Fixed point of $T(f) = \lambda \int_0^2 y^2 f(y)\,dy + 3 \lambda f(x/4) + x$ for $0 \leq x \leq 2?$ 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?
 A: 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 $\lambda \neq \frac{3}{17}$.
A: We are actually looking for a function f satisfying $T(f) = f$.
We have a functional equation and a bit like for differential equations, to look for a particular solution we see the precense of x which gives us the idea of looking for polynomial functions of degree <= 1 (of the form $f(x)=Ax+B$).
Replacing f in the equation, we obtain:
$\forall x, \frac16 \int_0^2 y^2(Ay+B)dy+\frac12 f\left(\frac{x}{4}\right)+x=Ax+B$
So $\forall x, \left(1-\frac{7}{8}a\right)x+\frac{2a}{3}-\frac{b}{18}$
I obtained $a=8/7$ and $b=96/7$
