The map $T: C[0,2] \to C[0,2]$, $T(f) = \lambda\int_0^2 y^2 f(y)\,dy + 3 \lambda f(x/4) + x$ is a contraction for $|\lambda| \leq \frac{3}{17}$ and $C[0,2]$ with the supremum metric is a complete metric space.
By the Contraction Mapping Theorem, I know $T$ has a unique fixed point. For fixed $\lambda = \frac{1}{6}$ how should I go about finding the fixed point?
In class notes, I have a similar example where they use the substitution $f(x) = Ax + B$ and compare coefficients. Why does this approach work (ie. why is the fixed point $f\in C[0,2]$ a linear function?
Suppose $Tf=f$. Then $$f(x)=\lambda A+3\lambda f\left(\frac{x}{4}\right)+x$$ where $A=\int_0^2 y^2f(y)dy.$ This shows that $$f(x)-3\lambda f\left(\frac{x}{4}\right)=\lambda A+x.$$ Since $f$ is continuous, if $|\lambda|<\frac13$, we see that $$f(x)=\sum_{k=0}^\infty (3\lambda)^k\Biggl(f\left(\frac{x}{4^k}\right)-3\lambda f\left(\frac{x}{4^{k+1}}\right)\Biggr)=\sum_{k=0}^\infty(3\lambda)^k\left(\lambda A+\frac{x}{4^k}\right).$$ Therefore $$f(x)= \frac{\lambda A}{1-3\lambda }+\frac{x}{1-\frac{3\lambda }{4}}.$$ That is $f$ is linear. Since $A=\int_0^2 y^2 f(y)dy$, we have $$A=\int_0^2 y^2\left(\frac{\lambda A}{1-3\lambda}+\frac{y}{1-\frac{3\lambda}{4}}\right)dy=\frac{8\lambda A}{3(1-3\lambda)}+\frac{4}{1-\frac{3\lambda}{4}}$$ so you can solve for $A$ to get $$A=\frac{48(1-3\lambda)}{(3-17\lambda)(4-3\lambda)}.$$ That is $$f(x)=\frac{48\lambda+4(3-17\lambda)x}{(3-17\lambda)(4-3\lambda)}$$ is the unique fixed point of $T$ for any $\lambda$ s.t. $|\lambda|<\frac13$ and $\lambda \neq \frac{3}{17}$.