Bounding Analytic Functions by derivatives

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SUMMARY

The discussion centers on the existence of an analytic function \( f \) that is unbounded in an infinite domain \( D \) while having a bounded first derivative \( f' \) and unbounded higher derivatives \( f'', f''', \) etc. Participants conclude that such a function cannot exist, referencing key concepts from complex analysis, including Cauchy's Integral bounds and Liouville's theorem. The consensus is that if \( f' \) is bounded throughout the complex plane, then it must be constant, leading to all higher derivatives being zero and thus bounded.

PREREQUISITES
  • Understanding of analytic functions in complex analysis
  • Familiarity with Cauchy's Integral theorem and bounds
  • Knowledge of Liouville's theorem and its implications
  • Concept of derivatives in the context of complex functions
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  • Study Cauchy's Integral theorem and its applications in complex analysis
  • Explore Liouville's theorem in detail and its consequences for bounded analytic functions
  • Investigate the properties of derivatives of analytic functions
  • Examine examples of unbounded analytic functions in complex domains
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Mathematicians, particularly those specializing in complex analysis, students studying advanced calculus, and anyone interested in the properties of analytic functions and their derivatives.

Hyperbolful
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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?
 
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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
 
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
 

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