Bizarre application of the mean value theorem

Question

In a book I am reading, if $|\xi|\le|x-y|^{-1}$, then the mean value theorem is used to estimate

$$|e^{-ix\cdot\xi}-e^{-iy\cdot\xi}|^{2}\le c|x\cdot\xi-y\cdot\xi|^{2}.$$

I am familiar with the mean value theorem for integrals and where we have an expression in the form of a quotient, but I don't see how it has been employed here.

I am solving an exercise in which I have a similar problem. That is, $|\xi|\le|y|^{-1}$ and I want to estimate

$$|e^{-i(x+2y)\cdot\xi}-2e^{-i(x+y)\cdot\xi}+e^{-ix\cdot\xi}|^{2}.$$

Answer 1

(I have assumed that $ x \cdot \xi ,y \cdot \xi$ are real here.)

Let $f(t) = e^{it}$, then $|f'(t)| = 1$ for all $t$ and so $|f(s)-(t)| \le |s-t|$ for all $s,t$.

In the above, $s=x \cdot \xi $, $t = y \cdot \xi$, hence $|e^{i x \cdot \xi} - e^{y \cdot \xi} | \le | x \cdot \xi - y \cdot \xi |$.

Addendum:

Since $f(s) -f(t)= \int_0^1 f'(t+\tau(s-t)) d \tau (s-t)$, we have $|f(s) -f(t)| \le \int_0^1 |f'(t+\tau(s-t))| d \tau |s-t| \le |s-t| $.

Answer 2

You know that $f: \mathbb{C} \to \mathbb{C}$ with $f(z) = e^z$ is an entire function with $f'(z) = e^z$. Thus, for any $u, v \in \mathbb{C}$ we have $$ \int_{u}^v e^z dz = e^v - e^u$$ and the absolute value of $e^z$ on the segment $[u, v]$ is bounded by $c_{u,v} = e^{\max(\text{Re} u,\text{Re} v)}$. We conclude that $$|e^v - e^u| \le c |u - v|.$$ Notice you can take $c$ as a constant independent of $u$ and $v$ if the numbers you are considering are bounded.