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I'd like to prove that the power sequence

$f_n(x) = x^n$

doesn't converges uniformly on $[0,1]$, but it does on $[0,a]$ if $a < 1$.

My textbook states that a sequence of functions converges uniformly to $f(x)$ if

$ \forall \, \epsilon > 0 \,\, \exists \, \overline{n}_{\epsilon}: \forall x \in A \quad |f_n(x) - f(x)| < \epsilon$

So I found that $f_n(x)$ converges pointwise to $f(a,b) = \begin{cases} \text{$0 \leq x < 1 \implies 0$}\\ \text{$x = 1 \,\,\,\, \,\,\,\,\,\,\implies 1$} \end{cases} $

So I tried to apply the definition of uniform convergence, but I can't understand how could I find an $\epsilon > 0: \forall \, \overline{n}, \,\,\exists \, x \in [0,1], \,\,\, \exists \, n \geq \overline{n}: |f_n(x) - f(x)| \geq \epsilon$

Did I misunderstand something about the definition? Thank you in advance

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  • $\begingroup$ How can I construct that sequence? $\endgroup$
    – Francesco Andreuzzi
    Dec 31, 2018 at 12:11
  • $\begingroup$ $f_n(1-1/2) = (1/2)^n$, $\,\, f(1-1/2) = 0$. But $(1/2)^2 < 1/2$ $\endgroup$
    – Francesco Andreuzzi
    Dec 31, 2018 at 12:18
  • $\begingroup$ I take $x_n=1-1/n$, not $1-1/2$. You have $f_n(1-1/n)=(1-1/n)^n\to 1/e$ and $f_n(1-1/n)\geq \frac{1}{4}$ for all $n$ (so take $\varepsilon =1/4$ instead of $1/2$). $\endgroup$
    – Surb
    Dec 31, 2018 at 12:28
  • $\begingroup$ Okay, so $x_n \to 1/e$ does the trick. Is it ok to take a value of $x$ for $f_n(x)$ that depends on $n$? $\endgroup$
    – Francesco Andreuzzi
    Dec 31, 2018 at 12:38
  • $\begingroup$ I don't understand your question : Is it ok to take a value of x for fn(x) that depends on n? Also, it's not correct that $x_n\to 1/e$. $\endgroup$
    – Surb
    Dec 31, 2018 at 12:39

1 Answer 1

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In the way that I think is more constructive to think about it, uniform convergence of $f_n$ to $f$ on $A$ is equivalent to $\underset{x\in A}{\sup} \vert f_n(x)-f(x)\vert \overset{n\rightarrow \infty}{\rightarrow} 0$. So if for all $n$ there exists $x_n\in A$ such that:

$\vert f_n(x_n)-f(x_n)\vert\equiv c>0$

then you would show that there is no uniform to $f$. You have to consider just one $f$, since uniform convergence also implies point-wise convergence.

You could have also gone another route. If a sequence of continuous functions converges uniformly to $f$, then $f$ is a continuous function, which you can see that $f$ is not.

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  • $\begingroup$ Could you please explain which $x_n$ should I consider? $\endgroup$
    – Francesco Andreuzzi
    Dec 31, 2018 at 12:45
  • $\begingroup$ For example if you want to show for $c=\frac{3}{4}$, you would want that $x_n^n-0=\frac{3}{4}$. So for example $x_n:=\Big( \frac{3}{4} \Big)^{\frac{1}{n}} $ would work. $\endgroup$
    – Keen-ameteur
    Dec 31, 2018 at 12:52
  • $\begingroup$ But I know that $f_n(x) = x^n$ converges on $[0,a]$ if $a < 1$. Your proof proves also that my sequence doesn't converge uniformly on that set $\endgroup$
    – Francesco Andreuzzi
    Dec 31, 2018 at 13:01
  • $\begingroup$ It won't since for any $1>a>0$, $x_n$ would eventually be larger than $a$, and thus not be relevant to the interval $[0,a]$. Furthermore for all $c>0$, choosing $x_n =\sqrt[n]{c}$ would mean that $x_n>a$ eventually. $\endgroup$
    – Keen-ameteur
    Dec 31, 2018 at 13:04
  • $\begingroup$ Okay, it makes sense. Thank you very much. Could you please check my last comment below my question, and tell me what you think? For instance, $lim_{n \to \infty} (3/4)^{1/n} = 1$, but this conflicts with the pointwise limit that I definied in my question $\endgroup$
    – Francesco Andreuzzi
    Dec 31, 2018 at 13:09

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