$f_n$ 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?

  • $\begingroup$ You've got the definitions wrong. One should define uniform convergence, but both claim to describe pointwise convergence. $\endgroup$
    – Thomas
    Feb 17, 2014 at 20:13
  • $\begingroup$ The second one is uniform continuity, ive just forgot to change the name (because i copied and modified from the pointwise version). Thanks anyway. $\endgroup$
    – Emo
    Feb 17, 2014 at 20:15
  • 1
    $\begingroup$ The same question is here. $\endgroup$
    – frabala
    Feb 17, 2014 at 20:53
  • $\begingroup$ Sorry, I haven't seen that! $\endgroup$
    – Emo
    Feb 17, 2014 at 20:55

3 Answers 3


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 $N\in\mathbb{N}$ such, that $\forall n\geq N$, $|f_n(x)-f(x)|<\epsilon$.

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 $\forall n\geq N$, $|f_n(x)-f(x)|<\epsilon, \forall x$.

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 $x \in\mathbb{R}$ 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 $\lim_{n\to\infty}f_n(x_n) \neq f(x)$ (if this limit exists at all). (Worse still, you can choose $x_n$ in such a way that $\lim_{n\to\infty}f(x_n) = \infty$. 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 $g:D\to \mathbb{C}$ 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 $\lim_{n\to\infty}f(x_n)$ 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 $x(n)=\frac{1}{\sqrt[n]{3}} \underset{n\to\infty}{\longrightarrow}1$ 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.

  • $\begingroup$ I don't understand why $\frac{1}{\sqrt[n]{3}}$ goes pointwise to 0, it goes to 1 for some $x$ as you say, right? Why does the uniform convergence break down, but the pointwise doesn't? $\endgroup$ Oct 9, 2019 at 10:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .