Application of Cauchy's integral formula.

Question

Let $f$ be an entire fundtion satisfying $|f^{\prime}(z)|\le 2|z|$ for any $z \in \Bbb C$. Then show that $f(z)=a+bz^2$ for some $a,b\in \Bbb C $ with $|b| \le 1$.

My trial : I tried to show that $f^{\prime\prime}(z)$ is bounded on $\Bbb C$.So, I tried to find relation between $f^{\prime}$ and $f^{\prime\prime}$. I mean, $|f^{\prime\prime}(z)|$ $\le$ {something with $f^{\prime}(z)$ product |z|} $\le R $ by using generalized Cauchy's integral formula. But, I failed... Further, I just thought it has to do with utilizing maximum modulus Theorem. But I had no idea of how to apply it.. Could anyone just give a few hints. it would be great help. Thansk!

Answer 1

Hint: What can you say about the function $$\mathbb{C} \backslash {0} \rightarrow \mathbb{C}: z \mapsto \frac{f'(z)-f'(0)}{z}?$$

Answer 2

Two versions ...

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Because $f(z)$ is entire, it has a Taylor expansion $$f(z)=a_0+\sum\limits_{n=1}a_nz^n \tag{1}$$ $$f'(z)=a_1+\sum\limits_{n=2}na_nz^{n-1} \tag{2}$$ From $$\left|f'(z)\right|\leq 2|z|, \forall z\in \mathbb{C} \Rightarrow |f'(0)|\leq 0$$ or $$0=|f'(0)|=|a_1| \Rightarrow a_1=0 \Rightarrow f(z)=a_0+\sum\limits_{n=2}a_nz^n \tag{3}$$ But then $$\left|f'(z)\right|\leq 2|z|, \forall z\ne0 \Rightarrow \left|\sum\limits_{n=2}na_nz^{n-1}\right|\leq 2|z| \Rightarrow |z|\left|\sum\limits_{n=2}na_nz^{n-2}\right|\leq 2|z|\Rightarrow\\ \left|\sum\limits_{n=2}na_nz^{n-2}\right|\leq 2, z\ne0$$ or $$\left|\sum\limits_{n=2}na_nz^{n-2}\right|\leq \max\{2,2|a_2|\}, \forall z \in \mathbb{C}$$ This means that $g(z)=\sum\limits_{n=2}na_nz^{n-2}$, which is entire, is also bounded. According to Liouville's theorem $g(z)$ is constant. But $f'(z)=z\cdot g(z)=Cz$ or $f(z)=a_0+\frac{C}{2}z^2$ and the result follows ...

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An alternative approach is to apply Cauchy's estimate to $(2)$ $$a_n=\frac{f^{(n)}(0)}{n!} \Rightarrow na_n=\frac{(f')^{(n-1)}(0)}{(n-1)!}=\frac{1}{2\pi}\int\limits_{C_R}\frac{f'(z)}{z^{n}}dz$$ leading to $$|na_n|\leq \frac{1}{2\pi}\int\limits_{C_R}\left|\frac{f'(z)}{z^{n}}\right||dz|\leq \frac{1}{2\pi}\int\limits_{C_R}\left|\frac{2}{z^{n-1}}\right||dz|=\frac{2}{R^{n-2}}$$ Taking the $\lim\limits_{R\rightarrow\infty}$ we have $a_n=0,\forall n\geq 3$. As a result, considering $(3)$ too $$f(z)=\sum\limits_{n=1}a_nz^n=a_0+a_2z^2=a+bz^2$$

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Last part, for $\forall z\ne 0$: $$|f'(z)|\leq 2|z| \Rightarrow |2bz|\leq 2|z| \Rightarrow |bz|\leq |z| \Rightarrow |b|\leq 1$$