I am studying uniform convergence, and I am confused about this:
Does $f_n:[0,0.5]\to[0,0.5], f_n(x)=x^n$ uniformly converge to $f:[0,0.5]\to[0,0.5], f(x)=0$?
My bet is that it does. I failed to see why it wouldn't. I suggested (to my friends with whom I am debating) $N=\left\lceil\frac{\log\epsilon}{\log x}\right\rceil$
As this must happen for all values of $x$, we choose the biggest $x=0.5$ for $N=\left\lceil\frac{\log\epsilon}{\log x}\right\rceil$ because the larger the $x$, the slower $x^n$ gets shrunken. (Someone on a chat has said: "yes, that is true; but you can come up with an $N$ that doesn't depend on $x$" ) If the $x$ that shrinks the slowest has shrunk until the difference is less than $\epsilon$, all other $x$ will have already shrunken even more, so all of them uniformly shrink, and the difference is less than $\epsilon$ for $n\ge N$ for all $x$
To this, I got the reply "You don't need to prove pointwise convergence here. The only way to solve this dispute is to ask on math exchange"
In order to prove that that sequence converges uniformly to the null function, you have to prove that, for every $\varepsilon>0$, there is some $N\in\Bbb N$ depending only on $\varepsilon$ such that$$n\geqslant N\text{ and }x\in\left[0,\frac12\right]\implies|x^n|<\varepsilon.\tag1$$But$$x\in\left[0,\frac12\right]\implies|x^n|\leqslant\frac1{2^n}.$$So, take $N\in\Bbb N$ such that $N>\log_2\left(\frac1\varepsilon\right)$ and then $(1)$ holds.