# Polynomials

## Big O Estimates

### Tie’s Extra Questions: Fall 2011, Fall 2009 (Polynomial upper bound, $$d=2$$) #complex/exercise/completed

Let $$f(z)$$ be entire and assume that $$f(z) \leq M |z|^2$$ outside some disk for some constant $$M$$. Show that $$f(z)$$ is a polynomial in $$z$$ of degree $$\leq 2$$.

Take a Laurent expansion at zero: \begin{align*} f(z) = \sum_{k\geq 0} c_k z^k,\qquad c_k = {1\over k!} f^{(k)}(0) = {1\over 2\pi i}\oint_{{\left\lvert {\xi} \right\rvert} = R} {f(\xi) \over \xi^{k+1}}\,d\xi .\end{align*} The usual estimate: \begin{align*} 2\pi i{\left\lvert {c_k} \right\rvert} \leq \oint_{{\left\lvert {\xi} \right\rvert} = R} R^{-(k+1)}{\left\lvert {f(\xi)} \right\rvert} \,d\xi &\leq \oint_{{\left\lvert {\xi} \right\rvert} = R}R^{-(k+1)} M R^2 \,d\xi\\ &= M R^{-(k-1)} \cdot 2\pi R \\ &= 2\pi M R^{-k+2} \\ &\overset{R\to\infty}\longrightarrow 0 ,\end{align*} provided $$-k+2<0 \iff k>2$$.

### Tie’s Extra Questions: Spring 2015, Fall 2016 (Polynomial upper bound, $$d$$ arbitrary) #complex/exercise/completed

• Let Let $$f:{\mathbb C}\rightarrow {\mathbb C}$$ be an entire function. Assume the existence of a non-negative integer $$m$$, and of positive constants $$L$$ and $$R$$, such that for all $$z$$ with $$|z|>R$$ the inequality \begin{align*}|f(z)| \leq L |z|^m\end{align*} holds. Prove that $$f$$ is a polynomial of degree $$\leq m$$.

• Let $$f:{\mathbb C}\rightarrow {\mathbb C}$$ be an entire function. Suppose that there exists a real number $$M$$ such that for all $$z\in {\mathbb C}, \Re(f) \leq M$$. Prove that $$f$$ must be a constant.

\begin{align*} {\left\lvert {f^{(n)}(z)} \right\rvert} &= {\left\lvert { {1\over 2\pi i} \oint_\gamma {f(\xi) \over (\xi - z)^{n+1}} \,d\xi} \right\rvert} \\ &\leq {1\over 2\pi i} \oint_\gamma { {\left\lvert { f(\xi) } \right\rvert} \over {\left\lvert {\xi - z} \right\rvert}^{n+1}} \,d\xi\\ &\leq {1\over 2\pi i } \oint_\gamma {LR^m \over R^{n+1} } \,d\xi\\ &= {L\over 2\pi i} R^{m-(n+1)} \cdot 2\pi R \\ &= LR^{m-n} \\ &\overset{R\to\infty}\longrightarrow 0 \qquad \iff m-n<0 \iff n>m ,\end{align*} so $$f$$ is a polynomial of degree at most $$m$$.

Now if $$f$$ is entire, $$g(z) \coloneqq e^{f(z)}$$ is entire and \begin{align*} {\left\lvert {g(z)} \right\rvert} = {\left\lvert {e^{f(z)}} \right\rvert} = e^{\Re(f)} \leq e^M ,\end{align*} so $$g$$ is an entire bounded function and thus constant by Liouville, making $$f$$ constant. Why this is true: \begin{align*} e^{f} = C \implies e^f \cdot f' = 0 \implies f'\equiv 0 ,\end{align*} since $$e^f$$ is nonvanishing, and $$f'\equiv 0$$ implies $$f$$ is constant.

### Asymptotic to $$z^n$$#complex/exercise/work

Suppose $$f$$ is entire and suppose that for some integer $$n\geq 1$$, \begin{align*} \lim_{z\to \infty} {f(z) \over z^n} = 0 .\end{align*}

Prove that $$f$$ is a polynomial of degree at most $$n-1$$.

