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Let $(M,d)$ be a compact metric space and for each $n$ in $\mathbb{N}$ let $f_n$ be a contraction mapping. Suppose $(f_n)$ converges uniformly to $f: M \rightarrow M$. Prove that $f$ has a fixed point which may not be unique.
My try:
$$d(f(x),f(y)) \le d(f(x),f_n(x)) + d(f_n(x),f_n(y)) + d(f_n(y),f(y))$$
$f_n$'s are contraction so $\exists \,\,\,\ 0 \leq k_n <1$ and since $f_n$ converges uniformly to $f$, we have that
$$d(f(x),f(y)) \le k_n d(x,y) \rightarrow d(f(x),f(y)) \le k d(x,y)$$
where
$0 \leq k \leq 1$. Since $k\leq1$, $f$ is non-expansive (not necessarily a contraction).
Question: How would one show that $f$ has a fixed point which may not be unique?
Best Answer
Sketch: First, an exercise. Suppose $g_1,g_2,\dots :M\to M$ are continuous and $g_n\to g$ uniformly on $M.$ Suppose $x_n\to x$ in $X.$ Then $g_n(x_n)\to g(x).$
In your problem we have for each $n$ that there exists $x_n\in M$ such that $f_n(x_n)=x_n.$ $M$ is compact, so there is a subsequence $n_k$ such that $x_{n_k}\to x.$ Now apply the exercise.
Nonuniqueness: On $M=[0,1]$ consider $f_n(x) = (1-1/n)x.$ These are contractions, but converge uniformly to $x.$