This is a question from a past exam I am trying to solve:
Let $C([a, b]; R)$ be the space of all continuous functions on $[a, b]$, $0 < a < b$ with the metric $∥f − g∥ = \max_{a≤x≤b} |f(x) − g(x)|$.
For each $g ∈ C([a, b]; R)$, define a map by: $$F(g(x)) = \frac{3}{(b^3-a^3)} \int_{a}^{x} t^3 g(t) dt$$ $∀x ∈ [a, b]$ Prove that there is a unique fixed point of F in the space $C([a, b]; R)$ (note $(C([a, b]; R), ∥ · ∥)$ is a complete metric space).
My attempt:
$$||F(f(x)) - F(g(x))| = \frac{3}{2(b^3-a^3)}|\int_{a}^{x}t^3 (f(x)-g(x))dt|$$ $$ \leq \frac{3(x-a)}{2(b^3-a^3)} *\max_{a≤t≤b}|t^3 (f(x)-g(x))|$$ $$ \leq \frac{3(b-a)}{2(b^3-a^3)} * b^3 *\max_{a≤t≤b}|f(x)-g(x)| $$ (this is as $x \leq b$) $$ \leq \frac{3(b-a)}{2(b^3-a^3)} * b^3 *||f-g||$$
so now I am stuck because I cant show that: $$0 \leq \frac{3(b-a)}{2(b^3-a^3)} * b^3 \leq 1$$
In fact I can see that it can be greater than 1.
Can someone point out my error?
Edit I've fixed up some of the typos