Ytukyg

$f_n$ converges pointwise to $f$ on $E$ if $forall x in E$ and $forall varepsilon > 0$, $exists N in mathbb N$, such that $forall n geq N$ we have $left|f_n(x) − f(x)right| < varepsilon$.

$f_n$ converges uniformly to $f$ on $E$ if $forall varepsilon > 0$, $exists N in mathbb{N}$ so that $forall n geq N$ we have $left|f_n(x) − f(x)right| < varepsilon$ which holds for all $x in E$.

I know the difference in definition, pointwise convergence tells us that for each point and each epsilon, we can find an $N$ (which depends from $x$ and $varepsilon$)so that ... and the uniform convergence tells us that for each $varepsilon$ we can find a number $N$ (which depends only from $varepsilon$) s.t. ... .

But it seems again to me a bit fuzzy. Can somebody explain the difference more 'deeper' or more 'philosophically'? Or may be illustrate it visually?

Look at this as a game between two people (practically, this can't happen but let our imagination flow). Each game actually is a "translation" of the corresponding definition.

You want to prove pointwise convergence. Lets make a game for it.

• 
  Step 1. You chose some $x$ from the domain.


• 
  Step 2. The opponent chooses some $epsilon >0$.


• 
  Step 3. You try to find an $Ninmathbb{N}$ such, that $forall ngeq N$, $|f_n(x)-f(x)|<epsilon$.

Additional Rules: Steps 2 and 3 must repeat until your opponent is convinced that for the particular $x$ you picked up, whatever $epsilon$ he tells you, you'll be able to find such an $N$. You will be able to select another $x$ from the domain, only after you have convinced your opponent. The games is over when you have done this for all $x$ in the domain.

As you may have already noticed, there is a precise correspondence between the definition and the game. Step 1 corresponds to binding the variable $x$. When $x$ takes a value, this value becomes fixed, so that we can range over $epsilon$, which corresponds to Step 2. After a value is chosen for $epsilon$, this value remains fixed. In Step 3 you find the proper $N$ within the context of already chosen values for $x$ and $epsilon$. This is why in pointwise convergence $N$ depends on both $x$ and $epsilon$.

Now, you want to prove uniform convergence. The game here changes.

• 
  Step 1. Your opponent chooses an $epsilon > 0$.


• 
  Step 2. You try to find an $N$ such that $forall ngeq N$, $|f_n(x)-f(x)|<epsilon, forall x$.

The difference here is that you don't check each $x$ separately. On the contrary, your opponent gives you an $epsilon$ and the $N$ you have to find refers to all $x$ in the domain, not just one you picked up. It is like considering the range of $x$ all at once. The game ends when your opponent is convinced that you'll find such an $N$, whatever $epsilon$ he tells you.

Let's look at an example. Consider the sequence of functions
$$
f_n(x) = begin{cases}
|x|-n &text{if }x in (-infty,-n)cup(n,infty)\
0 &text{if }x in [-n,n].
end{cases}
$$

Graph these functions. Without thinking about the definitions you have of different types of convergence, do you think that $f_n$ converges to some other function? If you said "yes", then what's its limit? $f(x) = 0$? But for each $M geq 0$ notice that, for each $n$, I can find an $x inmathbb{R}$ for which $|f_n(x) - f(x)| > M$. (Do you see an $x$ which works? Try $x = 2M + n$.) Doesn't that seem a bit weird, if ${f_n}$ converges to $f$? Indeed, this means that I can define a sequence ${x_n}$ of real numbers such that $|f_n(x_n) - f(x) > M$ for all $n$, and therefore $lim_{ntoinfty}f_n(x_n) neq f(x)$ (if this limit exists at all). (Worse still, you can choose $x_n$ in such a way that $lim_{ntoinfty}f(x_n) = infty$. I'll leave this up to you.)

Notice that $f_n to f$ pointwise. (Prove this.)

If we can agree that this is undesirable behavior to have for convergent sequences of functions, then you already know one of many reasons why we need a stronger definition of convergence $f_n to f$: pointwise convergence is too weak a property to be useful when discussing many properties of function sequences.

This is where uniform convergence comes into the picture. Can you see that ${f_n}$ does not converge to $f$ uniformly? (Try proving this from the definition.) Intuitively, uniform convergence encodes the idea that the entire function $f_n$ converges to $f$ at the same time (or same rate), not just that $f_n(x)$ moves towards $f(x)$ eventually at each point $x$. It's a global property of the functions $f_n$ and $f$ (i.e., a statement about $f_n$ involving the behavior of $f_n$ on its entire domain at once), not a local property like piecewise convergence (i.e., a property about $f_n$ that is equivalent to statements that individually only take into account small portions of the domain of $f_n$ (in this case, individuals points)). Indeed, a good exercise for you would be to show that if $f_n, f: mathbb{R} to mathbb{C}$ for each $n$, then $f_n to f$ uniformly if and only if $|f_n - f|_infty to 0$ as $n to infty$, where $|cdot|_infty$ is the sup-norm defined for $g:Dto mathbb{C}$ by

$$
|g|_infty = sup_{x in D}|g(x)|
$$

For ${f_n}$ and $f$ as in the above example, you can, in fact, show that $|f_n - f|_infty = infty$ for all $n$ (see the parenthetical about $lim_{ntoinfty}f(x_n)$ above), which gives you a second way to prove that $f_n$ does not converge uniformly to $f$.

Hopefully this clarifies the concepts some. I'd highly recommend trying different examples (both from a textbook and ones you construct on your own). Conveniently, good exercises aren't to terrible to come up with on your own for this concept; whenever you know $f_n to f$ pointwise, you can always ask whether or not $f_n to f$ uniformly, as well.

I understand why you find it fuzzy. I'll try to clarify it the way I see it:

I like to approach definitions by asking: Why were they thought of?

In this scenario, we have 2 very similar definitions that it's hard to visualize the difference between them.

I like to think of Uniform Convergence as the one that came first since it's the more intuitive of the two. In simple term, all it means is that $f_n(x)$ converges to $f(x)$ by only increasing $n$.

A simple example where it's easy to intuitively see the Uniform Converges is $f_n(x)=sqrt{x^2+frac{1}{n}}to sqrt{x^2}$.

Why do we need Point wise convergence? Because sometimes this definitions breaks down and doesn't work:

For example: Let $D=(0,1)$ and $f_n(x)=x^n$. At first, it seems like $f_n(x)$ Uniformly Converges to $0$. But upon a closer inspection we notice that if $x(n)=frac{1}{sqrt[n]{3}}
underset{ntoinfty}{longrightarrow}1$ then $f_n(x)$ doesn't converge to $0$. In other words, we have to limit ourselves to $x$s such that an $N$ exists where for $n>N$ it converges (in other words, $N$ is also a function of $x$).

Notice how the Uniform Convergence broke down on the edge case of $x(n)to 1$. This is the common theme for Pointwise convergence problems and is why (in my opinion) we have this definition.

To sum up: Uniform convergence is the intuitive definition. Pointwise Convergence was defined to handle the special cases where Uniform Convergence doesn't work.