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Say $f_n,f:\mathbb{R} \to \mathbb{R}$ are uniformly continous functions a $f_n\to f$ pointwise.

Is the convergence nessecarily uniform?

I couldn't find an appropriate counter-example for this case so I tried proving it :

We have $|f_n(x)-f_n(y)|<\epsilon/3$ for all $|x-y|<\delta_1$, also $|f(x)-f(y)|<\epsilon/3$ for $|x-y|<\delta_2$ and for all $n>N_x$ we have $|f_n(x)-f(x)|<\epsilon /3$.

Overall for any $|x-y|<\delta=min(\delta_1,\delta_2)$ and any $n>N_x$ we have $|f_n(y)-f(y)|\leq |f_n(y)-f_n(x)|+|f_n(x)-f(x)|+|f(x)-f(y)|<\epsilon $ for any $y\in (x-\delta,x+\delta)$ and for any $x\in A\subset\mathbb{R}$.

Now if $A$ is compact, we have a cover of $A$ and we can choose a finite subcover $U$, and take $N=max\{N_x|x\in U\}. $ and for all $n>N$ we have $|f_n(y)-f(y)|<\epsilon$ uniformly .

However, if my proof is correct I have only proven uniform convergence in compact sets, and not on the entire number line. Can my proof be modified so we can get uniform convergence on $\mathbb{R}$? Or can someone provide a counter example where it fails in this case?

Edit: Thanks to the counter-examples in the answers I can see my proof is false. Can anyone point out where my proof fails?

Sar
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  • This kind of like the Dini's theorem, in which the convergence is monotonic – xbh Aug 04 '18 at 15:39
  • This is true, The proof is inspired by Dini`s theorem proof, only with slightly different conditions. – Sar Aug 04 '18 at 15:40
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    A related concept is equicontinuity, which says that the same $\delta$ works for all values of $n$. – Teepeemm Aug 04 '18 at 16:32
  • Thank you @Teepeemm ! I didn`t know the term but I'll read into it since it seems like what I was looking for. – Sar Aug 04 '18 at 16:45

3 Answers3

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I'll give an example on $\mathbb{R}$. Take the sequence $f_n(x)=\frac{x}{n}$. All the functions are uniformly continuous, the limit function is the zero function which is also uniformly continuous. But I say there is no uniform convergence here. Take $\epsilon=1$. For any index $n_o\in \mathbb{N}$ you can take $n=n_0+1$ and $x=2n$ and you will get $|\frac{x}{n}-0|=2 \geq 1=\epsilon$. Hence there is no uniform convergence.

Mark
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The answer is no. To construct a counter-example, we consider $f_{n}(x)=\frac{nx}{1+(nx)^{2}}$ and $f(x)=0$ on interval [0,1]. Obviously they are all continuous functions defined on a compact subset of real axis, hence they are all uniform continuous. It’s also easy to verify that the series {fn(x)} converge point-wise to f(x). However, since for each n, fn(x) attains its maximum 1/2 at x=1/n, so the convergence is not a uniform one.

Shiningale
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  • Thank you. That`s seems like a good counter-example, however, Can you point out where my proof fails? – Sar Aug 04 '18 at 15:51
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    One of the problems I see in your proof is that you wrote that for a given $\epsilon$ there exists a unique $\delta_1$ that works for all $n \in \mathbb{N}$. This is not true in general. All the functions $f_n$ are uniformly continuous but $\delta$ is dependent on $n$. – Mark Aug 04 '18 at 16:00
  • @Mark Thank you for clarifying the same question for me at the same time! – xbh Aug 04 '18 at 16:05
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    @xbh If I remember correctly that's exactly the reason why in Dini's theorem we need monotone convergence. – Mark Aug 04 '18 at 16:08
  • @Mark Yes, the monotonicity allow us to pick one $N_x$ hence one $\delta$ and complete the proof. Thanks again! – xbh Aug 04 '18 at 16:12
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No. Let $f_n(x) = x^n$ on $[0,1]$. Then since continuous functions are uniformly continuous on compact sets, the $f_n$ are uniformly continuous, and converging pointwise to $\textbf{1}_{\{1\}}$, but the convergence is not uniform.

Daniel Xiang
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    OP also asked the limit function to be continuous. So your example is good if you remove the point $x=1$. – Mark Aug 04 '18 at 15:35
  • The limit function is $f=\begin{cases} 1\space\space, x=1 \ 0 \space\space,x\in[0,1)\end{cases}$ – Sar Aug 04 '18 at 15:35
  • But in your example, the limit function is not uniformly continuous, while the OP specifies the uniform continuity of $f$ – xbh Aug 04 '18 at 15:36
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    @xbh if you restrict to $[0, 1)$, then the limit is identically 0 and so is uniformly continuous. Restricting the domain of the $f_n$ doesn't affect their uniform continuity. So as Mark says the OP's example is good subject to a small edit. – Rob Arthan Aug 04 '18 at 15:40
  • @RobArthan Okay, thanks. – xbh Aug 04 '18 at 15:43
  • Notice that the limit function is not continuous over compact set $[0,1]$ – Mostafa Ayaz Aug 04 '18 at 15:44
  • @RobArthan ,When I was considering this counter-example in $[0,a]$ for $a<1$ we have $x^n$ monotonically decreasing (with $n$) and we can apply Dini`s theorem to get uniform convergence. Or am I missing something? – Sar Aug 04 '18 at 15:46
  • @Sar Yeah, in that case Dini's theorem is applicable. – xbh Aug 04 '18 at 15:50
  • @sar the proposed correction is to change the domain to the half-open interval $[0, 1) $. – Rob Arthan Aug 04 '18 at 17:11