Trying to remember what Theorem this is.

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Discussion Overview

The discussion revolves around the properties of analytic functions, specifically regarding a theorem related to the constancy of such functions on connected sets. Participants also explore the concept of a function being defined in a neighborhood of the unit disk, seeking clarification on its meaning and implications.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant proposes that if a function f is analytic on a connected set D and is constant on some nonempty open subset of D, then it should be constant on all of D, seeking confirmation of this statement as a specific theorem.
  • Another participant references a link to a resource discussing the uniqueness of analytic continuation, potentially related to the theorem in question.
  • A participant asks for clarification on what it means for a function to be defined in a neighborhood of the unit disk, expressing uncertainty about whether it refers to a neighborhood within the disk or one that contains an open ball encompassing the disk.
  • One participant shares their recollection of examples from a Calculus II class, suggesting that the neighborhood of the unit disk refers to points within the disk in a fixed plane, providing a specific interpretation based on their understanding.
  • Another participant echoes the previous comment about the unit disk but notes that their own instructor did not mention anything about it, indicating a lack of consensus on the interpretation of the term.

Areas of Agreement / Disagreement

The discussion contains multiple viewpoints regarding the theorem about analytic functions and the definition of a neighborhood of the unit disk. There is no consensus on the latter, as participants express differing interpretations and experiences.

Contextual Notes

Participants express uncertainty about the definitions and implications of terms related to analytic functions and neighborhoods, indicating potential limitations in their understanding or differing educational backgrounds.

Poopsilon
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I vaguely recollect that the following statement is true:

Let f be analytic on a connected set D, then if f is constant on some nonempty open subset of D then it is constant on all of D.

Can anyone confirm that this is true and is it a specific theorem? Thanks.
 
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Ah yes, thank you, and while I have your attention, what does it mean to say that a function is defined in a neighborhood of the unit disk? I know what it means for it to be defined in a neighborhood of a point, but I can't figure out what this means. I can't tell if it means some neighborhood inside of the unit disk, or some neighborhood which contains an open ball which contains the unit disk.
 
I recall my Calculus -II professor doing examples on the blackboard where the "neighbourhood" of the unit disk meant within the unit disk in a fixed plane. A unit disk is determined along two axes xy, yz, or xz. Well the function would be defined for all x<1 and y<1 where z is constant if the unit disk is defined along the xy plane. That is the function would not be defined on an open ball containing the unit disk but it means that if a unit disk us defuned along any two axes (xy, yz ,xz) then the function would be defined as long as those two variables ( x,y,z) are less than 1 and the third variable is constant.

Hope that makes sense.
 
hivesaeed4 said:
I recall my Calculus -II professor doing examples on the blackboard where the "neighbourhood" of the unit disk meant within the unit disk in a fixed plane. A unit disk is determined along two axes xy, yz, or xz. Well the function would be defined for all x<1 and y<1 where z is constant if the unit disk is defined along the xy plane. That is the function would not be defined on an open ball containing the unit disk but it means that if a unit disk us defuned along any two axes (xy, yz ,xz) then the function would be defined as long as those two variables ( x,y,z) are less than 1 and the third variable is constant.

Hope that makes sense.

Hmm, my Calc II teacher never mentioned anything about a unit disk.
 

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