Why this answer does not provide counter-example for $x^n\in C([0,1/2],||\cdot||_\infty)$
From:https://math.stackexchange.com/a/711371/342943
I have thought of a simpler answer, which I believe to be correct, it plays on the assumption of cauchy, and then letting $m=N$, and $n=N+1$. Then: $\|x^m-x^n\|_\infty= ||x^{N}-x^{N+1}\|_\infty = \|x^{N}(1-x)\|_\infty $ Now: $ ||x^{N}(1-x)\|_\infty = \sup_{x\in [0,1]}|x^{N}| |(1-x)| \geq \frac{1}{2}^N\times\frac{1}{2} = \frac{1}{2}^{N+1} $ Now let $$ \varepsilon =\frac{1}{4}^{N+1} $$, so $\|x^N-x^{N+1}\|_\infty\geq \frac{1}{2}^{N+1} \gt \frac{1}{4}^{N+1} = \varepsilon $
I know that $x^n\in C([0,1/2],||\cdot||_\infty)$ is uniformly convergent but not in $C([0,1],||\cdot||_\infty)$ But I cannot understand why the above answer from Ellya, is not valid in the case $x^n\in C([0,1/2],||\cdot||_\infty)$ so with this counter-exampe doesnot Ellya, show that $x^n\in C([0,1/2],||\cdot||_\infty)$ is not uniform convergent, as well?
That answer makes no sense, because in that answer the number $\varepsilon$ is $4^{-N}$; it should be a fixed number greater than $0$.