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!

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    $\begingroup$ 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

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 $E$.


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