1
$\begingroup$

So in my book I read two examples that the sequence of functions converges pointwise but not is an uniformly convergence. These sequences are:

$f_n : [0,1) \rightarrow [0,1)$ given by $f_n (x) = x^n \;\;\forall n \in \mathbb{N}$ and $g_n : [0,1) \rightarrow [0,1)$ given by $g_n (x) = \frac{x}{n} \;\;\forall n \in \mathbb{N}$.

So I could understand that both sequences above has a convergence pointwise. I know why $(f_n)$ is not uniformly convergence but I don't understand why $(g_n)$ is not uniformly convergence!

For me, $\forall \epsilon >0$, there is $n_0 = n_0(\epsilon) \in \mathbb{N}$ such that $d_{\infty}(g_n, 0) = \sup_{x \in [0,1]}|g_n(x)-0(x)| = \sup_{x \in [0,1]}|g_n(x)| = \sup_{x \in [0,1]}\Big| \frac{x}{n}\Big| = \frac{x}{n} < \frac{1}{n} < \epsilon. $ For this, just take $n_0 > \frac{1}{\epsilon}$. Thus, the sequence $(g_n)$ is uniformly convergente to function $g \equiv 0.$ But my book said that this sequence is not. Can you help me please?

$\endgroup$
3
  • 1
    $\begingroup$ Are you sure $g_n$ is not defined as $g_n\colon \mathbb{R}\to\mathbb{R}$? $\endgroup$
    – Clement C.
    Mar 30, 2019 at 18:03
  • $\begingroup$ I think that you are correct, because the author want to say that if you take out the compactness of the hypothesis, the Dini's theorem is not valid. $\endgroup$ Mar 30, 2019 at 18:12
  • $\begingroup$ That is good evidence, yes. $\endgroup$
    – Clement C.
    Mar 30, 2019 at 18:21

2 Answers 2

2
$\begingroup$

Of course you don't understand why $(g_n)_{n\in\mathbb N}$ doesn't converge uniformly. It does! And your proof is fine.

However, it doesn't converge uniformly in unbounded intervals (such as $\mathbb R$ or $[0,\infty)$). Perhaps that that's what your textbook says.

$\endgroup$
2
  • $\begingroup$ I think that you are correct, because the author want to say that if you take out the compactness of the hypothesis, the Dini's theorem is not valid. $\endgroup$ Mar 30, 2019 at 18:12
  • $\begingroup$ Indeed, it is not. $\endgroup$ Mar 30, 2019 at 18:14
0
$\begingroup$

You're book may be concern with unbounded intervals. In such case $\frac{x}{n}$ is not uniformly convergent.

The fuction is uniformly convergent if it uniformly approaches to its convergent point so the given sequence fails in to converge uniformly on unbounded intervals

$\endgroup$

You must log in to answer this question.

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