# Is a sequence of decreasing functions in $C^0$ pointwise convergent to $0$ implies the sequence is equicontinuous?

Suppose that $f_n \in C^0$, $C^0$ means the set of continuous functions from $[0,1]\to \mathbb R$, for each $x\in [a,b]$, $$f_1(x)\geq f_2(x)\geq f_3(x)\geq \ldots$$ And $$\lim_{n\to \infty}f_n(x)=0$$ Is the sequence equicontinuous? (Hint:Does $f_n$ converges uniformly to 0)

Here is my attempt, I want to show $f_n\to 0$ uniformly first.

Suppose not. Then $\exists \epsilon>0$ such that $\forall N\in \mathbb N$, $\exists$ some $n>N$ such that $\exists x\in [0,1]$ with $|f_n(x)|>\epsilon$.

Now let $k=N=1,2,3,\ldots$, then we can attain a sequence $x_k$ such that $|f_n(x_k)|>\epsilon$ for $n\leq k$ by the decreasing property. Since $[0,1]$ is compact then $x_k$ has a convergent subsequence, call it $y_n$ and suppose $y_n\to x_0$. I think at $x_0$, the limit of $f_n$ is not 0, but how to show it rigorously? And what role does decreasing play here because clearly without it the sequence may not converge uniformly.

Once we have uniform convergence, then $\exists N_0$ such that $n>N_0 \implies |f_n(x)|<\epsilon, \forall x\in [0,1]$, which means $|f_n(s)-f_n(t)|<\epsilon$ for any $s,t$ and $n>N_0$. Now only have finitely many $n$ to worry about, but they are all uniformly continuous, so we can pick the smallest $\delta$ that will make $f_n(s)-f_n(t)$ whenever $|s-t|<\delta$ for $n\leq N_0$.

Is this argument right?

Also it there any condition that can make pointwise convergence propagate to uniform converence? I know there's a Arzela-Ascoli propagation theorem, but it is quite hard to use it in practice.

Fix $\epsilon >0$

since $f_1(x)\geq f_2(x)\geq.....\geq 0$ (Check that)

$\implies |f_1(x)|\geq |f_2(x)|\geq.....\geq 0$(Check that)

$\implies \sup_{[0,1]}|f_1(x)|\geq \sup_{[0,1]}|f_2(x)|\geq.....\geq 0$

So, $\exists N(\epsilon)\in \mathbb N$ such that

$\sup_{[0,1]}|f_n(x)|\leq \epsilon\;\;\forall \,n\geq N(\epsilon)$

as $\sup_{[0,1]}|f_n(x)|\rightarrow 0$

Hence $f_n\rightarrow 0$ uniformly.

I think you argument following that are correct.

• Is there any hint for the second part of the question? That when will pointwise convergence propagate to uniform converence? – eeeethan1997 Feb 24 '15 at 17:44
• There is a theorem in measure theory where you can say that pointwise convergence is uniform convergence if restricted to certain subsets of domain. I cannot recall the name of theorem. – Harish Feb 24 '15 at 17:58