# Function in the Hardy space $\mathbb H^{2}$

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?! – Michelle Mar 28 '12 at 19:15
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 $\|f\|^{2}_{\mathbb H^{2}(\mathbb C^{+})}=2\pi\|g\|^{2}_{L^{2}}$. – Michelle Mar 29 '12 at 0:47
If you have that for all $y$ can't you just take the limit for $y \to \infty$? – Jonas Teuwen Mar 29 '12 at 9:27
I don't think so! this will mean that all functions in the Hardy space are zeros! – Michelle Mar 31 '12 at 23:57
Yes, so then your formula is incorrect, no? If this holds for all $y$ then also for very large $y$. – Jonas Teuwen Apr 1 '12 at 16:13