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For a part of my proof I need to establish that $e^{-\alpha x} \lt h(x)$, where $\alpha,x \gt 0, $ and $x,\alpha \in\mathbb{R}$. I thought for a while and couldn't find a function independent of $\alpha$ that fulfills my criteria. Any ideas?

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    $\begingroup$ Are you asking for an example of a function $h$ for which the inequality is true? $\endgroup$ – kimchi lover Jan 15 at 14:21
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    $\begingroup$ $h(x)=1$ fits the bill. $\endgroup$ – Adrian Keister Jan 15 at 14:31
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    $\begingroup$ If you don't want any specific $h$, there are a lot (and basically useless) functions that bound $e^{-ax}$ from above. For example given any strictly positive integer $n$, $h_n(x)=n$ is always strictly larger than $e^{-ax}$ and independent of $a$. $\endgroup$ – James Jan 15 at 14:36
  • $\begingroup$ I'm voting to close the question for lack of context. Without context, we cannot know what kinda of upper bound you want, and the question becomes too open-ended. $\endgroup$ – Simply Beautiful Art Jan 15 at 14:37
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    $\begingroup$ $$ \alpha,x>0\implies e^{-\alpha x}<1.$$ $\endgroup$ – Julián Aguirre Jan 15 at 15:15
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Also note that: $$e^{-\alpha x}=\sum_{n=0}^\infty\frac{(-1)^n\alpha^nx^n}{n!}$$ A polynomial of this expansion could be used, although take care to see that it is consistently greater as this is a series with alternating sign

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  • $\begingroup$ But a polynomial (of positive degree) gives a worse bound than the obvious $h(x)=1$. $\endgroup$ – David C. Ullrich Jan 15 at 17:41
  • $\begingroup$ The higher order used the closer this approximation is surely, especially at smaller values of $x$, since $h(x)=1$ is most accurate for $x\to\infty$ $\endgroup$ – Henry Lee Jan 15 at 18:53

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