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The $L_p$ norm $\|e^{-x/\epsilon}\|_p=(\frac{\epsilon}{p})^{1/p}(1-e^{-p/\epsilon})^{1/p}$, $1\leq p<\infty$ on the interval $(0,1)$. I am not sure why this norm is $O(\epsilon^{1/p})$ as $\epsilon$ approaches 0. Is it because as $\epsilon$ approaches 0, $e^{-p/\epsilon}$ vanishes leaving $(\frac{\epsilon}{p})^{1/p}$?

Also why is "$e^{-x/\epsilon}$ and zero function become indistinguishable with respect to the $L_p$ norm as $\epsilon$ approaches 0". Because if I take $\epsilon=0.01$ and $p=100$, $\epsilon^{1/p}\approx1$.

Michael Hardy
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Vaolter
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  • $p$ is fixed in "$e^{−x/\epsilon}$ and zero function become indistinguishable with respect to the $L^p$ norm as $\epsilon$ approaches $0$" (which is a clumsy way of saying $\lVert e^{-x/\epsilon}\lVert_p \to 0$ for $\epsilon \to 0$). – Daniel Fischer Jul 03 '13 at 15:39
  • Thanks, @DanielFischer, I get it now. Is my reasoning for the first part ok? – Vaolter Jul 03 '13 at 16:38

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