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**Complex exponential has $1$ as Lipschitz constant.**without wasting too much if your time.The question is published on by Tutorial Guruji team.

(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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