Choose $${\left\lvert {z} \right\rvert}$$ large enough so that $${\left\lvert {f(z)} \right\rvert}/{\left\lvert {z} \right\rvert}^n < {\varepsilon}$$. Then write $$f(z) = \sum_{k\geq 0} c_k z^k$$ and estimate \begin{align*} 2\pi {\left\lvert {c_k} \right\rvert} &\leq \oint_{{\left\lvert {z} \right\rvert} = R} {f(\xi) \over \xi^{k+1}}\,d\xi\\ &\leq \oint_{{\left\lvert {z} \right\rvert} = R} {{\varepsilon}{\left\lvert {\xi} \right\rvert}^n \over {\left\lvert {\xi} \right\rvert}^{k+1} } \,d\xi\\ &= {\varepsilon}R^{n-(k+1)} \cdot 2\pi R \\ &= {\varepsilon}C R^{n-k} \\ &\overset{{\varepsilon}\to 0}\longrightarrow 0 \end{align*} provided $$n-k \leq 0 \iff k\geq n$$, since $${\varepsilon}\to 0$$ forces $$R\to \infty$$.

### Spring 2021.3, Tie’s Extra Questions: Spring 2014, Fall 2009 (Polynomial lower bound, $$d$$ arbitrary) #complex/exercise/completed

Suppose $$f$$ is entire and there exist $$A, R >0$$ and natural number $$N$$ such that \begin{align*} |f(z)| \geq A |z|^N\ \text{for}\ |z| \geq R .\end{align*}

Show that

• $$f$$ is a polynomial and
• the degree of $$f$$ is at least $$N$$.

The easier version of this question: when $${\left\lvert {f} \right\rvert} \leq A{\left\lvert {z} \right\rvert}^N$$, $$f$$ is a polynomial of degree at most $$N$$ by Cauchy’s integral formula: \begin{align*} {\left\lvert {f^{(n)}(z)} \right\rvert} &= {\left\lvert { {1\over 2\pi i} \oint_\gamma {f(\xi) \over (\xi - z)^{n+1}} \,d\xi} \right\rvert} \\ &\leq {1\over 2\pi i} \oint_\gamma { {\left\lvert { f(\xi) } \right\rvert} \over {\left\lvert {\xi - z} \right\rvert}^{n+1}} \,d\xi\\ &\leq {1\over 2\pi i } \oint_\gamma {AR^N \over R^{n+1} } \,d\xi\\ &= {A\over 2\pi i} R^{N-(n+1)} \cdot 2\pi R \\ &= AR^{N-n} \\ &\overset{R\to\infty}\longrightarrow 0 \qquad \iff N-n<0 \iff n>N .\end{align*}

Now rearrange the given equality \begin{align*} {\left\lvert {f(z) \over z^N} \right\rvert} \geq A \qquad {\left\lvert {z} \right\rvert} \implies {\left\lvert {z^N\over f(z)} \right\rvert} \leq A^{-1} .\end{align*} A priori, $$f$$ is equal to its power series at $$z=0$$, so $$f(z) = \sum_{k\geq 0} c_k z^k$$. Since $${\mathbb{D}}_R$$ is compact, $$f$$ has finitely many zeros in this region, say $$\left\{{z_k}\right\}_{k\leq m}$$. This set must be finite, since an infinite subset of a compact set has a limit point, and being zero on a set with a limit point implies being identically zero by the identity principle.

Define \begin{align*} p(z) \coloneqq\prod_{1\leq k\leq m} (z-z_k) = z^m + { \mathsf{O}}(z^{m-1}) ,\end{align*} the product of these roots. Increase $$R$$ if necessary to ensure that \begin{align*} {\left\lvert {p(z)\over z^m} \right\rvert} < 1 \implies {\left\lvert {p(z)} \right\rvert} < {\left\lvert {z} \right\rvert}^m .\end{align*} Now define \begin{align*} G(z) \coloneqq{p(z) z^N \over f(z)} \implies {\left\lvert {G(z)} \right\rvert} = {\left\lvert {p(z) z^N\over f(z)} \right\rvert} = {\left\lvert {z^N\over f(z)} \right\rvert}\cdot {\left\lvert {p(z)} \right\rvert} \leq A^{-1}{\left\lvert {z} \right\rvert}^m .\end{align*}

