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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?

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    $\begingroup$ Intuitively, this is because $1^n=1$, so that the closer you get to $1$, the slower is convergence to $0$. Informally, $f_\infty(<1)=0$ and $f_\infty(1)=1$, the limit function is discontinuous. $\endgroup$
    – user65203
    Apr 27, 2015 at 13:34

6 Answers 6

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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 $N$.

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

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

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

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  • $\begingroup$ Clever! I like this answer a lot. $\endgroup$ Apr 27, 2015 at 13:52
  • $\begingroup$ Hi @N.S. it seems like you have shown that $f_n$ is not uniformly convergent on the half-open interval (0,1], and not on $(0,1)$, from showing that $f(x_n) \to \frac {1}{e} \ne 1 = f(\lim x_n)$. What do you think? Thanks, $\endgroup$
    – User001
    Dec 13, 2015 at 6:31
  • $\begingroup$ @User001 No, you don't use the limit at $1$. You just write the definition of uniform convergence on $(0,1)$ and use $x_n \in (0,1)$ to get a contradiction.... $\endgroup$
    – N. S.
    Dec 13, 2015 at 19:09
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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.

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  • $\begingroup$ Why is it impossible? $\endgroup$ Apr 8, 2023 at 23:46
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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 $(0,1)$.

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Let $0 < \epsilon < 1$. Put $$x=\frac{1+\epsilon}{2}$$ This gives $\epsilon < x < 1$. Then, $\forall n$, $$\begin{aligned} x^n = \bigl(\frac{1+\epsilon}2 \bigr)^n &=\frac{1}{2^n}(1+\epsilon)^n \\ &= \frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}\epsilon^k > \frac{1}{2^n} \epsilon \sum_{k=0}^n \binom{n}{k} =\frac{1}{2^n} 2^n \epsilon=\epsilon ~.\end{aligned}$$ Hence, there is no number $N$ such that $x^N < \epsilon$, when $0 < \epsilon < 1$. So, uniform convergence fails.

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