How can we show that $$\left|\frac{e^{-\lambda t}-1}t\right|\le|\lambda|$$ for all $t\in\mathbb R\setminus\left\{0\right\}$ and $\lambda\in\mathbb R$?
Clearly, $e^{-\lambda t}\in[0,1]$ for all $t,\lambda\ge0$, but that doesn't help.
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@JeanMarie This is not a duplicate of the above post. – Kavi Rama Murthy Mar 20 '19 at 23:09
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You are right. It is more or less a duplicate of https://math.stackexchange.com/questions/1683328/prove-that-left-frace-ht-1h-right-le-t-for-h0 – Jean Marie Mar 20 '19 at 23:31
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The inequality is false. See what happens as $\lambda \to -\infty$ with $t=1$ to see that it is false. [Indeed, when $\lambda=-n$ and $t=1$ the inequality says $|e^{n}-1| \leq n$ which is false as you can see from the series expansion of $e^{n}$]. It is true for $\lambda, t \geq 0$. For $\lambda =0$ this is obvious. For $\lambda > 0$ note that $\int_0^{1} e^{-\lambda tx} dx=\frac {1-e^{-\lambda t}} {\lambda t}$. Use the fact that the integrand is $\leq 1$. The inequality is also true when both parameters are negative.
Kavi Rama Murthy
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1It doesn't matter for the desired conclusion, but shouldn't it be $\int_0^1e^{-\lambda tx}:{\rm d}x=\frac{1-e^{-\lambda t}}{\lambda t}$? – 0xbadf00d Mar 20 '19 at 13:24
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@Oxbad00d Yes, that was a silly mistake. Thanks for pointing it out. – Kavi Rama Murthy Mar 20 '19 at 23:08