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Wrong corollary?

  1. Feb 27, 2006 #1

    quasar987

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    After stating the Weierstrass M-test for series of complex functions and the "[itex]f_n[/itex] continuous and uniformly convergeant to f on E ==> f continuous on E" thm, my teacher gives as a corollary that every power series [itex]\sum a_nz^n[/itex] is continuous on its disc of convergence D(0,R). And he doesn't give a proof, as if it's trivial.

    But I think the corollary is wrong. Am I right in thinking so?

    The convergence is absolute over all of D(0,R), but we only know for sure that the convergence is only uniform over [itex]\emptyset = \partial D(0,R) \cap U \subsetneq D(0,R)[/itex]. Hence, so is the continuity.
     
    Last edited: Feb 27, 2006
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  3. Feb 27, 2006 #2

    AKG

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    Corollary: A power series is continuous on its disc of convergence.
    Proof: On its disk, it converges absolutely and uniformly. In particular, it converges uniformly. What does it mean for a power series to converge uniformly? It means that the sequence of partial sums, as a sequence of functions, converges uniformly to the infinite sum. Each of these partial sums is a polynomial, hence continuous. So this sequence of polynomials is continuous and converges uniformly to the infinite series on the series' disk of convergence. Hence, on that disk, the power series is continuous, by the theorem.

    It seems you're getting confused about the ideas of uniform continuity of a (single) function, uniform convergence of a sequence of functions, and uniform convergence of a power series (which is a particular case of a sequence of functions being uniformly convergent).
     
  4. Feb 27, 2006 #3

    quasar987

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    Erm. As far as I know (i.e. according to what my book says and to what the teacher wrote on the blackboard), the convergence is only uniform on a closed disk [itex]\overline{D}(0,r)[/itex] of radius 0<r<R*. Not on the whole D(0,R). Is this not true?

    *I wrote this in a very complicated manner in the original post.
     
    Last edited: Feb 27, 2006
  5. Feb 27, 2006 #4

    HallsofIvy

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    Bit- given any point, p, in the disk of convergence, of radius R, there exist an r< R so that the point is inside the closed disk of radius r. Since convergence is uniform inside that closed disk, the function is continuous at p and therefore continous on the entire (open) disk.
     
  6. Feb 27, 2006 #5

    quasar987

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    I see. Thanks Halls.
     
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