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Why can't holomorphic functions be extended to a closed disc?

  1. Apr 5, 2012 #1
    If u is harmonic function defined on (say) the open unit disc, then it can be continuously extended to the closed unit disc in such a way that it matches any continuous function f(θ) on unit circle, i.e. the boundary of the disc. But my understanding is the same cannot be said of holomorphic functions, and this bothers me, because it seems like I can easily prove that it can be. Since u is harmonic it is the real part of some holomorphic function H=u+iv defined on the open unit disc, and v is harmonic. Then if F(θ) is some continuous function defined on the unit circle, we can extend u continuously to the closed unit disc so that it matches Re(F(θ)) on the unit circle, and we can extend v continuously to the closed unit disc so that it matches Im(F(θ)) on the unit circle. Thus we have successfully extended H continuously to the closed unit disc so that it matches F(θ) on the boundary. Where am I going wrong?

    Any help would be greatly appreciated.

    Thank You in Advance.
     
  2. jcsd
  3. Apr 5, 2012 #2
    consider the holomorphic function f(z)=1/(1-z), which cannot be extended to closed unit disk, neither can its real or imaginary part.
     
  4. Apr 5, 2012 #3
    But can't all harmonic functions defined on the open disc be extended continuously to the closed unit disc, so as to match any continuous function defined on the boundary?
     
  5. Apr 5, 2012 #4
    But in this case the boundary is not continuous, there's a pole at z=1.
     
  6. Apr 5, 2012 #5
    I think we're talking about different things. I'm talking about extending functions on the open disc to match arbitrary continuous functions on the boundary. You're talking about extending functions onto the boundary so that they're equal to what they're "supposed" to equal on the boundary.
     
  7. Apr 5, 2012 #6
    How wound you extend the harmonic function u(x,y)=0 on the open disk continuously to match the continuous boundary f(S1)=1?
     
  8. Apr 5, 2012 #7
    I'm sorry, I completely misstated the result. The actual result is the following: given a continuous function f defined on the boundary, there EXISTS a harmonic function defined on the open unit disc which can be extended continuously to the boundary to match f.

    So given this, why can't the same result be proven for holomorphic functions?
     
  9. Apr 5, 2012 #8

    mathwonk

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    try z-->zbar on the unit circle.
     
  10. Apr 5, 2012 #9
    OK, let u be a harmonic function on the open unit disc which can be continuously extended so as to equal cos(θ) on the unit circle, and let v be a harmonic function on the open unit disc which can be continuously extended so as to equal -sin(θ). Then H=u+iv can be continuously extended so as to equal zbar on the unit circle. Of course, since it does equal zbar, H will not be holomorphic on the unit circle, but that's not a problem, is it?
     
  11. Apr 5, 2012 #10


    I might be misunderstanding something basic here, but holomorphic functions are functions which are analytic on the whole complex plane, so how could one expect to extend such things to...somewhere?

    Of course, a holom. function is ALSO continuous everywhere, so also on the boundary and, thus, the result you mention applies to it...

    Perhaps you meant meromorphic functions? And even this: where a(some) pole(s) of the function are could make a difference...

    DonAntonio
     
  12. Apr 5, 2012 #11
    We can define holomorphic functions on any connected subset of the complex plane. Just like we can define a continuous real function on any open interval in the real line.
     
  13. Apr 5, 2012 #12
    You can't just take any old harmonic functions u and v and expect u+iv to be holomorphic. You have to check that these satisfy the Cauchy-Riemann equations, and the harmonic functions you refer to won't. Indeed, a simple application of the uniqueness theorem shows that u and v are continuous on the closed unit disk, harmonic on the interior, and u(x,y) + iv(x,y) = x-iy for all (x,y) on the boundary circle, then u(x,y) = x and v(x, y) = -y for all x and y. In other words, you just extended the function [itex]\overline{z}[/itex] on the unit circle to the function [itex]\overline{z}[/itex] on the closed unit disk, and that function is not holomorphic.

    Another way of proving that there are no continuous extensions of [itex]f(z)=\overline{z}[/itex] to the unit disk which are holomorphic on the interior is to use the Cauchy integral theorem is to use the Cauchy integral theorem. Specifically, if f is a continuous function on the disk holomorphic on the interior, then the integral of f around any closed circle centered at the origin and having radius r<1 must be zero (by the cauchy integral theorem). So taking the limit as [itex]r \rightarrow 1[/itex] and applying, say, the uniform convergence theorem, we also have that the integral of f around the unit circle must be 0 as well. But the integral of [itex]\overline{z}[/itex] around the unit circle is 2πi, not zero. So no such function can agree with [itex]\overline{z}[/itex] on the unit circle.
     
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