5
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(In the following, Lipschitz constant does not mean "best Lipschitz constant".)

I've just read this in a book that I highly regard:

Moreover, by mean value theorem, $u\to e^{iu}$ is Lipschitz continuous with Lipschitz constant $1$.

How does the author infer this from the mean value theorem ? The theorem applies only for real-valued functions.

The mean value theorem shows nontheless that $\sin$ and $\cos$ have Lipschitz constant $1$, and therefore $u\to e^{iu}$ has Lipschitz constant $2$.

How can this be lowered to $1$ ?

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Let $a \le b$.

$$|e^{ib} - e^{ia}| = \left|\int_a^b ie^{it}\,dt\right|\le \int_a^b |ie^{it}|\,dt = b -a.$$

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  • 1
    $\begingroup$ Nicely done! I don't know why the author referred to the mean value theorem. $\endgroup$ – Gabriel Romon Feb 25 '15 at 15:03

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