4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Is $\epsilon$ fixed, or does it depend on the value of $a$? in other words, are the functions uniformly contractionary? $\endgroup$ – Unwisdom Mar 5 '14 at 18:57
  • $\begingroup$ For the moment $\epsilon$ is fixed, i.e. it is uniform over $a$'s. But I would be also interested in the case where $\epsilon = \epsilon_a$. $\endgroup$ – user693 Mar 5 '14 at 18:58
  • 2
    $\begingroup$ 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 $\alpha_V>0$). $\endgroup$ – Etienne Mar 5 '14 at 19:41
  • $\begingroup$ Etienne, thanks for your commment. Can you please make it a formal answer (for the case of $a \mapsto \epsilon_a$ continuous)? $\endgroup$ – user693 Mar 5 '14 at 19:46
  • 1
    $\begingroup$ 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 $\frac{e\epsilon_{a}}{\epsilon_{b}}\leq 2e$. $\endgroup$ – Unwisdom Mar 5 '14 at 19:57
7
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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".)

$\endgroup$
  • $\begingroup$ 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 $kB$)? $\endgroup$ – user693 Mar 5 '14 at 19:29
  • $\begingroup$ The $k$ was a typo. (I had forgotten the $\epsilon$ denominator earlier and goofed the repair.) $\endgroup$ – copper.hat Mar 5 '14 at 19:35
  • $\begingroup$ Ok, thanks a lot again. $\endgroup$ – user693 Mar 5 '14 at 19:36
  • $\begingroup$ Unwisdom's proof below is very nice. $\endgroup$ – copper.hat Mar 5 '14 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.