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The following is a simple calculation I was trying to genearlize that I figured was just an application of l'hopitals rule.

Let $r >0$

1) Why is $$ \frac{\exp{\left( -\frac{1}{4y^{2r}} \right)}}{y^r} $$ bounded for $ y \in [0,1]$?

2) What if I have a different power of $y$ in the denominator $ s>0$ what are the conditions I need on $r,s$ to guarantee $$ \frac{\exp{\left( -\frac{1}{4y^{2r}} \right)}}{y^s} $$ is still bounded on $[0,1]$?

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We need no conditions. If $z$ is large positive, then $e^z>z^n/n!$. – André Nicolas Mar 29 '12 at 0:08
up vote 1 down vote accepted

Let $0<\delta<1$, then $$ f_{s,r}(y)=y^{-s}\exp\left(-\frac{1}{4y^{2r}}\right) $$ is continuous on $[\delta,1]$. So it is bounded by some constant $M_1>0$. Now we note that substitution $y=(4t)^{-\frac{1}{2r}}$ where $t\to+\infty$ (recall $r>0$) gives us that $$ \lim\limits_{y\to +0}f_{s,r}(y)= \lim\limits_{y\to +0}y^{-s}\exp\left(-\frac{1}{4y^{2r}}\right)= \lim\limits_{t\to +\infty}\frac{t^{\frac{s}{2r}}}{\exp(t)}=0 $$ The last limit is equal to $0$ because exponent grows faster then arbitrary power. Since $\lim\limits_{y\to +0}f_{s,r}(y)=0$, then $f_{s,r}(y)$ is bounded on $[0,\delta]$ by some $M_2>0$. Hence $f_{s,r}$ is bounded on $[0,1]$ by $\max(M_1,M_2)$

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Thank you am I correct in understanding this proof holds for any pair $r,s >0$? – user7980 Mar 29 '12 at 0:18
It holds for all $r\in\mathbb{R}_+$ and $s\in\mathbb{R}$ – Norbert Mar 29 '12 at 0:20

We can do this for any $r,x>0$. Writing $t=1/y$, we get $$ \lim_{y\to0^+}\frac{\exp{\left( -\frac{1}{4y^{2r}} \right)}}{y^s} =\lim_{t\to\infty}t^s\exp{\left( -\frac{t^{2r}}{4} \right)} =\lim_{t\to\infty}\frac{t^s}{\exp{\left(\frac{t^{2r}}{4} \right)}} =\lim_{t\to\infty}\frac{\exp{(s\log t)}}{\exp{\left(\frac{t^{2r}}{4} \right)}} =\lim_{t\to\infty}\exp\left(s\log t-\frac{t^{2r}}{4}\right)=0, $$ as $s\log t - t^{2r}/4\to-\infty$ as $t\to\infty$.

So the your function can be extended with continuity to zero. Thus we get a continuous function on the closed interval $[0,1]$ and so it is bounded.

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A variable change $\rm u=y^{-2r}$ puts expressions like this in the form

$$\rm u^k\exp(-au) ~~ \text{ as } ~~ u\to\infty.$$

Apply l'Hospital's $\rm \lceil k \rceil$ times (we assume $\rm k$ can be non-integer) and we obtain $0$.

We are assuming $\rm a,r>0$; otherwise we clearly have divergence.

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