Solving an equation of the form $f = T(f)$, Alternative to fixed point iteration

Question

Let $T$ be some integral operator and suppose $f:[0,\infty) \rightarrow [0,\infty)$ is a function which satisfies: $$f = T(f).$$

A popular method for finding $f$ is to take some arbitrary $f_0$ and then set $f_{n+1} = T(f_n)$ for each $n$ (a fixed point iteration). However, convergence does not always hold (especially when you start with a bad choice for $f_0$). Are there some alternative methods for solving these type of equations numerically?

Answer 1

It depends on the operator $T$ and the underlying space $X$.

For instance, if $X$ is a Hilbert space and $T$ is nonexpansive, then your iteration may not work: indeed, if $T$ is the negative identity, then you only bounce back and forth unless you start at the unique fixed point $0$.

If $T$ is nonexpansive in the Hilbert space $X$, a much better iteration is $$f_{n+1} = \tfrac{1}{2}f_n + \tfrac{1}{2}T(f_n).$$ This Krasnoselskii-Mann iteration will converge at least weakly to a fixed point provided there exists one.

Answer 2

To expand on Max's answers if you want more degrees of freedom. Any mean preserving smoothing operator will have stabler convergence properties.

For example convolution with the weighted mean: $$f_{n+1}= \beta T(f_n) +\displaystyle\sum_{k=0}^N \alpha_k f_{n-k}$$ where $$\displaystyle \beta + \sum_{k=0}^N \alpha_k = 1$$

Examples:

$\beta =1, \alpha_k = 0$ gives your fixed point iteration.

$\beta = 1/2, \alpha_0 = 1/2, \alpha_{>0} = 0$ gives Max's answer.