# Entire functions of order 1 and type 0

There is no entire function of order 1 and type 0 which is bounded on the real line. This result can for example be found in the book "Entire functions" by Boas.

I wonder under which weaker conditions of boundedness this is still true. For example, do there exist entire functions $f$ of order 1 and type 0, which in addition satisfy a bound like $|f(x)| < C\exp(c|x|^\rho)$ for real $x$, with some constants $c,C$ and $0<\rho<1$? Is the class of such functions non-emtpy, and maybe can even be characterized?

These questions would also benefit from explicit examples (no infinite products..) of functions of order 1 and type 0, of which I unfortunately don't know any. So if you know such a function, I'd be happy to learn it, too.

No, such functions do not exist. Let $u=\log|f|$. Your condition implies that $u(x)\leq O(|x|^\rho)$, where $0\leq \rho<1$, so the Poisson integral of $u^+$ is convergent, and one can obtain the formula $$u(z)=\frac{y}{\pi}\int_{-\infty}^\infty u(t)\frac{dt}{(t-x)^2+y^2}\quad\quad +ky+\log|B(z )|,$$ where $z=x+iy$, and $B$ is a Blaschke product for the upper half-plane. See, for example, theorems 7 and 5 in Levin, Distribution of zeros of entire functions, Chap V, sect 3. (Such functions are called of class $A$ in Levin's books, and of Cartwright's class in other books. It is a rather deep result of Cartwright that convergence of the poinsson integral of $u^+$ implies in this case the convergence of the Poisson integral of $u$).

Now the condition that $f$ is of zero type implies that $k=0$. (See Thm 6 in the same book, same chapter). Then $u$ is estimated from above by the Poisson integral since $\log|B|\leq 0$.

Inserting to the Poisoon integral $|x|^\rho$ instead of $u(x)$, we obtain an upper estimate for $u$. This estimate involvs an integral which can be computed, and computation shows that $u$ must be of order $\rho<1$.

So it cannot be of order $1$, minimal type.

On your other question (explicit examples of functions of order $1$ and type zero), of course the simplest answer is an infinite product, take zeros at the points $\pm [r\log r]$, for example, where $[.]$ is the integer part. If you don't like products then you can construst examples with explicit power series. Take the coefficients $a_n=n^{-n}(\log n)^{-n}$. Hope this is explicit enough.

EDIT. Here is another way to construct a function of minimal type order $1$. Functions $f$ of at most minimal type order 1 have the following description $$f(z)=\frac{1}{2\pi i}\; {\mathrm{res}}_{\zeta=0} F(1/\zeta)e^{z\zeta},$$ where $F$ is an arbitrary entire function with the property $F(0)=0$. This is called Leau's theorem (19 century).

Now if you want $f$ to be exactly of order $1$ (not "at most"), take $F$ of infinite order. For example $F(z)=\exp(\exp(z))$. So you have an explicit formula involving an integral.

• Thanks a lot for the answer. With "explicit examples" I was rather thinking of something like $\sin$, $\Gamma$, which one can easily estimate; but using Poisson integrals might be a good way out here. – Gandalf Lechner Dec 12 '12 at 11:39
• Elementary functions and most special functions have integer order, normal type. Gamma is an exception but it has maximal type. So you cannot have "explicit examples". Poisson integral is indeed a good tool to work with them, but it does not allow you to construct examples. – Alexandre Eremenko Dec 12 '12 at 14:35