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We know that the recurrence for $b>0$

(1) $f(0)=1$

(2) $f(z+1)=b{f(z)}$

has $f(z)=b^z$ as the only entire solution that is bounded on the strip $S=\{z: 0<\Re(z)\le 1\}$.

The image of $S$ under $b^z$ is an annulus for $b>0$ and so bounded. We know that for complex $b\not\ge 0$ the function $f(z)=b^z=\exp((\log(z)+2\pi i k)z)$ is not bounded on $S$ (the image is kinda infinite spiral) for any $k$. The question remains whether

Conjecture

There is no entire solution $f$ that satisfies (1) and (2) and is bounded on $S$ for complex $b\not\ge 0$.

How to prove this ?

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closed as too localized by Steve D, William, Andres Caicedo, J. M., Norbert Oct 3 '12 at 20:09

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Why close ? Maybe it is hard to add something that might be true. But closing sounds so bad. Or is that just my perspective ? –  mick Sep 17 '12 at 14:27
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1 Answer 1

up vote -2 down vote accepted

Remember that entire taylor series are coo everywhere. ( infinitely differentiable for all finite complex )

Let $f(z,1)$ be an entire periodic function with $f(0,1)=f(1,1)=1$ and period 1.

And $f(z,1)$ is not identically 1 for all $z$.

We will prove that for complex b with $arg(b) <> 0$ , the only solution to the equations is $f(z,1) * b^z$ and hence the proof follows.

Let k and n be positive integers.

$f(0) = 1$

$f(z+k) = b^k f(z)$

$f$ = entire

Then

Take the derivative of the equation $f(z+k) = b^k f(z)$ on both sides

$f ' (z+k) = b^k f ' (z)$ again $ f '' (z+k) = b^k f '' (z)$

And in general

f^(n) (z+k) = b^k f^(n) (z)

Hence because of taylors theorem we must conclude

$f(z) = f(0) * f(z,1) * b^z$ in the neighbourhood of 0.

But since f is entire it must be true everywhere and $f(0) = 1$ hence

$f(z) = f(z,1) b^z$

for all z.

If $arg(b) <> 0$ then the period of $b^z$ does not have $Re <> 0$ and hence $b^z$ is unbounded on the strip.

If $f(z)$ needs to be bounded and $b^z$ is not bounded , this implies that $f(z,1)$ needs to be bounded.

But this is impossible since $f(z,1)$ has a real period and is entire , it must be unbounded on the strip.

( remember $f(z,1) =/= 1$ everywhere by definition )

The product of two functions unbounded in the same region must be unbounded in that region.

QED

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