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Let the following function: \begin{equation} \pi(x) = \frac{e^x}{1 + e^x},\quad \textrm{for all } x \in \mathbb{R}. \end{equation} I want to know if it is Lipschitz? Here my proof and I like to know if it is true?

\begin{equation} \pi'(x) = \frac{e^x}{(1 + e^x)^2}. \end{equation}

Using the fact that for all $x \in \mathbb{R},$ we have $ (1 + e^x)^2 > 1+e^x > e^x,$ then I can conclude that for all $x \in \mathbb{R},$ we have $|\pi'(x)| < 1.$ Hence, the derivative of $\pi(\cdot)$ is bounded. Then, I conclude that the function $\pi(\cdot)$ is $k-$lipschitz for some $k\leq 1.$

Best regards,

mathdatastatsml
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  • It's correct (but not enough), you need to use the mean value theorem to conclude. – idm Feb 02 '15 at 12:17
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    @idm I supose the OP already knows how to prove a function is Lipschitz if the derivative is bounded (MVT), otherwise hardly would he have calculated the derivative and etc. Under this assumption, I think the above is correct. +1 – Timbuc Feb 02 '15 at 12:19

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