Beurling's majorant is defined as the unique entire function $B(z)$ such that the Fourier transform of $B(x)$ is compactly supported in the interval $[-1;1]$, $B(x) \geq \text{sgn}(x)$ and $B(x)$ minimizes the integral, $$ \int_{-\infty}^{\infty} B(x) - \text{sgn}(x) \text{d} x=1 $$


$$ B(z) = \left ( \frac{\sin (\pi z)}{\pi} \right )^2 \cdot \left ( \frac{2}{z} + \sum_{n = 0}^{\infty} \frac{1}{(z-n)^2} - \sum_{n=1}^{\infty} \frac{1}{(z+n)^2} \right ) $$

Note that by Paley-Wiener the property that the Fourier transform of $B(x)$ is compactly supported in $[-1;1]$ is equivalent to the property that $B(z)$ is a function of exponential type, with $B(z) = O(e^{2 \pi |\Im z|})$.

In general $B(x)$ is also a very good point-wise approximation to $\text{sgn}(x)$ and this makes it a very valuable function when one needs optimal numerical constants. The need for such arises sometimes in analytic number theory, for example.

Seulberg modified Beurling's function, and considered,

$$ S_{+}(z) = \frac{1}{2} B(\delta (z - a)) + \frac{1}{2} B(\delta ( b - z)) $$

This is a very good majorant for the characteristic function of the interval $[a;b]$. Here $\delta$ is a parameter that regulates the quality of the approximation. The price to pay for a larger $\delta$ is that $\hat{S_{+}}(x)$ vanishes when $x$ is larger, i.e when $|x| > \delta$.

Selberg's majorant has the properties that $S_{+}(x) \geq 1$ when $a \leq x \leq b$ and $S_{+}(x) \geq 0$ otherwise. Furthermore $\hat{S_{+}}(x) = 0$ when $|x| \geq \delta$ and finally $$ \int_{-\infty}^{\infty} S_{+}(x) \text{d} x = b - a + \frac{1}{\delta} $$ When $\delta(a-b)$ is an integer this is in fact the optimal majorant satisfying these constraints, but in general it is quite good.

After this long introduction, my question is the following: Is there a known construction of an optimal majorant $M_{+}(x)$ with the following properties,

1) $M_{+}(x) \geq 1$ when $a \leq x \leq b$, and $M_{+}(x) \geq 0$ otherwise

2) The fourier transform of $M_{+}(x)$ is compactly supported in $[-\delta;\delta]$

3) The difference

$$ \int_{-\infty}^{\infty} |M_{+}(x)|^2 \text{d} x - (b-a) $$

is as small as possible? How small can it be, in terms of $\delta$ ? (EDIT: I am particularly interested in upper bounds for the above quantity).

I will be grateful for any insights, references concerning this question. I know that Valeer has done quite some work with Beurling and Selberg's majorants but I haven't been able to locate anything relevant to my question in his publications.

You can view the Selberg majorant approximating the unit interval with $\delta = 3$ at http://www.freeimagehosting.net/xffsu . (If one of the mods could link the image into my question I will be grateful).


1 Answer 1


Given such a $M$ note that $F(x)=M^2(x)$ will be (1) non-negative, (2) majorize $1_{[a,b]}$, and (3) has $\hat{F}$ supported in $[-2\delta,2\delta]$. Thus it will be a (possibly not optimal) solution to the standard $2\delta$ Beurling-Selberg problem. This implies that $$\int |M(x)|^2dx - (b-a) \geq \frac{1}{2\delta}$$ (at least in the cases when the standard Beurling-Selberg construction is known to be sharp).

  • $\begingroup$ Thank you. This certainly places a limitation on how well we can do. I should have added in my question that I am particularly interested in upper bounds for the above quantity. The quality of the constants is crucial ! I believe (based on some numerical experiments) that it is possible to obtain an upper bound of the form c/delta with c < 1. $\endgroup$
    – boinkboink
    Apr 3, 2012 at 4:08
  • $\begingroup$ I did figure out an answer to my question in a case of interest, and so would like to close this question, and accept your answer. $\endgroup$
    – boinkboink
    Apr 3, 2012 at 5:55
  • 4
    $\begingroup$ What constant did you get? What is your construction? $\endgroup$
    – Mark Lewko
    Apr 3, 2012 at 6:11

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