Showing Entire Function is Bounded

Question

Let $f$ be an entire function, such that $$ \lvert\, f(x+yi)\rvert\le \frac{1}{\lvert y\rvert}, $$ for all $x$ and all $y\ne 0$.

Show that $f$ is bounded.

Answer 1

Let $$ f(z)=\sum_{n=0}^\infty a_nz^n. $$ If $z=x+iy$, then $y=\dfrac{z-\overline{z}}{2i}$. We are given that $$ \lvert y\,f(x+iy)\rvert\le 1, $$ and hence $$ \left|\sum_{n=0}^\infty a_nz^{n+1}-\lvert z\rvert^2\sum_{n=0}^\infty a_nz^{n-1}\right\rvert=\left\lvert(z-\overline{z})\sum_{n=0}^\infty a_nz^n\right\rvert\le 2 $$ Thus, for real $R>0$, and integer $m\ge 0$, $$ \frac{1}{2\pi i}\int_{\lvert z\rvert=R}\left(\sum_{n=0}^\infty a_nz^{n+1}-\lvert z\rvert^2\sum_{n=0}^\infty a_nz^{n-1}\right)\,\frac{dz}{z^m}=\frac{1}{2\pi i}\int_{\lvert z\rvert=R}\left(\sum_{n=0}^\infty a_nz^{n+1}-R^2\sum_{n=0}^\infty a_nz^{n-1}\right)\,\frac{dz}{z^m} \\ = a_{m-2}-R^2a_m, $$ while $$ \left|\frac{1}{2\pi i}\int_{\lvert z\rvert=R}\left(\sum_{n=0}^\infty a_nz^{n+1}-\lvert z\rvert^2\sum_{n=0}^\infty a_nz^{n-1}\right)\,\frac{dz}{z^m}\right|\le \frac{1}{2\pi}\cdot 2\pi R\cdot \frac{2}{R^m}=\frac{1}{R^{m-1}}. $$ Therefore, for all $R>0$. $$ \lvert a_{m-2}-R^2a_m\rvert\le \frac{1}{R^{m-1}}. $$ Letting $R\to\infty$, we obtain that $a_m$ has to be equal to zero, and so $a_{m-2}$. Hence $f\equiv0$.

Answer 2

By the given condition, we may bound $f$ on the strip $$ S = \{(x,y) \in \mathbb{C} : |y| \geq 1 \}. $$ Without using some of the heavier machinery of complex analysis, I do not see an easy way to bound $\mathbb{C} - S$.

However, the maximum modulus principle informs us that on $\mathbb{C}-S$, $|f|$ must achieve its maximum on the boundary of this set. The boundary of this set are the lines $y = 1$ and $y = -1$. But on both of these we have our easy bound on $S$, thus $f$ must be bounded.

Here is an alternative short proof relying on the identity theorem:

Assuming that in your statement above that $x$ and $y$ are both real variables, note that $f(x)$ must be $0$ for any real $x$, since $$ |f(x)| = \lim_{n \rightarrow \infty} \left|f(x+i/n) \right| \leq \lim_{n \rightarrow \infty} 1/n = 0. $$

Then since $f$ agrees with the function $0$ on all of $\mathbb{R}$, which has a limit point in $\mathbb{C}$, $f$ must be in fact equal to $0$ everywhere. Therefore, $f$ is bounded.