Suppose a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is locally Lipschitz. Prove $f$ is Lipschitz on $[a,b]$.

Here is what I have so far: Let $[a, b]$ be some closed, bounded interval. Since f is locally Lipschitz, for each $x\in[a; b]$, we may find some $U_x$ and some $M_x$ such that $|f(y)-f(z)| < M|x-y|$. Let $\mathcal{U}$ denote the collection of all such open neighborhoods $U_x$. Then $\mathcal{U}$ is an open cover of $[a, b]$. By the Heine-Borel Theorem, $[a, b]$ is compact, so $\mathcal{U}$ has a finite subcover, \mathcal{V}. Label the members of $\mathcal{V}$ as $U_{x_1}, U_{x_2},\ldots,U_{x_n}$. Then $[a, b]$ and for each $U_{x_k}$, we associate a corresponding $M_{x_k}$ such that if $y, z\in U_{x_k}$ , then $|f(y)-f(z)|< M_{x_k}|y-z|$. Let $M = \max\{M_{x_1},\ldots,M_{x_n}\}$. Let y and z be some points in $[a, b]$ with z < y. If both y and z lie in the same neighborhood in $\mathcal{V}$, then we are done.

This is as far as I got. I do not know how to handle the case when $y$ and $z$ are in different neighborhoods.

  • $\begingroup$ You're missing some dollar signs around \mathcal{V}. I'd've just edited it to fix it, but two-character edits by other people aren't allowed :( $\endgroup$ – Ben Millwood Jun 6 '12 at 16:01

Very good work so far.

By the lebesgue covering lemma, there exists $\delta >0$ s.t. if $x,y$ satisfy $|x-y|<\delta$ then there is $k$ s.t. $x ,y \in U_{x_k}$.

Let $x<y$. Choose $x=x_1\leq x_2 \leq \dots \leq x_n = y$ s.t. $|x_i - x_{i+1}| < \delta$. Then we have $|f(x)-f(y)| \leq |f(x_1) -f(x_2)| + |f(x_2) -f(x_3)|+\dots + |f(x_{n-1}) - f(x_n)| \leq \sum_i M |x_i-x_{i+1}| = M|x-y|$

  • $\begingroup$ why is $\sum_i M|x_{i+1}-x_i|=M|x-y|$? $\endgroup$ – john Jun 5 '12 at 21:17
  • $\begingroup$ Because the points are chosen to be increasing, from x to y, so the sum of their distances is just the distance from x to y. $\endgroup$ – Seth Jun 5 '12 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.