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Trying to remember what Theorem this is.

  1. Apr 20, 2012 #1
    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.
  2. jcsd
  3. Apr 20, 2012 #2
  4. Apr 20, 2012 #3
    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.
  5. Apr 21, 2012 #4
    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.
  6. Apr 23, 2012 #5
    Hmm, my Calc II teacher never mentioned anything about a unit disk.
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