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Proposition. Let $f : \mathbb R^n \to \mathbb R$ be a function with the following property: every point in $\mathbb R^n$ has an open neighborhood where $f$ is Lipschitz. Then it is Lipschitz on every closed and bounded subset of $\mathbb R^n$.

I think it's fairly easy to prove this directly: let $K \subset \mathbb R^n$ be closed and bounded. Define $K'$ as a compact set that is the minimal convex set containing $K$. By hypothesis, for every $x \in K'$ there exists an open neighborhood $\Omega_x$ such that $\forall y,z \in \Omega_x$ the Lipschitz condition holds: $$|f(y) - f(z)| \leq L_x|y-z| $$ for some $L_x > 0$ depending on $x$. The family $$\mathcal F \doteq \{\Omega_x\ |\ x \in K' \} $$ is an open cover of $K'$; since $K'$ is compact, one can select a finite number of points $x_1,\dots,x_k$ such that $$K' \subseteq \bigcup_{j=1}^k \Omega_{x_k} $$ In each of these neighborhoods, $f$ will be Lipschitz with constant $L_{x_k}$; but if $f$ is Lipschitz with constant say $L$, then it is Lipschitz with constant $L'$, for all $L' \geq L$. This means we can define $$L \doteq \max_k \{L_{x_k}\} $$ and $f$ will be Lipschitz with constant $L$ on $K'$, therefore on $K$.


However, I was wondering if this result could be proven by contradiction. Suppose $f$ is not Lipschitz on $K$. Then $$\forall L > 0\quad \exists x,y \in K \qquad |f(x)- f(y)| > L|x-y| $$ Let $\{L_k\}$ be a sequence of strictly positive real number. Then for all $k$ we can find $x_k,y_k \in K$ such that $|f(x_k) - f(y_k)| > L_k |x_k - y_k|$.

From here on, I'm stuck. I'm not sure whether I should be arriving at a negation of the compactness (possibly through sequential compactness, since we're dealing with sequences) of $K$, or of the "locally Lipschitz" condition in the hypothesis. Any suggestions?

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  • $\begingroup$ In your argument by contradiction, where are you using the fact that $f$ is Lipschitz in some open neighborhood of any point? Without this assumption, the result is false. $\endgroup$ – Francesco Polizzi Jun 25 '17 at 14:08
  • $\begingroup$ Well, I know that should come up at some point in the proof, but I'm stuck. Precisely, I don't know whether to assume compactness and arrive at a negation of the property you say, or alternatively to assume it and arrive at a negation of compactness. $\endgroup$ – giobrach Jun 25 '17 at 14:11
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Proof by contradiction: I'll be brief (ask if you have questions). First note that the local Lipschitz condition implies $f$ is continuous everywhere.

If the result fails for some compact $K,$ then for every $m\in \mathbb N,$ there exist $x_m,y_m\in K$ such that

$$\tag 1|f(y_m)- f(x_m)| > m|y_m-x_m|.$$

Because $K$ is compact, there exists a subsequence $m_k$ such that both $y_{m_k},x_{m_k}$ converge to some $x,y\in K$ respectively.

If $y\ne x,$ then $|y-x|>0.$ Hence for large $k$ we have $|y_{m_k}-x_{m_k}| > |y-x|/2.$ Thus for large $k$ we see

$$|f(y_{m_k})-f(x_{m_k})| > m_k|y_{m_k}-x_{m_k}|> m_k|y-x|/2\to \infty.$$

That's a contradiction, because $f$ is continuous on $K,$ hence is bounded on $K.$ Therefore we must have $y=x.$

But if $y=x$ we can invoke the local Lipschitz condition at $x$ to see that $(1)$ cannot happen, contradiction. This proves the result.

Added later: Note that this result doesn't depend on convexity at all: If $X,Y$ are arbitrary metric spaces, and $f:X\to Y$ is locally Lipschitz, then $f$ is Lipschitz on each compact subset of $X.$ The above proof will work nearly word for word in this setting.

