Mean Value Property to show that entire function is a constant

Question

Let $f(z)$ be an entire function so that,

$$ \int \frac{|f(z)|}{1 + |z|^3} dA(z) < \infty$$

where the integral is taken over the entire complex plane. Show that $f$ is a constant.

I believe that the idea is to use the mean value property; that is:

$$f(z) = \frac{1}{\pi\delta^2}\int_{D(z, \delta)}f(w)dA(w)$$

and then do some manipulation to relate the two integrals. But I'm not sure otherwise how to proceed. Can anyone help?

Answer 1

You can use Cauchy's integral formula to show that $f^{(2)}=0$. Given $w\in \mathbb C$, take $r\geq\max\{1,2|w|\}$. Let $C_r$ be the circle with radius $r$ centered at $0$. Then $$f^{(2)}(w)=\frac{1}{\pi i}\int_{C_r}\frac{f(z)}{(z-w)^3}\,dz,$$

so $$|f^{(2)}(w)|\leq \frac1\pi \int_0^{2\pi}\frac{|f(re^{i\theta})|}{r^3-|w|^3}r\,d\theta\leq \frac3\pi\int_{0}^{2\pi}\frac{|f(re^{i\theta})|}{1+r^3}r\,d\theta.$$

The last inequality follows because $r\geq1$ and $r\geq2|w|$, so that

$$r^3-|w|^3\geq\frac78r^3>\frac13(r^3+r^3)\geq\frac13(1+r^3).$$

By integrating with respect to $r$ from $c=\max\{1,2|w|\}$ to $\infty$, $$\int_c^\infty |f^{(2)}(w)|\,dr\leq \frac3\pi\int_c^\infty\int_0^{2\pi}\frac{|f(re^{i\theta})|}{1+r^3}r\,d\theta\,dr=\frac3\pi\int_{|z|\geq c}\frac{|f(z)|}{1+|z|^3}\,dA(z)<\infty,$$

which implies $f^{(2)}(w)=0$, and $w$ was arbitrary.

Once you know $f^{(2)}=0$ you have $f(z) = az+b$ for constants $a$ and $b$, and $a$ must be zero for $\displaystyle{\int_\mathbb{C}\dfrac{|az+b|}{1+|z|^3}\,dA(z)}$ to converge.

Answer 2

Fix any $z_0\in \Bbb C$. $\forall r > 0$, we have by Mean Value Theorem, $$~~~~~~~~~~f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta})\, d\theta$$

$$\implies |f(z_0)| \leq \frac{1}{2\pi} \int_0^{2\pi} |f(z_0 + re^{i\theta})|\, d\theta$$

For every $R>0$, we have, $$\int_0^{R} \lvert f(z_0)\rvert~ \dfrac{r}{1+r^3}\, dr \le \frac{1}{2\pi}\int_0^R \int_0^{2\pi} \lvert f(z_0 + re^{i\theta})| ~\dfrac{r}{1+r^3}\, d\theta\, dr $$

which means, for $$A_R=\int_0^R \dfrac{rdr}{1+r^3}$$

we obtain, $$\lvert f(z_0)\rvert A_R \le \frac{1}{2\pi}\iint_{D(0;R)} \dfrac{|f(z)|}{1+|z|^3}\,dA$$

Therefore, letting $R\to\infty$, for some constants $C$ and $M$,

$$C |f(z_0)|\le \iint_{\mathbb{C}} \dfrac{|f(z)|}{1+|z|^3}\, dx\, dy=M$$

So $f(z_0) \leq M/C$. Since $z_0$ was arbitrary, $f$ is constant.

Answer 3

Suppose $f$ is nonconstant. Then there is a constant $c>0$ such that

$$\tag 1 \frac{1}{2\pi}\int_0^{2\pi} |f(re^{it})|\, dt > cr$$

for large $r.$

Proof: We can write $f(z) = f(0) + z^ng(z)$ for some $n\in \mathbb N$ and some entire $g$ such that $g(0) \ne 0.$ We then have $(1)$ bounded below by

$$ \frac{1}{2\pi}\int_0^{2\pi} r^n|g(re^{it})|\, dt - |f(0)|.$$

This is at least $r^n|g(0)| - |f(0)|$ as a consequence of the mean value property for circles applied to $g.$ This gives the result.

So if $f$ is nonconstant, then

$$\int_{\mathbb C} \frac{|f(z)|}{1 +|z|^3}\, dA(z) = \int_0^\infty \frac{r}{1+r^3}\int_0^{2\pi} |f(re^{it})|\,dt\,dr.$$

By $(1),$ the last integral is $\infty.$ Since in our problem we are given an $f$ for which this integral is finite, such an $f$ must be constant.