# 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 $g_n\colon \mathbb{R}\to\mathbb{R}$? Commented Mar 30, 2019 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. Commented Mar 30, 2019 at 18:12
• That is good evidence, yes. Commented Mar 30, 2019 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.