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Question. Is there an entire function $F$ satisfying first two or all three of the following assertions:

  • $F(z)\neq 0$ for all $z\in \mathbb{C}$;
  • $1/F - 1\in H^2(\mathbb{C}_+)$ -- the classical Hardy space in the upper half-plane;
  • $F$ is bounded in every horizontal half-plane $\{z\colon \text{Im}(z) > \delta\}$?

Thoughts. Let $G= 1/F$. Then we have $G(z) = 1 + \int_0^{\infty} h(x)e^{izx}\, dx$ for some $h\in L^2[0,\infty)$ and all $z\in \mathbb{C}_+$. For nice functions $h$ (e.g., for super-exponentially decreasing) this integral representation can be extended to the whole complex plane and probably the example can be constructed in terms of $h$. However, I don't know if it is possible to find $h$ such that $G$ is non-zero for every $z$.

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    $\begingroup$ Can you explain, what is a "horizontal half-plane" ? Are half-planes $\{ z=x+iy: y\geq 0\}$ and $\{ x+iy: y\leq 0\}$ both horizontal? $\endgroup$ Commented Feb 10, 2023 at 14:40
  • $\begingroup$ @AlexandreEremenko, thank you for your correction, I meant to consider only the upper half-plane. It is fixed now $\endgroup$ Commented Feb 10, 2023 at 15:28

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There is a zero-free entire function bounded in every left half-plane, and such that $f-1$ is in $H^2$ in every left half-plane.

Let $\gamma$ be the boundary of the region $$D=\left\{ x+iy: |y|<2\pi/3, x>0\right\} .$$ Consider the function $$g(z)=\int_\gamma \frac{\exp e^\zeta}{\zeta-z}d\zeta,\quad z\in {\mathbf{C}}\backslash D.$$ The integral evidently converges and $g(z)=O(1/z)$ in ${\mathbf{C}}\backslash D.$ Now, $g$ has an analytic continuation to an entire function: deforming the contour to $\partial D_t$, where $D_t=\{ x+iy:|y|<2\pi/3, x>t\},\; t>0$ does not change $g$ in $D$, and shows that $g$ has an analytic continuation to ${\mathbf{C}}\backslash D_t$, and this is for every $t>0$, so $g$ is entire. Now $f(z)=e^{g(z)}$ is the desired function. If you want upper half-planes take $f(iz)$.

Remark. You can improve the estimate $g(z)=O(1/z)$. Evidently, $g$ has infinitely many zeros $z_1,z_2,\ldots$. Then $g_k(z)=g(z)/((z-z_1)\ldots(z-z_k))$ satisfies $g_k(z)=O(z^{-k-1})$ as $z\to\infty$ outside $D_t$.

Remark 2. This construction is standard in the theory of entire functions, see, for example, Entire function bounded at every line Sometimes this $g$ is called the Mittag-Leffler function.

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  • $\begingroup$ Thanks, this is a very useful example! Will follow the links $\endgroup$ Commented Feb 10, 2023 at 16:19
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Here is a slightly different construction, from this answer by @reuns , the function built this way has a $\exp(\exp(|z|))$ growth compared to triple exponent in the wonderful example by @AlexandreEremenko.

For $z\in\mathbb{C}$, consider $$ g(z) = \int_{i\infty}^z\frac{e^{is}(e^{is} - 1)}{s}ds = - \frac{e^{iz}(e^{iz} - 2)}{2iz} + \int_{i\infty}^z\frac{e^{is}(e^{is} - 2)}{2is^2}ds, $$ where the equality is from integration by parts. For $z\neq 0$, let the contour in the last integral consist of two components: the vertical half-line and the arc of the circle with center in $0$ (as in the picture).contour

After integration by parts the integral over the half-line is exponentially small and the integral over the arc is $O(1/|z|)$ in every upper half-plane.

It follows that $|g(z)| = O(1/|z|)$ as $z\to\infty$ uniformly in every upper half-plane and $f(z) = e^{g(z)}$ will be the answer to the initial question.

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