# 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 $\langle u,x\rangle =u_1x_1+\cdots+u_nx_n$.

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Try \langle and \rangle, the result is cute (and more TeX-canonical). – Did Sep 23 '13 at 17:45
also note the geometric intuition which compares the chord length with the corresponding arc length also. – Evan Sep 23 '13 at 17:50
@evan which part is chord length and which is arc length? i see $e^{i\langle u,x\rangle}-e^{i\langle u,y\rangle}$ is the chord lenth in complex space $\mathbb{C}$, while $u,x,y$ are vectors in $\mathbb{R}^n$ space... looks one is orange and the other is apple... – athos Sep 24 '13 at 2:06
@athos well noting many proofs first proceed by bounding $|e^{ix}-e^{iy}|\leq |x-y|$ this part is what I meant to be interpreted geometrically. The former is chord length and the latter is arc length. – Evan Sep 25 '13 at 4:31
@evan aha i got it! $\alpha := \langle u, x \rangle$, $\beta := \langle u, y \rangle$, then $\alpha$, $\beta$ is the angle, also the length of arc (from x axis). so $|\alpha-\beta|$ is the length of arc between angle $\alpha$ and $\beta$, while $|e^{i\alpha}-e^{i\beta}|$ is the chord length. thank you! – athos Sep 25 '13 at 4:40

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.

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Don't we need to say something about complex-differentiability (since the function in question isn't scalar-valued, when viewed as a function between Euclidean spaces)? – Jonathan Y. Sep 23 '13 at 17:45
No we do not, since $h$ is defined on $[0,1]$ (with values in a 2D vector space but the MVT I linked to takes care of this). The only difference is that one cannot exhibit $t$ such that $h(1)-h(0)=h'(t)$, fortunately the proof only uses $|h(1)-h(0)|\leqslant\sup|h'|$. – Did Sep 23 '13 at 17:48
Oh, thanks, I forgot we don't actually need that specific $t$ to show the inequality. – Jonathan Y. Sep 23 '13 at 18:30

$${{\rm d}{\rm e}^{{\rm i}\mu\left\langle u, x - y\right\rangle} \over {\rm d}\mu} = {\rm i}\left\langle u, x - y\right\rangle {\rm e}^{{\rm i}\mu\left\langle u, x - y\right\rangle}$$

$${\rm e}^{{\rm i}\mu\left\langle u, x - y\right\rangle} - 1 = {\rm i}\left\langle u, x - y\right\rangle \int_{0}^{1}{\rm e}^{{\rm i}\mu'\left\langle u, x - y\right\rangle}\,{\rm d}\mu'$$

$${\rm e}^{{\rm i}\mu\left\langle u, x\right\rangle} - {\rm e}^{{\rm i}\mu\left\langle u,y\right\rangle} = {\rm i}\left\langle u, x - y\right\rangle {\rm e}^{{\rm i}\mu\left\langle u,y\right\rangle} \int_{0}^{1}{\rm e}^{{\rm i}\mu'\left\langle u, x - y\right\rangle}\,{\rm d}\mu'$$

$$\color{#ff0000}{\large% \left\vert{\rm e}^{{\rm i}\mu\left\langle u, x\right\rangle} - {\rm e}^{{\rm i}\mu\left\langle u,y\right\rangle}\right\vert} = \left\vert\left\langle u, x - y\right\rangle\right\vert \left\vert \int_{0}^{1}{\rm e}^{{\rm i}\mu'\left\langle u, x - y\right\rangle}\,{\rm d}\mu' \right\vert \color{#ff0000}{\large% \leq \left\vert u\right\vert\left\vert x - y\right\vert}$$

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From $\left|{d\over dt}e^{it}\right|=1$ for real $t$ it follows that $$\left|e^{i\langle u,x\rangle}-e^{i\langle v,x\rangle}\right|\leq \left|\langle u,x\rangle-\langle u,y\rangle\right|\leq|u|\ |x-y|\ .$$

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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.

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Taking $x = y$ shows the inequality cannot be strict in all cases. – Robert Lewis Sep 23 '13 at 18:21
@RobertLewis, thank you. Indeed, $\langle u,y+t(x-y)\rangle$ is constant where $x-y$ is perpendicular to $u$, and in particular when $x=y$ (but that's an equivalency; the inequality is strict whenever $\langle u,x-y\rangle\neq 0$). – Jonathan Y. Sep 23 '13 at 18:28
My pleasure, sir! – Robert Lewis Sep 23 '13 at 18:34

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

$e^{i\langle u, x \rangle} = \cos \langle u, x \rangle + i \sin \langle u, x \rangle \tag{1}$

$\nabla e^{i\langle u, x \rangle} = \nabla (\cos \langle u, x \rangle + i \sin \langle u, x \rangle) = (-\sin \langle u, x \rangle + i\cos \langle u, x \rangle)u, \tag{2}$

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

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