If $f$ is entire on $\mathbb{C}$ and $|f(z)| \leq 100 \log_{e}|z|$ for each $z$ in $|z| \geq 2$ ,
If $f(i) = 2i$ then $f(1) = ?$,
I thought of applying ML inequality , Cauchys integral formula but i could not proceed with these tools in hand , i have read Contour integrations , ML inequality , Cauchy Goursat theorem .If you can cite any simple method for a begineer it would be good.
Actually i thought of using Liouville's theorem but due to the $\geq$ sign in modulus of $z$ i am unable to prove anything !
Thanks!
The proof will closely follow that of Louisville's theorem.
Fix some $z\in \Bbb C$, and let $R>|z|+2$. Thus we have that $B(0;2)\subset B(z; R)$. Furthermore, we have that for all $\zeta \in \partial B(z; R)$, since $|\zeta |>2$, $f(\zeta)\leq 100\log_e|\zeta|$. Since $\log_e $ is an increasing function, it then follows that $f(\zeta)\leq 100\log_e(R+|z|)$.
We thus have that \begin{eqnarray} |f'(z)| &&=\Big|\frac{1}{2\pi i}\int_{\partial B(z;R)} \frac{f(\zeta)}{(\zeta - z)^2}d\zeta\Big|\\ &&\leq \frac{1}{2\pi} \int_{\partial B(z;R)} \frac{|f(\zeta)|}{|(\zeta - z)^2|}|d\zeta|\\ &&\leq \frac{1}{2\pi} \int_{\partial B(z;R)} \frac{100\log_e(R+|z|)}{R^2}|d\zeta|\\ &&= 100\frac{\log_e(R+|z|)}{R} \end{eqnarray}
Since for a fixed $z$ this is true for all $R$ large enough, we may take the limit to see that $f'(z)=0$. Since $z$ was arbitrary, $\frac{d}{dz}f$ is identically zero, so $f$ is constant.