Bounding Analytic Functions by derivatives

  • #1
Ok my last post was trivial, but it led to this question

Assume f is unbounded and analytic in some domain D, and f' is bounded in D

does there exist a function for which the above holds and f'',f''',... are all unbounded in D?
 

Answers and Replies

  • #2
D must be an infinite domain, else, if f was unbounded, f' would also be unbounded.

now, suppose that f'' is unbounded, that means that the area under it will grow unboundedly fast,
which implies that f' wil be unbounded

there is not such function
 
  • #3
By D infinite do you mean an unbounded domain? I'm referring to subsets of the complex plane.

So yes if f is analytic, then f' is analytic, and if f' is bounded on all of C then by louiville's theorem f' is constant so then the rest of the derivatives are zero and hence bounded.

Sorry, I meant my question to more complex analysis based. What it amounts to is a function unbounded on an unbounded subset of the complex plane, that isn't the whole plane, and who's derivative is bounded in the same domain, but who's other derivatives are all unbounded in that domain as well.

I think it still doesn't exist as a consequence of Cauchy Integral bounds though
 

Related Threads on Bounding Analytic Functions by derivatives

Replies
3
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
554
Replies
7
Views
2K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
13
Views
4K
  • Last Post
Replies
4
Views
10K
Top