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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 $\|f\|_\infty = \sup|f(x)|$ .

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    I'd just compute $\|g_n-g_m\|_1$ and $\|g_n-g_m\|_\infty$.2012-04-19

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It is clear that $g_n$ converges pointwise to the function g(x)=\begin{cases}1&\text{if }0 Check that $\lim_{n\to\infty}\int_{-1}^1|g(x)-g_n(x)|\,dx=0$ and deduce that $\{g_n\}$ is convergent in the $L^1$ norm, and hence Cauchy.

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.