# Show uniformly convergent sequence of contraction maps has a fixed point but is not unique?

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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?

• Compactness implies completeness. – zhw. Aug 17 '19 at 3:11
• @zhw: how would that help? – Saeed Aug 17 '19 at 3:16
• I was just correcting your second sentence. – zhw. Aug 17 '19 at 3:16

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 $$x.$$
• for the exercise, can we say since $g_n$ are contraction so they are Lipschitz, then they are continuous. Since each $g_n$ is continuous and uniformly converges to $g$, $g$ is continuous. Am I correct? – Saeed Aug 17 '19 at 4:04
• I didn't assume they were contractions; that's why i used the notation $g_n.$ Lipschitz has nothing to do with it. Continuous + uniformly convergent implies the limit function is continuous – zhw. Aug 17 '19 at 4:09
• But I have not assume that $f_n$'s are continuous? – Saeed Aug 17 '19 at 4:11