For the sequence $g_n \in C[-1,1]$ defined by $g_n = x^\frac{1}{2n-1}$ how would you show that $g_n$ is Cauchy under $\|\cdot\|_1$ but not under $\|\cdot\|_\infty$?
$\|f\|_1 = \int_a^b |f|$ and $\|f\|_\infty = \sup|f(x)|$ .
For the sequence $g_n \in C[-1,1]$ defined by $g_n = x^\frac{1}{2n-1}$ how would you show that $g_n$ is Cauchy under $\|\cdot\|_1$ but not under $\|\cdot\|_\infty$?
$\|f\|_1 = \int_a^b |f|$ and $\|f\|_\infty = \sup|f(x)|$ .
It is clear that $g_n$ converges pointwise to the function g(x)=\begin{cases}1&\text{if }0
On the other hand, if $\{g_n\}$ were Cauchy in the $L^\infty$ norm, since $C[-1,1]$ is complete for that norm, its limit would be continuous. But $g$ is not continuous.