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Let $(X,d)$ be a complete metric space and let $(\Lambda,d_\Lambda)$ be a metric space. Suppose we have a continuous map $F:X \times \Lambda \to X$ such that there exist $0<k<1$ satisfying $$d(F(x,\lambda),F(y,\lambda)) \leq k d(x,y)$$ for all $x,y \in X$ and all $\lambda \in \Lambda$.

By the Banach fixed point for each $\lambda$ there is a unique $x(\lambda)\in X$ satisfying $F(x(\lambda),\lambda) = x(\lambda)$.

I have to prove that the map $\Lambda \ni \lambda \to x(\lambda)\in X$ is continuous. How do I prove this map is continuous?

$\textbf{My attempt:}$ If we define $f_\lambda(x) = F(x,\lambda)$ we have that for an arbitrary point $p\in X$ the identity $$x(\lambda) = \lim_{n\to \infty} f_\lambda^n(p)$$ holds by the Banach fixed point. Therefore:

$$d(x(\lambda),x(\lambda_0)) = \lim_{n\to \infty}d(f_{\lambda}^n(p),f_{\lambda_0}^n(p)).$$

Essentially I have to prove the limits $\lim_{\lambda \to \lambda_0} $ and $\lim_{n\to \infty}$ commutes on the identity above, but I do not know how.

Hugo
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1 Answers1

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Suppose for contradiction that the mapping $\lambda \mapsto x(\lambda)$ is not continuous at $\lambda_0$. Then there exists an $\epsilon > 0$ such that for all $\delta > 0$, there exists a $\lambda_\delta$ such that $$d(\lambda_\delta, \lambda_0) < \delta \ \ \ {\rm but } \ \ \ d(x(\lambda_\delta), x(\lambda_0)) > \epsilon.$$

Now observe that $$d\left( x(\lambda_\delta), F(x(\lambda_0), \lambda_\delta)\right) = d(F(x(\lambda_\delta), \lambda_\delta), F(x(\lambda_0), \lambda_\delta))\leq kd(x(\lambda_\delta), x(\lambda_0)),$$ by the contraction property.

Furthermore, we have $$ d(F(x(\lambda_0), \lambda_\delta), F(x(\lambda_0), \lambda_0) = d(F(x(\lambda_0), \lambda_\delta) , x(\lambda_0)), $$ and, by the triangle inequality, we obtain, \begin{multline} d(F(x(\lambda_0), \lambda_\delta) , x(\lambda_0))\geq d(x(\lambda_\delta), x(\lambda_0)) - d( x(\lambda_\delta) , F(x(\lambda_0), \lambda_\delta))\\ \geq (1-k)d(x(\lambda_\delta), x(\lambda_0)) \geq(1-k)\epsilon > 0.\end{multline}

Thus for every $\delta > 0$, we have $$d((x(\lambda_0), \lambda_\delta), (x(\lambda_0), \lambda_0)) = d(\lambda_\delta, \lambda_0) < \delta$$ but $$d(F(x(\lambda_0), \lambda_\delta), F(x(\lambda_0), \lambda_0) ) \geq(1-k)\epsilon.$$

This contradicts the assumption that $F$ is continuous at $(x(\lambda_0), \lambda_0)$.

Kenny Wong
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