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Can the function $(-1)^n$, $n=1,2,...$ be extended to an analytic function $f(z)$ defined on the right half complex plane satisfying the growth condition $$|f(x+iy)|\le C e^{Px+A|y|},$$ with $A<\pi$ and $C,P\in\mathbb{R}$?

I know that such an extension would be unique by Carlson's theorem and I showed that the obvious "power function" extension from complex analysis does not work for any choice of branch. I'm not sure where to go from here.

zbrads2
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    Readers should note that the restriction $A<\pi$ is crucial in light of the function $\cos \pi z = \frac12(e^{i\pi z}+e^{-i\pi z})$. – Greg Martin May 19 '22 at 23:48
  • Does the conventional definition $(-1)^z = e^{z \ln(-1)}$ not work for you? (And yes, my ambiguity in $\ln(-1)$ is deliberate here.) – Dan May 20 '22 at 00:02
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    @Dan Unfortunately, no. Writing $-1=e^{\pi i}$, this becomes $e^{z(2k+1)\pi i}=e^{-(2k+1)\pi y}e^{(2k+1)\pi ix}$ for some choice of $k$. Let's take $x=0$. Letting $y>0$, the growth condition then implies $k>-1$. Taking $y<0$, the growth condition implies $k<0$. Thus, no choice of $k$ satisfies this constraint. – zbrads2 May 20 '22 at 00:25

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If $f$ is holomorphic in the right half-plane with $f(n) = (-1)^n$ for all positive integers $n$ and with the given growth restrictions, then $$ g(z) = f(z+1)+f(z) $$ satisfies $|g(x+iy)| \le C e^{Px+A|y|}$ for some constant $C$, and $g(n) = 0$ for all positive integers $n$. Carlson's theorem can be applied to $g$, and it follows that $g$ is identically zero.

It follows that $f(z+2) = f(z)$ for all $z$ in the right half-plane. This periodicity allows to extend $f$ to an entire function.

Finally we can apply Entire periodic function with bounded growth is constant to $F(z) = f(2z)$ and conclude that $f$ is constant.

Therefore such a function $f$ does not exist.

Martin R
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