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The German Wikipedia is full of articles like this one saying one function falls slower than another. I wrecked google, but could not find the definition or sufficient conditions. Is it:

$ \lim_\limits{x \to \alpha }\frac{f(x)}{g(x)} = 0$, or

$\displaystyle{\frac{d}{dx}\frac{f(x)}{g(x)} < 0} $ (in appropriate region)

Or is any of these sufficient? I guess the first one, since it doesn't require differentiability.

JohnColtraneisJC
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Ludi
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    If you want to use the definition for non-differentiable function I would go with the limits definition of yours. This is usually written as $f\in \mathrm o(g)$ and means that $f$ grows much slower than $g$, or in your case, that $f$ falls much faster than $g$, assuming that they are falling at all. See big-O-notation. – M. Winter Sep 18 '17 at 13:21
  • @M.Winter I just want to clarify it, because there seems to be an error in a proof of mine and I suspected this assumption I made. Is the second also correct, if differentiable? – Ludi Sep 18 '17 at 13:23
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    The second definition is at least not the same. Take $f(x)=1+x$ and $g(x)=x$. You have $(f/g)'<0$ but $f/g\to1$ for $x\to\infty$. It includes cases like $f\in\Theta(g)$. – M. Winter Sep 18 '17 at 13:26

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Let's restrict this discussion to those stretched exponentials in the article you linked to, since they have the property that their limit is zero as $t \to +\infty$, which makes talking about the concept of "falling" relatively straightforward.

If we have two such functions, $f$ and $g$, then to say "$g$ falls more slowly than $f$" means that $g$ approaches $0$ more slowly than $f$ does. In other words, $f$ falls (decreases) more quickly than $g$ does. So it is the case that $\displaystyle\lim_{t\to +\infty} \frac{f(t)}{g(t)} = 0$, because the numerator approaches $0$ more quickly than the denominator does. There are two ways of stating this here: $$ \lim_{t \to+\infty} \frac{f(t)}{g(t)} = 0 \qquad \text{and} \qquad \lim_{t\to+\infty} \frac{g(t)}{f(t)} = +\infty$$

In general, $g(t) = e^{-t^\beta}$ falls more slowly than $f(t) = e^{-t}$ if we have $\beta < 1$. That's just from the link you provided. Consider: $$ \lim_{t\to +\infty} \frac{g(t)}{f(t)} = \lim_{t\to +\infty} \frac{e^{-t^\beta}}{e^{-t}} = \lim_{t\to+\infty} \frac{e^{-t^\beta} \cdot e^t}{e^{-t} \cdot e^t} = \lim_{t\to+\infty} e^{t-t^\beta}$$

Since $\beta < 1$ and $t \to +\infty$, then $t^\beta \to 0$. Thus $t - t^\beta \to +\infty$ as $t \to +\infty$, and so $\displaystyle \lim_{t\to+\infty} e^{t-t^\beta} = +\infty$. This paragraph isn't exactly rigorous, but I'm going for clarity here.

A similar argument shows the other limit mentioned above: $\displaystyle\lim_{t\to+\infty} \frac{f(t)}{g(t)} = 0$

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    @Ludi, which case? Note that I modified my post fairly heavily since I didn't like the way I originally phrased it, then I realized that I had to change a lot to match it. If you're asking about the formal definition of falling more slowly, I think the limit definition is the way to go. This is related to the concept of little-oh notation as M. Winter mentioned above, but I think those concepts are more generally used for functions that grow without bound rather than approach $0$. –  Sep 18 '17 at 13:39
  • Oh, I had misunderstood something. Perhaps in the middle of revision – Ludi Sep 18 '17 at 13:42