de Branges' positivity condition

Let $E(z)$ be an entire function satisfying $\vert E(\bar z)\vert<\vert E(z)\vert$ 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

\begin{displaymath}\vert F(z)\vert^2\leqslant \Vert F\Vert^2_{\mathcal H(E)} K(z, z) \end{displaymath}

for all complex $z$, where the inner product of the space is given by

\begin{displaymath}\langle F(z), G(z)\rangle_{\mathcal H(E)}=\int_{-\infty}^\infty
{F(x)\bar G(x)\over \vert E(x)\vert^2}dx\end{displaymath}

for all elements $F, G\in\mathcal H(E)$ and where

\begin{displaymath}K(w, z)={E(z)\bar E(w)-\bar E(\bar z)E(\bar w)\over
2\pi i(\bar w-z)}\end{displaymath}

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 [ MR 87m:11050] and [ MR 93f:46032] proved the following beautiful

Theorem . Let $E(z)$ be an entire function having no real zeros such that $\vert E(\bar z)\vert<\vert E(z)\vert$ 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 $\vert E(x+iy)\vert$ 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 $\vert E(x-iy)\vert<\vert E(x+iy)\vert$ for $y>0$, and that $\vert E(x+iy)\vert$ 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

\begin{displaymath}\Re \langle F(z), F(z+i)\rangle_{\mathcal H(E)}\geqslant 0 \end{displaymath}

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 [ MR 2001h:11114] that this condition is not satisfied.

Back to the main index for The Riemann Hypothesis.