If I remember correction from my abstract algebra course, if $f(x)$ is defined for all x and is bounded, then composition mapping $f\cdot g$ is also bounded, and so should $g\cdot f$ since the range of $f(x)$ will become the domain for $g(x)$.
I was given this question. I understand what it is asking for. I just go not know how to prove it. The question reads:
Suppose $f(x)$ and $g(x)$ are defined for all x. Prove if true; give a counterexample if false:
(a) $f(x)$ bounded $\rightarrow$ $f(g(x))$ bounded.
(b) $f(x)$ bounded $\rightarrow$ $g(f(x))$ bounded.
I want to go about the proof by saying, let $f(x)$ be defined for all x $\in$ $R$ and bounded above and below by some inf, $i$, and sup, $s$. Let $g(x)$ be defined for all x $\in$ $R$ and not bounded.
Part A. $f \cdot g$ implies the range of $g$ will become the domain of $f$. Since $f(x)$ is bounded for all x $\in$ $R$, and the range of $g(x)$ is the $R$. Then $f(R)$ is also bounded.
Part B. $g \cdot f$ implies the range of $f$ will become the domain of $g$. Since $f(x)$ is bounded above and below by some inf and sup, such that the interval, $I$, becomes (inf $\leq$ f(x) $\leq$ sup), and $g(x)$ is not bounded. Then $g(f(x))$ implies it is also bounded.
This is my idea for out to prove it. This is not the final proof. More of a rough draft of a proof.
Thanks for taking the time to read this, and thanks in advanced for your feedback.
You are thinking of continuous functions, as in the second case you argue that a bounded set $[i,s]$ is mapped by $g$ to a bounded set $g([i,s])$. But if you do not require $g$ to be continuous, the second implication need not to be true: Let $g \colon \def\R{\mathbb R}\R\to \R$ given by $$ g(x) = \begin{cases} \frac 1x & x \ne 0 \\ 0 & x= 0\end{cases} $$ and $\def\abs#1{\left|#1\right|}f(x) =\exp(-\abs x)$ for all $x \in \mathbb R$, so $f$ is bounded by $1$. Then for each $x \in \R$,
$$(g \circ f)(x) = \frac 1{\exp(-\abs x)} = \exp(\abs x), $$ which is unbounded. $(g \circ f)(\R) = [1,\infty)$ to be explicit.
For the other case, your argumentation is fine, as $(f \circ g)(\R) \subseteq f(\R)$ and the latter is bounded, the former is also bounded.