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Let $E$ be a path-connected subset of $\mathbb{R}^n$ and $f$ a differentiable function on $E$. Prove or disprove:

for any $x,y\in E$, there exists $z\in E$ such that $f(x)-f(y)=\nabla f(z)\cdot (x-y)$.

How to solve this? Thanks.

I have tried but failed:

(1). Suppose $E$ is convex. Let $g(t)=f(x+t(y-x))$. Then $g'(t)=\nabla f(x+t(y-x))\cdot (y-x)$. By mean value theorem, there exists $t_0\in (0,1)$ such that $g'(t_0)=g(1)-g(0)=f(y)-f(x)$. We choose $z=x+t_0(y-x)$.

(2). In general, $E$ is only assumed path-connected. If I let $\gamma$ be a curve such that $\gamma(0)=x$ and $\gamma(1)=y$, and $g(t)=f(\gamma(t))$, then $g'(t)=\nabla f(\gamma(t))\cdot \gamma'(t)$. However, $\gamma'(t)\neq y-x$. Hence I cannot solve...

Shiquan
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2 Answers2

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Hint: consider $t\to (\cos t, \sin t)$.

mookid
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  • $f$ is not allowed to be vector-valued here. Otherwise the mean-value theorem even fails in dimension $1$. – PhoemueX Jun 26 '14 at 15:24
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The claim is not true in general if the domain is not assumed to be convex.

As a counterexample, note that validity of your estimate would in particular imply

$$ |f(x) - f(y)| = |\nabla f(z) \cdot (x-y)| \leq \Vert \nabla f \Vert_\infty \cdot |x-y|, $$

if the gradient of $f$ is bounded on $\Omega$, i.e. $f$ would be Lipschitz in that case. In the example below, $f$ will have bounded gradient, but will not be Lipschitz, so that your claim is not true for that $f$.

Take

$$ \Omega = [((-1,0) \cup (0,1)) \times (0,1)] \cup [(-1,1) \times (1/2, 1)] $$ and note that $\Omega$ is indeed bounded, open and path-connected.

The idea is now to take $f \equiv 1$ on the "lower" part of the "left half" of $\Omega$ and $f \equiv 0$ on the "lower" part of the "right half" of $\Omega$ and to let $f$ vanish on the "upper" part of $\Omega$ to make everything smooth.

More precisely, take a smooth (at least $C^1$) function $g : \Bbb{R} \to \Bbb{R}$ with $g \equiv 0$ on $(-\infty, -2)$ and on $(1/4, \infty)$ and with $g \equiv 1$ on $(0, 1/8)$.

Define $f : \Omega \to \Bbb{R}$ by $f(x,y) = g(y)$ for $x \in (-1,0)$ and $f(x,y) = 0$ for $x \notin (-1,0)$. Note that $f$ with this definition is smooth with $\nabla f(x,y) = (0, g'(y))$ for $x \in (-1,0)$ and $\nabla f(x,y) = 0$ for $x \notin (-1,0)$, so that $\nabla f$ is bounded.

Now note

$$f(-\varepsilon, 1/16) = 1 \text{ and } f(\varepsilon, 1/16) = 0$$

for all $\varepsilon \in (0,1)$ which shows that $f$ is not Lipschitz.

PhoemueX
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  • This seems to contradict the mean value inequality. See: https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorem_in_several_variables – Bombyx mori Dec 30 '14 at 10:45
  • @BombyxMori: Yes, it seems that way but actually it does not, because the Wikipedia article assumes the following: "Fix points $x,y\in G$ such that the interval ${tx + (1-t)y \mid t\in [0,1]}$ lies in $G$ ". This is not the case here in general, and this is exactly the part where the shape of $\Omega$ enters. The idea is to have points which are very close together, but such that the shortest path connecting them is still quite long. – PhoemueX Dec 30 '14 at 11:13