Let $f$ be an entire function s.t. $|f(z)|≤A|z|$ for all $z$, where $A$ is a fixed positive number. Show that $f(z)= a_1\cdot z$ where $a_1$ is a complex constant.
Here's my attempt:
Using Cauchy's inequality, we have $f(z)≤ \dfrac{o! M_R}{R^0} = M_R < A(|z_0| + R)$
Of course that is far from being the right answer but that all i have
Because $f(z)$ is entire, it has a Taylor expansion $$f(z)=a_0+\sum\limits_{n=1}a_nz^n \tag{1}$$ From $$\left|f(z)\right|\leq A|z|, \forall z\in \mathbb{C} \Rightarrow |f(0)|\leq 0$$ or $$0=|f(0)|=|a_0| \Rightarrow a_0=0 \Rightarrow f(z)=\sum\limits_{n=1}a_nz^n \tag{2}$$ But then $$\left|f(z)\right|\leq A|z|, \forall z\ne0 \Rightarrow \left|\sum\limits_{n=1}a_nz^n\right|\leq A|z| \Rightarrow |z|\left|\sum\limits_{n=1}a_nz^{n-1}\right|\leq A|z|\Rightarrow\\ \left|\sum\limits_{n=1}a_nz^{n-1}\right|\leq A, z\ne0$$ or $$\left|\sum\limits_{n=1}a_nz^{n-1}\right|\leq \max\{A,|a_1|\}, \forall z \in \mathbb{C}$$ This means that $g(z)=\sum\limits_{n=1}a_nz^{n-1}$, which is entire, is also bounded. According to Liouville's theorem $g(z)$ is constant. But $f(z)=z\cdot g(z)$ and the result follows ...
An alternative approach is to apply Cauchy's estimate (like I did here) to $(1)$ $$a_n=\frac{f^{(n)}(0)}{n!}=\frac{1}{2\pi}\int\limits_{C_R}\frac{f(z)}{z^{n+1}}dz$$ leading to $$|a_n|\leq \frac{1}{2\pi}\int\limits_{C_R}\left|\frac{f(z)}{z^{n+1}}\right||dz|\leq \frac{1}{2\pi}\int\limits_{C_R}\left|\frac{A}{z^{n}}\right||dz|=\frac{A}{R^{n-1}}$$ Taking the $\lim\limits_{R\rightarrow\infty}$ we have $a_n=0,\forall n\geq 2$. As a result, considering $(2)$ too, $f(z)=\sum\limits_{n=1}a_nz^n=a_1z$
It is easy to see that $|f'(0)|\le A$. Define $$ F(z)=\left\{\begin{array}{ll}\frac{f(z)}{z}&\text{ if }z\neq0,\\ f'(0)&\text{ if }z=0.\end{array}\right. $$ Then $F(z)$ is an entire function and $|F(z)|\le A$ and hence $F(z)$ is a constant. $F(z)=a_1$ or $f(z)=a_1z$.
