How to prove $\sup(f \circ g(x))$ is less than $\sup(g(x))$

Question

I am being asked to show that $g: \mathbb{R} \to B$ is a function whose range $B$ is a bounded subset of $\mathbb{R}$, and that $f: \mathbb{R} \to \mathbb{R}$ is monotone increasing. If we let $u = \sup\{g(x): x \in \mathbb{R}\}$, how do we show that $\sup\{f \circ g)(x): x \in \mathbb{R}\} \leq f(u)$?

I have seen how to prove $$ \sup\bigl(f(x) + g(x)\bigr) \leq \sup\bigl(g(x)\bigr) + \sup\bigl(f(x)\bigr) $$ but not sure what to do here really. Anything helps!

Answer 1

Let $x\in\mathbb{R}$ be given. Since $g(x)\leq u$, $u<\infty$, and $f$ is increasing, we have $f(g(x))\leq f(u)$ and hence $(f\circ g)(x)\leq f(u)$. The arbitrariness of $x\in\mathbb{R}$ yields $\sup_{x\in\mathbb{R}}(f\circ g)(x)\leq f(u)$.