Welcome to our community

Be a part of something great, join today!

Pointwise convergence implies uniform convergence

Siron

Active member
Jan 28, 2012
150
Hi,

I have to prove the following theorem:

Let $f_n:[0,1] \to \mathbb{R}, \forall n \geq 1$ and suppose that $\{f_n|n \in \mathbb{N}\}$ is equicontinuous. If $f_n \to f$ pointwise then $f_n \to f$ uniformly.

Before I start the proof I'll put the definitions here:
$f_n \to f$ pointwise if and only if $\forall \epsilon>0, \forall x \in X, \exists N \in \mathbb{N}$ such that $\forall n \geq N: |f_n(x)-f(x)|<\epsilon$
$f_n \to f$ uniformly if and only if $\forall \epsilon>0, \exists N \in \mathbb{N}, \forall x \in X: |f_n(x)-f(x)|<\epsilon$
$\{f_n|n \in \mathbb{N}\}$ equicontinuous if and only if $\forall \epsilon>0, \exists \delta>0, \forall n \in \mathbb{N}: |x-y|<\delta \Rightarrow |f_n(x)-f_n(y)|<\epsilon$
Attempt:
Let $\epsilon>0$ and define the sets $U_k=\{x \in [0,1]||f_n(x)-f(x)|<\epsilon, \forall n \geq k\}$ then I claim $U_k$ are open sets and hence because of the pointwise convergence of $f_n \to f$ they form an open covering for $[0,1]$. Since $[0,1]$ is compact there exists a finite subcover, that is, $\exists k_1,\ldots,k_n \in \mathbb{N}$ such that $[0,1] \subset \bigcup_{j=1}^{n} U_{k_j}$. We can assume that $k_1\leq k_2\leq \ldots \leq k_n$ and since $U_{k_j} \subseteq U_{k_1}, \forall j \geq 1$ I conclude $[0,1] \subseteq U_{k_1}$, i.e $\exists k_1 \in \mathbb{N}: \forall x \in X: |f_n(x)-f(x)|<\epsilon, \forall n \geq K$ which means $f_n \to f$ uniformly.

Is this proof correct? If yes, I'll show my proof for the claim: $U_k$ is open which follows by the equicontinuity.

Is there a way to show the statement by using Ascoli's theorem?


Thanks in advance!
Cheers.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
From mathwonk:

This seems to be a standard application of ascoli's theorem, as explained in wikipedia.

one remark, the questioner had misstated the definition of equicontinuous, giving instead the definition of uniformly equicontinuous.