Consider the following functions defined on the interval $[0,1]$:
$f_k(x)=x^k$ and $f(x)=\left\{\begin{array}{rcl} 0 & \mbox{if} & 0\leq x < 1, \\ 1 & \mbox{if} & x = 1. \end{array} \right.$
I need to find $\lim_{k \to \infty} \sup_{x \in [0,1]}|f_k(x)-f(x)|$.
My Attempt:
If $x=0$, then $|f_k(x)-f(x)|=|0-0|=0$ for all $k$, so, the $\sup$ is not reached at $x=0$.
If $x=1$, then $|1^k-1|=|1-1|=0$ for all $k$, so, the $\sup$ is not reached at $x=1$ either.
So, the $\sup$ must be reached in an $x \in (0,1)$, but I can't find out which $x$ is it (and I don't really need to, I just need to find the $\sup$ and then take the limit, but I don't know what else I could do to find it.)
Thank you in advance for your help.
Note that the $x^k$ are continuous, and $1^k=1$. Thus very near, but not at $x=1$ and for any $k=1,2,\ldots$ you have numbers of the form $1-\epsilon$, $0<\epsilon<1$ arbitrarily small, yet $\lim f_k=0$. To put it easily: can you find $$\sup_{x\in(0,1)}|x^k|?$$