0
$\begingroup$

Theorem $7.12$ (Rudin): If $\{f_n\}$ is a sequence of continuous functions on $E$ and if $f_n \to f$ uniformly, then $f$ is continuous on $E$.

Given the sequence, $f_n(x) = 3-(x/4)^n$, use Theorem $7.12$ to show that $f_n(x)$ does not converge uniformly on $[0,4]$.

So, we have that $E=[0,4]$ and I thought to use the contrapositive of $7.12$ to show that $f_n$ does not converge uniformly. Now, we have $\lim_{n\to\infty} f_n(x) = \lim_{n\to\infty}3- \lim_{n\to\infty}(x/4)^n = 3 - \frac{1}{1-x} = f(x)$ for $x<4$. This limit is undefined at $x=1$ so $f(x) \equiv f$ is not continuous on $E$ since $1 \in E$. So, $f$ not continuous on $E$ $\implies$ $f_n \not\to f$ uniformly. Is this the right path? Any suggestions would be greatly appreciated. Thanks!

$\endgroup$
  • 2
    $\begingroup$ Where did you get the idea that $\lim_{n \to \infty} (x/4)^n=\frac{1}{1-x}$? That limit is $0$ if $|x|<4$ and $1$ when $x=4$. That is, the pointwise limit of your $f_n$ is $3$ on $[0,4)$ and $2$ at $4$. $\endgroup$ – Ian Feb 23 '15 at 1:49
  • $\begingroup$ Oh, I realize I mixed up the geometric series with this limit. $\endgroup$ – RXY15 Feb 23 '15 at 1:54
3
$\begingroup$

You are right to use the contrapositive, but you didn't find $f$ correctly. Note that for $x \in [0,4)$, $\lim_{n \to \infty}3-\left(\frac{x}{4}\right)^n = 3$. But for $x=4$ then $\lim_{n \to \infty}3-\left(\frac{4}{4}\right)^n =2$. So you have some jump discontinuity at the end of the interval of $E$. Hence, $f$ is not continuous on $E$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.