I have problem showing this:
Suppose $ f: X \to X $ is a continuous surjective function and there exists a sequence in $ \mathbb{N} $ such as $ \lbrace n_{k} \rbrace_{k \geq 1} $ such that $ n_{k} \to \infty$ and $ f^{n_{k}} \to I_{X} $ uniformly . How can we show that any power of $f^{n_{k}}$ converges to $ I_{X} $ uniformly ?
Notice that $ d $ is a metric on $ X $ and :
\begin{align*}
D(g,h) := \sup_{x \in X } d(g(x) , h(x))
\end{align*}
Is a metric on $C(X,X)$.
Asked
Active
Viewed 30 times
1
Reza Yaghmaeian
- 1,055
-
You have some notation which is not so clear. What do you mean by $f^{n^k}$ if $X$ is not necessarily $\mathbb{R}$ , what is $I_x$? Is $X$ only a topological space? – Keen-ameteur Sep 05 '19 at 12:00
-
The sequence is the $n_k$ fold concatenations $f^{\circ n_k}$? And you are concerned with the convergence of the sequence $f^{\circ (mn_k)}$ for any fixed $m$ as "power" of the original sequence? – Lutz Lehmann Sep 05 '19 at 12:04
-
@LutzL Yes exactly. – Reza Yaghmaeian Sep 05 '19 at 12:07
-
@Keen-ameteur X is a compact metric space, and $ d $ is a metric on that. $I_{x}$ is identity function. – Reza Yaghmaeian Sep 05 '19 at 12:08
1 Answers
2
If $D(f^{∘n_k},I_X)<\varepsilon$, then $D(f^{∘(mn_k)},I_X)<mε$.
On the point level for any $x_0\in X$ and with $x_j=f^{∘(jn_k)}(x_0)$ you get $$ d(f^{∘(mn_k)}(x_0),x_0)=d(x_m,x_0) \le \sum_{j=0}^{m-1}d(x_{j+1},x_j)=\sum_{j=0}^{m-1}d(f^{∘n_k}(x_j),x_j)<mε $$ Now take the supremum on the left side to get the claim.
With some minimal additional boilerplate the uniform convergence of the function sequence $f^{∘(mn_k)}$ is thus established.
Lutz Lehmann
- 126,666