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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 functiuon. 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.

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.

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 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 functiuon. 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.

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 functiuon. 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.

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 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 functiuon. 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.

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 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 functiuon. 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.