# Why doesn't any pointwise convergent imply uniform?

Roughly speaking, the difference between the pointwise convergence and the uniform convergence is the N's dependence on x. So for pointwise convergence in domain $$E$$:

$$\exists N \text{ s.t. } |f_n(x)-f(x)|<\epsilon \text{ for } n>N, x \in E, \epsilon >0$$

For uniform convergence to work, we add the condition $$\forall x$$. ie:

$$\exists N \text{ s.t. } |f_n(x)-f(x)|<\epsilon \text{ for } n>N, \forall x \in E, \epsilon >0$$

In the former, N can depend on x, while in the latter, it can only depend on epsilon. The thing is, can't we choose the greatest of all Ns in the former? E.g: if:

$$|f_n(x_1)-f(x_1)|<\epsilon \implies N\geq 2$$ $$|f_n(x_2)-f(x_2)|<\epsilon \implies N\geq 6$$ $$|f_n(x_3)-f(x_3)|<\epsilon \implies N\geq 3$$

can't we choose $$N \geq 6$$, such that all of the following statements are true? Hence implying uniform convergence?

• You're observing that if you have finitely many $x$ in your domain, then pointwise convergence implies uniform convergence (by choosing the largest $N$). The problem is, if you have infinitely many $x$ in your domain, then there may be no largest $N$ to choose.
– jl00
Commented Dec 30, 2022 at 20:43
• A good counterexample is the sequence $$f_n : x\in [0,1] \mapsto x^n \in \mathbb{R}$$ Commented Dec 30, 2022 at 20:45
• It's the quantifiers, mainly. For pointwise read as "for each $x,$ there exists $N$ such that..." versus for uniform "there exists $N$ such that for each $x, \ldots$ Commented Dec 30, 2022 at 20:45
• @oliverjones which x in particular should I pick? Commented Dec 30, 2022 at 20:47
• Well every $x$ should be considered, I suggest playing with the sequence of functions I gave to test the two modes of convergence in question. Commented Dec 30, 2022 at 20:55

If $$E$$ consists of finitely many points, then what you're saying works. But generally you will be interested in infinite domains, in which case the maximum you're considering need not exist.
For example, if $$E=(0,1]$$, and
$$|f_n(1)-f(1)|<\epsilon \implies N\geq 1$$ $$|f_n(1/2)-f(1/2)|<\epsilon \implies N\geq 2$$ $$|f_n(1/3)-f(1/3)|<\epsilon \implies N\geq 3$$ $$\dots$$ $$|f_n(1/k)-f(1/k)|<\epsilon \implies N \geq k$$
then there is no single $$N$$ you can choose that will work for all $$x$$.