Is this right?
If $f_{n}(x):C \rightarrow \mathbb{R}$ is continuous, $C$ is compact, and $f_{n}(x)$ converges to $f(x)$ pointwisely for each $x$ in $C$, and $f(x)$ is continuous, then $f_{n}(x)$ converges to $f(x)$ uniformly on $C$.
I thought that since $C$ is compact, and each $f_{n}, f$ is continuous, by extreme value theorem, there exists $c$ in $C$ s.t $f_{n}(c)-f(c)=\sup|f_{n}(x)-f(x)|$ and $f_{n}(c) \rightarrow f(c)$ as $n \rightarrow \infty$, so we get $\limsup|f_{n}-f|=0$.
I know that if I add a condition that $f_{n}$ is monotone, then this is Dini's theorem, but I don't know why we need that condition. Where am I wrong?

  • $\begingroup$ @Davide Giraudo: You ruined the question there with your $\mathbb C$'s! $\endgroup$ – TonyK Oct 7 '14 at 9:07
  • $\begingroup$ It is not my $\mathbb C$: there were here before. $\endgroup$ – Davide Giraudo Oct 7 '14 at 9:09
  • $\begingroup$ Oh, sorry. It was Pravisha John (approved by mt_ and idm). $\endgroup$ – TonyK Oct 7 '14 at 9:14
  • $\begingroup$ why is $f_n(c)-f(c)=\sup |\cdot|$? what does $\limsup$ in the last equation mean? There are argumetns of $f_n$ and $f$ missing. I also edited that to the correct meaning but somehow this is again cancelled :( Who downvoted this question? I think, the typesetting is not optimal but to my mind its absolutely reasonable. $\endgroup$ – Quickbeam2k1 Oct 7 '14 at 9:15
  • $\begingroup$ Let g(x)=f_n(x)-f(x) then by EVT, there exists c in C s.t |g(c)|=sup|g(x)|. and since f_n goes to f pointwise, g(c) also goes to 0. this implies the uniform convergence. this is my argument. $\endgroup$ – ooooo Oct 7 '14 at 9:29

Probably the most prominent counterexample is the following. Let $f_n:[0,1]\to\mathbb{R}$ be such that

$$f_n(x)=\begin{cases}n^2x& x\in[0,1/n]\\ 2n-n^2x& x\in[1/n,2/n]\\0& \text{otherwise} \end{cases}$$

Then $f_n$ converges poitnwise to $f=0$ which is continuous but $||f_n-f||_\infty=n$ hence we do not obtain uniform convergence!

Edit: As you see here, the existence of $c$ u claimed is precsiley the point $c_n=1/n$. Particularly, $c_n$ depends on $n$, hence the argument with the pointwise convergence for a fixed $c$ fails.

  • $\begingroup$ I mean, pointwise is also one of my assumptions. I wonder if I get rid of just monotone condition, it converges uniformly. $\endgroup$ – ooooo Oct 7 '14 at 9:23
  • $\begingroup$ No, you can't. The example only violates the monotonicity assumption. As stated above. your argument, that $c$ is independent of $n$ is false. $\endgroup$ – Quickbeam2k1 Oct 7 '14 at 9:41
  • $\begingroup$ does it converge to 0 pointwise? at x=1/n, f_n(1/n)=n. why does it converge to 0?? $\endgroup$ – ooooo Oct 7 '14 at 9:54
  • $\begingroup$ How is pointwise convergence defined? For $x=0$ we have $f_n(x)=0$ for every $n$. Hence $f_n\to 0$ in $x=0$. For $x>0$. There is an $n_0(x)$ such that $f_n(x)=0$ for every $n>n_0(x)$. Thus $f_n$ converges to zero everywhere on $[0,1]$. $x=1/n$ is not a fixed point. Hence it is irrelevant for pointwise convergence! $\endgroup$ – Quickbeam2k1 Oct 7 '14 at 10:08

Dini's theorem requires that the sequence $\{f_n(x)\}$ is monotone! (it.wikipedia.org/wiki/Lemma_di_Dini)


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