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Doing the limit we can see that in the open interval it converges pointwise to the constant function $f(x) = 0$. In the closed interval it doesn't converge uniformly because in $x=1$ $f(x) =1$ and when $0<x<1$ then $f(x)=0$. It isn't a continuous function although $f_n(x)$ is continuous for all $n$. So it can't be uniform.

However,let $0<ε<1$, we know the function increases as $x$ increases. So,

$|f_n(x) - f(x)| \leq ε^{n}$

And ε^(n) goes to 0 as n tends to infinity. Doesn't it mean that the convergence is uniform?

José Carlos Santos
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Marco Villalobos
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  • You might also want to see these pages: https://math.stackexchange.com/questions/1254285/why-is-f-nx-xn-not-uniformly-convergent-on-0-1 and https://math.stackexchange.com/questions/1773464/is-xn-uniformly-convergent-on-0-1. – Minus One-Twelfth Apr 23 '19 at 10:17

2 Answers2

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No, it doesn't mean that. Yes, $\lim_{n\to\infty}\varepsilon^n=0$, but that is not relevant here. Just note that$$(\forall n\in\mathbb N):\left(\sqrt[n]{\frac12}\right)^n=\frac12.$$But then $(f_n)_{n\in\mathbb N}$ doesn't converge uniformly to the null function. And, since it does converge pointwise to the null function, you can conclude that it doesn't converge uniformly at all.

José Carlos Santos
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  • I don't get why it isn't relevant. Isn't uniform convergence about finding something greater than |fn(x) - f(x) | that doesn't depend on x and which gets closer to 0 as n tends to infinity? – Marco Villalobos Apr 23 '19 at 09:34
  • It's about, given a number $\varepsilon>0$, to prove that $\bigl\lvert f(x)-f_n(x)\bigr\rvert<\varepsilon$ when $n$ is large enough and for every $x$ in the domain. – José Carlos Santos Apr 23 '19 at 09:45
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    Using your $|f_n(x)-f(x)|\le \varepsilon^n$ argument, you are effectively showing (via the squeeze theorem) that if $x\le \varepsilon$, then $f_n(x)\to f(x)$ as $n\to \infty$. But to show uniform convergence, you would essentially have to show that for any $\epsilon > 0$, there exists an $N>0$ (which can depend on $\epsilon$ but not on $x$) such that for all $x\in[0,1)$, whenever $n > N$, we have $|f_n(x)-f(x)| < \epsilon$. Your $|f_n(x)-f(x)|\le \varepsilon^n$ argument didn't show this. – Minus One-Twelfth Apr 23 '19 at 10:27
  • ε^(n) is greater than any x in the domain [0,ε) because it's an increasing function. – Marco Villalobos Apr 23 '19 at 10:40
  • Why do you write $\varepsilon^{(n)}$ instead of $\varepsilon^n$? – José Carlos Santos Apr 23 '19 at 10:41
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    Even though $\varepsilon > x$ for all $x\in [0,\varepsilon)$, this will not imply uniform convergence on $[0,1)$. You might want to see this question: https://math.stackexchange.com/questions/1773464/is-xn-uniformly-convergent-on-0-1. – Minus One-Twelfth Apr 23 '19 at 10:56
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No, $x^n$ does not converge to zero uniformly on $[0,1)$, because

$$\lim_{n\to\infty}\sup_{0\leq x<1}|x^n|=1\neq 0$$

uniquesolution
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  • Where does that expression comes from? – Marco Villalobos Apr 23 '19 at 09:37
  • It comes from the definition of uniform convergence. Have you seen it in your studies? A sequence $f_n$ converges uniformly to $f$ on a set $A$ if and only if $\lim_{n\to\infty}\sup_{x\in A}|f_n(x)-f(x)|=0$. The proof is easy. – uniquesolution Apr 23 '19 at 09:39
  • I was trying to apply the definition to solve the original problem – Marco Villalobos Apr 23 '19 at 09:44
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    Go right ahead and apply the definition. Take $\varepsilon=\frac{1}{2}$. Now I give you some positive integer $N$. Now you find $n>N$ sufficiently large and $x$ sufficiently close to $1$ such that $x^n>\frac{1}{2}=\varepsilon$.
    1. Can you do that?
    2. Do you understand why it proves that there is no uniform convergence? 3. Do you see that this process is equivalent to the one line I wrote in my answer?
     – uniquesolution Apr 23 '19 at 09:48
  • @Marco Villalobos, "the definition" is ambiguous thing unless you write it down. To me, uniform convergence in some function space is convergence in sup norm, precisely what uniquesolution used in their answer. – Ennar Apr 23 '19 at 09:53
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    @uniquesolution all right, I see your point, but what if we only study the interval [0,ε]. Applying the same reasoning, x^(n) can't be greater than ε. Moreover, i get points 1 and 2, but i don't see why 3 is equivalent – Marco Villalobos Apr 23 '19 at 09:56
  • We do have uniform convergence on $[0,\varepsilon]$ for any constant $\varepsilon$ with $0 < \varepsilon < 1$. But this does not imply uniform convergence on $[0,1)$. – Minus One-Twelfth Apr 23 '19 at 10:54
  • @minusOne-Twelfth That's where i was trying to get! But ε can be arbitrarily close to 1, so why not in the open interval? – Marco Villalobos Apr 23 '19 at 10:59
  • See this question maybe, it's asking a similar thing: https://math.stackexchange.com/questions/1773464/is-xn-uniformly-convergent-on-0-1. – Minus One-Twelfth Apr 23 '19 at 10:59
  • The problem is, even though $\varepsilon$ can be arbitrarily close to $1$, you have to fix the value of $\varepsilon$. Once you fix it, no matter how close to $1$ it is, given any $N > 0$, there will exist an $n \ge N$ and an $x\in [0,1)$ such that $x^n \ge 1/2$ (say); by definition, this implies that we don't have uniform convergence. Such $x$ may well have to be extremely close to $1$, even greater than $\varepsilon$! But the key is that $\varepsilon$ is fixed, so no matter how close to $1$ it is, we will be able to get $x$ even closer to $1$ and have $x^n \ge 1/2$. – Minus One-Twelfth Apr 23 '19 at 11:04
  • By the way, you may also like to see the answer here to possibly gain some intuition for uniform convergence in general: https://math.stackexchange.com/questions/679951/the-difference-between-pointwise-convergence-and-uniform-convergence-of-function. – Minus One-Twelfth Apr 23 '19 at 11:06