Composite bounded functions

Question

Prove $f(x)$ is bounded $\rightarrow$ that $f(g(x))$ is bounded. For all x in $f(x)$ ang $g(x)$.

To my understanding, suppose $f(x)$ is bounded, then do we need to show that the composition function of $f(g(x))$ is bounded by proving $g(x)$ is bounded as well.

Answer 1

No, $g(x)$ does not need to be bounded. If $f(x)$ is bounded on its entire domain, then restricting its domain to $g(x)$ does not 'unbound' the function.

Answer 2

Not at all. In fact, the boundedness (or not) of $g$ has nothing to do with anything. You need to show that there is some $M$ such that for all $x$ in the domain of $f\circ g,$ we have $|f(g(x))|\le M$. Use the fact that $f$ is bounded to prove it.

Answer 3

With $Im()$ and $Dom()$ I will mean the domain of a function. Obviously $Im(g)\cap Dom(f)\subseteq Dom(f)$, so if $|f(x)|\le M$ for all $x\in Dom(f)$, then $|(f\circ g)(x)|=|f(g(x))|\le M$.