This question is from Textbook Wayne patty page 161 and I am struck on following question.
For each n $\in \mathbb{N}$ , let $f_n : [0,1] \to \mathbb{R}$ be the function defined by $f_n(x) = x^n$and let F = {$f_n : n\in \mathbb{N}$}.
Then Show that : (a) Show that F is closed subset of (C ([0,1], $\mathbb{R}),\rho)$.
(b) Show that F is not equicontinuous.
$\rho$ here is sup metric on $ C(X,\mathbb{R}^n)$ is defined by $\rho$(f,g)= max { d((f(x), g(x)) : x$\in X$} where d is usual metric on $\mathbb{R}^n$.
For (a) I think the limit point will be f = 0 if 0$\leq $ x< 1 and 1 if x=1.
Clearly this immediately implies that the set is closed.
For equicontinuity, definition: Let (X,T) be a topological space, let (Y,d) be a metric space, let F$\subseteq C(X,Y)$, and let $x_0 \in X$ . Then F is equicontinous at $x_0$ if for each $\epsilon >0$ there exists a neighbourhood U of $x_0$ such that if f $\in F$ and x$\in U$ , then $d(f(x),f(x_0))< \epsilon$. If F is equicontinuous at each point of X , then it is said to be equicontinuous.
I have to find a point $x_0$ which doesn't satisfies the defintion that is $d(f(x), f(x_0))\geq\epsilon$ .but I am unable to construct such a point and needs help.
For (b), consider the point $x_0 = 1$. Intuitively, the functions $f_n$ get steeper and steeper around that point, so it will be impossible to have a neighbourhood in which all functions are $\epsilon$-close to $f_n(x_0) = 1$. Let me know if you would like me to expand on how to show this rigorously.