An entire function $f$ is said to be of exponential type if there exist constants $c$ and $k$ such that $|f(z)|\leq c e^{k |z|}$.

A famous result of Polya says if $f$ is an entire function of exponential type strictly less than $\log 2$ and take integer values at $\mathbb{Z_+}$ (the set of non-negative integers) then $f$ must be a polynomial.

A generalization of this result was used by Gelfond to prove his famous result $\alpha$ and $\beta$ are algebraic numbers with $\alpha \neq 0,1$ and if $\beta$ is not a rational number, then any value of $\alpha ^\beta = \exp(\beta \log \alpha)$ is a transcendental number.

It is well known that the theory of entire functions of exponential type is closely related to that of Harmonic analysis.

If some one can guide me to other results/method of proofs in transcendence theory where techniques or results related to harmonic analysis are used I would be very grateful.

A Question:-

Maybe the result of Polya can be seen as a manifestation of the Uncertainty Principle? In the sense that a function and its fourier transform both cannot be small.

Since, the function of exponential type is the Fourier transform of a compactly supported distribution, thus support is small there.

Now, a transcendental entire function of exponential type takes all values (with the possible exception of one) infinitely often (thus the zero set of such a function is usually infinite i.e., big and hence the support is small).

What Polya's result says is with the extra arithmetic condition on the function, that is it takes integer value at non negative integers, the function no longer can not be transcendental type (infinitely many zeros, so support is small) and hence a polynomial (only finitely many zeros i.e., support is big). (Of course this is heuristic, can this be made precise ? )

  • $\begingroup$ You question would be much improved if you used the topic of your last sentence as your main query. $\endgroup$
    – S. Carnahan
    Feb 8, 2012 at 9:50
  • $\begingroup$ Thankyou, Prof. Carnahan for your suggestion. I will try to modify my question as per your suggestion. $\endgroup$
    – Vagabond
    Feb 8, 2012 at 10:20
  • $\begingroup$ I have changed the question as per S. Carnahan's suggestion. $\endgroup$
    – Vagabond
    Feb 8, 2012 at 12:24


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