Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

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).

share|improve this answer
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. –  boinkboink Apr 3 '12 at 4:08
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. –  boinkboink Apr 3 '12 at 5:55
What constant did you get? What is your construction? –  Mark Lewko Apr 3 '12 at 6:11
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.