# Need some help on a non-example of equicontinuity

In an attempt to better understand the definition of an equicontinuous family of continuous functions, I want to find a simple non-example.

My intuition says that the family $\{f_n\colon[0,1]\to\mathbb R\}_{n\in\mathbb N}$ given by $f_n(x)=x^n$ is not equicontinuous, but I do not know how to show this.

Any help is appreciated.

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Can you elaborate on your "intuition?" Where do you think this family family fails to be equicontinuous? – Alex R. Jul 6 '12 at 15:18
Could this also be tagged functional-analysis? – Rudy the Reindeer Jul 6 '12 at 15:38
@MattN. In my opinion, not really. While the concept of equicontinuity plays a role in a functional analysis, the question here asks for the explicit construction of a counterexample, giving the question a much more real-analytic flavor. – user12014 Jul 6 '12 at 17:55
@PZZ Ok. Thanks for the comment! – Rudy the Reindeer Jul 6 '12 at 17:57

We claim that the point $x=1$ ruins the family's chance at equicontinuity. We choose $\epsilon = \frac{1}{2}$ and let $\delta>0$ be given. Pick any $y\in(1-\delta,1)$ you want. Letting $n\to \infty$ we have $y^n \to 0,$ so we may choose an $n$ large enough that $y^n < \frac{1}{2}$, then we would have $$|f_n(1)-f_n(y)| = |1-y^n| \geq \frac{1}{2}.$$

Hence the family is not equicontinuous at $x=1$.

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I think you should mention that you chose a $y \in (1 - \delta, 1)$. – el_tenedor Apr 4 '15 at 15:58

According to Wikipedia (link given in OP): "...the limit of an equicontinuous pointwise convergent sequence of continuous functions $f_n$ on either metric space or locally compact space is continuous. ..."

In your example the pointwise limit of $f_n$ is not continuous. It is $0$ everywhere except at $1$ (where it's $1$). So we may conclude that $f_n$ are not equicontinuous.

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I hope I'm not missing anything. – Rudy the Reindeer Jul 6 '12 at 15:35

The problem is at $1$. Indeed, assume that $\{f_n\}$ is equi-continuous at $1$. Then we can find a $\delta>0$ such that if $|1-x|\leq \delta$ and $0\leq x\leq 1$ then for each integer $n$, $|f_n(x)-f_n(1))|\leq 1/3$, hence $|x^n-1|\leq 1/3$ and $x^n\geq 2/3$. If $1/{n^2}\leq \delta$ then we should have $\left(1-\frac 1{n^2}\right)^n\geq 2/3$. It's impossible, since the RHS converges to $0$.

In fact, we know by Arzela-Ascoli-'s theorem that a uniformly bounded sequence of equi-continuous functions on a compact interval admit a converging subsequence for the uniform norm (and this limit is continuous). We can check that the pointwise limit of $\{f_n\}$ is not continuous.

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Why can we simply check that $f_n$ is non-uniformly convergent? Doesn't the Arzela-Ascoli theorem you quoted refer to subsequences? I mean, even if we show that $f_n$ is non-uniformly convergent, does this necessarily mean that all its subsequences are too? – Ryan Nov 11 '13 at 9:44
The pointwise limit of subsequences is non-continuous (as it's the case for the whole sequence) so the argument applies. – Davide Giraudo Nov 11 '13 at 9:50