How will fixed point change if the mapping changes? Generally, the question is we have a contraction mapping $M:\mathbb{R}^n\rightarrow\mathbb{R}^n$. Define a new mapping $\tilde{M}:\mathbb{R}^n\rightarrow\mathbb{R}^n$ where $\tilde{M}:Mx+c$ where $c$ is a constant vector. We know $\tilde{M}$ is also a contraction mapping. Thus, we know there exists a unique fixed point $x^*\in\mathbb{R}^n$ that solves $Mx^*=x^*$ and a unique $\tilde{x}\in\mathbb{R}^n$ such that $\tilde{M}\tilde{x}=\tilde{x}$. How the fixed point $x^*$ will differ from $\tilde{x}$? How the change of $c$ affect the change of the fixed point.
To be more specific, in reinforcement learning, we define $Q$ functions where $Q\in\mathbb{R}^{S\times A}$. 
The optimal $Q$ function denoted by $Q^*$ has to satisfy the Bellman equation given as follows:
$$
Q^*(i,a)=c(i,a) + \beta \sum_j p(i,j,a)\min_b Q^*(j,b),\ \ i\in \mathcal{S}, a\in \mathcal{A},
$$
where $c:\mathcal{S}\times\mathcal{A}\rightarrow \mathbb{R}$ and $p(i,j,a)$ is the probability that given state $i$ and action $a$, the next state is $j$.
Define $F:\mathbb{R}^{S\times A}\rightarrow \mathbb{R}^{S\times A}$ by 
$$
[F(Q)]_{i,a}=c(i,a) + \beta \sum_j p(i,j,a)\min_b Q^*(j,b)
$$
for $i\in\mathcal{S},a\in\mathcal{A}$.
We know that $F$ is a contraction mapping. Thus, we can apply iterative algorithm to find the fixed point $Q^*$ which is unique.
But I was wondering how the fixed point $Q^*$ would change as we change $c(i,a)$ a little bit. Is the mapping $c\mapsto Q^*$ continuous? If we want the fixed point $Q^*$ to be the values that we desire, how should we select $c$.
I don't know what theories I can reply on to solve my problems. Do you know any papers that provide tools for solving my problems?
 A: If $c$ is a small perturbation, then the fixed point will be a small perturbation of the original fixed point, whose size is in fact quite explicitly controlled by $c$. This is a feature known as stability that is observed in many problems that are resolved using contraction mapping. To see this, let $0 < L < 1$ be the constant in the contraction mapping definition of $M$, so that $\|Mx - My\| \leq L\|x-y\|$. Let $x$ be a fixed point of $M$, so that $x = Mx$, and let $y$ be a fixed point of $\tilde{M}$, so that $y = My + c$. Then the difference is
$$
y-x = (My - Mx) + c.
$$
By the triangle inequality and the contraction hypothesis,
$$
\|y-x\| \leq L\|y-x\| + \|c\|.
$$
Rearranging,
$$
(1-L)\|y-x\| \leq \|c\|,
$$
and since $0<L<1$ the constant factor is positive, so we can divide through to conclude that
$$
\|y-x\| \leq \frac{1}{1-L}\|c\|.
$$
Therefore if $c$ is small, then $y-x$ is also small. In fact, taking this argument just slightly further, one concludes that the fixed points $x_c$ of the maps $x\mapsto Mx + c$ depend in a Lipschitz continuous way on the perturbation $c$.
