Pointwise vs. Uniform Convergence

Question

This is a pretty basic question. I just don't understand the definition of uniform convergence.

Here are my given definitions for pointwise and uniform convergence:

Pointwise convergence: Let $X$ be a set, and let $F$ be the real or complex numbers. Consider a sequence of functions $f_n$ where $f_n:X\to F$ is a bounded function for each $n\in \mathbb N$. $f:X\to F$ is the pointwise limit of $f_n$ if for every $x \in X$, $$\lim_{n\to \infty}f_n(x)=f(x).$$

Uniform convergence: Let $f_n$ be a sequence of functions in the set of all bounded functions from $X$ to $F$ where $F$ is the real or complex numbers. The sequence is said to converge uniformly to a bounded function $f:X \to F$ if, given $\epsilon>0$, there exists an $N\in \mathbb N$ s.t. $\sup\{|f_n(x)-f(x)| : x \in X \}<\epsilon$ for $n\ge N$

I'm sorry I don't have a more specific question. I just don't see the exact relation/difference between the two definitions. I've asked two different professors to explain this to me but neither of their explanations helped.

Edit: Attempting to show that uniform convergence implies pointwise convergence if $f_n$ converges uniformly to f, then $\sup\{|f_n(x)-f(x)| : x\in X \}$ for $n\ge N$. Thus, $|f_n(x)-f(x)|<\epsilon$ for $n\ge N$, which is the definition of pointwise convergence.

Answer 1

It may help if you unfold the definition of limit in pointwise convergence.

Then pointwise convergence means that for each $x$ and $\epsilon$ you can find an $N$ such that (bla bla bla). Here the $N$ is allowed to depend both on $x$ and $\epsilon$.

In uniform convergence the requirement is strengthened. Here for each $\epsilon$ you need to be able to find an $N$ such that (bla bla bla) for all $x$ in the domain of the function. In other words $N$ can depend on $\epsilon$ but not on $x$.

The latter is a stronger condition, because if you have only pointwise convergence, it may be that some $\epsilon$ will require arbitrarily large $N$ for some $x$s.

For example, the functions $f_n(x)=\frac{x}{n}$ converge pointwise to the zero function on $\mathbb R$, but do not converge uniformly. For example, if we choose $\epsilon=1$, then the convergence condition boils down to $N>|x|$. For each $x\in\mathbb R$ we can find such an $N$ easily, but there's no $N$ that works simultaneously for every $x$.

Answer 2

$f_n\to f$ pointwise on $(a,b)$ if for each fixed $x\in(a,b)$, $|f_n(x)-f(x)|\to 0$ as $n\to\infty$. Notice this is a pointwise (local) criterion.

On the other hand, $f_n\to f$ uniformly on $(a,b)$ if $\sup_{a< x<b}|f_n(x)-f(x)|\to 0$ as $n\to\infty$. This is a "global" criterion in that is requires the maximum of all the pointwise errors on $[a,b]$ to tend to zero.

As an example, $f_n(x)=x^n$, $0\le x\le 1$ converges pointwise to $f(x)=\begin{cases} 0, &0\le x<1,\\ 1, &x=1.\end{cases}$, because the first condition above holds, but the convergence is not uniform since the second condition does not hold.

Answer 3

The pointwise convergence depends on each $x$, that is you need to count and fix every $x$ before passing to the limit. On the other hand, uniform convergence is independent of $x$. Hence, the pointwise convergence is enough when your space $X$ is a countable set and not enough when you are working with an uncoutable set like $[0,1]$, in such case you need a uniform convergence to generalise your convergence on $[0,1]$.

Answer 4

Let $(f_n)$ be a sequence of functions defined on a subset $E$ of $\mathbb R$, and suppose that $f:E\to\mathbb R$ has the property that for all $x\in E$, $$ f(x)=\lim_{n\to\infty}f_n(x) \, . $$ Then, $f$ is called the limit function of the sequence $(f_n)$, and we say that $(f_n)$ converge pointwise to $f$.

It is natural to ask the question: "to what extent do the functions $f_n$ end up resembling the function $f$"? If no further hypotheses are added, then the resemblance may be weak. For instance, the limit of a sequence of continuous functions might be discontinuous, the limit of a sequence of integrable functions might be non-integrable, and so on.

Why does this alarming state of affairs occur? Well, if we unfold the meaning of the statement "for all $x\in E$, $f(x)=\lim_{n\to\infty}f_n(x)$", it becomes $$ (\forall x\in E)(\forall\varepsilon>0)(\exists N\in\mathbb N)(\forall n\in\mathbb N)(n\ge N\implies |f_n(x)-f(x)|<\varepsilon) \, . $$ This formidable looking statement holds the answer to our question. Note that the clause "$(\forall x\in E)$" appears before the clause "$(\exists N\in\mathbb N)$", and so the value of $N$ is allowed to depend on the value of $x$. Therefore, there is a possibility that no value of $N$ works for every value of $x$.

It is easiest to understand this with an example. If $f_n(x)=x^n$ on $[0,1]$, then $(f_n)$ converges pointwise to the function $f(x)$ that equals $0$ when $0\le x<1$ and equals $1$ when $x=1$. This is trivial to see for $x=0$ and for $x=1$; if $0<x<1$, we can proceed as follows: given an $\varepsilon>0$, we can choose an $N\in\mathbb N$ satisfying $N>\log(\varepsilon)/\log(x)$. Then, if $n\ge N$, it follows that $$0<x^n\le x^N<x^{\log(\varepsilon)/\log(x)}=\varepsilon \, .$$

On the other hand, in the above example, it is impossible to choose an $N\in\mathbb N$ which does not depend on the value of $x$. In other words, it is not the case that $$ (\forall\varepsilon>0)(\exists N\in\mathbb N)(\forall n\in\mathbb N)(\forall x\in E)(n\ge N\implies|f_n(x)-f(x)|<\varepsilon)\tag{*}\label{*} $$ To prove this, we can have to demonstrate the negation of $\eqref{*}$, which is $$ (\exists\varepsilon>0)(\forall N\in\mathbb N)(\exists n\in\mathbb N)(\exists x\in E)(n\ge N\text{ and }|f_n(x)-f(x)|\ge\varepsilon) \, . $$ Take $\varepsilon=1/2$, and consider that if $x=\sqrt[N]{1/2}$ then $x^N=1/2$.

What this shows is that no matter how big $N$ is, there is a value of $x$ such that $|f_N(x)-f(x)|=1/2$, and consequently the function $f_N$ looks noticeably different to $f$ at this point.

If, on the other hand, the condition $\eqref{*}$ holds, then for any $\varepsilon>0$ thrown at you, there is an $N\in\mathbb N$ such that the functions $f_N,f_{N+1},f_{N+2},\dots$ are "within $\varepsilon$" of $f$: at no point does the value of $f$ differ from one of these functions by a margin greater than $\varepsilon$. Consequently, the functions $f_n$ really do end up visually resembling the function $f$, and many of the properties that the functions $f_n$ have will end up being shared by $f$.

The condition $\eqref{*}$ is precisely what uniform convergence means.