# Showing that a function is constant using Liouville's theorem

Let $$α ∈ (0, 1)$$ be a fixed exponent, let $$C > 0$$, and let $$f$$ be an entire function satisfying $$|f(z)| ≤ C|z|^α$$

Show that $$f$$ is constant.

My attempt:

I'll show that using Liouville's theorem.

Let $$z=x+iy$$, $$(x, y) \in \mathbb R^2$$, then $$|z| = {(x^2+y^2)}^{1/2}$$ and so $$|z|^α = {(x^2+y^2)}^{α/2}$$.

Thus, $$|f(z)|≤ C{(x^2+y^2)}^{α/2}$$. And since $$\mathbb C$$ is endowed with $$\mathbb R^2$$, we can rewrite the above as $$|f(x, y)|≤ {(x^2+y^2)}^{α/2}$$, which shows that $$f$$ is continuous on $$\mathbb C$$ for all $$z$$. And we know that continuity implies boundness. So $$f$$ is also bounded.

And since it is entire, by Liouville's, it's constant.

Is my attempt correct?

• This is not correct. First of all, minor thing, $|z|^\alpha=(x^2+y^2)^{\alpha/2}$. But more importantly, all you've shown is that $|f(z)|\leq M(z)$. You have not shown $f$ is bounded since $M$ depends on $z$. – zugzug Mar 26 at 16:36
• Okay I'll correct it. Check my edit on few minutes – JOJO Mar 26 at 16:38
• maybe think about $\frac{f(z)-f(0)}{z}$; note that continuity doesn't imply boundness on an unbounded set like the plane – Conrad Mar 26 at 16:45
• Why do you believe continuity implies boundedness? $f(z) = z$ is continuous and unbounded on $\Bbb{C}$. – Eric Towers Mar 26 at 16:46
• Your edit is still not correct. Why does the inequality show $f$ is continuous? Why do you need to show $f$ is continuous when it's entire (therefore continuous)? Why are continuous functions bounded on $\mathbb{C}$? (e.g. $f(z)=e^z$). – zugzug Mar 26 at 16:47

Hint: $$\frac{f(z)-f(0)}{z}$$ is an entire function that goes to zero at infinity by hypothesis; what can you conclude?

Much like the proof of Liouville's Theorem, we use the Cauchy estimates

$$|\frac{f^{(n)}(0)}{n!}|\leq\frac{1}{2\pi}\int_\gamma\frac{|f(\zeta)|}{R^{n+1}}d\zeta$$

where $$\gamma$$ is a circle of radius $$R$$ centred at 0.

If $$|f(\zeta)|\leq C|\zeta|^{\alpha}$$, we get the following estimate

$$|\frac{f^{(n)}(0)}{n!}|\leq\frac{1}{2\pi}\int_\gamma CR^{-n-1+\alpha} d\zeta = CR^{ -n+\alpha}$$

which tends to 0 as $$R$$ tends to $$\infty$$ for $$\alpha\in(0,1)$$ and $$n\geq 1$$. Hence $$\frac{f^{(n)}(0)}{n!}=0$$ and so, considering the Taylor series of $$f$$, we have $$f(z)=f(0)$$

• I did not get this part: "and so, considering the Taylor series of $f$, we have $f(z)=f(0)$" – JOJO Mar 26 at 17:05
• Nice answer, upvote. Very minor point: when you apply the estimates, shouldn't you use $|d\zeta|$ rather than $d\zeta$? Otherwise, it doesn't make sense. – zugzug Mar 26 at 17:14
• @JOJO By Taylor expanding $f$ around $0$, we know it is given by $f(0) +f^{(1)}(0)x+\frac{1}{2}f^{(2)}(0)x^2+\cdots$. Since all the derivatives vanish, we must that that the Taylor series is just $f(0)$. – Aidan Apr 4 at 14:46

Another way. $$|f(0)| \leq C|0|^\alpha = 0$$, so $$f(z)/z$$ is also an entire function. It satisfies, $$\left| \frac{f(z)}{z} \right| \leq C|z|^{\alpha - 1} \text{.}$$ $$|z|^{\alpha -1}$$ is a decreasing function on $$|z| \geq 1$$, with maximum $$1$$ attained on $$|z| = 1$$. $$f(z)/z$$ is continuous on the compact set $$|z| \leq 1$$, so attains a maximum, $$M$$, on the closed unit disk. Therefore, $$f(z)/z$$ is an entire function bounded by $$\max \{C,M\}$$ on $$\Bbb{C}$$ and by Liouville's theorem, $$f(z) / z$$ is a constant, $$K$$.

So we have $$f(z) = Kz$$ and we are given $$|Kz| \leq C |z|^\alpha$$. Then $$|K| \leq C|z|^{\alpha - 1}$$ for all $$z \neq 0$$. Taking the limit as $$|z| \rightarrow \infty$$, we find $$K = 0$$. Therefore, not only is $$f$$ constant, it is the zero function.

Every entire function can be represented by a power series about any $$a\in\Bbb C$$

$$f(z) = \sum_{n\geqslant 0}c_n{(z-a)}^n$$

that converges for all $$z\in\Bbb C$$. Pick $$a=0$$ and consider what Cauchy's Integral Formula tells you about $$c_n$$ for $$n\geqslant 1$$. (Hint: You can pick a circle as large you like in the integral formula.)