how to prove $|e^{i \langle u,x \rangle}-e^{i \langle u,y \rangle }|<|u|\cdot|x-y|$? I'm reading Bernt Oksendal's "Stochasticc Differential Equations" and this is one of the result that I don't see the proof.
This is from Appendix A, page 309 (sixth edition):
$$\large \lvert e^{i \langle u,x \rangle}-e^{i \langle u,y \rangle}\rvert < \lvert u \rvert \cdot \lvert x-y\rvert$$
Here $u,x,y\in \mathbb{R}^n$,   $i\in \mathbb{C}$ is the imaginary unit and $\langle u,x\rangle =u_1x_1+\cdots+u_nx_n$.
 A: By noting that
$h(t)=\exp(\mathrm i\langle u,(1-t)x+ty\rangle)$ is such that $h(0)=\mathrm e^{\mathrm i\langle u,x\rangle}$, $h(1)=\mathrm e^{\mathrm i\langle u,y\rangle}$ and, for every $t$ in $[0,1]$, $|h(t)|=1$ and $h'(t)=\mathrm i\cdot\langle u,y-x\rangle\cdot h(t)$. Thus, $|h'(t)|=|\langle u,y-x\rangle|$ for every $t$ in $[0,1]$. 
By the mean value theorem for functions of several variables, $|h(1)-h(0)|\leqslant|\langle u,y-x\rangle|$. Since $|h(1)-h(0)|=|\mathrm e^{\mathrm i\langle u,y\rangle}-\mathrm e^{\mathrm i\langle u,x\rangle}|$ and, by Cauchy-Schwarz inequality, $|\langle u,y-x\rangle|\leqslant\|u\|\cdot\|x-y\|$, we are done.
A: $$
{{\rm d}{\rm e}^{{\rm i}\mu\left\langle u, x - y\right\rangle} \over {\rm d}\mu}
=
{\rm i}\left\langle u, x - y\right\rangle
{\rm e}^{{\rm i}\mu\left\langle u, x - y\right\rangle}
$$
$$
{\rm e}^{{\rm i}\mu\left\langle u, x - y\right\rangle} - 1
=
{\rm i}\left\langle u, x - y\right\rangle
\int_{0}^{1}{\rm e}^{{\rm i}\mu'\left\langle u, x - y\right\rangle}\,{\rm d}\mu'
$$
$$
{\rm e}^{{\rm i}\mu\left\langle u, x\right\rangle}
-
{\rm e}^{{\rm i}\mu\left\langle u,y\right\rangle}
=
{\rm i}\left\langle u, x - y\right\rangle
{\rm e}^{{\rm i}\mu\left\langle u,y\right\rangle}
\int_{0}^{1}{\rm e}^{{\rm i}\mu'\left\langle u, x - y\right\rangle}\,{\rm d}\mu'
$$
$$
\color{#ff0000}{\large%
\left\vert{\rm e}^{{\rm i}\mu\left\langle u, x\right\rangle}
-
{\rm e}^{{\rm i}\mu\left\langle u,y\right\rangle}\right\vert}
=
\left\vert\left\langle u, x - y\right\rangle\right\vert
\left\vert
\int_{0}^{1}{\rm e}^{{\rm i}\mu'\left\langle u, x - y\right\rangle}\,{\rm d}\mu'
\right\vert
\color{#ff0000}{\large%
\leq
\left\vert u\right\vert\left\vert x - y\right\vert}
$$
A: From $\left|{d\over dt}e^{it}\right|=1$ for real $t$ it follows that
$$\left|e^{i\langle u,x\rangle}-e^{i\langle v,x\rangle}\right|\leq \left|\langle u,x\rangle-\langle u,y\rangle\right|\leq|u|\ |x-y|\ .$$
A: One way would be to use complex path integration:
$$\left|e^{i\langle u,x\rangle}-e^{i\langle u,y\rangle}\right| = \left|\int_\gamma e^{z}dz\right|,$$
where $\gamma:[0,1]\to\mathbb{C}$ is defined $\gamma(t) = (1-t)i\langle u,y\rangle + ti\langle u,x\rangle$.
To evaluate this integral, simply note
$$\left|\int_\gamma e^{z}dz\right| = \left|\int_0^1 e^{i((1-t)\langle u,y\rangle + t\langle u,x\rangle)}i\langle u,x-y\rangle dt\right| \leq \int_0^1 |\langle u,x-y\rangle|dt =\\
= |\langle u,x-y\rangle| \leq \|u\|\|x-y\|.$$
Strict inequality can be achieved by noting that the integrand doesn't have a constant angle.
A: Won't this work?
First, note that taking $x = y$ shows the inequality cannot be strict.  What is wanted is
$\vert e^{i\langle u, y \rangle} - e^{i\langle u, x \rangle} \vert \le \vert u \vert \vert y - x \vert. \tag{0}$
Having said that, use 
$e^{i\langle u, x \rangle} = \cos \langle u, x \rangle + i \sin \langle u, x \rangle \tag{1}$
and take the gradient:
$\nabla e^{i\langle u, x \rangle} = \nabla (\cos \langle u, x \rangle + i \sin \langle u, x \rangle) = (-\sin  \langle u, x \rangle + i\cos  \langle u, x \rangle)u, \tag{2}$
then write the line integral along the path $\gamma:[0, 1] \to \Bbb R^n$, $\gamma(s) = (1 - s)x + sy$, so that $\gamma(0) = x$ and $\gamma(1) = y$, noting that $\gamma'(s) = y - x$ for all $s \in [0, 1]$:
$e^{i\langle u, y \rangle} - e^{i\langle u, x \rangle} = \int_0^1 \nabla e^{i\langle u, \gamma(s) \rangle} \cdot \gamma'(s)ds = \int_0^1 \nabla e^{i\langle u, \gamma(s) \rangle} \cdot (y - x)ds, \tag{3}$
and take the norm of both sides:
$\vert e^{i\langle u, y \rangle} - e^{i\langle u, x \rangle} \vert = \vert \int_0^1 \nabla e^{i\langle u, \gamma(s) \rangle} \cdot (y - x)ds \vert \le \vert y - x \vert \, \vert \int_0^1 \nabla e^{i\langle u, \gamma(s) \rangle}ds \vert$ 
$\le \vert y - x \vert \,  \int_0^1 \vert\nabla e^{i\langle u, \gamma(s) \rangle}\vert ds  \le \vert u \vert \vert y - x \vert, \tag{4}$
by virtue of (2), which easily is seen to imply
$\vert \nabla e^{i\langle u, x \rangle} \vert = \vert (-\sin  \langle u, x \rangle + i\cos  \langle u, x \rangle)u \vert = \vert u \vert. \tag{5}$
Cheers, and
Fiat Lux
