Here's the problem I'm working on:
Let $(M,d)$ be a metric space and $f_n:M\rightarrow M$ be a sequence of contraction mappings. Suppose $(f_n)$ converges uniformly to $f: M \rightarrow M$. 1) Show that $f$ is Lipschitz. 2) Let M be complete. Show that $f$ has a fixed point. 3) Provide an example of $f_n$ and $(M,d)$ that the above fixed point may not be unique.
Here's my reasoning:
NTS: $\exists K \ge 0$ such that $\forall x,y \in M, d(f(x),f(y)) \le K d(x,y).$
By the triangle inequality, we have that $$d(f(x),f(y)) \le d(f(x),f_n(x)) + d(f_n(x),f_n(y)) + d(f_n(y),f(y)).$$ Since $f_n$ is a contraction for all $n \in N$ (N = Natural numbers), we have that $\exists K_0 \in [0,1)$ such that $$d(f(x),f(y)) \le d(f(x),f_n(x)) + K_0 d(x,y) + d(f_n(y),f(y)).$$ Since $f_n$ converges uniformly to $f$, we have that for $S_n = sup \{ \,d(f_n(x),f(x)) \, | x \in M\}$, $S_n \rightarrow 0$ as $n \rightarrow \infty.$
Hence $$d(f(x),f(y)) \le S_n + K_0 d(x,y) + S_n,$$ and as $n \rightarrow \infty,$ we have that $d(f(x),f(y)) \le 0 + K_0 d(x,y) + 0.$ Thus we have that $f$ is Lipschitz, and since $K_0 \in [0,1),$ $f$ is a contraction.
Then if $(M,d)$ is complete, by the contraction mapping theorem, there exists a unique fixed point.
This result is somewhat confusing because the original question asks for an example where there may be more than one fixed point. Did I do something wrong?