So I know of three possible ways to try and prove something is not uniformly convergent and I have tried all three but not one of them has given me peace about a definitive answer.
I know for $[0,1)$ this function converges uniformly to $f(x)=1$
Now for the interval $[0,1]$ the limit function $f(x)= \begin{cases} 1 & 0\le x< 1 \\ \frac{1}{2} & x=1 \end{cases} $
I thought I could use this definition I have, to prove it is not UC but I am not sure if it works right.Let me know what you think:
The definition states:
The function sequence $\{f_{n}\}$ fails to converge uniformly on $D$ to the function $f(x)$ provided that there exists a positive number $\epsilon$ that for infinitley many $n\in \mathbb{N}$ there is a number $x_n$ in $D$ satisfying $|f_n(x_n)-f(x_n)| \ge \epsilon.$
So say I choose $\epsilon =\frac{1}{6}$ and I let $x_n =(\frac{1}{2})^{1/n}$
Obviously $x_n$ is in $D$ and $x_n\le 1$for infinitley many $n\in \mathbb{N}$
So now we evaluate our inequality and notice
$|\frac{1}{1+((\frac{1}{2})^{1/n})^n}-1|=| \frac{2}{3}|\ge\frac{1}{6}$ and $|\frac{1}{1+((\frac{1}{2})^{1/n})^n}-\frac{1}{2}|=| \frac{1}{6}|\ge\frac{1}{6}$
Thus we can conclude that $f_n(x)$ does not converge uniformly on $f(x)$
Does this proof look correct? if not can you suggest some other method or point me in the right direction?
Note that $f_n({1 \over \sqrt[n]{3}}) = {3 \over 4}$ for all $n$.
Hence $|f_n({1 \over \sqrt[n]{3}}) -f({1 \over \sqrt[n]{3}})| = {1 \over 4}$ for all $n$.