I already proved that the space $C_b ([0,\infty)) = \{ f: [0,\infty) \to \mathbb{R} | f $ is continuous and limited $ \} $ is a Banach space with the norm $\|f\| = \sup |f| $. Now I need to prove that $\forall g \in C_b ([0,\infty))$ there exists an unique $f \in C_b ([0,\infty))$ such that:
$$ f(t) = g(t) + \int_{0}^{\infty} \frac{ f(s|\cos t|)}{1+s^2} ds, \forall t \in [0,\infty) $$
Here's my attempt: I'll show that the map $(Tf)(t) = g(t) + \int_{0}^{\infty} \frac{f(s|\cos t|)}{1+s^2} ds $ takes the space $C_b([0,\infty))$ in itself and it's a contraction. For the second part, my teacher gave the hint for using a theorem that I already proved saying that it is sufficient to prove that there is an $n$ such that $T^n$ is a contraction.
$(i)$ Let $f \in C_b ([0,\infty))$, show that $(Tf) \in C_b ([0,\infty))$
Let $B = \sup |f| $
Since $g \in C_b([0,\infty)) $ and the sum of limited functions is limited, then it is sufficient to show $\int_{0}^{\infty} \frac{f(s| \cos t|)}{1+s^2} ds$ is limited:
$\int_{0}^{\infty} \frac{f(s| \cos t|)}{1+s^2} ds \leq \int_{0}^{\infty} \frac{B}{1+s^2} ds = B \int_{0}^{\infty} \frac{1}{1+s²} ds = B \frac{\pi}{2}$
Then $(Tf) \in C_b([0,\infty))$.
$(ii)$ Prove $(Tf)$ is a contraction.
$d(Tf_1,Tf_2) = \| \int_{0}^{\infty} \frac{f_1(s| \cos t|)}{1+s^2} ds - \int_{0}^{\infty} \frac{f_2(s| \cos t|)}{1+s^2} ds \| = \| \int_{0}^{\infty} \frac{f_1(s| \cos t|) - f_2(s| \cos t|)}{1+s^2} ds \| $
But I'm not sure how to get the above to $ < \| f_1 - f_2 \|$ neither how to show $T^n$ is contraction for some $n$, which is probably easier.
Any help would be appreciated. Thanks.