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Analytic on open strip

  1. Jun 26, 2008 #1
    Let f be analytic on {z : 0 < Im z < 1} and continuous on the closure of this set.

    Suppose that f(z) = 0 if z is real. Show that f is identically zero.

    Any help please?
  2. jcsd
  3. Jun 26, 2008 #2

    matt grime

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    Well, what (standard?) results do you know that might help? Any that tell you when an analytic function is zero? I can think of 2, one of which gives you the answer.
  4. Jun 26, 2008 #3
    Well, I know the identity theorem, but we can't apply it here :(

    [because f is not zero inside the region where f is analytic]
  5. Jun 26, 2008 #4

    matt grime

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    Can you think of anyway to extend f to be analytic on the region where it is zero?
  6. Jun 27, 2008 #5


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    Of course the fact that f is continuous on the closure of the set is crucial- you will have to use that.
  7. Jun 27, 2008 #6
    I once asked about what kind of things can happen that prevent analytic continuation. I think it turned out that there can be boundaries (lines or other one dimensional manifolds) over which analytic continuation cannot be carried out, and was it the reason, that the function approaches zero or infinity at the boundary? If the claim in opening post is correct, I'm probably remembering something wrong then?

    Or does the boundary of continuation arise so that the function does not have any limits on it?
  8. Jun 27, 2008 #7
    My mistake. I was thinking about this: http://en.wikipedia.org/wiki/Domain_of_holomorphy

    But it doesn't give a counter example to the claim in the opening post.
  9. Jun 27, 2008 #8

    matt grime

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    I would hope it doesn't give a counter example: the reflection principle is one of the most elementary results in complex analysis - it's a simple consequence of Morera's theorem.
  10. Jun 27, 2008 #9


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    jostpuur - having zeros which accumulate at the boundary is different from saying that it goes to zero at the boundary.
    eg, sin(1/z) is defined for z !=0 and has zeros which accumulate at z=0, but it doesn't go to zero as z->0.
  11. Jun 27, 2008 #10
    I had understood this by the time of my previous post, where I said that the function described by the Wikipedia isn't a counter example.

    When writing post #6, I thought I had seen something that could be a counter example, but then I found the Wikipedia page, took a closer look, and noticed that I remembered it wrong, and then wrote the post #7.
    Last edited: Jun 27, 2008
  12. Jun 27, 2008 #11
    Define [itex]g_y : \mathbb{R} \rightarrow \mathbb{C}[/itex] so that [itex]g_y(x) = f(x+iy)[/itex]. Cauchy's theorem plus continuity of [itex]f[/itex] at the boundary imply that

    [tex]\int_{-a}^a (g_y(x)+g_y(-x))dx = 0[/tex]

    (taking a symmetric rectangular contour with base arbitrarily close to the real line). The continuity of [itex]g_y[/itex] gives that

    [tex]\frac{1}{2\epsilon} \int_{\epsilon}^{\epsilon} (g_y(x)+g_y(-x))dx \Rightarrow g_y(0)[/tex]

    So [itex]f(iy)=0[/itex], and you can apply the identity theorem.
    Last edited: Jun 27, 2008
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