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  • $\begingroup$ Nice. But how can you conclude that $f$ is continuous on $K$? $\endgroup$ – giobrach Jun 25 '17 at 16:30
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    $\begingroup$ @giobrach To prove continuity at $x$ just note that for $y\in\Omega_x, $ $|f(y)-f(x)|\le L_x|y-x|.$ $\endgroup$ – zhw. Jun 25 '17 at 16:34
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It does not suffice to cover $K$ with open balls.

To understand this, say $n=1$ and $K=[0,1]\cup [2,3]$. Then the maximum of the Lipschitz constants $L_{[0,1]}$ and $L_{[2,3]}$ does not necessarily work for $K$, since $f$ can have a huge derivative in the interval $[1,2]$.

It order to take care of this issue, first define $$ C=\mathrm{convex\,hull\,\, of\,\,}K=\{t_1x_1+\cdots+t_{n+1}x_{n+1}:t_i\ge 0,\,\sum t_i=1,\, x_i\in K, \forall i=1,n+1\}. $$ Indeed, $C$ is uniquely defined.

Observe that $C$ is also compact, and then cover $C$ with finitely many open balls.

If now $x_1,x_2\in K\subset C$, then $$ [x_1,x_2]=\{(1-t)x_1+tx_2: t\in [0,1]\}\subset C, $$ and if $B_1,\ldots,B_k$ the ball which cover $C$, then there exist $y_1,\ldots,y_m\in [x_1,x_2]$, such that $$ y_1,\ldots,y_m\in[x_1,x_2], $$ and $$ [x_1,y_1]\subset B_{i_1}, [y_1,y_2]\subset B_{i_2}, \ldots, [y_m,x_2]\subset B_{i_{m+1}}, $$ and $$ |f(x_1)-f(x_2)|\le |f(x_1)-f(y_1)|+|f(y_1)-f(y_2)|+\cdots+|f(y_m)-f(x_2)| \le L(|x_1-y_1|+|y_1-y_2|+\cdots+|y_m-x_2|)=L|x_1-x_2|, $$ since $x_1,y_1,\ldots,y_m,x_2$ lie in a line in that order.

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  • $\begingroup$ Wow. So if I require that $K$ itself be convex by hypothesis, the proof holds? $\endgroup$ – giobrach Jun 25 '17 at 14:36
  • $\begingroup$ P.S. Is such a set $C$ unique? If so, how does one prove it? (I corrected the proof in the question. Thank you for pointing this out.) $\endgroup$ – giobrach Jun 25 '17 at 14:51
  • $\begingroup$ See my edited answer $\endgroup$ – Yiorgos S. Smyrlis Jun 25 '17 at 17:23
  • $\begingroup$ So $C$ is the set of all possible linear combinations of $n+1$ elements of $K$ such that the coefficients are nonnegative and add up to $1$? Very interesting. But I don't understand what this has to do with convexity. Can you please point me to some sources that clarify this? $\endgroup$ – giobrach Jun 25 '17 at 17:51
  • $\begingroup$ @giobrach I wrote the answer a bit more in detail. $\endgroup$ – Yiorgos S. Smyrlis Jun 25 '17 at 18:26
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Hint: I think it is easier to work with the definition of open covers of $K \subset \mathbb{R^n}$ instead of sequences. Since $K$ is compact, you can find for any covering of open supsets a finite subset of that covering $\{A_1,\dots,A_k \}$ that already covers $K $.
Now choose $L:=max\{L_1,\dots,L_k\}$, where $L_i$ is the Lipschitz constant for $f|A_i$, $i=1,\dots,k$.
Use this $L$ as your Lipschitz constant for $f|K$.

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  • $\begingroup$ well... I might overlook something but it seems to me that this is exactly what OP did. $\endgroup$ – Surb Jun 25 '17 at 14:23
  • $\begingroup$ true that, did not read it carefully enough. @Surb $\endgroup$ – Simonsays Jun 25 '17 at 15:01

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