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Given a function $f(z)$, $z=x+iy, x,y\in \mathbb R$, which belongs to $\mathbb H^{2}(\mathbb C^{+})$, where $\mathbb C^{+}$ is the upper half plane Im$(z)>0$ and $f(a_{n})=0$, for all $n\in \mathbb Z$, where $a_{n}'s$ are all real numbers. What can we say about this function $f$? should it be the zero function!? or there is something else that we can say about?

Edit: $f$ is continuous on the real line, and has singularities in the lower half plane, and the sequence $\{a_{n}\}$ has no accumulation point.

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    I just got this: since $f\in \mathbb H^{2}(\mathbb C^{+})$ , then $$|f(z)|\leq \sqrt{\frac{2}{\pi y}}\|f\|_{2}$$ for all $y>0$. Does this help?!2012-03-28
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    Also, for $f\in \mathbb H^{2}(\mathbb C^{+})$, there is $g\in L^{2}(0, \infty)$ such that $$f(z)=\frac{1}{2\pi}\int_{0}^{\infty}g(t)e^{itz}dt$$ and $\|f\|^{2}_{\mathbb H^{2}(\mathbb C^{+})}=2\pi\|g\|^{2}_{L^{2}}$.2012-03-29
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    If you have that for all $y$ can't you just take the limit for $y \to \infty$?2012-03-29
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    I don't think so! this will mean that all functions in the Hardy space are zeros!2012-03-31
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    Yes, so then your formula is incorrect, no? If this holds for all $y$ then also for very large $y$.2012-04-01
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    This theorem might be helpful. Theorem (Fatou):Let $K$ be a closed set of Lebesgue measure zero on the unit circle. Then there exist a function which is continuous on closed disk $\bar{\mathbb{D}}$ and holomorphic on open unit disk $\mathbb{D}$ and vanishes precisely on $K$.2015-05-09

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