Continuity of Fixed Point For all $a \in \mathbb{R}$, let $f_a: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be continuous and contractive, that is, there exists $\epsilon \in (0,1)$ such that $\left\| f_a(x)-f_a(y) \right\| \leq (1-\epsilon) \left\| x-y\right\|$ for all $x,y \in \mathbb{R}^n$.
Assume that for all $x \in \mathbb{R}^n$, the mapping $a \mapsto f_a(x)$ is continuous.
Now let $x_0 \in \mathbb{R}^n$ be fixed. 
For all $a \in \mathbb{R}$, define the sequence $\{ x_a^0, x_a^1, x_a^2, \cdots \} := \{ x_0, \ f_a(x_0), \ f_a( f_a(x_0) ), \ \cdots \}$, that is, $x_a^k := f_a^{(k)}(x_0)$ for all $k \geq 0$. Define $ \bar{x}_a := \lim_{k \rightarrow \infty} x_a^k$.
Under what additional conditions the mapping $a \mapsto \bar{x}_a$ is continuous?
 A: Under the assumption that $\epsilon$ is constant across $a$, then no additional assumptions are needed. 
Suppose that $x_a$ is the fixed point of $f_a$, and choose $e>0$.  Then there is a $\delta$ such that for all $b$ within $\delta$ of $b$, you have $e\epsilon\geq \vert f_b(x_a)-f_a(x_a)\vert=\vert f_b(x_a)-x_a\vert$. 
Now, repeatedly apply $f_b$ to $x_a$.  If $x_b$ is the fixed point of $f_b$, then 
$$\vert x_b-x_a\vert\leq(e\epsilon)\sum_{i=0}^{\infty}(1-\epsilon)^{i}=\frac{e\epsilon}{\epsilon}=e.   $$
A: We can view $\bar{x}_a$ as a minimizer of the continuous function $x\mapsto \|f_a(x)-f_a(f_a(x))\|$. If $f$ is jointly continuous as a function of $\mathbb{R}^n\times\mathbb{R}$, then the argmin correspondence that maps $a$ to the set of minimizers of this functions is upper hemi-continuous by Berge's maximum theorem (one has to show that locally all solutions lie in some compact set, but that's not hard to do). Since there is a unique minimzer for each $a$, the function $a\to\bar{x}_a$ is continuous. 
A: You can use the contraction mapping estimates directly.
You have the estimate
$\|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} \|f_a(x_0) - x_0\|$, so we can see that if we let $B = \sup_{a \in B(\hat{a},1)} \|f_a(x_0) - x_0\|$, then
$\|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} B$ for all $a \in B(\hat{a},1)$.
So, if $a,a' \in B(\hat{a},1)$,
we have the estimate
\begin{eqnarray}
\|\bar{x}_a - \bar{x}_{a'} \| &\le& \|\bar{x}_a - f_a^{(k)}(x_0)\| +
\| f_a^{(k)}(x_0) - f_{a'}^{(k)}(x_0) \|
+\|\bar{x}_{a'} - f_{a'}^{(k)}(x_0)\| \\
&\le& 2{(1-\epsilon)^k \over \epsilon} B + \| f_a^{(k)}(x_0) - f_{a'}^{(k)}(x_0) \|
\end{eqnarray}
Now let $\epsilon>0$ and choose $k$ such that $2{(1-\epsilon)^k \over \epsilon} B < {1 \over 2} \epsilon$, then since $a \mapsto f_a^{(k)}(x_0)$ is continuous at $\hat{a}$, we can find a $\delta\le 1$ such that $\| f_a^{(k)}(x_0) - f_{a'}^{(k)}(x_0) \| < {1 \over 2} \epsilon$ for all $a,a' \in B(\hat{a},\delta)$, and so
$\|\bar{x}_a - \bar{x}_{a'} \| < \epsilon$.
(I think the first time I saw these estimates was in Kantorovich & Akilov's "Functional analysis".)
