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I try to prove Corollary 2.5.5:

Corollary 2.5.5. Let $f: {\bf R}^n \to {\bf R}^n$ be a $C^1$ mapping and let $K \subset {\bf R}^n$ be compact. Then the restriction $f|_K$ of $f$ to $K$ is Lipschitz continuous.

Obviously, I somehow need to use this theorem:

Theorem 2.5.3 (Mean Value Theorem). Let $U$ be a convex open subset of ${\bf R}^n$ and let $f: U \to {\bf R}^p$ be a differentiable mapping. Suppose that the derivative $Df: U \to \operatorname{Lin}({\bf R}^n,{\bf R}^p)$ is bounded on $U$, that is, there exists $k>0$ with $\lVert Df(\xi)h\rVert \leq k\lVert h\rVert$, for all $\xi \in U$ and $h \in {\bf R}^n$, which is the case if $\lVert Df(\xi)\rVert_{\text{Eucl}} \leq k$. Then $f$ is Lipschitz continuous on $U$ with Lipschitz constant $k$, in other words

$$ \lVert f(x) - f(x')\rVert \leq k\lVert x-x'\rVert \qquad (x,x' \in U). \tag{2.17} $$

Here are my thoughts about this problem:

If $x,x'\in U(x,x')$ where $U(x,x')$ is a convex open subset of $ℝ^n$ and the derivative $Df$ is bounded on $U(x,x')$, then $\|f(x)-f(x') \| ≤ k(x,x') \|x-x'\|$.

Surely, I can find for every $x,x'\in K$ such an $U(x,x')$ and because $K$ is compact I can even find finite many of them, so that I can take $k$ as the maximum of $k(x,x')$.

The problem in my reasoning is that if make the open cover $∪U(x,x')$ finite, then I have for single $x\in K$ a $U(x,x')$, but I don't think I can find for every pair $(x,x')$ where $x,x'\in K$ a set $U(x,x')$ in my finite cover that contain both.

Edit: The answer of @Umberto P. makes sense to me, but when I look at the official solution, I don't see why they would make it so difficult then:

http://mathematics.discoursehosting.net/uploads/db1409/484/12492d22695e2e3a.png

Kasper
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  • just use the definition and some fact and it is easy – Rosa Maria Gtz. Oct 04 '13 at 19:54
  • @Knight According to my teacher, this is not an easy exercise. You may have overlooked some difficulties. – Kasper Oct 04 '13 at 19:56
  • @kahen Thanks ! – Kasper Oct 04 '13 at 19:59
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    Needless complication. Since $f$ is globally defined, you can simply use the convex hull of $K$, which is again compact. – Daniel Fischer Oct 04 '13 at 20:31
  • @DanielFischer Hm.. so would you conclude that the author had a bad day writing this ? Or is there an other reason to make it so complicated. I don't know much of the complex hull of $K$. But using a ball as Umberto P. suggested, would suffice as well right ? – Kasper Oct 04 '13 at 20:46
  • Bad day, probably. Maybe the author thought of something different that requires that type of argument. Yes, any convex set containing $K$ on which the derivative is bounded works. The convex hull of $K$ is the smallest such. – Daniel Fischer Oct 04 '13 at 20:51
  • @DanielFischer Hm.. I may write him letter then. But is the convex hull open ? Because theorem 2.5.3 needs an open set right ? – Kasper Oct 04 '13 at 20:55
  • The convex hull of a compact set is compact, that's one of the nice things. The theorem is formulated to use an open $U$, but that's not necessary for the conclusion. It's convenient for the theorem because in that, the function is defined only on $U$, and properly defining differentiability of a function in points that are not interior points is a tad more complicated. If you use a ball, you still need to use the fact that its closure is compact (and $Df$ defined and continuous there) to conclude the boundedness of $Df$ on the ball. – Daniel Fischer Oct 04 '13 at 21:04
  • @DanielFischer Ah, I understand, thanks for explaining. – Kasper Oct 04 '13 at 21:16
  • @DanielFischer I emailed the author of the book. His reaction was in dutch, and quite long, but to summarize (if I understand it correctly) he says: "If you take for $K$ the cantor set. Then it is not true that you can find for every element $x$ in $K$ a ball with center $x$ with the property that this ball is still in $K$. " – Kasper Oct 12 '13 at 23:28
  • @Kasper That's right, but irrelevant. The Cantor set is not convex. Theorem 2.5.3 holds also for convex compact sets, and corollary 2.5.5 follows from that variant of 2.5.3 if we use the convex hull of $K$. – Daniel Fischer Oct 12 '13 at 23:36
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    I've emailed Mr. Kolk, let's see what he says. – Daniel Fischer Mar 30 '14 at 13:49
  • The link to the image of the book page is broken. If you still have it on your computer, could you upload the image to i.stack.imgur? – Daniel Fischer Jan 22 '16 at 15:55

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If $K$ is compact it is in particular bounded. You can cover $K$ with a single convex open set $U$ (for instance a ball).

Umberto P.
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  • I've edited my post. I don't understand why the official solution make things so difficult. – Kasper Oct 04 '13 at 20:23