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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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    Are you considering here $||\cdot ||_{1}$ and $||\cdot ||_{\infty}$ as the $L^{1}$ and $L^{\infty}$ norms?2012-04-19
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    Ah, I've edited in what the norms are defined as.2012-04-19
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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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