de Branges space

Remarks

$\mathcal{H}(E)$ is a Hilbert space and a reproducing kernel Hilbert space. For $w \in \C$ the reproducing kernel is $$K_w(z) = \frac{\overline{E(w)} E(z) - \overline{E^\#(w)} E^\#(z)}{2\pi i (\bar w - z)},$$ so that $F(w) = \ip{F}{K_w}$ for all $F \in \mathcal{H}(E)$. The condition on $E$ implies that $E$ has no zeros in $\mathbb{H}$, so condition 1 makes sense.

Example: $E(z) = e^{-iaz}$ with $a > 0$. For $z = x + iy$ with $y > 0$ one has $\abs{E(z)} = e^{ay} > e^{-ay} = \abs{E(\bar z)}$. Here $E^\#(z) = e^{iaz}$, and the kernel becomes $K_w(z) = \frac{\sin(a(\bar w - z))}{\pi (\bar w - z)}$. $\mathcal{H}(E)$ is the Paley–Wiener space of entire functions of exponential type at most $a$ that are square integrable on $\R$.

Non-example: $E(z) = 1$ gives $\abs{E(\bar z)} = \abs{E(z)}$, so the strict inequality fails and $E$ does not define a de Branges space.