Application of Banach's theorem

Question

For the past few days I've been studying analysis/metric spaces and right now I'm busy with Banach's theorem. I was making an old exam question which I have no clue how to begin solving it. It goes as follows: Using Banach's theorem we need to proof that there exists a unique bounded (and continuous) function $f: \mathbb{R} \rightarrow \mathbb{R}$ for which the following holds, $$\forall x \in \mathbb{R}: f(x) = \frac{1}{2+|f(x+1)|} +3cos(x)$$ I do not have any clue how to start with this problem, the only thing I did was come up with a condition for a fixed point since this is a crucial part of Banach's theorem. Does anyone know how to start with this problem, if so any help would be greatly appreciated :))

Answer 1

Let $B := B(\mathbb{R}, \mathbb{R})$ be the space of bounded functions equipped with the sup norm, and let $\mathcal{C_b} := \mathcal{C}_b(\mathbb{R}, \mathbb{R})$ be the space of continuous bounded functions equipped with the sup norm as well.
Consider then $T_1 : B \to B$ and $T_2 : \mathcal{C}_b \to \mathcal{C}_b$ defined by: $$T_k(f) = \left(x \mapsto \frac{1}{2 + |f(x+1)|} + 3\cos x\right)$$ for $f$ in the respective domains of $T_1$ or $T_2$.
Both $B$ and $\mathcal{C_b}$ are complete w.r.t. their norm. Moreover, due to the continuity and boundedness of $\cos$, as well as the fact that $0 \leq \frac{1}{2 + |f(x+1)|} \leq \frac{1}{2}$ and that $t \mapsto \frac{1}{t}$ is continuous on $\{t \geq 2\}$, $T_1$ and $T_2$ are well-defined, so there only remains to check that they are contractions.

Let $(f,g) \in B^2$. Then, for all $x \in \mathbb{R}$ : $$\begin{split}\left|T_k(f)(x) - T_k(g)(x)\right| &= \left|\frac{1}{2 + |f(x+1)|} - \frac{1}{2 + |g(x+1)|}\right|\\ &= \frac{\Big||g(x+1) - |f(x+1)|\Big|}{(2 + |f(x+1)|)(2 + |g(x+1)|)}\\ &\leq \frac{|g(x+1) - f(x+1)|}{4}\\ &\leq \frac{{\|f - g\|}_\infty}{4}\end{split}$$ where the second-to-last step uses the second triangle inequality (or reverse triangle inequality? I do not remember how it's called in English) and the fact that the denominator is bigger than $(2 + 0)(2 + 0)= 4$.
Hence, taking the sup: $${\left\|T_k(f) - T_k(g)\right\|}_\infty \leq \frac{1}{4} {\|f - g\|}_\infty$$ Thus both $T_1$ and $T_2$ are contractions as $\frac{1}{4} < 1$, and satisfy the conditions of the Banach-Picard fixed point theorem, meaning that both of them admit respective unique fixed points $f_1 \in B$ and $f_2 \in \mathcal{C}_b$.
But since $\mathcal{C_b} \subset B$, $f_2$ is also a fixed point of $T_1$, hence by uniqueness $f_1 = f_2 =: f^*$, which means that $f^*$ is unique among all bounded functions, while being continuous.

You'll notice that any continuous bounded function could have replaced $3\cos$, as it did not play a role in determining whether $T_2$ was a contraction other than having these properties.