Analytic on open strip

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.

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.

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]

matt grime
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Can you think of anyway to extend f to be analytic on the region where it is zero?

HallsofIvy
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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.

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?

When n = 1, then every open set is a domain of holomorphy: we can define a holomorphic function which has zeros which accumulate everywhere on the boundary of the domain

But it doesn't give a counter example to the claim in the opening post.

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.

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.

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.

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.

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Define $g_y : \mathbb{R} \rightarrow \mathbb{C}$ so that $g_y(x) = f(x+iy)$. Cauchy's theorem plus continuity of $f$ at the boundary imply that

$$\int_{-a}^a (g_y(x)+g_y(-x))dx = 0$$

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

$$\frac{1}{2\epsilon} \int_{\epsilon}^{\epsilon} (g_y(x)+g_y(-x))dx \Rightarrow g_y(0)$$

So $f(iy)=0$, and you can apply the identity theorem.

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