I have some function from $R^n$ to $R^n$ defined as: $$T(v) = u + f(v) F$$ where u is some vector, f is some non-linear function and F is a n,n matrix. i want to prove that there is a unique fixed point. i know the following is true:
- the spectral radius of is stricly less that 1 ($\rho(F) < 1$)
- every entry of F is between 0 and 1 and the sum of each row is <= 1
- for every $i$, $(f(v))_i \leq v_i$
- $f(f(v)) = f(v)$, meaning that $f$ sends v to some area where $f$ acts as identity.
- $|Df(v)|_{\ell_\infty} \leq 1$, $|Df(v)|_{\ell_1} < 2$ where $Df(v)$ is the jacobian of $f$
The problem is cant show that T is a contraction since $F$ is contraction iin defferent norm than $f$. can anyone give me any ideas?
Edit:
To understand how $f$ looks like, we can say the following the gives us more information about $f$. For every vector $v$ (of size $n$) we can split $v$ to vectors of size $k$, and we have some consts $c_1, ..., c_{n/k}$. now we can split $f$ to n/k functions $f_1, ..., f_{n/k}$, and v to $n/k$ vectoros of size $k$: $v_1, ..., v_{n/k}$. So for every $i\in[n/k]$: $\sum_{j=1}^k(f_i(v_i))_j \leq c_i$. if this sum before aplying $f$ was less than $c_i$, f is the identity. Meaning that $f$ reduces the sum (if needed) to $c_i$.
That actually means that v can be splited to smaller vectors of equall sizes, such that f reduces the sum of each of these sub-vectors to be $c_i$, and it does this by reducing each of the components differently. We can further state that $f_i$ multiply each of its components by some weight $0 \leq w_i(v_i) \leq 1$, and each of the weights is a function of the subvector and the index. i.e $(f_i(v_i))_j=(w_i(v_i))_j*(v_i)_j$
This explain (4), since after applying $f$, the $f(v)$ is in the area where each of the sub vector's sums is $\leq$ the capcity $c_i$
