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Suppose that $f$ is an entire function satisfying $|f(z)| \leq 1+|z|^\frac{3}{2}$ for all $z \in \mathbb{C}$. Show that f must be a linear function, that is, $f(z)=az+b$ for some $a,b \in \mathbb{C}$.

I have not much insight on where to go with this, but here is what I have thought up already:

As $f$ is entire it is analytic everywhere in the complex plane, this means that it is complex differentiable everywhere in $\mathbb{C}$. We also have that $|f(z)| \leq 1+|z|^\frac{3}{2}$. Allowing $z=a+bi$ we then have $|f(z)| \leq 1+|a+bi|^\frac{3}{2}$. We can also simplify $|a+bi|^\frac{3}{2}$ to be $(a^2 +b^2)^\frac{3}{4}$. But this approach seems wrong, or that it won't lead anywhere. Where can I take this? Or is where I'm starting wrong?

Joey
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  • Since $f$ is entire, it's power series at point $z=0$ has infinite radius of convergence. Try to show that $a_m = 0$ for $m \ge 2$ (where $f(z) = \sum_{n=0}^\infty a_nz^n$). Are you familiar with formulas for $f^{(n)}(0)$ in terms of integral? – Presage Oct 19 '20 at 23:13

2 Answers2

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Hint: Let $f(z)=\sum a_kz^{k}$. Let $g(z)=a_2+a_3z+a_4z^{2}+...$. Then $g$ is also an entire function and $f(z)=a_0+a_1z+z^{2}g(z)$. Show that $g$ is bounded using the hypothesis. By Liouville's Theorem $g$ is a constant and this gives $f(z)=a_0+a_1z+a_2z^{2}$. Show now (using the hypothesis) that $a_2$ must be $0$.

Kavi Rama Murthy
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Let $f(z) = \sum_{n=0}^\infty a_nz^n$. Note that $a_n = \frac{f^{(n)}(0)}{n!}$. Recall that for any $R > 0$ you can write $$ |a_n| = |\frac{f^{(n)}(0)}{n!}| = \left|\frac{1}{2\pi i} \int_{\partial B(0,R)} \frac{f(\omega)}{\omega^{n+1}}d\omega \right| \le \frac{1}{2\pi}\int_{\partial B(0,R)}\frac{|f(\omega)|}{|\omega|^{n+1}}d\omega = \frac{1}{2\pi}\int_{\partial B(0,R)} \frac{|f(\omega)|}{R^{n+1}}d\omega$$

Using assumption, we'll get for any $R>0$ that $$ |a_n| \le \frac{1}{2\pi} \int_{\partial B(0,R)} \frac{1+|\omega|^{\frac{3}{2}}}{R^{n+1}} = \frac{1 + R^{\frac{3}{2}}}{R^n}$$

Clearly for $n \ge 2$ if we take $R \to \infty$ we'll get $a_n = 0$. Hence $f(z) = \sum_{n=0}^\infty a_nz^n = \sum_{n=0}^1 a_nz^n = a_0 + a_1z$

Presage
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  • Why is $|f(w)|$ $\leq$ $1+|w|^\frac{3}{2}$ when the assumption is for z? Similarly why is $w^{n+1} \geq R^{n+1}$. I really follow the rest of the solution, except for those parts. – Joey Oct 19 '20 at 23:43
  • What do you mean the assumption is for $z$? You have pointwise inequality $|f(z)| \le 1+|z|^{\frac{3}{2}}$. It's the same as you would have function $h:\mathbb C \to \mathbb C$ given by $h(z) = z$. Then $h(\omega) = \omega$, too, it's just a symbol used. For the second part, note that we're integrating the boundary of $B(0,R)$, in other words ${\omega \in \mathbb C : |\omega|= R }$ so in fact we have equality $|\omega|^{n+1} = R^{n+1}$ on that curve we're integrating (I'll edit solution with one step more, just for better understanding) – Presage Oct 19 '20 at 23:47
  • Gotcha okay, I guess I completely overlooked that, that part makes sense. – Joey Oct 19 '20 at 23:50
  • Oh, last question on this, why does $R^{n+1}$ just become $R^n$ – Joey Oct 20 '20 at 00:12
  • Cause after inequalities, we're integrating $\frac{1+R^{\frac{3}{2}}}{R^{n+1}}$ over $\partial B(0,R)$ so that it becomes $\frac{1+R^{\frac{3}{2}}}{R^{n+1}} \cdot 2\pi R$ (we'reintegrating constant function ) – Presage Oct 20 '20 at 00:16
  • ohh gotcha yes yes that makes sense ! – Joey Oct 20 '20 at 00:20