# Uniform convergence of power sequence

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

• How can I construct that sequence? Dec 31, 2018 at 12:11
• $f_n(1-1/2) = (1/2)^n$, $\,\, f(1-1/2) = 0$. But $(1/2)^2 < 1/2$ Dec 31, 2018 at 12:18
• 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 $1/2$).
– Surb
Dec 31, 2018 at 12:28
• 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 $n$? Dec 31, 2018 at 12:38
• 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 $x_n\to 1/e$.
– Surb
Dec 31, 2018 at 12:39

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.

• Could you please explain which $x_n$ should I consider? Dec 31, 2018 at 12:45
• 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. Dec 31, 2018 at 12:52
• 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 Dec 31, 2018 at 13:01
• 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. Dec 31, 2018 at 13:04
• 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 Dec 31, 2018 at 13:09