Convergence of fixed points of a sequence uniformly convergent.

Solution 1:

Let $k_m$ be the contraction constant of $f_m$. Then \begin{align} d(x_m,x_n)&=d(f_m(x_m),f_n(x_n)) \\ &\leq d(f_m(x_m),f_m(x_n))+d(f_m(x_n),f_n(x_n)) \\ &\leq k_md(x_m,x_n)+d'(f_m,f_n) \end{align} for all $m,n\in\mathbb{N}$. Since $k_m<1$, we may rearrange to get \begin{align} d(x_m,x_n)\leq\frac{d'(f_m,f_n)}{1-k_m}. \end{align} We are done if $\limsup k_m<1$. Otherwise counterexamples exist: Consider contraction mappings on $[0,1]$ defined by \begin{align} f_n=\begin{cases} (1-1/n)x &\text{if $n$ is odd}, \\ (1-1/n)x+1/n &\text{if $n$ is even}. \end{cases} \end{align} Then $x_n=0$ if $n$ is odd and $x_n=1$ if $n$ is even, so $(x_n)$ does not converge. However, $f_n$ converges uniformly to the identity function on $[0,1]$.