Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

5 Answers 5

up vote 4 down vote accepted

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.

share|improve this answer
    
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
add comment

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

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

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

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

share|improve this answer
add comment

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

share|improve this answer
add comment

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.

share|improve this answer
    
Taking $x = y$ shows the inequality cannot be strict in all cases. –  Robert Lewis Sep 23 '13 at 18:21
1  
@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
1  
My pleasure, sir! –  Robert Lewis Sep 23 '13 at 18:34
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.