Locally Lipschitz function is bounded

Hoi, I'm trying to prove that in a banach space: If f is locally lipschitz, then it maps bounded sets onto bounded sets.

Sounds fair right?

I started an argument as follows, but I couldn't finish it...

Choose $\displaystyle x_1\in B$, and let $\displaystyle U_1:=U(x_1)$ be the largest neighborhood in $\displaystyle B$ containing $\displaystyle x_1$, such that $\displaystyle f|_{U_1}$ is Lipschitz with constant $\displaystyle L_1$. For arbitrary $\displaystyle n$, if $\displaystyle B_n:=B\setminus \bigcup_{i<n} U_{i}$ is nonempty, choose $\displaystyle x_n\in B_n$ and define $\displaystyle U_n$ similarly. Repeat indefinately. If we find a finite sequence $\displaystyle U_1,\cdots, U_n$ we're done since $\displaystyle f(B) = \bigcup_{i\leq n} f(U_i)$ is clearly bounded since all $\displaystyle f(U_i)$ are bounded. If we find an infinite sequence $\displaystyle U_1,U_2,\cdots$ we have that $\displaystyle L_n\to \infty$, and we must show that $\displaystyle f(B)$ remains bounded. Clearly, since $\displaystyle B$ is bounded we have diam($\displaystyle U_n)\to 0$...

, and I wanted to use this. But, now I'm thinking that my argument wouldn't work because $\displaystyle f(B)$ could still be unbounded, if the diam($\displaystyle f(B_i)$) doesn't go to 0 fast enough...

what can I use here?? I dont see how this would especially work in a banach space. Maybe, the fact that we are working in a banach space is not even necessary?

Re: Locally Lipschitz function is bounded

Maybe you need the Banach space because of compacity.

I don't know if it is correct, but consider the following:

In a metric space, a set is closed and bounded $\displaystyle \Leftrightarrow $ it is compact.

So choosing a sequence of $\displaystyle (x_n)_{n \in \mathbb{N}} $ and the corresponding neighborhoods in which

the function f restricted to these neighborhoods is Lipschitz with constants $\displaystyle L_i $, you can always

pick a finite sequence to still cover the whole bounded (and closed) set.

If the set is open, wasn't there some theorem that one can, for a continuous function, extend it to the boundary continuously?

Well, with closed bounded sets it works fine - find some argument for open ones. ;-)

Then space doesn't need to be Banach though, it only has to be a metric space.

Re: Locally Lipschitz function is bounded

Thanks for the response, but that doesn't work. Banach-property only means that the metric d, is induced by some norm. $\displaystyle \left\|\cdot\right\|$

The compactness property does not hold in general Banach-spaces, as they may be infinite dimensional. What you say is true in finite dimensional metric spaces.

So, i wish i could pick such finite sequence...But i think I found an argument, considering a line-segment through 2 points x, and y in B. This is something we can do, because we're working in a normed space. And then showing that |f(x)-f(y)| must be finite. Then on this line-segment we can make a sequence on covers U(x) where we can show that the diameter can not converge to 0, and must therefore be a finite sequence...bla bla. And then the argument works.

If i wrote up the full argument...if i still agree with myself by then (LOL), I'll post it :)

Re: Locally Lipschitz function is bounded

Still, the argument does not depend on any dimension.

The theorem of Heine-Borel applies to ANY metric space.

No dimension needed!

Re: Locally Lipschitz function is bounded

Lol no, please...:P

Closed and bounded is equivalent with compact, only in complete finite dimensional metric spaces....

For arbitrariy (also complete) metric spaces this is different: compact is equivalent with closed and 'totally bounded'

totally bounded means that for any cover u can find a finite subcover...etc. It's not the same as being bounded.

In finite dimensional spaces, bounded is the same as totally bounded. Not in infinite dimensional spaces, like many banach spaces....

please check this out....

Re: Locally Lipschitz function is bounded

Ah yeah, you're right!

Sorry, it was very late when I answered last. ;-)