Issue: this might not be entire? There could be poles at the zeros of $$f$$ outside of $${\mathbb{D}}_R$$

By the previous result, $$G$$ is a polynomial of degree at most $$m$$. Now consider leading terms: on one hand, \begin{align*} f(z) G(z) = p(z) z^N \sim \qty{z^m + \cdots }\cdot z^N = z^{N+m} + \cdots .\end{align*} On the other hand, \begin{align*} f(z) G(z) &= f(z) \qty{z^m + \cdots} \\ &\sim \sum_{k\geq 0} c_k z^{k+m} + z^{m-1}f(z) + \cdots \\ &= (z^m + \cdots + c_{N}z^{N+m} + \cdots) + z^{m-1}f(z) + \cdots ,\end{align*} and by the previous expression, this must be a polynomial of degree at most $$N+m$$. This forces $$c_k = 0$$ for all $$k> N$$, otherwise these would contribute higher order terms.

Note: maybe not quite right!

Alternatively, note that the inequality can be rewritten as \begin{align*} {\left\lvert {G(z)} \right\rvert} \leq A^{-1}{\left\lvert {z} \right\rvert}^m \implies {\left\lvert {p(z)\over f(z)} \right\rvert} \leq A^{-1}{\left\lvert {z} \right\rvert}^{m-N} .\end{align*}

• If $$m-N = 0$$, then $$p/f$$ is an entire bounded function and thus constant, making $$p(z) = \lambda f(z)$$ and $$f$$ is a polynomial of degree exactly $$N$$. -If $$m-N>0$$, then $$p/f$$ is a polynomial of degree at most $$m-N$$ by the previous result. But $$p/f$$ is a polynomial with no zeros, since $$Z_p = Z_f$$, and the only nonvanishing polynomial is a constant, so again $$p = \lambda f$$.
• If $$m-N<0$$, then use the inequality \begin{align*} {\left\lvert {z^{N-m}p(z) \over f(z)} \right\rvert} \leq A^{-1} ,\end{align*} so the LHS is an entire bounded function and thus constant, so $$z^{N-m}p(z) = \lambda f(z)$$. But the LHS is evidently a polynomial of degree $$(N-m)+m = m$$.

Note that the analogue of this problem where $${\left\lvert {f(z)} \right\rvert} \leq A {\left\lvert {z} \right\rvert}^N$$ implies $$f$$ is a polynomial of degree at most $$N$$ is easy by the Cauchy estimate: \begin{align*} {\left\lvert {f(z)} \right\rvert} ={\left\lvert {\sum_{k\geq 0} c_k z^k } \right\rvert} \implies {\left\lvert {c_n} \right\rvert} = {\left\lvert {f^{(n)}(0)} \right\rvert} &= {\left\lvert {{n!\over 2\pi i }\int_\gamma {f(\xi) \over (\xi-a)^{n+1} } \,d\xi} \right\rvert} \quad \text{ at } a=0\\ &\leq {n!\over 2\pi }\int_\gamma {{\left\lvert {f(\xi)} \right\rvert} \over {\left\lvert {\xi} \right\rvert}^{n+1} } \,d\xi\\ &\leq {n!\over 2\pi }\int_\gamma {A {{\left\lvert {\xi} \right\rvert}^N } \over {\left\lvert {\xi} \right\rvert}^{n+1} } \,d\xi\\ &= {A n!\over 2\pi }\int_\gamma {{R ^N } \over R^{n+1} } \,d\xi\\ &= {An!\over 2\pi} \cdot {2\pi R \over R^{n+1-N}} \\ &= {An! \over R^{n-N}} \\ &\overset{R\to\infty}\longrightarrow 0 \quad \iff n-N>0 \quad\iff n>N ,\end{align*} so $$f(z) = \sum_{0\leq k\leq N} c_k z^k$$.

For the case at hand, a solution I liked from MSE:

• Write $$g(z) \coloneqq f(1/z)$$, so $$g$$ has a singularity at $$z=0$$. The claim is that this is a pole.

• It can’t be removable: \begin{align*} {\left\lvert {g(z)} \right\rvert} \geq A {\left\lvert {1\over z} \right\rvert}^n \to\infty \quad \text{ for } {\left\lvert {1/z} \right\rvert} \geq R \,\, (\iff {\left\lvert {z} \right\rvert} < 1/R) ,\end{align*} so $$g$$ is unbounded near $$z=0$$.

