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I've got two problems on this issue:

If $X,Y$ are metric spaces, $X$ compact, suppose $f:X\to Y$ is locally Lipschitz continuous on $X$, then is it also globally Lipschitz continuous?

My guess is yes. I tried using the finite subcover property (as was used in the proof for uniform continuity on a compact set implied by ordinary continuity): suppose for $x\in X$, there exists an open ball $B(x,r(x))$ with Lipschitz constant $L(x)$, we can adjoin all such open balls to obtain an open cover and hence a finite subcover $B(x_1,r(x_1)),\cdots, B(x_n,r(x_n))$ with Lipschitz constants $L(x_1),\cdots,L(x_n)$. But things were not the same as in uniform continuity proof, because I can no longer confine the two points in comparison to a small neighborhood. I can only find an uniform $L$ such that $$d_Y(f(\tilde{x}),f(\hat{x}))\le L(d_X(\tilde x, x_i)+d_X(x_i,x_j)+d_X(x_j,\hat x))$$ So I think I'm on the wrong track. Or worse yet, my initial guess is false itself?

The second problem is

If $E$ is a compact subset of $\Bbb R^2$, $f(x,y)$ is locally Lipschitz continuous with regard to $y$ on $E$, that is, for each $(x,y)\in E$, there exists an open ball $B((x,y),r(x,y))$ with Lipschitz constant $L(x,y)>0$ such that $(x_0,y_1),(x_0,y_2)\in B((x,y),r(x,y))$ implies $$|f(x_0,y_1)-f(x_0,y_2)|\le L(x,y)|y_1-y_2|$$ then is $f$ also globally Lipschitz continuous with regard to $y$ on $E$? In other words, can we choose an uniform $L$ on $E$?

I don't know whether this one is right, let alone any idea to prove/disprove it. So I need your help here. Thanks in advance.

Vim
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  • Global continuity is clear (continuity is a local property, and $f$ is continuous at every $x$), maybe you are asking about uniform continuity? – Henno Brandsma Nov 20 '15 at 07:05
  • @HennoBrandsma I'm asking about global Lipschitz continuity, sorry. – Vim Nov 20 '15 at 07:06
  • http://math.stackexchange.com/a/154753/4280 has a proof for 1, there are links in the whole question to other answers. – Henno Brandsma Nov 20 '15 at 07:11
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    As a side note, it's generally frowned upon to ask two different questions in the same post as a matter of the way the search engines work. It's better practice to ask them in separate questions (You can include a link to each from the other if you feel the need to connect them) – Alan Nov 20 '15 at 07:13
  • @HennoBrandsma Thanks for your sources! It really helps. – Vim Nov 20 '15 at 07:15
  • @Alan Thanks for your suggestion. But since in this post the two problems are so closely related, I feel it might be acceptable to put them together. – Vim Nov 20 '15 at 07:18
  • Closely related, and the same proof works for both; hence, a duplicate. –  Dec 16 '15 at 07:15

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