Vtyjkyuk

I'm stuck in an analysis problem, it states:

Let $lbrace f_n:[0,1)rightarrow mathbbRrbrace_nin mathbbN$, with $f_n(x)=x^n$. Prove that:

$i)$ The familiy is pointwise equicontinuous (i.e. for every $x_0in[0,1)$.

$ii)$ The familiy is NOT uniformly equicontinuous.

My attempt:

Por $i)$, we need to prove that for every $x_0in[0,1)$, and $forall varepsilon>0 quad exists delta=delta(varepsilon;x_0)>0$ such that if $|x-x_0|<delta$, then $|f_n(x)-f_n(x_0)|<varepsilon$, for all $nin mathbbN$

So working with the last inequality, and using the mean value theorem, it exists $cin (0,1)$, such that $f_n(x)-f_n(x_0)=f_n^'(c)(x-x_0)$

$|f_n(x)-f_n(x_0)|=|f_n^'(c)(x-x_0)|=|nc^n-1||x-x_0|<n|c|^n-1delta<ndelta<varepsilon$

So, by this kind of estimation, $delta=varepsilon/n$. But this choice of $delta$ clearly depends on $n$. And it shouldn't!

And for part 2 I don't have clue. Any help or advise will be appreciated.

Let $varepsilon>0$ be given, and let $K$ be a nonempty compact subset of $[0, 1)$. Set $s = max K $. Either $s=0 $ or $s>0$. The former case is trivial ($K=0$), and so we assume there is a number $r>0$ such that $s=frac11+r$. By induction we obtain
beginalign displaystylefrac1(1+r)^n leq frac11+nr
endalign
for all positive integers $n$ (Bernoulli's Inequality). We may find a positive integer $N$ such that $varepsilon N > frac1r$ (Archimedean Principle). Thus
beginalign displaystyle left| ,f_n (s) right| < varepsilon , text whenever , ngeq N ,.
endalign

Since $K$ was given as an arbitrary compact subset of $[0,1)$, we have shown that the sequence of functions $, f_n_n=1^infty$ converges uniformly on every compact subset of $[0, 1)$. In other words, the sequence of functions $,f_n_n=1^infty$ is normal in $[0,1)$. I think knowing some Italians would be a good thing to have to your avail at this point.

If you really want to hold my feet to a fire, I can explain how to further show that the family is equicontinuous.

For part ii): Suppose $f_n$ is uniformly equicontinuous on $[0,1).$ Let $epsilon= (e^-1/2 - e^-1)/2.$ Then there exists $delta > 0$ such that $x,y in [0,1),$ $|y-x| < delta,$ implies $|f_n(y)-f_n(x)| < epsilon.$ Now for large $n,$ $1/2n< delta.$ So for large $n,$

$$tag 1 |f_n(1-1/n + 1/2n) -f_n(1-1/n)| < epsilon.$$

But the left side of $(1)$ converges to $e^-1/2 - e^-1.$ Thus for large $n,$ $(1)$ fails. We have a contradiction, proving $f_n$ is not uniformly equicontinuous on $[0,1).$