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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?

Any other answers?

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    $\begingroup$ 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 $z$. $\endgroup$ – zugzug Mar 26 at 16:36
  • $\begingroup$ Okay I'll correct it. Check my edit on few minutes $\endgroup$ – JOJO Mar 26 at 16:38
  • $\begingroup$ maybe think about $\frac{f(z)-f(0)}{z}$; note that continuity doesn't imply boundness on an unbounded set like the plane $\endgroup$ – Conrad Mar 26 at 16:45
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    $\begingroup$ Why do you believe continuity implies boundedness? $f(z) = z$ is continuous and unbounded on $\Bbb{C}$. $\endgroup$ – Eric Towers Mar 26 at 16:46
  • $\begingroup$ 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. $f(z)=e^z$). $\endgroup$ – zugzug Mar 26 at 16:47
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Hint: $\frac{f(z)-f(0)}{z}$ is an entire function that goes to zero at infinity by hypothesis; what can you conclude?

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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 $f(z)=f(0)$

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  • $\begingroup$ I did not get this part: "and so, considering the Taylor series of $f$, we have $f(z)=f(0)$" $\endgroup$ – JOJO Mar 26 at 17:05
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    $\begingroup$ 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. $\endgroup$ – zugzug Mar 26 at 17:14
  • $\begingroup$ @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 $f(0)$. $\endgroup$ – Aidan Apr 4 at 14:46
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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.

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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.)

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