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 , and let be the largest neighborhood in containing , such that is Lipschitz with constant . For arbitrary , if is nonempty, choose and define similarly. Repeat indefinately. If we find a finite sequence we're done since is clearly bounded since all are bounded. If we find an infinite sequence we have that , and we must show that remains bounded. Clearly, since is bounded we have diam( ...

, and I wanted to use this. But, now I'm thinking that my argument wouldn't work because could still be unbounded, if the diam( ) 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?