I would like to ask about, does there exists an entire function which is bounded on every line parallel to $x$ - axis , but unbounded on the $x$ - axis.
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3$\begingroup$ somewhat related: mathoverflow.net/questions/29734/… $\endgroup$– WolfgangCommented Dec 16, 2014 at 11:04
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4 Answers
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Yes, there are such functions. Take a very narrow region $D$ containing the positive ray, with nice boundary and such that $D$ intersects every any horizontal line other than the real line by a bounded interval. Let $g$ be a conformal map of $D'$ onto the right half-plane, where $D'\subset D$ is another similar region. Then with appropriate choice of $D$ and $D'$ $$f(z)=\int_{\partial D'} \frac{e^{g(\zeta)}}{\zeta-z}d\zeta$$
will converge for $z$ outside $D$ and the function $f$ will be bounded outside $D$ and extend to an entire function (by a deformation of the contour).
For the details, see for example MR2753600 or MR0545054.
EDIT. The method is very flexible of course. Taking $D$ to be a half-strip and $g(z)=e^z$ one obtains a Mittag-Leffler function. Replacing it by $f(z+4i)$ you obtain a function that is bounded on every line from the origin. But to obtain a function as you ask, a half-strip $D$ will not work, so the function is less elementary.
With the same method one can also construct functions which tend to zero on every line: just replace $f$ by $f(z)/(z-z_0)$ where $f(z_0)=0$. Existence of infinitely many zeros of the original $f$ is easy to prove. Repeating this you can find a function which tends to $0$ on every line faster than any polynomial.
answered Dec 16, 2014 at 15:55
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1$\begingroup$ My article "Hyperbolic entire functions with full hyperbolic dimension and approximation by Eremenko-Lyubich functions" also treats this Cauchy integral method in quite some generality. (arxiv.org/abs/1106.3439 , ams.org/mathscinet-getitem?mr=3214678 ). $\endgroup$ Commented Dec 17, 2014 at 10:00
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$\begingroup$ Sorry for the noise, but how do you define $g$? It is defined on $D' \subset D$ but we evaluate it on $\partial D$ in the integral?... $\endgroup$ Commented Jul 8, 2018 at 20:59
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$\begingroup$ @barto: D' must be sufficiently nice so that the conformal map has an analytic continuation to a slightly bigger domain. For details see the references that I cited. $\endgroup$ Commented Jul 9, 2018 at 21:15
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Yes. It is even better ! See the friendly paper MR2290290 David H. Armitage, Entire functions that tend to zero on every line. Amer. Math. Monthly 114 (2007), no. 3, 251–256.
Notice that such an example shows that there exist non-trivial functions (any directional derivative of the entire function) in the plane, whose Radon transform vanishes identically. Whence the necessity of rather strong assumptions when inverting the Radon transform. If you think that this inversion is instrumental in the medical scanner, you see that accuracy of mathematical statements can be crucial for human health!
answered Dec 16, 2014 at 16:00
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3$\begingroup$ The question was not about entire functions tending to zero on EVERY line, but such functions are also traditionally (since approx 1900) constructed by the general method I described. $\endgroup$ Commented Dec 16, 2014 at 19:20
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6$\begingroup$ Another funny twist of this same problem is to show that every entire function is a sum of an entire function bounded on every horizontal strip and an entire function bounded on every vertical strip. $\endgroup$– fedjaCommented Dec 17, 2014 at 0:12
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$\begingroup$ @fedja. Do you have a reference for that ? $\endgroup$ Commented Dec 17, 2014 at 6:37
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$\begingroup$ Alas, no, only a proof, which I can post if you want. :-) Avner Kiro showed me this puzzle when visiting me a few months ago and I have no idea where he took it from. $\endgroup$– fedjaCommented Dec 17, 2014 at 11:19
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3$\begingroup$ @fedja, is it too late to ask for that proof? $\endgroup$– LSpiceCommented Aug 5, 2020 at 11:28
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This should be a comment but it is too long... It is also possible to construct a function satisfying these demands using Taylor series only. For instance, the function
$$F(z)=\sum_{n\geq 0} \frac{z^n}{(\log(\log(n+1+e^e)))^n}$$ is unbounded on the positive real ray, but it is bounded outside the domain $\{x+iy\;:\; x>0,\; |y|<\pi e^{-x}\}$.
To see it, just write $$F(z)=\int_{-1/2-i\infty}^{-1/2+i\infty}\frac{z^s}{(\log\log(s+1+e^e))^s}\frac{ds}{e^{2\pi i s}-1}$$ and shift the contour.
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$\begingroup$ The series you wrote is not an entire function: its radius of convergence is 1. $\endgroup$ Commented Aug 10, 2020 at 1:37
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See Rudin: Real and Complex Analysis sect. 12.7 p274
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16$\begingroup$ For those who don't have the book, perhaps you could say what is there and how it answers the question. $\endgroup$ Commented Dec 12, 2015 at 0:08