# Uniform convergence- Point-wise convergence. Doubt regarding the difference

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.

• I see you edited your question to have $N^*$ as the $\sup$, this is now correct notationally but the problem is fundamental. – TSF May 15 '19 at 7:10

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.
• 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 – RScrlli May 15 '19 at 7:11
• 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 $\epsilon$. – TSF May 15 '19 at 7:12