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<h1 align=center> de Branges' positivity condition
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Let $E(z)$ be an entire function satisfying $|E(\bar z)|<|E(z)|$ for
$z$ in the upper half-plane.  Define a  Hilbert space of entire functions
$\mathcal H(E)$ to be the set of all entire functions $F(z)$ such that
$F(z)/E(z)$ is square integrable on the real axis and such that
$$|F(z)|^2\leqslant \|F\|^2_{\mathcal H(E)} K(z, z)  $$
for all complex $z$, where the inner product of the space is
given by
$$\langle F(z), G(z)\rangle_{\mathcal H(E)}=\int_{-\infty}^\infty
{F(x)\bar G(x)\over |E(x)|^2}dx$$
for all elements $F, G\in\mathcal H(E)$ and where 
$$K(w, z)={E(z)\bar E(w)-\bar E(\bar z)E(\bar w)\over
2\pi i(\bar w-z)}$$
is the reproducing kernel function of the space $\mathcal H(E)$,
that is, the identity
$$
F(w)=\langle F(z), K(w, z)\rangle_{\mathcal H(E)} 
$$
holds for every complex $w$ and for every element $F\in\mathcal H(E)$.

 de Branges [
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<a href="http://www.ams.org/msnmain?fn=130&s1=87m:11050&pg1=MR">MR 87m:11050</a>
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<a href="http://www.ams.org/msnmain?fn=130&s1=93f:46032&pg1=MR">MR 93f:46032</a>
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proved the following beautiful

 \noindent{\bf Theorem .}
{\it    Let $E(z)$ be an entire function having no real zeros
such that $|E(\bar z)|<|E(z)|$ for $\Im z>0$, such that $\bar E(\bar z)=\epsilon E(z-i)$
for a constant $\epsilon$ of absolute value one, and such that
$|E(x+iy)|$ is a strictly increasing function of $y>0$ for each fixed real $x$.  
If  $\Re\langle F(z), F(z+i)\rangle_{\mathcal H(E)}\geqslant 0$ for every
element $F(z)\in\mathcal H(E)$ with $F(z+i)\in\mathcal H(E)$, then the zeros 
of $E(z)$ lie on the line $\Im z=-1/2$, and 
$\Re\{\bar{E^\prime}(w) E(w+i)/2\pi i\}\geqslant 0$
when $w$ is a zero of $E(z)$. }

 

Let $E(z)=\xi(1-iz)$.  Then it can be shown that 
$|E(x-iy)|<|E(x+iy)|$ for $y>0$, and that
$|E(x+iy)|$ is a strictly increasing function of $y$ on $(0, \infty)$ for
each fixed real $x$.
Therefore, it is natural to ask whether the Hilbert
space of entire functions $\mathcal H(E)$ satisfies the condition that
$$\Re \langle F(z), F(z+i)\rangle_{\mathcal H(E)}\geqslant 0  $$
for every element $F(z)$ of $\mathcal H(E)$ such that $F(z+i)\in\mathcal H(E)$,
because if so, then the Riemann Hypothesis would follow.

It is shown in [
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<a href="http://www.ams.org/msnmain?fn=130&s1=2001h:11114&pg1=MR">MR 2001h:11114</a>
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<p> <br>
<hr>
<p> <br>
Back to the 
<a href="http://www.aimath.org/WWN/rh/"> main index</a> 
for The Riemann Hypothesis.
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