Ascoli-Arzelà example and equicontinuity

Question

My teacher did an example of Ascoli-Arzelà Theorem that I really don't understand, above all how he show that a series of functions is equicontinuous.

Let $f_n:=x^n \quad , \quad x\in [0,1] \quad , \quad f_n\subset C([0,1],\Bbb{R})$

$\forall x, \quad \exists \lim_{x\to +\infty}f_n(x)=f(x) \quad with \quad f(x):= \begin{cases} 0 & \text{if $0\le x<1$} \\ 1 & \text{if $x=1$} \end{cases}$

$f(x)$ is not continuous $\to$ $f_n$ doesn't uniformly converge to $f$

Now: $f_n(0)=0 \quad \forall n$

And $f_n$ isn't uniformly equi-continuous: $\quad$ $f_n'(x)=(x^n)'=nx^{n-1} \qquad f_n'(1)=n \longrightarrow +\infty$

I really don't understand this last part: Why did he calculated the derivate in $1$? And why can he say that $f_n$ isn't uniformly equi-continuous in this way? And does a correlation exist between derivates and equicontinuity?

Answer 1

I don't think the behavior of the derivative at the point $x=1$ proves (without much effort) that the sequence is not uniformly equi-continuous . So your teacher's argument is not complete and not a good approach for this.

If the sequence is uniformly equicontinuous then there exists $\delta >0$ such that $|f_n(x)-f_n(y)| <1-\frac 2 e$ for all $n$ whenever $|x-y| \leq \delta$. In particular $|f_n(1-\frac 1 n)-f_n(1)| <1$ for all $n$ sufficiently large. This lead to the contradiction $1-\frac 1 e < 1-\frac 2 e$

Answer 2

Just to complement Kavi Rama Murthy's answer to you question.

Consider a family of functions $\mathcal{F}\subset C[0,1]$ such that every $f\in\mathcal{F}$ is differentiable on $[0,1]$, and $\sup_{x\in [0,1]}|f'(x)|\leq M$ for some constant $M>0$. An application of the mean value theorem gives $$|f(x)-f(y)|=|f'(c_{x,y})||x-y|\leq M|x-y|$$ where $c_{x,y}$ is a point between $x$ and $y$. From this one infers that $\mathcal{F}$ is equicontinuous. If in addition, $\mathcal{F}$ is uniformly bounded (i.e. $\sup_{f\in\mathcal{F}}\|f\|_u<\infty$), then $\mathcal{F}$ is relatively compact by the aforementioned theorem.

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The uniform bound of the derivatives, when it happens, is a quick way to suspect compactness of a family of functions (it gives the equicontinuity part). The other condition (uniform boundedness or compactness of the orbits $\mathcal{F}_x=\{f(x):f\in\mathcal{F}\}$ for each $x\in[0,1]$) is usually easier to check.

Answer 3

We can have uniform equicontinuity even though $f_n'(1)\to \infty:$ Define

$$f_n(x) = \frac{x^{n^2}}{n},\,\,x\in [0,1].$$

Then $f_n\to 0$ uniformly. A uniformly convergent sequence of continuous functions on $[0,1]$ is uniformly equicontinuous. However $f_n'(1)= n\to \infty.$