Does there exist an unique continuous bounded $f$ such that $f(x)= \frac{\sin(f(x))}{2+x^2} - \frac{\cos^2(x)}{1+e^x}$?

Question

Does there exist an unique continuous bounded $f$ such that $f(x)= \frac{\sin(f(x))}{2+x^2} - \frac{\cos^2(x)}{1+e^x}$?

I wanted to prove this by proving this is a strict contraction and than applying the fixed point theorem of Banach.

EDIT: My calculations so far: $|Tf(x) - Tg(x)| = |\frac{\sin(f(x))}{2+x^2} - \frac{\cos^2(x)}{1+e^x} - \frac{\sin(g(x))}{2+x^2} + \frac{\cos^2(x)}{1+e^x}| = |\frac{\sin(f(x))}{2+x^2} - \frac{\sin(g(x))}{2+x^2} | \leq |sin(f(x)) - sin (g(x))| $

But here i get stuck. I can see that because of the continuity of the sinus function : $|f(x)-g(x)|< \delta \rightarrow |sin(f(x)) - sin (g(x))|<\epsilon$. But I do not see how this can help me to make $|sin(f(x)) - sin (g(x))| \leq c |f(x) - g(x)|$.

Answer 1

You want to apply the Banach fixed point theorem to the space $C_b(\Bbb R)$ of bounded continuous functions with the supremum norm $\Vert \cdot \Vert_\infty$ and the operator $T: C_b(\Bbb R) \to C_b(\Bbb R)$ defined by $$ Tf(x) = \frac{\sin(f(x))}{2+x^2} - \frac{\cos^2(x)}{1+e^x} \, . $$

For $f, g \in C_b(\Bbb R)$ and all $x \in \Bbb R$ is $$ |Tf(x) - Tg(x)| = \frac{|\sin(f(x))-\sin(g(x))|}{2+x^2} \le \frac 12 |\sin(f(x))-\sin(g(x))| \\ \le \frac 12 |f(x) -g(x)| \le \frac 12 \Vert f-g \Vert_\infty \, . $$ It follows that $$ \Vert Tf - Tg \Vert_\infty \le 12 \Vert f-g \Vert_\infty $$ so that $T$ is a strict contraction.