I'm trying to prove the convergence of a new method with a fixed point iteration equation: $$Y_{n+1} = f(Y_n)$$ And I was able to obtain the Jacobian matrix of function $f$ at the fixed point $Y^*$ as below:
Jac = $$ \begin{pmatrix} (1-X_1)(1-\alpha_1) & -(1-X_2)\alpha_1 & ... & -(1-X_n)\alpha_1 \\ -(1-X_1)\alpha_2 & (1-X_2)(1-\alpha_2) & ... & -(1-X_n)\alpha_2 \\ ... & ... & ... & ... \\ -(1-X_1)\alpha_n & -(1-X_2)\alpha_n & ... & (1-X_n)(1-\alpha_n) \\ \end{pmatrix} $$
where $${1\over4} < X_i<2, i=1,2,...,n$$ $$ 0<\alpha_i <1, i=1,2,...,n $$ $$ \sum_{i=1}^n \alpha_i=1 $$
In order to show the fixed point iteration converge, I need to show the spectral radius of the Jacobian matrix < 1. I have run millions of numerical experiments with random numbers. For all tested cases, the spectral radius of this matrix is < 1, so I think it might be possible to prove the convergence mathematically. I learned a lot reading posts and answers on this forum and hopefully, somebody can help me with my question. :)
Do you know whether $f$ is continuous and the sequence $Y_n$ converges?
If so, then let $Y^* = \lim Y_n$.
Since $\lim_{Y \rightarrow Y^*} f(Y^*) = f(Y)$, we have $\lim f(Y_n) = f(Y)$.
So $Y = \lim f(Y_{n+1}) = \lim f(Y_n) = f(Y)$.