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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?

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Is $\epsilon$ fixed, or does it depend on the value of $a$? in other words, are the functions uniformly contractionary? – Unwisdom Mar 5 '14 at 18:57
For the moment $\epsilon$ is fixed, i.e. it is uniform over $a$'s. But I would be also interested in the case where $\epsilon = \epsilon_a$. – Adam Mar 5 '14 at 18:58
In fact, all the proofs in the answers below show that if $\varepsilon_a$ depends continuously on $a$, then $\overline x_a$ still depends continuously on $a$. This is because any $a\in\mathbb R$ has a neighbourhood $V$ such that the function $v\mapsto \varepsilon_v$ is bounded below on $V$ (by some constant $\alpha_V>0$). – Etienne Mar 5 '14 at 19:41
Etienne, thanks for your commment. Can you please make it a formal answer (for the case of $a \mapsto \epsilon_a$ continuous)? – Adam Mar 5 '14 at 19:46
To adapt my argument to show this, just choose $\delta$ so that in addition to the stated requirement, $\epsilon_b$ does not dip below $\epsilon_a/2$. The upper bound for $\vert x_b-x_a \vert$ will become $\frac{e\epsilon_{a}}{\epsilon_{b}}\leq 2e$. – Unwisdom Mar 5 '14 at 19:57
up vote 4 down vote accepted

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 $$\vert x_b-x_a\vert\leq(e\epsilon)\sum_{i=0}^{\infty}(1-\epsilon)^{i}=\frac{e\epsilon}{\epsilon}=e. $$

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+1: Nice & succinct. – copper.hat Mar 5 '14 at 19:24

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.

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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".)

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Thanks for your answer. There still is one thing I am not clear with. From the definition of $B$, could be also get $\| \bar x_a - f_a^{(k)}(x_0) \| \leq \frac{(1-\epsilon)^k}{\epsilon} B $ (instead of $kB$)? – Adam Mar 5 '14 at 19:29
The $k$ was a typo. (I had forgotten the $\epsilon$ denominator earlier and goofed the repair.) – copper.hat Mar 5 '14 at 19:35
Ok, thanks a lot again. – Adam Mar 5 '14 at 19:36
Unwisdom's proof below is very nice. – copper.hat Mar 5 '14 at 19:50

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