Why Does Choosing z=1/n Demonstrate Non-Uniform Convergence?

[Fermat1]

on page 4, example 9 in this link, http://www.personal.psu.edu/auw4/M401-notes1.pdf, they show a sequence of functions is not uniformly convergent. To show this, you need to show that for some epsilon, there is no 'universal' N.

But they didn't pick a particular value of $z$, they chose $z=1/n$, which is a function of $n$. Can anyone explain why this proves that the sequence is not uniformly convergent

 

[Opalg]

on page 4, example 9 in this link, http://www.personal.psu.edu/auw4/M401-notes1.pdf, they show a sequence of functions is not uniformly convergent. To show this, you need to show that for some epsilon, there is no 'universal' N.

But they didn't pick a particular value of $z$, they chose $z=1/n$, which is a function of $n$. Can anyone explain why this proves that the sequence is not uniformly convergent

If you could find a particular (fixed) value of $z$ then the functions would not converge pointwise (because they would not converge at that particular value of $z$). The aim of Example 9 is to provide a sequence of functions that converges pointwise but not uniformly. That necessarily means that the value of $z$ is going to have to depend on $n$.

 

[Fermat1]

If you could find a particular (fixed) value of $z$ then the functions would not converge pointwise (because they would not converge at that particular value of $z$). The aim of Example 9 is to provide a sequence of functions that converges pointwise but not uniformly. That necessarily means that the value of $z$ is going to have to depend on $n$.

So basically, they hare saying that whatever $N$ is chosen, that $N$ 'won't do' for $z=1/N$ ?

 

[Deveno]

Pick an $\epsilon < \dfrac{1}{2}$.

For example, say $\epsilon = \dfrac{1}{4}$.

Suppose, for the sake of showing a contradiction, that $\{f_n\}$ was uniformly convergent on $(0,\infty)$.

We then could find a natural number $N$ (having chosen $\epsilon$, this is now a FIXED natural number) such that:

$|f_n(x)| < \dfrac{1}{4}$ for all $n > N$ and all $x \in (0,\infty)$.

Let $x = \dfrac{1}{N+1}$.

We have:

$|f_{N+1}(N+1)| = \dfrac{1}{2} > \dfrac{1}{4}$, contradiction.

So $\{f_n\}$ must not be uniformly convergent, there is no such $N$.

What actually happens here, is we can find an $N$ that works on $(1/n,\infty)$, but no matter how big $N$ gets, there's still a little bit of $f$ for which the convergence is "too slow", even though that "little bit" winds up getting closer and closer to 0.

This example is closely related to the behavior of $f(x) = \dfrac{1}{x}$ near 0. Clearly, as we approach 0, $f$ approaches $\infty$, but no matter "how close" we get to 0, if we are not AT it, $f$ is still finite. We can bound $f$ on any interval:

$[\epsilon,1]$

but we cannot bound $f$ on $(0,1]$, even though it is defined everywhere on this interval.

This is essentially "built-in" to the real numbers, we can keep creeping closer to the cliff, and we're never actually "forced" to jump off. Even a tiny bit of "wiggle room" is enough to allow functions to do some very odd things.