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
On
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.
On
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.
On
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.
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.