# Does $f_n$ converge uniformly?

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?

• Are you sure $g_n$ is not defined as $g_n\colon \mathbb{R}\to\mathbb{R}$? – Clement C. Mar 30 at 18:03
• 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. – Thiago Alexandre Mar 30 at 18:12
• That is good evidence, yes. – Clement C. Mar 30 at 18:21

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