Why is $f_n(x) = x^n$ not uniformly convergent on $(0, 1)$?

Question

Definition of uniform convergence:

For all $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $d(f_n(x), f(x)) < \epsilon$ for all $n > N \in \mathbb{N}$ and all $x \in (0,1)$.

It's easy to see that for all $x$, $f_n(x) \to 0$ on $(0, 1)$ which means that $f_n$ is pointwise convergent, but since $f_n$ converges pointwise to $0$ for all $x$, I don't see any counter-examples that we can take $x$ to be in order for $f_n$ to converge to any other limit besides $0$. I also don't see how we can use the definition in order to prove that $f_n$ is not uniformly convergent. In this case, what should I do?

Answer 1

Note that $\lim_{x\to1} f_n(x) = 1$ for all $n$. This breaks uniform convergence because we can get close enough to $1$ such that $f_N(x) > \frac12$ for any fixed $N$.

Answer 2

Hint. You have $$\sup_{x\in(0,1)}f_n(x)=1.$$

Answer 3

We can see this directly. If it were uniformly convergent (to $0$), for any $\varepsilon>0$, we could find $N_0$ such that $0<x^n<\varepsilon$ for all $n\ge N_0$ and all $x\in(0,1)$. In particular, $x^{N_0}<\varepsilon$, which is the same as $\,0<x<\varepsilon^{\frac1{N_0}}$ for all $x\in(0,1)$. That is impossible.

Answer 4

Hint If $x_n =1-\frac{1}{n}$ then $f_n(x_n) \to \frac{1}{e}$.

Use this to contradict $d(f_n(x_n), f(x_n))<\epsilon$.

Answer 5

What exponent is needed for $(.9)^n$ to be less than $.1$? Answer: $$n > \ln(.1) / \ln(.9) $$

What exponent is needed for $(.99)^n$ to be less than $.1$? Answer: $$n > \ln(.1) / \ln(.99) $$

What exponent is needed for $(.999)^n$ to be less than $.1$? Answer: $$n > \ln(.1) / \ln(.999) $$

What is the limit of these exponents $n$ as the number of $9$'s increases to $+\infty$? Answer: $$\lim \, n = \lim_{x \to 1^-} \ln(.1) / \ln(x) = +\infty $$

Therefore, the sequence of functions $f_n(x) = x^n$ does not converge uniformly to zero on the interval $(0,1)$.