I)}$ is bounded. Then $g(f(x))$ is bounded.

If $x>0$ then both $e^{x}$ and $1/x$ are unbounded. Can you compose them to get something bounded?

Answer 2

Define $f$ and $g$ by

$$ \begin{align*} f(x)) &=\left\{ \begin{array}{lr} x & \mbox{ if $x \ge 0$} \\ 0 & \mbox{ if $x < 0$} \end{array}\right. \\ g(x)) &=\left\{ \begin{array}{lr} 0 & \mbox{ if $x \ge 0$} \\ x & \mbox{ if $x < 0$} \end{array}\right. \end{align*} $$ Then neither $f$ nor $g$ is bounded, but $f \circ g$ is identically $0$.