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I know the definition of "pointwise convergence" and "uniform convergence", nevertheless I have some difficulties understanding the difference between those two concepts.

My book defines Uniform convergence as follows:

Let be $f_n$ a sequence of functions on $A\subseteq \mathbb{R}$. Then $(f_n)$ converges uniformly on $A$ to a limit function $f$ defined on $A$ if, for every $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $|f_n(x)-f(x)|<\epsilon$ whenever $n\geq N$ and $x\in A$.

Now I do understand that in the case of uniform convergence we can choose $N$ irrespectively of the point $x$ while in the case of Pointwise convergence this is not the case (something like the difference between continuity and uniform continuity).

If the sequence of functions converge pointwise to $f$ it means that $\forall \epsilon>0, x\in A$ there exists $N$ such that $|f_n(x)-f(x)|<\epsilon$ whenever $n\geq N$, right? In this case $N$ depends both on $\epsilon$ and $x$.

My question is, given $\epsilon$ couldn't we pick an $N^*= \sup_{x\in A}{N(x,\epsilon)}$ (maybe this notation is a little bit messy).

Then we know that for this $\epsilon$ $, |f_n(x)-f(x)|<\epsilon$ whenever $n\geq N^* \forall x\in A$.

I know that in some cases this cannot be done, but I cannot see whether this is always false.

Would you mind to spot my error here? Thanks in advance.

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  • $\begingroup$ I see you edited your question to have $N^*$ as the $\sup$, this is now correct notationally but the problem is fundamental. $\endgroup$ Commented May 15, 2019 at 7:10

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The value $N^*$ might not exist in $\mathbb{N}$ (i.e. it's $\infty$). You write $\max$ but really you should only write $\max$ when the underlying set (in this case, $A$) is finite. If it is not you must write $\sup$ and the $\sup$ is not always attained. If the $\sup$ is attained, i.e. $N^*(\epsilon)$ is finite, for all $\epsilon$ then there is no issue with choosing $N^*$ and you have uniform convergence.

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  • $\begingroup$ Okey, so we can claim that if the set $A$ is finite pointwise and uniform convergence are both verified. Instead if $A$ is not finite then I cannot state anything in the general case $\endgroup$
    – Chaos
    Commented May 15, 2019 at 7:11
  • $\begingroup$ Well more precisely, regardless of the set $A$, we have that if $N^*(\epsilon)$ is finite for all $\epsilon$ then you have uniform convergence (by the argument you used, just take $N^*$) but it must hold for all values of $\epsilon$. $\endgroup$ Commented May 15, 2019 at 7:12
  • $\begingroup$ It makes completely sense, I'll accept the answer as soon as the system allow me. Thanks sir. $\endgroup$
    – Chaos
    Commented May 15, 2019 at 7:14

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