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Suppose that if $f(z)$ is an entire function such that $\dfrac{f(z)}{z^n}$ is bounded for $|z|\ge R$ then $f(z)$ must be a polynomial of degree at most $n$

This same question has already been asked and solved on this website but the solution relies on power series and this question is asked in my text before power series has been introduced.

My book gives the hint that I should use Cauchy's estimates for $f^{(m+1)}(z)$ on a disk $|z-z_0|<R$ and then let $R \to \infty$ to obtain that $f^{(m+1)}(z_0)=0$

I proved the hint as follows. $f$ is analytic on $|z|<R$ and also $|f(z)|<k|z|^n$ when $|z|\ge R$ so by Cauchy's estimate on $|z|<R$, $f^{(m)}(0)\le \dfrac{m!}{R^m}k|z|^n=m!kR^{n-m}\to0$ when $m>n$ and so $f^{(m)}(0)=0 \; \forall \;m>n$.

Now let $z_0\in \mathbb C$ and choose $R'$ big enough such that $B_R(0) \subset B_{R'}(z_0)$, hence $|z|>R$ when $z \in \partial B_{R'}(z_0).$ $f$ being entire is then analytic on $B_{R'}(z_0)$ and so again by Cauchy's estimates we see that $f^{(m)}(z_0)\le \dfrac{m!}{(R')^m}k|\hat z|^n$ for $\hat z\in \partial B_{R'}(z_0)$ such that $|\hat z| \ge |z|$ for any $z \in \partial B_{R'}(z_0)$. Note that $|\hat z|=|z_0|+R'$ since for any $z \in \partial B_{R'}(z_0)$ we have $|z|\le |z-z_0|+|z_0|= R' +|z_0|$. So $|\hat z| \le R' + |z_0|$ but we see that $z_0 + R'\dfrac{z_0}{|z_0|}$ is on $\partial B_{R'}(z_0)$ and $|z_0 + R'\dfrac{z_0}{|z_0|}|=|z_0|+R'$ and this means that $|\hat z| = R' +|z_0|$. Using this we now see that $f^{(m)}(z_0)\le \dfrac{m!}{(R')^m}k(|z_0|+R')^n$ and if we let $m>n$ then we see that $f^{(m)}(z_0) \to 0$ and $R' \to \infty$.

Now we can say that $f^{(m)}(z)=0$ whenever $m>n$.

Now I do not know how to continue? Why does this mean that $f$ must be a polynomial? I see how it is true if we can use power series but thats not available to me now. Can I use induction as follows?

I want to show that if $f$ is entire and $f^{(m)}(z)=0$ for $m>n$ means that $f$ is a polynomial with degree at most $n$.

For the base case suppose $f^{m}(z)=0$ whenever $m>0$. Hence $f'(z)=0$ which means $f$ is a constant function, or in other words a polynomial with degree $0$. So the base case is true.

Now suppose that $f^{m}(z)=0$ whenever $m>k$ means that $f$ is a polynomial if at most degree $k$. Suppose that $f^{m}(z)=0$ whenever $m>k+1$. Let $g(z)=f'(z)$ which is also entire since $f$ is entire. Then we see that $g^{m}(z)=0$ whenever $m>k$ which implys that $g$ is a polynomial of degree at most $k$ and therfore $f'(z)$ is a polynomial of degree at most $k$ which means that $f(z)$ is a polynomial of degree at most $k+1$ which proves the result.

Is this all correct?

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  • $\begingroup$ $\sin(z)$ is not a bounded function when the domain is $\mathbb C$ $\endgroup$ – alpastor Mar 29 at 20:19
  • $\begingroup$ @PeterForeman, are you sure $\sin(z)$ is bounded in the entire complex plane? Or, did I misunderstand what you were saying? $\endgroup$ – avs Mar 29 at 20:19
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The statement that $$ f^{(n)}(0) = 0 \mbox{ for } n \geq m $$ (which you already proved) means that $f$ is a polynomial in a neighborhood $D$ of the origin. Let us show that this polynomial cannot be the restriction of a non-polynomial entire function to $D$.

Since $f$ is entire, so are all of its derivatives. Therefore, a derivative $g_{k}(z) = f^{(k)}(z)$ of $f$ of order $k \geq m$ ($m$ or higher) is also entire. And this function coincides in the whole interior of $D$ with the function $f^{(k)} \equiv 0$. Regard $f^{(k)}$ as the restriction to $D$ of the entire function constantly equal to zero.

Since two holomoprhic functions defined in the same domain and equal on a set that has a limit point in that domain are identical (this is the key here; see Cartan), it follows that $g_{k}(z)$ is identically equal to the entire function that is constantly zero.

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  • $\begingroup$ I mentioned in the above question that the power series representation of entire functions has not been introduced yet in my text book (Gamelin's Complex Analysis) so I can't use power series to answer this question $\endgroup$ – alpastor Mar 29 at 20:46
  • $\begingroup$ Have you covered the Cauchy integral formula? $\endgroup$ – avs Mar 29 at 21:01
  • $\begingroup$ Yes i have covered that $\endgroup$ – alpastor Mar 29 at 21:17
  • $\begingroup$ Well, you've used that by using the Cauchy estimates. Anyway, I edited the proof, removing the part about Taylor series. $\endgroup$ – avs Mar 29 at 22:05
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    $\begingroup$ I doubt that if the OP's course has not yet covered series that it's covered the identity theorem. $\endgroup$ – Mark Viola Mar 29 at 23:00

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