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I am having trouble understanding the intuition behind the difference between point-wise convergence of functions and uniform convergence of functions once revisited. I am quoting the following definition from a book:

Theorem: Let$\{f_n\}$ be defined on $S$. Then (a) $\{f_n\}$ converges point wise to $F$ on $S$ if and only if there is, for each $\epsilon>0$ and $x\in S$ an integer $N$(which may depend on $x$ as well as $\epsilon$) such that: $|F_n(x)-F(x)|<\epsilon\:\:\:\:\:\:\text{if}\:\:\:\:\:n\geqslant N$ (b)$F_n$ converges uniformly to $F$ on $S$ if and only if there is for each $\epsilon>0$ an integer $N$(which depends only on $\epsilon$ and not on any particular $x$ in $S$) such that: $|F_n(x)-F(x)|<\epsilon\:\:\:\:\:\:\text{for all}\:x\in S\:\text{if}\:n\geqslant N$

So uniform convergence means that after the $N$ the $f_n$ is going to approach $f$ at the same pace $\epsilon$ for all $x$? While point wise convergence means that a function converges on a given point $x$ that is why $N$ depends on $x$, right?

Could someone provide me examples of both cases, functions that converge point wise and functions that converge uniformly?

Thanks in advance!

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  • $\begingroup$ By "theorem" you must mean "definition" ;). $\endgroup$ – Megadeth Apr 8 '17 at 15:55
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You can at least try to see that uniform convergence implies pointwise convergence, which follows directly from the definitions.

For a function sequence converging pointwisely but not uniformly, a handy example that came to my mind is this: $$ f_{n}: x \mapsto \frac{x}{n}, \mathbb{R} \to \mathbb{R} $$ for all $n \in \mathbb{N}$. For every $x \in \mathbb{R}$, the sequence $(f_{n}(x))$ of real numbers converges to $0$ apparently. However, for every $l \neq 0$, there is some $\varepsilon > 0$, say $\varepsilon := |l|/2$, such that for every $N \in \mathbb{N}$ there are some $n \geq N$ and some $x \in \mathbb{R}$, say $n := N$ and $x := 2Nl$, such that $$ |f_{n}(x) - l| = |\frac{x}{n} - l| = |l| \geq \varepsilon. $$ For $l = 0 $, take $\varepsilon := 1$, for example, to see that it still holds true that for every $N \in \mathbb{N}$ there are some $n \geq N$ and some $x \in \mathbb{R}$, say $n := N$ and $x := 2N$, such that $$ |f_{n}(x) - l| = |\frac{x}{n}-l| = 2 \geq \varepsilon. $$ This proves the negation of the definition of uniform convergence; so $(f_{n})$ is not uniformly convergent.

An example of uniformly convergent function sequence is even easier. Let $$ g_{n}: x \mapsto \frac{1}{n}, \mathbb{R} \to \mathbb{R} $$ for all $n \in \mathbb{N}$. I assert that $g_{n} \to 0$ uniformly. Note that for every $\varepsilon > 0$, there is some $N \in \mathbb{N}$, say $N := \lfloor \frac{1}{\varepsilon} \rfloor + 1$, such that $$ |g_{n}(x) - 0| = \frac{1}{n} < \varepsilon $$ for all $n \geq N$. We are done.

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