Prove that: $f$ is a polynomial of degree $\le m$. I have a problem:
Suppose $f \in \mathcal{H}(U,F)=\left \{ f: U \to F,~ \text{f is holomorphic mapping} \right \}$.
Where $E,F$ are two complex Banach spaces, $U$ is an open set in $E$.
We assume $\exists m \in \mathbb{N}_0=\{0,1, \ldots\}$ and $c>0$ such that 
$$\left \|f(x)  \right \| \le c (\left \| x \right \|^m+1),~ \forall x \in E$$.
Prove that: $f$ is a polynomial of degree $\le m$.
Ps: I need your help. Thanks.
 A: I thought I had made a few mistakes here.
We don't need step 1, we only need to show step 2 in proof. Which means:
Since $||a_n||=||\dfrac{1}{2\pi i} \int_{| \zeta|=r} \dfrac{f(a+\zeta  t) \mathrm{d} \zeta}{\zeta^{n+1}} ||$ 
$\le \dfrac{1}{2 \pi} \int_{|\zeta|=r} \dfrac{||f(a+ \zeta t)|| \cdot | \mathrm{d}\zeta |}{| \zeta |^{n+1}}$
$=\dfrac{1}{2 \pi} (2 \pi r)\dfrac{C(|\zeta|^{m}+1)}{|\zeta|^{n+1}}$
$=\dfrac{C(r^{m+1}+r)}{r^{n+1}}$, $|\zeta|=r$, $\forall n>m, \mathrm{C}>0$.
$=\dfrac{\mathrm{C}}{r^{n-m}}+\dfrac{\mathrm{C}}{r^{n}}$
Let $r \to +\infty$ that I will obtain $||a_n||=\|P^m f(a)(t)\|\le \left(\dfrac{\mathrm{C}}{r^{n-m}}+\dfrac{\mathrm{C}}{r^{n}} \right)  \underset{r \to \infty}{\longrightarrow} 0$.
Therefore, $a_n=0,~~~ \forall n>m$.
Do you think so? Patrick. :)
A: Thank you very much! Patrick. 
With Patrick's hints, I will rewrite as follows:
We show that $a_n=0$, for $n>m$. 
Firstly, we show that: $||f(x)|| \le \mathrm{C}||x||^m$.
$||x||^m+1 \le 2||x\||^m,~~ \forall ||x|| \ge 1$ 
Hence, $\implies ||f(x)|| \le 2C||x||^m=\mathrm{C}||x||^m$ 
Step 2: We'll have to get the bound $\||f(x)\|| \le \dfrac{ \mathrm{C}}{r^{n-m}}$,$ \forall n>m$.
Since $||a_n||=||\dfrac{1}{2\pi i} \int_{| \zeta|=r} \dfrac{f(a+\zeta  t) \mathrm{d} \zeta}{\zeta^{n+1}} ||$ 
$\le \dfrac{1}{2 \pi} \int_{|\zeta|=r} \dfrac{||f(a+ \zeta t)|| \cdot | \mathrm{d}\zeta |}{| \zeta |^{n+1}}$
$=\dfrac{1}{2 \pi} (2 \pi)\dfrac{C(|\zeta|^{m+1}+1)}{|\zeta|^{n+1}}$
$=\dfrac{C(|\zeta|^{m+1}+1)}{|\zeta|^{n+1}}$
$=\dfrac{C(|r|^{m+1}+1)}{r^{n+1}}$
$\le \dfrac{\mathrm{C}}{r^{n-m}}$
Let $r \to +\infty$ that I will obtain $||a_n||=\|P^m f(a)(t)\|\le \dfrac{\mathrm{C}}{r^{n-m}}  \underset{r \to \infty}{\longrightarrow} 0$.
Therefore, $a_n=0$, with $n>m$.
It means $f$ is a polynomial of degree $\le m$. :P
*. I'm trying in case:
If $||x|| \le 1$ then $||f(x)|| \le 2c=\mathrm{C}$. It means $m=0$
$\implies||a_n|| \le \dfrac{\mathrm{C}}{r}$.
Is it right? Patrick. 
Can you help me? Thanks.   Prove that: $\sup_{z \in \overline{D}} |f(z)|=\sup_{z \in \Gamma} |f(z)|$
