# Locally Lipschitz function is bounded

• May 12th 2012, 06:56 AM
Dinkydoe
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 $x_1\in B$, and let $U_1:=U(x_1)$ be the largest neighborhood in $B$ containing $x_1$, such that $f|_{U_1}$ is Lipschitz with constant $L_1$. For arbitrary $n$, if $B_n:=B\setminus \bigcup_{i is nonempty, choose $x_n\in B_n$ and define $U_n$ similarly. Repeat indefinately. If we find a finite sequence $U_1,\cdots, U_n$ we're done since $f(B) = \bigcup_{i\leq n} f(U_i)$ is clearly bounded since all $f(U_i)$ are bounded. If we find an infinite sequence $U_1,U_2,\cdots$ we have that $L_n\to \infty$, and we must show that $f(B)$ remains bounded. Clearly, since $B$ is bounded we have diam( $U_n)\to 0$...

, and I wanted to use this. But, now I'm thinking that my argument wouldn't work because $f(B)$ could still be unbounded, if the diam( $f(B_i)$) 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?
• May 12th 2012, 11:32 AM
mastermind2007
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 $\Leftrightarrow$ it is compact.
So choosing a sequence of $(x_n)_{n \in \mathbb{N}}$ and the corresponding neighborhoods in which
the function f restricted to these neighborhoods is Lipschitz with constants $L_i$, 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.
• May 12th 2012, 12:18 PM
Dinkydoe
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. $\left\|\cdot\right\|$

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 :)
• May 12th 2012, 07:31 PM
mastermind2007
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!
• May 12th 2012, 10:51 PM
Dinkydoe
Re: Locally Lipschitz function is bounded

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....