Let $\Lambda=\{\lambda_n\}_1^\infty$ a set of points on the real line. We denote by $\bar{n}(r)$ the largest number of points in any interval $[x,x+r]$, $r>0$. Define the upper uniform density (fist done by Beurling \begin{equation*} \delta(\Lambda):=u.u.d.(\Lambda)=\lim_{r\rightarrow \infty}\frac{\bar{n}(r)}{r} \end{equation*} Without loss of generality assume that $0\in\Lambda$.

Now define \begin{equation*} f(z)=\lim_{R\rightarrow\infty}\bigg\{\prod_{0<|\lambda|<R}(1-\frac{z}{\lambda})\bigg\} \end{equation*} Furthermore, fix $a>\delta(\Lambda)$. I am interested in calculating a sharp bound of the form below (with $C$ and $m$ as sharp as possible)

\begin{equation*} |f(x+iy)|\le C(|x+iy|+1)^m e^{a|y|} \end{equation*} Without the sharp bounds this has been studied by Arne Beurling and has implications for interpolation of deltas type functions (note that f is 1 at 0 and 0 on other entries in $\Lambda$) using entire functions.

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    $\begingroup$ Bound under what restrictions? You realize that the condition involving only the limit allows you to chose absolutely any $\lambda_j$ and make the function as large as you wish in any finite domain, don't you? $\endgroup$ – fedja Oct 24 '13 at 23:17

Grepstad has carried out the details of Beurling's paper in her Master thesis, see


For the question you ask, see formula (3.6). I do not know whether the exponential type obtained is sharp...

  • $\begingroup$ Thanks, The argument carried out here is exactly the same as Beurling's argument (as mentioned). However, this is a really good reference for clarifying what I want. I'm interested in bounding $g(z)$ on page 14 on the real line in infinity norm. For this purpose we have $\endgroup$ – mohi Oct 28 '13 at 3:34
  • $\begingroup$ $|g(z)|\le max_j|w_j|\sum_j|g_j(z)|\le max_j|w_j|\tilde{C}$. I'm interested in knowing when is it that I can have $\tilde{C}=1$. That is, what is the smallest ratio $a/\delta(\Lambda)$ under which it is possible to have $\tilde{C}=1$. $\endgroup$ – mohi Oct 28 '13 at 3:38
  • $\begingroup$ Thanks for the reference Gustavo. Also, Grepstad is actually a woman. $\endgroup$ – Tony S.F. Jun 16 '16 at 20:48

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