# Application of Cauchy's integral formula.

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!

• You can definitely replicate this answer – rtybase Mar 29 at 13:33

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 ...
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 $$f(z)=\sum\limits_{n=1}a_nz^n=a_0+a_2z^2=a+bz^2$$
Last part, for $$\forall z\ne 0$$: $$|f'(z)|\leq 2|z| \Rightarrow |2bz|\leq 2|z| \Rightarrow |bz|\leq |z| \Rightarrow |b|\leq 1$$
Hint: What can you say about the function $$\mathbb{C} \backslash {0} \rightarrow \mathbb{C}: z \mapsto \frac{f'(z)-f'(0)}{z}?$$