• It can’t be essential: if so, take the neighborhood of $$z=0$$ given by $$U\coloneqq D_{1\over R}(0)\setminus\left\{{0}\right\}= \left\{{z{~\mathrel{\Big\vert}~}0< {\left\lvert {z} \right\rvert} < {1\over R} }\right\}$$. Then $$g(U) \subseteq {\mathbf{C}}$$ would be dense by Casorati-Weierstrass, but note that $$g(z) = w\in g(U) \implies {\left\lvert {w} \right\rvert} \coloneqq{\left\lvert {g(z)} \right\rvert} \geq A{\left\lvert {1/z} \right\rvert}^n$$ since $${\left\lvert {z} \right\rvert}<1/R$$, so $$g(U) \subseteq ({\mathbf{C}}\setminus D_{A\over R^n}(0))$$ and in particular does not intersect the interior of $$D_{A\over R^n}(0)$$.

• Since $$z=0$$ is a pole, it has some finite order $$m$$, so write \begin{align*} g(z) = \qty{c_{-m}z^{-m} + \cdots + c_{-1}z^{-1}} + \qty{c_0 + c_1 z + \cdots} \coloneqq p(1/z) + h(z) ,\end{align*} where $$p$$ is polynomial of degree exactly $$m$$ (since $$c_{-m} \neq 0$$) and $$h$$ is entire. In particular, $$z=0$$ is not a singularity of $$h$$.

• Now \begin{align*} g(z) = p(1/z) + h(z) \implies f(z) = p(z) + h(1/z) .\end{align*}

• Then \begin{align*} f(z) - p(z) = h(1/z) \overset{{\left\lvert {z} \right\rvert}\to \infty}\longrightarrow c_0 \coloneqq h(0) ,\end{align*} since holomorphic functions are continuous.

• Then $$h$$ is an entire function with a finite limit $$L$$ at $$\infty$$. $$h$$ is bounded by $$c_0$$ in a neighborhood $$U_\infty$$ of $$\infty$$ and takes on a maximum on $$U_\infty^c$$ by compactness and the maximum modulus principle. So $$h$$ is bounded on all of $${\mathbf{C}}$$, and thus constant by Liouville, and thus $$h(1/z) = L$$ for all $$z$$.

• So \begin{align*} f(z) &= p(z) + h(1/z) = p(z) + c_0 \\ \implies f(z) &= (c_{-1}z + \cdots + c_{-m}z^m) + c_0 ,\end{align*} which is a polynomial of degree exactly $$m\coloneqq\deg p$$.

• Why $$m \geq N$$: if not, $$m<N$$ so $$N-m > 0$$. Then for large $$z$$, \begin{align*} A \leq {\left\lvert {f(z) \over z^N} \right\rvert} &= {\left\lvert {c_0 + c_{-1}z + \cdots + c_{-m}z^m \over z^N} \right\rvert}\\ &= {\left\lvert { {c_0 \over z^N} + {c_{-1} \over z^{N-1}} + \cdots + {c_{-m} \over z^{N-m}} } \right\rvert} \\ &\overset{{\left\lvert {z} \right\rvert}\to\infty}\longrightarrow 0 ,\end{align*} since every term has a factor of $$z$$ in the denominator. This contradicts $$A>0$$. $$\contradiction$$

## Misc

### Spring 2021.4 #complex/qual/completed

Let $$f = u + iv$$ be an entire function such that $$\Re(f(x+iy))$$ is polynomial in $$x$$ and $$y$$. Show that $$f(z)$$ is polynomial in $$z$$.

To clear up notation: write $$f(z) = u(x, y) + iv(x, y)$$, here we’re assuming that $$u$$ is polynomial in $$x$$ and $$y$$.

If $$u$$ is polynomial in $$x,y$$, then so is $$v$$.

