Complex exponential has $1$ as Lipschitz constant. Code Answer

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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, $uto e^{iu}$ 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 $uto e^{iu}$ has Lipschitz constant $2$.

How can this be lowered to $1$ ?

Answer

Let $a le b$.

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

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