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I want to bound $f_h(y) = h^{-1}|e^{hy}-1|$ where $h\in(0,1)$ and $y\in\mathbb R$ with something independent on $h$ and growing as slow as possible with $y\to \pm\infty$.

Can I do better than $f_h(y) \ \leq e^{|y|}$?

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@Mercy. $f_h(x)\leq |x|$ is not true. Take $y=\frac{1}{h^2}$ and $h$ large (say h=100). Note that also $f_h(x)\leq |x|^N$ is not true for any fixed $N\in\mathbb N$. – Hans Oct 27 '12 at 9:13
In the previous comment I mean $h$ small of course ($h=\frac{1}{100}$)... – Hans Oct 27 '12 at 16:22
up vote 2 down vote accepted

For $y\rightarrow -\infty$ its possible, but for $y\rightarrow \infty$ it is not possible, because

\begin{eqnarray} \frac{|e^{hy}-1|}{h} &\geq& \frac{e^{hy}}{h}-\frac{1}{h} \nonumber \end{eqnarray}

From the last inequality, we see that for fixed $h$, your function grows equal or even more than an exponential in the set $\{(h,y):\ y>0\}$.

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