This follows from the fact that $$u$$ is a harmonic conjugate of $$v$$, and the explicit process computing the conjugate will result in a polynomial. Gamelin describes this process in detail, see Ch.2 Section 5 on Harmonic functions where he proves the formula \begin{align*} v(x, y) = \int_{y_{0}}^{y} \frac{\partial u}{\partial x}(x, t) \,dt -\int_{x_{0}}^{x} \frac{\partial u}{\partial y}\left(s, y_{0}\right) \,ds+ C .\end{align*}

Since $$f(x, y)$$ is a polynomial in $$x, y$$, $$f(z)$$ must be a polynomial in $$z$$.

Since $$f$$ is entire, it’s equal to its Laurent series everywhere, so \begin{align*} f(z) = \sum_{k\geq 0} c_k z^k, \qquad c_k = {f^{(k) }(0) \over k!} = {1\over 2\pi i} \int_{S^1} {f(\xi) \over \xi^{k+1} } \,d\xi .\end{align*} Thus $$f$$ will be a polynomial if $$c_{N} = 0$$ for all $$N$$ large enough, which will be true if $$f^{(N)}(z) = 0$$ for large enough $$N$$. But we can write \begin{align*} {\frac{\partial }{\partial z}\,} f(z) = {\frac{\partial }{\partial x}\,} f(x, y) \implies 0 = \qty{{\frac{\partial ^N}{\partial x^N}\,}} f(x, y) = \qty{{\frac{\partial ^N}{\partial z^N}\,}} f(z) \coloneqq f^{(N)}(z) ,\end{align*}

### Spring 2019.4 (Eventually bounded implies rational) #complex/qual/completed

Let $$f$$ be a meromorphic function on the complex plane with the property that $$|f(z)| \leq$$ $$M$$ for all $$|z|>R$$, for some constants $$M>0, R>0$$.

Prove that $$f(z)$$ is a rational function, i.e., there exist polynomials $$p, q$$ so that $$f=\frac{p}{q}$$.

Note that $$f$$ must have finitely many poles – either $$z=\infty$$ is a pole or a removable singularity, and since poles are isolated, there is some $$R\gg 0$$ such that all other poles of $$f$$ are in $${\left\lvert {z} \right\rvert} \leq R$$. The set $$P_f$$ of poles is a closed set and $$\overline{{\mathbb{D}}_R}$$ is compact, so if $$P_f$$ is infinite it has an accumulation point, contradicting that poles are isolated.

So enumerate $$P_f$$ as $$\left\{{p_k}\right\}_{k\leq N}$$, define $$g(z) \coloneqq\prod_{k\leq N}(z-p_k)$$, and set $$F(z) \coloneqq g(z) f(z)$$. Then $$F$$ is an entire function, and the claim is that $$F$$ is bounded and thus constant by Liouville. Proving the bound: take $${\left\lvert {z} \right\rvert} > R$$, then \begin{align*} {\left\lvert {G(z)} \right\rvert} &= {\left\lvert {f(z)} \right\rvert} {\left\lvert {g(z)} \right\rvert} \\ &\leq M C {\left\lvert {z} \right\rvert}^N ,\end{align*} using that $$g$$ is a polynomial of degree $$N$$, so $${\left\lvert {g(z)\over z^N} \right\rvert}\to 1$$ as $${\left\lvert {z} \right\rvert}\to \infty$$ since $$g$$ is monic. So after possibly increasing $$R$$, we can choose $${\left\lvert {z} \right\rvert}$$ large enough so that $${\left\lvert {g(z)\over z^N} \right\rvert} < C$$ for, say, some constant $$C<2$$. In any case, by a common qual question, if $${\left\lvert {G} \right\rvert} \in { \mathsf{O}}({\left\lvert {z} \right\rvert}^N)$$ for $${\left\lvert {z} \right\rvert}$$ large enough then $$G$$ is a polynomial of degree at most $$N$$. Then $$f(z) \coloneqq G(z)/g(z)$$ is a rational function.

### Spring 2020 HW 3.5, Tie’s Extra Questions: Fall 2015 #complex/exercise/completed

Let $$f$$ be entire and suppose that $$\lim_{z \rightarrow \infty} f(z) = \infty$$. Show that $$f$$ is a polynomial.

Note that $$f$$ has finitely many zeros: since $$f$$ is unbounded, there is some $$R$$ such that $$f({\mathbb{D}}_R^c) \subseteq {\mathbb{D}}^c$$, so in particular $$f$$ is nonvanishing on $${\mathbb{D}}_R^c$$. So $$Z_f$$ is a closed subset of a compact set, so is either finite or has an accumulation point. In the latter case, $$f\equiv 0$$ by the identity principle, so suppose not.

