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.

Let $U \subset \mathbb{R}^d$ be open and path-connected. Let $f: U \to \mathbb{R}^m $ be differentiable on $U$ and suppose there exists a real $M$ such that $|| D_f(x) || \leq M $ for all $x \in U$. then,

$$ \|f(b) - f(a) \| \leq M\|b-a\|.$$

for all $a,b \in U$.

Is this result true. Can someone show me how to prove it? thanks

share|improve this question
1  
This is not true for non-convex $U$: if you put $U={\bf R}^2\setminus [0,\infty)\times \{0\}$, then the angle $f\colon U\to (0,2\pi)$ is a counterexample. –  tomasz Mar 8 at 15:29
    
In a nutshell, even though $U$ is path-connected, when $d \gt 1$ the length of short paths from $a$ to $b$ need not be bounded by a constant times $||b-a||$. –  hardmath Mar 8 at 15:35
    
This can also be generalised to arbitrary Riemannian manifolds. But in any case, you need to consider the geodesic distance. –  tomasz Mar 8 at 15:35

4 Answers 4

up vote 2 down vote accepted

First of all, it is clearly true if $m=1$. Indeed, in such case $f : U\to\mathbb R$, and setting $$ g(t)=f\big((1-t)a+tb\big), $$ then $f(b)-f(a)=g(1)-g(0)=\int_0^1 g'(t)\,dt$, or equivalently $$ f(b)-f(a)=\int_0^1 \frac{d}{dt} f\big((1-t)a+tb\big)\,dt= \int_0^1 \nabla f\big((1-t)a+tb\big)\cdot (b-a)\,dt, $$ and hence \begin{align} \lvert f(b)-f(a)\rvert&\le \int_0^1 \lvert\nabla f\big((1-t)a+tb\big)\cdot (b-a)\rvert\,dt \\&\le \| b-a\| \int_0^1\big\|\nabla f\big((1-t)a+tb\big)\big\|\,dt \le M\,\| b-a\|. \end{align} In general now, if $m>1$, let $w\in\mathbb R^m$ arbitrary, and set $g(x)=f(x)\cdot w$. Then $g: U \to \mathbb R^m$, and hence \begin{align} \big|\big(\,f(b)-f(a)\big)\cdot w\big|&=\lvert f(b)\cdot w-f(a)\cdot w\rvert=\lvert g(b)-g(a)\rvert\le \|b-a\|\sup_{x\in U} \|Dg(x)\| \\&=\|b-a\|\sup_{x\in U} \|Df(x)\cdot w\|\le \|b-a\|\cdot \|w\|\cdot\sup_{x\in U} \|Df(x)\|. \end{align} In particular, for $w=f(b)-f(a)$, the above inequality yields $$ \big|\big(\,f(b)-f(a)\big)\cdot \big(\,f(b)-f(a)\big|\le\|b-a\|\cdot \|f(b)-f(a)\|\cdot\sup_{x\in U} \|Df(x)\|, $$ or $$ \|f(b)-f(a)\|^2\le\|b-a\|\cdot \|f(b)-f(a)\|\cdot\sup_{x\in U} \|Df(x)\|, $$ which implies that $$ \|f(b)-f(a)\|\le\|b-a\|\cdot\sup_{x\in U} \|Df(x)\|. $$

Note. This argument works in the case in which the segment $[a,b]$ lies in $U$. In general, this argument hold whenever $U$ is convex.

share|improve this answer
    
Why $|f(b) \cdot w - f(a) \cdot w | \leq ||b-a|| \sup ||D g(x)|| $ ? –  user130448 Mar 8 at 15:09
    
@ThomasMann: See updated version. –  Yiorgos S. Smyrlis Mar 8 at 15:12
    
The OP asked about $\mathbb{R}^m$ valued $f$ with source $n$ dimensional. Your reasoning works only for $f$ defined on one dimensional space.The inequality $|g(b)-g(a)| \le..$ makes use of this assumption. –  Thomas Mar 8 at 15:19
    
@Thomas This $$ f(b)-f(a)=\int_0^1 \frac{d}{dt} f\big((1-t)a+tb\big)\,dt= \int_0^1 \nabla f\big((1-t)a+tb\big)\cdot (b-a)\,dt, $$ works for every dimension. –  Yiorgos S. Smyrlis Mar 8 at 15:20
2  
only if the path is in the domain of definition of $f$. ($U$ is open and path connected, not convex) –  Thomas Mar 8 at 15:21

I doubt you can do that if $U$ is not convex or satisfies some similar condition. You will have to replace $||b-a||$ by the length of the shortest curve in $U$ from $b$ to $a$. If $U$ is, e.g., convex, this is straightforward, since then the line from $a$ to $b$ is contained in $U$ and then $$f(b)-f(a) = f(a+t(b-a))|_{t=1} - f(a+t(b-a))|_{t=0} = Df(\xi)(b-a)$$ for some $\xi =a +t_0(b-a)$ by the (one dimensional) MVT. The estimate then follows directly. The general case follows by joining $a$ an $b$ by, e.g., piecewise linear curves.

share|improve this answer
1  
You can also just join $a$ and $b$ with a smooth, arc-length parametrised curve and compose it with $f$ and just apply the MVT directly. For a simple counterexample for non-convex $U$, just consider $U={\bf R}^2\setminus [0,\infty)$ and put let $f$ be the angle. –  tomasz Mar 8 at 15:27

Counterexample. Let $$ f(x,y)={\mathcal {Im}}\log (x+iy), $$ where the complex logarithm is defined in $\mathbb C\smallsetminus (-\infty,0]$, and hence for $x<0$, we have $$ \lim_{y\to 0^+} f(x,y)=\pi \quad\text{whereas}\quad \lim_{y\to 0^-} f(x,y)=-\pi. $$ If we set as $U$ the following domain $$ U=\{(x,y): x^2+y^2>1\}\smallsetminus\{(x,0): x<0\}, $$ then, due to Cauchy-Riemann equations $$ f_x(x,y)={\mathcal Im}\frac{d}{dz}\log z=\frac{y}{x^2+y^2}, $$ and $$ f_y(x,y)=-{\mathcal Re}\frac{d}{dz}\log z=-\frac{x}{x^2+y^2}, $$ and hence $\|Df(x,y)\|\le 1$ in $U$.

At the same time, for $b=(-2,\varepsilon)$ and $a=(-2,-\varepsilon)$, we have $$ \|b-a\|=2\varepsilon \quad\text{and}\quad \lvert f(b)-f(a)\rvert\approx 2\pi. $$ So the inequality $$ \lvert f(b)-f(a)\rvert\le \|Df(x,y)\|\cdot\|b-a\| $$ DOES NOT hold.

share|improve this answer

It can be disproved by the following physical experiment. Consider a large flat capacitor with the voltage $V$. Inside the capacitor, the electric field is constant and equal to $\frac{V}{d}$ ($d$ is the distance between the plates). It's the largest possible electric field in the system. If you move a unit charge that was outside but near a plate through the capacitor to other similar point outside, the work will be equal to $V$.

But you can choose a long way around: Starting from the initial point of the charge where the electric field is almost $0$, you move farther and farther from the plates and strength of the electric field will reach some small value (because from the charge's view they don't look like infinite anymore) and then you make a way around and approach from the other side.

The electric field was definitely smaller than $\frac{V}{d}$ and when you multiply the supremum of its module by $d$, it will be smaller than the difference in potentials.

share|improve this answer

Your Answer

 
discard

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