The inverse continuity as a continuous dependence of a fixed point in terms of a parameter

Question

Just came across the following:

Let $X$ be a complete metric space and B a metric space. Consider a family of functions $$f_b: X \rightarrow X$$ indexed by $b \in B$. These functions are contractions with constant $K \in (0, K_*)$ such that $K_* < 1$ regardless of $b$.

Suppose the function $b \in B \mapsto f_b:X \rightarrow X$ is continuous in the following sense:

$$\forall \epsilon > 0 \hspace{1mm} \exists\delta >0 \hspace{1mm} \forall b_1, b_2 \in B: d(b_1, b_2) < \delta \Longrightarrow \sup_x d(f_{b_1}(x), f_{b_2}(x)) < \epsilon$$

Then the fixed point $x(b)$ varies continuously with $b$.

I think I understood the proposition, but couldn’t prove it. Can somebody help me on that?

Answer 1

Well, the Banach fixed point theorem says:

If $T:X\to X$ is a map so that there exists $c\in[0,1)$ so that $d(Tx,Ty)\leq cd(x,y)$, then there exists a unique $x_o\in X$ such that $Tx_o=x_o$.

But it's not just that. The constant point $x_o$ has the following property (look at the proof of the fixed point theorem): if $x\in X$ is any point, then $T^nx\to x_o$.

Now to our question. First of all I will denote the fixed point of $f_b$ by $x_b$. We have the maps $f_b:X\to X$ so that the mapping $b\mapsto f_b$ is continuous when we regard the space of the functions $X\to X$ with the uniform norm.

Now let $\varepsilon>0$. Then we may find $\delta>0$ so that whenever $b_1,b_2\in B$ satisfy $d(b_1,b_2)<\delta$, then $\sup_{x}d(f_{b_1}(x),f_{b_2}(x))<\varepsilon.$

As we said, for any point $x\in X$ we have that $x_{b_1}=\lim_{n\to\infty}f_{b_1}^n(x)$ and that $x_{b_2}=\lim_{n\to\infty}f_{b_2}^n(x)$. Now we have that $$\require{cancel}\cancel{d(x_{b_1},x_{b_2})=\lim_{n\to\infty}d(f_{b_1}^n(x),f_{b_2}^n(x))\leq d(f_{b_1}(x),f_{b_2}(x))<\varepsilon} $$ and we are done.

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Edit

As pointed out in the comments, there was a mistake in the striked out part of my answer from 3 years ago; I apologize. There is an easy fix however; The functions $\{f_b\}_{b\in B}$ are all contractions with common constant $\gamma\in(0,1)$, by assumption. Let $\varepsilon>0$. Find $\delta>0$ such that, for all $b_1,b_2\in B$, if $d(b_1,b_2)<\delta$, then $\sup_{x\in X}d(f_{b_1}(x),f_{b_2}(x))<\varepsilon$.

Let $b_1,b_2\in B$ be such that $d(b_1,b_2)<\delta$. I claim that, for all $x\in X$ and for all $n\in\mathbb{N}$ we have $$d(f_{b_1}^n(x),f_{b_2}^n(x))\le(1+\gamma+\gamma^2+\dots+\gamma^{n-1})\varepsilon.$$

Indeed, we prove this by induction. For $n=1$ there is nothing to show. Assume that this is true for some $n\in\mathbb{N}$. Then $$d(f_{b_1}^{n+1}(x),f_{b_2}^{n+1}(x))\le d(f_{b_1}^{n+1}(x),f_{b_1}(f_{b_2}^n(x)))+d(f_{b_1}(f_{b_2}^n(x)),f_{b_2}^{n+1}(x))=$$ $$=d(f_{b_1}(f_{b_1}^{n}(x)_,f_{b_1}(f_{b_2}^n(x)))+d(f_{b_1}(f_{b_2}^n(x)),f_{b_2}(f_{b_2}^{n}(x)))\le\gamma\cdot d(f_{b_1}^n(x),f_{b_2}^n(x))+\varepsilon\le$$ $$\le \gamma\cdot(1+\gamma+\dots+\gamma^{n-1})\cdot\varepsilon+\varepsilon=(1+\gamma+\dots+\gamma^n)\varepsilon$$ as we wanted.

Therefore, we have $$d(x_{b_1},x_{b_2})=\lim_{n\to\infty}d(f_{b_1}^n(x),f_{b_2}^n(x))\le\lim_{n\to\infty}(1+\gamma+\dots+\gamma^{n-1})\varepsilon=\varepsilon\cdot\sum_{j=0}^\infty\gamma^j=\frac{\varepsilon}{1-\gamma}$$

To conclude: given $\varepsilon>0$, we found $\delta>0$ (this $\delta=\delta(\varepsilon)$ depends on $\varepsilon$ only) such that $d(b_1,b_2)<\delta\implies d(x_{b_1},x_{b_2})<\frac{\varepsilon}{1-\gamma}$. So given $\varepsilon>0$, take $\delta=\delta((1-\gamma)\cdot\varepsilon)$ to obtain a quantity small enough so that $d(b_1,b_2)<\delta\implies d(x_{b_1},x_{b_2})<\varepsilon$.