Using the Banach fixed-point theorem, I can prove that there exists exactly one continuous function $f:[0,1]→\mathbb{R}$ that satisfies $$f(x) = \frac{1}{2}f(3x)+\cos x.$$
I currently don't know how to solve this $f$ precisely. My attempt is similar to Picard iteration. Define $$Tf = \frac{1}{2}f(3x)+\cos x.$$ Let $f_0 = 2$, and $f_{n+1} = Tf_n$. I can calculate that $$f_n = \frac{1}{2^{n-1}}\cos 3^{n-1}x + \cdots + \frac{1}{2}\cos 3x+ \cos x + \frac{1}{2^{n-1}}.$$ But what is the limit of this $f_n$? Thank you for any solutions or hints to this problem.
Edit: Adding the plots of $f_1,f_3$ and $f_5$ in the interval $x\in[-\pi,\pi]$, JL