Write $$Z_f = \left\{{z_k}\right\}_{k\leq n}$$ for the $$n$$ many zeros of $$f$$, included with multiplicity, and set \begin{align*} \Phi(z) \coloneqq\prod_{k\leq n} (z-z_k), \qquad F(z) \coloneqq{\Phi(z) \over f(z) } .\end{align*} Now $$F$$ is a nonvanishing entire function.

$$F$$ is bounded on $${\mathbf{C}}$$.

Choose $$R\gg 1$$ so that all of $$z_k$$ are in $${\mathbb{D}}_R$$, so $${\left\lvert {\xi - z_k} \right\rvert} < R$$ for all $$\xi \in {\mathbb{D}}_R$$ and all $$k$$. By Cauchy’s integral formula, \begin{align*} {\left\lvert {F(z)} \right\rvert} &\leq {1\over 2\pi} \oint_{{\left\lvert {\xi} \right\rvert} = R} {\left\lvert {F(\xi) \over \xi} \right\rvert} \,d\xi\\ &={1\over 2\pi} \oint_{{\left\lvert {\xi} \right\rvert} = R} {\left\lvert {\Psi(\xi) \over f(\xi) \cdot \xi} \right\rvert} \,d\xi\\ &\leq {1\over 2\pi} \oint_{{\left\lvert {\xi} \right\rvert} = R} {R^m \over {\left\lvert { f(\xi) } \right\rvert} R } \,d\xi\\ &\leq {1\over 2\pi} \oint_{{\left\lvert {\xi} \right\rvert} = R} R^{m-1} \,d\xi\\ &= R^m ,\end{align*} where $$R$$ is increased if necessary to ensure $${\left\lvert {1\over f(z)} \right\rvert} < 1$$, which can be done since $${\left\lvert {f(z)} \right\rvert}\to \infty$$ as $$R\to \infty$$. So $${\left\lvert {F(z)} \right\rvert} \leq C{\left\lvert {z} \right\rvert}^m$$ in $$\overline{{\mathbb{D}}_R}^c$$ for $$R$$ large enough, making $$F$$ a polynomial of degree at most $$m$$. Since $$F$$ is continuous in $$\overline{{\mathbb{D}}_R}$$ which is compact, $$F$$ is bounded in here as well, making $$F$$ bounded on all of $${\mathbf{C}}$$.

Given this, $$F$$ is entire and bounded and thus constant by Liouville. So $$F(z) = c$$, and as a result $$f(z) = c\Phi(z)$$ which is a polynomial of degree $$n$$.

### Spring 2020 HW 2, SS 2.6.13 #complex/exercise/completed

Suppose $$f$$ is analytic, defined on all of $${\mathbf{C}}$$, and for each $$z_0 \in {\mathbf{C}}$$ there is at least one coefficient in the expansion $$f(z) = \sum_{n=0}^\infty c_n(z-z_0)^n$$ is zero. Prove that $$f$$ is a polynomial.

Hint: use the fact that $$c_n n! = f^{(n)}(z_0)$$ and use a countability argument.

Write $$Z_n \coloneqq\left\{{z\in {\mathbf{C}}{~\mathrel{\Big\vert}~}f^{(n)}(z) = 0 }\right\}$$, then by hypothesis $$\bigcup_{n\geq 0} Z_n = {\mathbf{C}}$$. A version of the Baire category theorem is that if $$X$$ is a complete metric space and $$X$$ is a countable union of closed sets, then at least one such set has a nonempty interior. Thus some $$Z_n$$ has an interior point $$z_0$$, and as a result there is some disc $${\mathbb{D}}_{\varepsilon}(z_0)$$ on which $$f^{(n)}(z_0) \equiv 0$$. This implies that $$f^{(k)}(z_0) \equiv 0$$ on $${\mathbb{D}}_{\varepsilon}(z_0)$$ for every $$k\geq n$$, so $$f$$ is a polynomial of degree at most $$n$$.

#complex/exercise/completed #complex/exercise/work #complex/qual/completed